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  1. #1 Hydrogen atom 
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    I read that the new QM model about the H-atom says the electron has no speed, is that true, but all other parameter of classical model are valid (quantum numbers, orbits, spin, transition values etc?
    Could someone explain what that concretely means and what are the advantages of the new model?
    In particular, I do not see how the electron, if it doesn't move, can resist the nucleus attraction.
    Thanks


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    Quote Originally Posted by whizkid View Post
    I read that the new QM model about the H-atom says the electron has no speed, is that true, but all other parameter of classical model are valid (quantum numbers, orbits, spin, transition values etc?
    Could someone explain what that concretely means and what are the advantages of the new model?
    In particular, I do not see how the electron, if it doesn't move, can resist the nucleus attraction.
    Thanks
    Are you quite sure you have read the electron has no speed, and do you have a reference for that? It doesn't sound right to me. Surely it is rather that its position and momentum cannot simultaneously be defined, due to quantum indeterminacy. The electron does in fact have speed, because the model assumes a constant interchange between potential and kinetic energy of the electron as it moves within the potential well represented by nuclear attraction. I wonder if perhaps what you may be thinking of is that, in the ground state (1s orbital) of the hydrogen atom, the electron has no angular momentum. (i.e. the azimuthal quantum number, l, = 0). It is impossible for a classical "orbit" to have zero angular momentum of course: it would mean the electron was stationary and would fall into the nucleus. The QM model allows for this to be possible.

    It follows from this that are no "orbits" in the modern QM model of the atom. There are orbitals, but these are not much like classical orbits, partly for the reason I've given. They are "clouds" of electron probability, spread out in 3D, rather than the neat, 2D, geometrical ellipses of classical mechanics.


    Electrons have spin, certainly - which gives them a magnetic moment, the effects of which can be seen in appropriate circumstances.

    As for the advantages of all this, you are really asking for a series of tutorials on the structure of the atom and how this explains various atomic properties. This is a longish haul. Suggest we deal first with any questions you may have on what I've said so far and then we can take it from there.


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    Thanks, you have been very kind and clear.
    I do not wish to learn QM or QM model of the atom. I have read that that classical model is fine for light atoms, except for angular momentum, which should be zero.
    I would like to understand only why the old vale of h/2pi is a problem for the economy of the atom
    and , if any, the other differences in values or Physical laws, for example if the value for mu (the magnetic moment) remains h/4pi.

    As to QM, as you seem so competent and willing to oblige, could you tell whether the cloud refers only to the probability of finding the electron or something more complex and theoretical.
    Thanks a lot, again
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    Quote Originally Posted by whizkid View Post
    Thanks, you have been very kind and clear.
    I do not wish to learn QM or QM model of the atom. I have read that that classical model is fine for light atoms, except for angular momentum, which should be zero.
    I would like to understand only why the old vale of h/2pi is a problem for the economy of the atom
    and , if any, the other differences in values or Physical laws, for example if the value for mu (the magnetic moment) remains h/4pi.

    As to QM, as you seem so competent and willing to oblige, could you tell whether the cloud refers only to the probability of finding the electron or something more complex and theoretical.
    Thanks a lot, again
    Well I think the classical model is not exactly fine if you are a chemist, as the distribution of the electron clouds in space is key to atomic size and chemical bonding. But it works to a point, so long as you are prepared to just park the bits that don't work and accept you will learn about how QM deals with all that later on.

    I don't follow what you mean by the "economy" of the atom, though. Economy seems a strange word to use in this context. Can you explain what you are getting at?

    I'll need to look up the values of magnetic moment etc, as I don't carry these in my head (my QM was learnt 40 years ago - but I do dip into the books now and again.)
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    Quote Originally Posted by exchemist View Post

    Well I think the classical model is not exactly fine if you are a chemist, as the distribution of the electron clouds in space is key to atomic size and chemical bonding.

    I don't follow what you mean by the "economy" of the atom, though. Economy seems a strange word to use in this context. Can you explain what you are getting at?
    The size for H1 was bohr radius: 0.5292*10^-8 cm, what is it in QM?, I read that ionization energy remains 13.6 eV so, can you confirm that even in QM speed is C/137?
    I meant the general structure of the atom, why was angular momentum = h/2pi a problem? it helps stabilize the orbit

    I got the info in this thread The maximum number of elements, and I could not understand why the electron has no speed and is not a material object. Moderators are usually academics, but you seem to disagree with their statements?
    Last edited by whizkid; March 9th, 2014 at 08:11 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post

    Well I think the classical model is not exactly fine if you are a chemist, as the distribution of the electron clouds in space is key to atomic size and chemical bonding.

    I don't follow what you mean by the "economy" of the atom, though. Economy seems a strange word to use in this context. Can you explain what you are getting at?
    The size for H1 was bohr radius: 0.5292*10^-8 cm, what is it in QM?, I read that ionization energy remains 13.6 eV so, can you confirm that even in QM speed is C/137?
    I meant the general structure of the atom, why was angular momentum = h/2pi a problem? it helps stabilize the orbit

    I got the info in this thread The maximum number of elements, and I could not understand why the electron has no speed and is not a material object. Moderators are usually academics, but you seem to disagree with their statements?
    Aha, now I see what you have read. What this says is that the electron has no well-defined speed. That's important, as it is referring to the quantum indeterminacy that I previously mentioned. It has a speed alright (because it has kinetic energy), but it is not well-defined (i.e you can't find out what it is, as a function of time or position). I don't think there's any inconsistency between what I'm saying and what this person is saying.

    As for this about the electron not being a "material object", I think one needs to be careful how one interprets what one reads. In the QM description, an electron behaves in some ways like a distinct particle and in some ways like a smeared-out "probability wave". But both aspects are "material", seeing as an electron is a constituent of what we call "matter". It is just that the concept of what is "material" has to be broadened to encompass behaviour that is not intuitive to us from our everyday observation of macroscopic things. It is true that, to the extent that a wave is not an "object" but a disturbance, it is hard to see the wave aspect of the electron as an "object". But it is material. At least, that is how I would look at it.

    As for Bohr radius, no that doesn't apply in QM, as the electron is not in an orbit at any fixed distance from the nucleus. It can go right up to the nucleus and can spend time quite far way from it. In the ground 1s state, it occupies a spherically symmetrical cloud of probability around the nucleus, which diminishes in intensity as the radius increases, a bit like the way the atmosphere thins as you go to higher altitudes.

    The ionisation energy is an experimental fact, so does not depends on what model we use.

    It is true that in Bohr's orbit model you can calculate the kinetic energy and hence the orbital speed of the electron from this, but it doesn't work in QM, because the electron is not in an orbit at a fixed distance. So this means that its speed is presumed to vary and that the proportion of its energy that is in the form of potential energy and the proportion that is in the form of kinetic energy will be unknown at any given instant - and so too will be its speed.
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    Quote Originally Posted by exchemist View Post
    It has a speed alright (because it has kinetic energy), but it is not well-defined (i.e you can't find out what it is, as a function of time or position).

    As for Bohr radius, no that doesn't apply in QM, as the electron is not in an orbit at any fixed distance from the nucleus. It can go right up to the nucleus and can spend time quite far way from it. In the ground 1s state, it occupies a spherically symmetrical cloud of probability around the nucleus, which diminishes in intensity as the radius increases, a bit like the way the atmosphere thins as you go to higher altitudes.

    the electron is not in an orbit at a fixed distance. So this means that its speed is presumed to vary and that the proportion of its energy that is in the form of potential energy and the proportion that is in the form of kinetic energy will be unknown at any given instant - and so too will be its speed.
    Thanks, exchemist, that is amazingly clear, I understand the theory now, but I have a few doubts: it seems that all the rules of physics do not apply here:
    As the engine of the electron (electrostatic force) is costant,
    -how can the speed be incostant, erratic?
    - how can it go and even stay by the nucleus and then bounce away , as PE far away is lessser and lesser?
    - as the motion is anyway sort of circular, ho can it have no angular momentum, and yet have magnetic moment?

    The Bohr model can be adapted to explain the cloud: if the orbit is elliptict an rotates in 3-D, the speed varies a lot and also the distance fron the nucleus, and as it rotates trillions of times a second , it appears like a cloud, no more-no less than a stone rotating on a string of variable length that disappears in a cloud of probability.
    Why this is not possible or even plausible?
    Thanks for your in valuable help.
    If you have no more time to spend do you know of a link where I can find all these details?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    It has a speed alright (because it has kinetic energy), but it is not well-defined (i.e you can't find out what it is, as a function of time or position).

    As for Bohr radius, no that doesn't apply in QM, as the electron is not in an orbit at any fixed distance from the nucleus. It can go right up to the nucleus and can spend time quite far way from it. In the ground 1s state, it occupies a spherically symmetrical cloud of probability around the nucleus, which diminishes in intensity as the radius increases, a bit like the way the atmosphere thins as you go to higher altitudes.

    the electron is not in an orbit at a fixed distance. So this means that its speed is presumed to vary and that the proportion of its energy that is in the form of potential energy and the proportion that is in the form of kinetic energy will be unknown at any given instant - and so too will be its speed.
    Thanks, exchemist, that is amazingly clear, I understand the theory now, but I have a few doubts: it seems that all the rules of physics do not apply here:
    As the engine of the electron (electrostatic force) is costant,
    -how can the speed be incostant, erratic?
    - how can it go and even stay by the nucleus and then bounce away , as PE far away is lessser and lesser?
    - as the motion is anyway sort of circular, ho can it have no angular momentum, and yet have magnetic moment?

    The Bohr model can be adapted to explain the cloud: if the orbit is elliptict an rotates in 3-D, the speed varies a lot and also the distance fron the nucleus, and as it rotates trillions of times a second , it appears like a cloud, no more-no less than a stone rotating on a string of variable length that disappears in a cloud of probability.
    Why this is not possible or even plausible?
    Thanks for your in valuable help.
    If you have no more time to spend do you know of a link where I can find all these details?
    Well, that's just QM, I'm afraid. The Bohr model has lots wrong with it, as was very quickly realised at the time. One enormous problem with it is that an electric charge in orbit, according to Maxwell's by then well-established theory of electromagnetic radiation, should constantly radiate, losing energy and eventually spiralling into the nucleus. Which manifestly does not happen! Instead, what actually happens is that electrons in an atom can absorb (or emit, if they are not already in the ground state) distinct frequencies only of radiation, and thereby "jump" from one "orbit" to another. This also cannot be explained by Bohr's model.
    Your idea of a precessing orbit, though ingenious, would suffer from the same problems, I'm afraid.

    The presumed variation in speed of the electron can to some degree be thought of as like a pendulum swinging, or a vibrating spring, or some other form of "harmonic" motion. In harmonic motion, energy is continuously moving from potential to kinetic and back. At the extremes of the motion, where the object reverses direction, its speed is zero, so its kinetic energy is zero, and its potential energy is at a maximum. Conversely, at the centre of the motion, the object is moving at its fastest and the potential energy is at a minimum.

    Notice I use the analogy of "harmonic" motion. The electron in an atom, having a wavelike character, can only exist in a set number of "resonant" states, corresponding to spherical "standing wave" patterns. The various orbitals are in effect the fundamental and successive harmonics.

    There is more about all this here: Atomic orbital - Wikipedia, the free encyclopedia

    The magnetic moment of the electron is something else, attributed to what is called "spin". It can be thought of as spinning on its own axis (it's a particle, now: confusing, eh?!). A spinning charge creates a magnetic field. Needless to say, being a QM phenomenon, it is not exactly like that, but for most purposes in chemistry - my former discipline - that analogy is good enough. So the magnetic moment of the electron itself is not a consequence of orbital motion about the nucleus of the atom: it is intrinsic to the electron itself.

    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.

    There is more about this here: Electron magnetic dipole moment - Wikipedia, the free encyclopedia
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    Quote Originally Posted by YellowKazooie3 View Post
    The Hydrogen atom has 1 proton and 1 electron
    Well done, darling!
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    Quote Originally Posted by whizkid View Post
    Thanks, you have been very kind and clear.
    I do not wish to learn QM or QM model of the atom. I have read that that classical model is fine for light atoms, except for angular momentum, which should be zero.
    I would like to understand only why the old vale of h/2pi is a problem for the economy of the atom
    and , if any, the other differences in values or Physical laws, for example if the value for mu (the magnetic moment) remains h/4pi.

    As to QM, as you seem so competent and willing to oblige, could you tell whether the cloud refers only to the probability of finding the electron or something more complex and theoretical.
    Thanks a lot, again
    Just realised I left this question about the probability cloud dangling. It is just a probability density, i.e. if you integrate it over a certain volume of space, it tells you the proportion of the electron's time that is spent in that volume of space. You often see pictures of orbitals, which show things such as the envelope in which the electron spends 95% of its time, or something.

    Formally, the "resonance" or "standing wave" states of the electron in the atom are solutions to the time-independent Schroedinger equation. These are mathematical functions which , when multiplied by their complex conjugate (sort of equivalent to squaring them, for something involving complex numbers) give the probability density. So the electron wave can be thought of as a wave representing the square root of the probability, more or less. Don't spend too much time worrying about these details though. The main thing is to realise the wave nature of the electron results in a model of the atom in which the only stable states are standing wave patterns.

    (Transitions between different states are caused by the wave of electromagnetic radiation coupling to these standing wave patterns and changing one to another, with absorption or emission of energy as appropriate. This process is described by the time-dependent Schroedinger equation, since these transitions occur over a very short but finite time interval. So QM can explain the "jumps" between states, unlike Bohr's model, which had to postulate implausible instantaneous jumps, with no explanation of what was going on.)
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    [QUOTE=exchemist;536563]
    Quote Originally Posted by whizkid View Post
    Thanks, you have been very kind and clear.
    I do not wish to learn QM or QM model of the atom. I h. So QM can explain the "jumps" between states, unlike Bohr's model, which had to postulate implausible instantaneous jumps, with no explanation of what was going on.)
    Thanks for your masterly explanations, exchemist.
    You just forgot to tell me what is the current extinmate for the radius of a not-exiced H atom, where the probability is 95%, and the average speed of the electron and the magnetic moment: the magnetic field interacting with the spin is (q=1 *)vr/2 , so if QM says it stays h/4pi they have a definite idea of v and r
    Also , if you can spare some more time:
    if there is a magnetic moment, shouln't there be also an angular momentum?
    What makes the standing wave precess , according to QM?
    finally
    couldn't the jumps be explained simply as Brehmstralung?
    Last edited by whizkid; March 10th, 2014 at 05:43 AM.
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    [QUOTE=whizkid;536743]
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by whizkid View Post
    Thanks, you have been very kind and clear.
    I do not wish to learn QM or QM model of the atom. I h. So QM can explain the "jumps" between states, unlike Bohr's model, which had to postulate implausible instantaneous jumps, with no explanation of what was going on.)
    Thanks for your masterly explanations, exchemist.
    You just forgot to tell me what is the current extinmate for the radius of a not-exiced H atom, where the probability is 95%, and the average speed of the electron and the magnetic moment: the magnetic field interacting with the spin is (q=1 *)vr/2 , so if QM says it stays h/4pi they have a definite idea of v and r
    Also , if you can spare some more time:
    if there is a magnetic moment, shouln't there be also an angular momentum?
    What makes the standing wave precess , according to QM?
    finally
    couldn't the jumps be explained simply as Brehmstralung?
    Ah yes. The thing is, as I've been saying, the electron cloud thins out with no fixed boundary, just like the Earth's atmosphere. There is no calculated radius, really, from QM. One could calculate a 95% confidence interval envelope, but it would have little physical significance. When this sort of thing is done it is to give a qualitative idea of the shape of the orbitals, rather than to define atomic size. With size, what matters in practice is the effective size of the atom in the compounds it forms. As you may know, there are tables of atomic radii, derived from experimentally measured interatomic distances, in the various compounds that hydrogen forms.

    Here's a link I found to a visualisation program for orbitals, which uses the 95% interval, just to show what's involved.
    http://nbviewer.ipython.org/urls/raw...0working.ipynb

    I was interested to note they point out when you square the wavefunction (or multiply by complex conjugate) you lose the phase, because the -ve phase of the wave gets squared to +ve of course.
    Typically when these pictures are made the +ve and -ve phases are indicated, though, as this is important for understanding it as a standing wave pattern - and for seeing what will happen when orbitals overlap to form chemical bonds in compounds.

    I've searched for a value for the size of the 95% confidence level envelope for the 1s orbital but don't seem to be able to find one for you. However, you might be interested to read this: Bohr radius - Wikipedia, the free encyclopedia

    which talks about the relationship between the expectation value of the electron-nucleus distance, in relation to the radius of the Bohr orbit.

    I have been able to find for you the average potential and kinetic energy of the electron in the 1s of hydrogen, as follows:
    The average potential energy is -me4/(ħ2n2) and the average kinetic energy is me4/(2ħ2n2).
    (m is the electron mass, e is the charge on the proton, the charge on the electron being -e.).

    I'll leave you to get the mean speed of the electron from this, if you want.

    As to your other questions, I haven't time to get into the magnetic moment stuff just now, but bremstrahlung doesn't sound very relevant to the mechanism for emission and absorption of radiation. That is a process producing a continuous spectrum rather than the line spectrum we observe from atoms - which was one of the things Bohr was trying to account for with his orbits. Do you know of a reverse bremstrahlung process to account for absorption? And moreover, absorption at only specific frequencies?



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    Thanks, exchemist, you have been of great help
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    [QUOTE=whizkid;536743]
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by whizkid View Post
    Thanks, you have been very kind and clear.
    I do not wish to learn QM or QM model of the atom. I h. So QM can explain the "jumps" between states, unlike Bohr's model, which had to postulate implausible instantaneous jumps, with no explanation of what was going on.)
    Thanks for your masterly explanations, exchemist.
    You just forgot to tell me what is the current extinmate for the radius of a not-exiced H atom, where the probability is 95%, and the average speed of the electron and the magnetic moment: the magnetic field interacting with the spin is (q=1 *)vr/2 , so if QM says it stays h/4pi they have a definite idea of v and r
    Also , if you can spare some more time:
    if there is a magnetic moment, shouln't there be also an angular momentum?
    What makes the standing wave precess , according to QM?
    finally
    couldn't the jumps be explained simply as Brehmstralung?
    On the magnetic moment issue, orbital magnetic moment is (-eh/4pi.m)√(l(l+1)), where l is the azimuthal quantum number. This shows it is zero for an s orbital.

    Spin magnetic moment is, I think, approx eh/4pi.m, though I tend to get lost in unscrambling the h bar, Bohr Magneton and other factors commonly used. But have a look here at both: Electron magnetic dipole moment - Wikipedia, the free encyclopedia.

    I can't quite follow your comment about magnetic field interacting with the spin, etc. I can't seem to relate it to the expressions I've just given above. What is "q"?

    (By the way, if we go much further we are likely to get to the bottom of what I feel I can recall with confidence and I may have to hand you off to someone with a more recent recall of the quantum physics involved. I am a mere chemist so I'm better on the qualitative principles than the rigour of the maths behind it.)
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    I agree with excemist, this guy knows what he is talking about
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    Quote Originally Posted by exchemist View Post
    One enormous problem with it is that an electric charge in orbit, according to Maxwell's by then well-established theory of electromagnetic radiation, should constantly radiate, losing energy and eventually spiralling into the nucleus.
    Could you please expand on that, why should it lose KE energy?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    One enormous problem with it is that an electric charge in orbit, according to Maxwell's by then well-established theory of electromagnetic radiation, should constantly radiate, losing energy and eventually spiralling into the nucleus.
    Could you please expand on that, why should it lose KE energy?
    An accelerating electric charge is what you have in a radio transmission antenna. The resulting radio waves have energy, which comes from the power input of the transmitter. An orbiting electron (which is, as you know, in a state of continuous acceleration due to its constantly changing direction) should do the same, but in its case there is no energy source other than its own KE. So the KE should decline progressively.

    There is a qualitative explanation of how an accelerating charge gives rise to an EM wave here: electromagnetism - How & Why does accelerating charges radiate electromagnetic radiation? - Physics Stack Exchange
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    Quote Originally Posted by exchemist View Post
    An orbiting electron (which is, as you know, in a state of continuous acceleration due to its constantly changing direction) should do the same, but in its case there is no energy source other than its own KE.
    Why should it accelerate? and the energy to change direction comes from the proton, doesn'it?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    An orbiting electron (which is, as you know, in a state of continuous acceleration due to its constantly changing direction) should do the same, but in its case there is no energy source other than its own KE.
    Why should it accelerate? and the energy to change direction comes from the proton, doesn'it?
    Whoa, steady! You know your Newtonian mechanics, don't you? At least I had assumed you do, in view of the sophistication of your earlier questions.

    Anything in circular or elliptical motion is constantly accelerating, because it is not moving in a straight line at constant speed. Right?

    And there is no "energy" involved in the changing of direction. That is caused by the force of electrostatic attraction. Force, not energy.

    Similarly, an orbiting sateillite is constantly accelerating (it is in free fall under the influence of the force of the Earth's gravity) but the kinetic energy of its motion comes from the rocket that got it up there and gave it its initial tangential velocity. That kinetic energy remains constant and it stays up there indefinitely (unless it is in low orbit and encounters friction from the residual upper atmosphere.)
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    Quote Originally Posted by exchemist View Post
    And there is no "energy" involved in the changing of direction.
    I know mechanics, I think that is debatable: changing direction cannot take place without energy /work, if it were so we could move around a planet like Jupiter just placing a rocket precisely in line with its center of mass. Couldnn't we?.
    Last edited by whizkid; March 12th, 2014 at 08:54 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    And there is no "energy" involved in the changing of direction.
    I know mechanics, but I donot think that changing direction can take place without energy /work.
    Oh dear.

    Work is force x distance gone in the direction of that force.

    In the case of the circular orbit, the direction is always perpendicular to the (radial) line of action of the force. So no work is done.

    With an elliptical orbit, some work is done as the satellite gains KE as it moves from apogee to perigee, but then reverse work is done as it swings out to apogee and slows down again, getting you back to where you started.

    In either case, no net energy change occurs during a complete orbit.

    Check it out for yourself with a textbook or on line.
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    Quote Originally Posted by exchemist View Post
    Check it out for yourself with a textbook or on line.
    I know what the textbook says, I ask:
    A body is moving in a straight line => +x, if we push it perpendicularly => +/-y and make it change direction, we need to apply what force?
    A boat is speeding on a lake, suppose a slave chained to it in the water swims perpendicularly making it rotate and describe a circle (if the effor is constant), does the boat accelerate? does the slave do any work, does he burn any energy?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    Check it out for yourself with a textbook or on line.
    I know what the textbook says, I ask:
    A body is moving in a straight line => +x, if we push it perpendicularly => +/-y and make it change direction, we need to apply what force?
    A boat is speeding on a lake, suppose a slave chained to it in the water swims perpendicularly making it rotate and describe a circle (if the effor is constant), does the boat accelerate? does the slave do any work, does he burn any energy?
    Right, if you push a body with a force perpendicular to its direction of motion it will be deflected in a curve but no work is done. But that implies that, as the body starts to deviate, the angle of the applied force changes, so as to maintain its perpendicularity to the direction of motion. Which means it will no longer act along the Y axis as deviation starts. This is what happens with motion in a circle (orbit, conker on a string etc).

    The speedboat and swimmer are terrible analogies, as water creates huge amounts of frictional drag. Try a skater on ice or something, tied to a stick.

    But I have to say we are deviating rather disappointingly from the subject of this thread. The Mods may ask that we take this excursion into classical mechanics onto a separate thread, for the sake of other readers.
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    Quote Originally Posted by exchemist View Post
    Right, if you push a body with a force perpendicular to its direction of motion it will be deflected in a curve but no work is done.

    that we take this excursion into classical mechanics onto a separate thread, for the sake of other readers.
    If you are bang on center of mass no matter the direction, the shape, the medium etc.. you'll always be perpendicular, or give a short push (say 10^-43: sec Plancks time)
    If you are interested I can open a new thread, this issue has always fascinated me.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    Right, if you push a body with a force perpendicular to its direction of motion it will be deflected in a curve but no work is done.

    that we take this excursion into classical mechanics onto a separate thread, for the sake of other readers.
    If you are bang on center of mass no matter the direction, the shape, the medium etc.. you'll always be perpendicular, or give a short push (say 10^-43: sec Plancks time)
    If you are interested I can open a new thread, this issue has always fascinated me.
    Yes, why don't you? Then we may find more people contribute.

    This hydrogen atom thing is close to my own area of interest, but it seems we are almost the only players. I'm happy to continue the hydrogen atom discussion too, but if there are problems of classical mechanics to sort out first, let's do that on your new thread and then revert.
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    Quote Originally Posted by exchemist View Post
    This hydrogen atom thing is close to my own area of interest, but it seems we are almost the only players.
    There are a few spectators but your answers were so good I had nothing to add
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    Quote Originally Posted by PhDemon View Post
    Quote Originally Posted by exchemist View Post
    This hydrogen atom thing is close to my own area of interest, but it seems we are almost the only players.
    There are a few spectators but your answers were so good I had nothing to add
    You're too kind. Clearly RPW and PWA did a decent job then, back in 1972-6. Seriously though, good to know you are watching out for any blunders.
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    PWA was still going strong in my time ('94-'98), now sadly retired...
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    Exchemist, I'm just starting to explore chemistry and atomic physics and I must tell you this thread is fascinating!
    Your posts are inspiring me to imagine the configuration of electron shells in a whole new way. Thank you and the OP
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    Quote Originally Posted by RobinM View Post
    Exchemist, I'm just starting to explore chemistry and atomic physics and I must tell you this thread is fascinating!
    Your posts are inspiring me to imagine the configuration of electron shells in a whole new way. Thank you and the OP
    That makes it all worthwhile! As you can maybe tell, I found it mind-expanding 40 years ago at university and still do today.
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    Quote Originally Posted by exchemist View Post
    What this says is that the electron has no well-defined speed. That's important, as it is referring to the quantum indeterminacy that I previously mentioned. It has a speed alright (because it has kinetic energy), but it is not well-defined (i.e you can't find out what it is, as a function of time or position)..
    We'll sort out if an orbiting body accelerate in an onther thread.
    but here, is "no well defined" means we just that we cannot find out but it is constant or that spped varies? Because: if speed is variable you surely havethat "Maxwell problem" here, since the electron would be constantly accelerating and decelerating.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    What this says is that the electron has no well-defined speed. That's important, as it is referring to the quantum indeterminacy that I previously mentioned. It has a speed alright (because it has kinetic energy), but it is not well-defined (i.e you can't find out what it is, as a function of time or position)..
    We'll sort out if an orbiting body accelerate in an onther thread.
    but here, is "no well defined" means we just that we cannot find out but it is constant or that spped varies? Because: if speed is variable you surely havethat "Maxwell problem" here, since the electron would be constantly accelerating and decelerating.
    Indeed! If it was a classical particle, that would be so. But this is the thing about quantum indeterminacy. It is a tricky and mysterious area. I am not an expert on the various interpretations of QM but I'll stick my neck out and say the way I look at it.

    In the other thread we discuss the ability of individual electrons, fired one at a time through a slit to interfere with themselves to build up, dot by dot, the interference pattern we expect from waves. How the hell can that be? It seems to me it might be that the "underlying reality" of the electron's nature is that it is a kind of matter wave that explores space, but that it can only be detected in quantised units - which we call electrons.

    Some physicist may well jump in at this point and tell me I am speaking ex ano, in which case I'll have learnt something.

    But if this could be so, then maybe the "speed" of the electron in the orbital is "undefined" in the same mathematical sense as a quantity divided by zero - a meaningless thing.
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    Quote Originally Posted by exchemist View Post
    In the other thread we discuss the ability of individual electrons, fired one at a time through a slit to interfere with themselves to build up, dot by dot, the interference pattern .
    In that thread I ventured to guess that it may depend on phase, we'll see there.
    Now, about the H atom, I have discovered that my guess about orbit precessing was not ingenious nor wild, it is called in classical model 'gyromagnetic ratio'. I had never heard of it, physics is just a hobby, in some aspects I am an expert in others just a novice.

    I cannot figure out what it is in QM as you said that orbital angular momentum is 0 , is there a gyromagnetic ratio? Here Gyromagnetic ratio - Wikipedia, the free encyclopedia , they say g=2
    in classical physics one would expect the g-factor to be g=1. however in the framework of relativistic quantum mechanics, g= 2
    Last edited by whizkid; March 13th, 2014 at 08:22 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    In the other thread we discuss the ability of individual electrons, fired one at a time through a slit to interfere with themselves to build up, dot by dot, the interference pattern .
    In that thread I ventured to guess that it may depend on phase, we'll see there.
    Now, about the H atom, I have discovered that my guess about orbit precessing was not ingenious nor wild, it is called in classical model 'gyromagnetic ratio'. I had never heard of it, physics is just a hobby, in some aspects I am an expert in others just a novice.

    I cannot figure out what it is in QM as you said that orbital angular momentum is 0 , is there a gyromagnetic ratio? Here Gyromagnetic ratio - Wikipedia, the free encyclopedia , they say g=2
    in classical physics one would expect the g-factor to be g=1. however in the framework of relativistic quantum mechanics, g= 2
    OK, firstly, let's make sure we are not confusing the orbital magnetic moment with the spin magnetic moment. All electrons, being fermions, are spin 1/2 particles. So, being charged, they have an intrinsic magnetic moment due to their spin. Secondly, an electron in an orbital for which l, the azimuthal quantum number, is >0, will have an orbital angular momentum, giving rise to an orbital magnetic moment. Electrons in s orbitals, e.g. in the ground state of hydrogen, have l=0, i.e. no angular momentum and thus no orbital magnetic moment.

    My understanding is the "g factor" for electron spin is a "fudge factor" to account for what is experimentally observed and applies only to spin magnetic moment. A value of 2, i.e. double what is classically expected, was derived by Dirac, using, I think, relativistic considerations, but the actual value is not exactly 2, it is 2.0023. I gather from Atkins' book (which you have forced me to get off the shelf) that QED accounts for this extra bit, but you will need a proper physicist to tell you how.

    N.B. "Magnetogyric ratio", as I was taught to call it, appears to be synonymous with the "gyromagnetic ratio" as you and Wiki call it, i.e. γ = -e/2m. This is not the "g factor". The g factor is as I say the fudge factor that gives the right answer for spin magnetic moment.

    As to what you say about precession of a classical orbit, precession is a phenomenon, not the value of a physical constant, as γ is. Please clarify what you mean.
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    Quote Originally Posted by exchemist View Post
    N.B. "Magnetogyric ratio", as I was taught to call it, appears to be synonymous with the "gyromagnetic ratio" as you and Wiki call it, i.e. γ = -e/2m. This is not the "g factor". The g factor is as I say the fudge factor that gives the right answer for spin magnetic moment.

    As to what you say about precession of a classical orbit, precession is a phenomenon, not the value of a physical constant, as γ is. Please clarify what you mean.
    Wiki says:
    ..."In physics, the gyromagnetic ratio of a particle or system is the ratio of its magnetic dipole moment to its angular momentum, and it is often denoted by the symbol γ, gamma...."gyromagnetic ratio" is sometimes used[1] as a synonym for a different but closely related quantity, the g-factor. The g-factor, unlike the gyromagnetic ratio, is dimensionless. ...Any free system with a constant gyromagnetic ratio, such as a rigid system of charges, a nucleus, or an electron, when placed in an external magnetic field B(measured in teslas) that is not aligned with its magnetic moment, will precess at a frequency f (measured in hertz), that is proportional to the external field"

    :
    I was referring to that, what is not clear to me :
    - what is the direction of the precession, is it on the plane of the orbit or normal to it , or what?
    - what happens if the ratio is 1? is there no precession?
    - I studied in school that classical angular momentum of H1 (mvr) is "L" h/2pi , not 0 , 0 is the angular momentum quantum number "l".

    As to precession , consider a standing wave n -lambda(here n=8, I could not find a picture of n=1, but the principle is the same)
    File:Circular Standing Wave.gif - Wikipedia, the free encyclopedia
    if precession is in the plane of the orbit, then the picture you get when you probe the atom is a diffused cloud as in QM , and not an orbit.
    That was my guess that you defined ' ingenious'
    Have I made myself clear now?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    N.B. "Magnetogyric ratio", as I was taught to call it, appears to be synonymous with the "gyromagnetic ratio" as you and Wiki call it, i.e. γ = -e/2m. This is not the "g factor". The g factor is as I say the fudge factor that gives the right answer for spin magnetic moment.

    As to what you say about precession of a classical orbit, precession is a phenomenon, not the value of a physical constant, as γ is. Please clarify what you mean.
    Wiki says:
    ..."In physics, the gyromagnetic ratio of a particle or system is the ratio of its magnetic dipole moment to its angular momentum, and it is often denoted by the symbol γ, gamma...."gyromagnetic ratio" is sometimes used[1] as a synonym for a different but closely related quantity, the g-factor. The g-factor, unlike the gyromagnetic ratio, is dimensionless. ...Any free system with a constant gyromagnetic ratio, such as a rigid system of charges, a nucleus, or an electron, when placed in an external magnetic field B(measured in teslas) that is not aligned with its magnetic moment, will precess at a frequency f (measured in hertz), that is proportional to the external field"

    :
    I was referring to that, what is not clear to me :
    - what is the direction of the precession, is it on the plane of the orbit or normal to it , or what?
    - what happens if the ratio is 1? is there no precession?
    - I studied in school that classical angular momentum of H1 (mvr) is "L" h/2pi , not 0 , 0 is the angular momentum quantum number "l".

    As to precession , consider a standing wave n -lambda(here n=8, I could not find a picture of n=1, but the principle is the same)
    File:Circular Standing Wave.gif - Wikipedia, the free encyclopedia
    if precession is in the plane of the orbit, then the picture you get when you probe the atom is a diffused cloud as in QM , and not an orbit.
    That was my guess that you defined ' ingenious'
    Have I made myself clear now?
    Yes, you've made yourself clear now, but no, it isn't the same as what you had been suggesting. What this is all about the precession of magnetic dipoles in the presence of an external, repeat, external, magnetic field.

    Your idea was not that, it was the idea that an electron in a classical orbit might have that orbit precess or something (it was I who attached the label precession to your description) as an putative explanation for how a 2D orbit could produce a 3D "probability cloud". So clearly it would have had to work in general, not just when an external field was applied. If you have a model in which the electron cloud should collapse to a classical orbit when you turn off an exterior laboratory magnet, your model has, shall we say, some way to go.

    The precession is of the magnetic dipole, which is aligned with the angular momentum vector. As you may know, in classical mechanics that vector projects at right angles to the plane of the motion, along the orbital axis. So it is like a precessing spinning top or gyroscope. So I afraid I can't follow what you mean by precession "in the plane of the orbit". Precession is motion about an axis.
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    Quote Originally Posted by exchemist View Post
    Your idea was not that, it was the idea that an electron in a classical orbit might have that orbit precess or something (it was I who attached the label precession to your description) as an putative explanation for how a 2D orbit could produce a 3D "probability cloud".
    Yes my idea was rotating in a direction normal to the orbit, but also a rotation in the plane of the orbit, I suppose just a change of phase, would have a similar effect, when you are scanning an atom.
    As you know the vector of the mechanical angular momentum points in a direction opposite to the one of the orbital magnetic momentum, I thought that the unbalance of the two vectors (or the interaction with spin angular momentum) would cause a rotation of the plane of orbit, is this impossible?

    I found an animation of what I suppose is a good picture od a standing wave in H1 (1s) orbit.
    As the picture changes 10^16 times a second, I think we get a probability cloud anyway, the same as a stone rotating on a string will disappear at a certain speed.
    If there is any reason in the world that mahes it rotate in the normal direction than the result would be even more striking. What is wrong with my guess?

    Thanks for your help.

    P.S I found out that the radius of H is 79 pm, and in water 32 pm
    Last edited by whizkid; March 15th, 2014 at 04:26 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    Your idea was not that, it was the idea that an electron in a classical orbit might have that orbit precess or something (it was I who attached the label precession to your description) as an putative explanation for how a 2D orbit could produce a 3D "probability cloud".
    Yes my idea was rotating in a direction normal to the orbit, but also a rotation in the plane of the orbit, I suppose just a change of phase, would have a similar effect, when you are scanning an atom.
    As you know the vector of the mechanical angular momentum points in a direction opposite to the one of the orbital magnetic momentum, I thought that the unbalance of the two vectors (or the interaction with spin angular momentum) would cause a rotation of the plane of orbit, is this impossible?

    I found an animation of what I suppose is a good picture od a standing wave in H1 (1s) orbit.
    As the picture changes 10^16 times a second, I think we get a probability cloud anyway, the same as a stone rotating on a string will disappear at a certain speed.
    If there is any reason in the world that mahes it rotate in the normal direction than the result would be even more striking. What is wrong with my guess?

    Thanks for your help.

    P.S I found out that the radius of H is 79 pm, and in water 32 pm
    OK, I don't follow much of this, I have to confess.

    I'm not sure what you mean about the "mechanical" angular momentum vector pointing in the opposite direction to the "orbital" angular momentum. Can you provide a reference to help me understand what you are thinking of?

    Secondly, the last sentence of your main para, about rotation in the "normal" direction, is very obscure. What is your "guess"? Can you explain more fully what you have in mind?

    Lastly, what do you mean about the radius of H? Its covalent radius is of the order of 30pm +/-5pm or so. This is from averaging interatomic distances when it forms various compounds. Where do you get 79pm from?
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    Quote Originally Posted by exchemist View Post
    I'm not sure what you mean about the "mechanical" angular momentum vector pointing in the opposite direction

    Secondly, the last sentence of your main para, about rotation in the "normal" direction, is very obscure.
    Where do you get 79pm from?
    wiki:
    According to the modern, quantum-mechanical understanding of the hydrogen atom, the average distance −its expectation value− between electron and proton is ≈1.5a0, (1.5*52.9= 79)
    ,Interactive Periodic Table: Hydrogen

    i
    F it is so, it fits the animation where the oscillation is half the radius, so the cloud would extend from 26 to 79 pm
    Last edited by whizkid; March 16th, 2014 at 02:30 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    I'm not sure what you mean about the "mechanical" angular momentum vector pointing in the opposite direction

    Secondly, the last sentence of your main para, about rotation in the "normal" direction, is very obscure.
    Where do you get 79pm from?
    wiki:
    According to the modern, quantum-mechanical understanding of the hydrogen atom, the average distance −its expectation value− between electron and proton is ≈1.5a0, (1.5*52.9= 79)
    ,Interactive Periodic Table: Hydrogen

    i
    F it is so, it fits the animation where the oscillation is half the radius, so the cloud would extend from 26 to 79 pm
    Ah I see. But of course the expectation value of the distance doesn't define the atomic radius in any commonly used sense. And in fact, the cloud in s orbitals actually goes right up to the nucleus. It doesn't stop at 26pm.
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    Quote Originally Posted by exchemist View Post
    the cloud in s orbitals actually goes right up to the nucleus. It doesn't stop at 26pm.
    All right, exchemist, I thank you for the invaluable help, I hope you know a link where I can find details of the H1-cloud and how they get its picture.

    I have some questions about the H-atom fine structure, I suppose it's better to start a fresh thread.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    the cloud in s orbitals actually goes right up to the nucleus. It doesn't stop at 26pm.
    All right, exchemist, I thank you for the invaluable help, I hope you know a link where I can find details of the H1-cloud and how they get its picture.

    I have some questions about the H-atom fine structure, I suppose it's better to start a fresh thread.
    It's all here, at your very fingertips, you know, including pictures: Hydrogen atom - Wikipedia, the free encyclopedia

    I see you have started a new one for the fine structure questions. I'll leave the physicists to run with that one for now.

    If you ever get round to a thread on your problem with motion in a circle, I'll be interested.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by whizkid View Post
    the H1-cloud and how they get its picture..
    I'll leave the physicists to run with that one for now.

    If you ever get round to a thread on your problem with motion in a circle, I'll be interested.
    I am looking for details of the technique used to get the picture of the atom

    - I hope you can join soon the new discussion

    -maybe a new thread is not necessary, just think you are out in space on an asteroid/planet,
    ...if you push a body with a force perpendicular to its direction of motion it will be deflected in a curve but no work is done.
    if it took no work, you could push a whole planet around just jumping in a normal direction?
    Last edited by whizkid; March 17th, 2014 at 10:02 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by whizkid View Post
    the H1-cloud and how they get its picture..
    I'll leave the physicists to run with that one for now.

    If you ever get round to a thread on your problem with motion in a circle, I'll be interested.
    I am looking for details of the technique used to get the picture of the atom

    - I hope you can join soon the new discussion

    -maybe a new thread is not necessary, just think you are out in space on an asteroid/planet,
    Rigif you push a body with a force perpendicular to its direction of motion it will be deflected in a curve but no work is done.
    if it took no work, you could push a whole planet around just jumping in a normal direction?
    If you start a new thread on this I'll have a go.
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    Quote Originally Posted by exchemist View Post
    the cloud in s orbitals actually goes right up to the nucleus. It doesn't stop at 26pm.
    Hi, exchemist, I thought this picture might be of interest
    The First Image Ever of a Hydrogen Atom's Orbital Structure

    A
    n additional question why is transition in H in the original link (hyperphysics) from 3s to 2p and not to 2s?
    Last edited by whizkid; March 21st, 2014 at 01:45 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    the cloud in s orbitals actually goes right up to the nucleus. It doesn't stop at 26pm.
    Hi, exchemist, I thought this picture might be of interest
    The First Image Ever of a Hydrogen Atom's Orbital Structure

    A
    n additional question why is transition in H in the original link (hyperphysics) from 3s to 2p and not to 2s?
    Yes indeed. It seems to have 2 spherical nodes and no angular ones, which would make it a 3s orbital, I think. Though the drawing they make looks a bit wrong, so I'm not sure.

    As to why the transition is from 3s to 2p and not 2s, this is due to a very important principle in spectroscopy called the "selection rules". These were originally empirical observations that spectral lines only seemed to arise from transitions in which l, the azimuthal quantum number, changed by 1. So an s - s transition, or a p-p one, is apparently "forbidden".

    The reason why is due to something I mentioned earlier, which is that (most) electronic transitions occur by a coupling of the electric vector of the photon (which is an oscillating dipolar field and is thus spatially antisymmetric) with the orbitals of the atom. This has the effect that a photon interacting with a symmetrical orbital (such as an s or d orbital) must turn it into an antisymmetric one (such as a p or f one) , or conversely, if it interacts with an antisymmetric orbital it turns it into a symmetric one.

    Once more, you will note this is a direct consequence of the wave nature of matter - and not something that the Bohr model could predict.

    There are other ways for orbitals to couple with the EM field and these allow spectra lines that disobey this rule to appear, albeit weakly. Furthermore, "l" is itself not always "well-defined". In multi-electron atoms, the angular momenta and spin couple to each other in various complex ways and, due to quantum indeterminacy, there are limits on what properties remain "well defined" in various scenarios. These effects give rise to a change in the effective selection rules, depending on what is going on.

    But that is another fascinating chapter in the story………....

    Note added later: I've just realised there is another way to look at this, which is that photons have spin angular momentum and so there has to be a change of one unit of angular momentum when they are emitted or absorbed. So l has to change by +/- 1. But again this rule can be broken in some circumstances………….
    Last edited by exchemist; March 21st, 2014 at 05:26 AM. Reason: footnote added; typos
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    Quote Originally Posted by exchemist View Post
    As to why the transition is from 3s to 2p and not 2s, this is due to a very important principle in spectroscopy called the "selection rules". These were originally empirical observations that spectral lines only seemed to arise from transitions in which l, the azimuthal quantum number, changed by 1. So an s - s transition, or a p-p one, is apparently "forbidden".
    ….
    Thanks, is the 2p a circular or elliptical orbit?
    Why in transitions do we consider only the value of B of the orbit od arrival? during the transition the electron is influenced by a varying magnetic field, why don't we integrate?

    I read that in the yellow emission of sodium the predicted B field is of over 18 Tesla, how can you ever get such a strong magnetic field in a 3s orbit?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    As to why the transition is from 3s to 2p and not 2s, this is due to a very important principle in spectroscopy called the "selection rules". These were originally empirical observations that spectral lines only seemed to arise from transitions in which l, the azimuthal quantum number, changed by 1. So an s - s transition, or a p-p one, is apparently "forbidden".
    ….
    Thanks, is the 2p a circular or elliptical orbit?
    Why in transitions do we consider only the value of B of the orbit od arrival? during the transition the electron is influenced by a varying magnetic field, why don't we integrate?

    I read that in the yellow emission of sodium the predicted B field is of over 18 Tesla, how can you ever get such a strong magnetic field in a 3s orbit?
    I think you need to read up some of the references I have already given you, and re-read some of our previous correspondence.

    p orbitals are dumbell-shaped, with a pair of lobes either side of the nucleus, one of +ve phase and one of -ve. I do not understand why, after all we've been through, you are still banging on about circular or eliptical "orbits". THEY ARE NOT ORBITS.

    Your second question makes no sense, in relation to what I have told you. I told you transitions occur due to the electric vector of the radiation coupling to the electron's standing wave. Magnetism did not enter the discussion at any stage. I do not understand why you keep trying to calculate magnetic field strength.

    I continue to be very happy to discuss all this further, but would ask that you please try to take in what I am saying to you, when you formulate your replies.
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    Quote Originally Posted by exchemist View Post
    . Magnetism did not enter the discussion at any stage. I do not understand why you keep trying to calculate magnetic field strength. .
    I am following the classical model as a template and trying to learn how QM deals with the concepts.
    Does QM not account for electron spin and magnetic moment? How do you deal with the fact that the fine structure split in sodium is a huge 0.0021 eV? where does that energy come from?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    . Magnetism did not enter the discussion at any stage. I do not understand why you keep trying to calculate magnetic field strength. .
    I am following the classical model as a template and trying to learn how QM deals with the concepts.
    Does QM not account for electron spin and magnetic moment? How do you deal with the fact that the fine structure split in sodium is a huge 0.0021 eV? where does that energy come from?
    OK, I now understand what you re doing. But please understand my answers will always be in terms of QM, not the Bohr model.

    Yes QM does account for spin and magnetic moment, as we discussed that some time ago. But that is not what we were talking about just now, which was the selection rules for transitions between orbitals, accompanied by absorption or emission of EM radiation. This process is not (usually) a magnetic phenomenon.

    Now, re sodium, you are evidently talking about the splitting between the doublet lines in the spectrum. The pair of D lines is (in emission) caused by transitions from 3p to 3s. The splitting of what one might think should be one line due to this transition indeed arises due to a magnetic effect. This is that, in the upper, p, state, there are two possible alignments of the orbital angular momentum with the spin angular momentum of the electron. If the alignment is opposed you have the 3p 1/2 state, and if they are aligned you have the 3p 3/2 state, the latter being of slightly higher energy, due to the stronger magnetic field. So you two energy states, not one, both of which can undergo transition to the 3s ground state, leading to a pair of lines. This magnetic interaction between spin and orbital angular momenta is called spin-orbit coupling and can get quite hairy in multi-electron atoms, as there will be numerous combinations of alignments, all with slightly differing energy levels.

    The sodium case is described here (if you have not already read it): Sodium doublet with and without magnetic field.

    I'm not sure I follow what you mean by "Where does that energy come from?". The state in which the 2 magnetic moments are aligned - leading to a stronger net field - is of slightly higher energy than that in which they are opposed, which seems reasonable, doesn't it?

    By the way here is another description: http://chemistry.illinoisstate.edu/s...tomicterms.pdf

    which also explains the term symbols for the states, just in case this is where we end up going next.
    Last edited by exchemist; March 24th, 2014 at 07:15 AM. Reason: phrase added for clarity
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    Quote Originally Posted by exchemist View Post
    The sodium case is described here (if you have not already read it): Sodium doublet with and without magnetic field.

    I'm not sure I follow what you mean by "Where does that energy come from?". The state in which the 2 magnetic moments are aligned - leading to a stronger net field - is of slightly higher energy than that in which they are opposed, which seems reasonable, doesn't it?
    That is the site I was referring to, the energy of the split , as you can see, is 0.0021 eV which is a lot of energy if you reckon it's coming only from the orientation of the spin , in classical model it is explained by an interaction with a huge B field of 18 T, how does QM explain that, is there a magnetic field of the orbital or not, what are its parameters, what is the formula that allows to calculate the value of B in various n levels etc....?
    B = 18 TeslaThis is a very large magnetic field by laboratory standards. Large magnets with dimensions over a meter, used for NMR and ESR experiments, have magnetic fields on the order of a Tesla.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    The sodium case is described here (if you have not already read it): Sodium doublet with and without magnetic field.

    I'm not sure I follow what you mean by "Where does that energy come from?". The state in which the 2 magnetic moments are aligned - leading to a stronger net field - is of slightly higher energy than that in which they are opposed, which seems reasonable, doesn't it?
    That is the site I was referring to, the energy of the split , as you can see, is 0.0021 eV which is a lot of energy if you reckon it's coming only from the orientation of the spin , in classical model it is explained by an interaction with a huge B field of 18 T, how does QM explain that, is there a magnetic field of the orbital or not, what are its parameters, what is the formula that allows to calculate the value of B in various n levels etc....?
    B = 18 TeslaThis is a very large magnetic field by laboratory standards. Large magnets with dimensions over a meter, used for NMR and ESR experiments, have magnetic fields on the order of a Tesla.
    I think the key phrase is "by laboratory standards". Don't forget, we are speaking of tiny charges exerting a field in concomitantly tiny areas of space. All you need is the region of space to be a bit tinier than you might (arbitrarily) think, in relation to the charge, and hey presto you have a field strength that looks "huge" by macroscopic laboratory standards. The magnitude of the splitting of the two energy levels is extremely modest by the standards of electron energy levels in an atom.

    Re whether or not there is a magnetic field of the orbital or not, sure there is, if it has angular momentum and consequently an orbital magnetic moment (the formula for which I've given you earlier, here: http://www.thescienceforum.com/physi...ost536797.html )

    I have found in Atkins a formula for the orbital magnetic field, which is B = -(1/c²r)(dV/dr)l, in which l is the angular momentum and V(r) is the electric potential constraining the electron.

    But, Health Warning, I have to tell you this is all very rusty knowledge on my part. If you want to go further into the maths I'd advise you to do you own reading, rather than relying on my recollections. You can find a lot of this stuff on the web. After a gap of 40 years I am a lot stronger on the concepts than on the maths.
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    Quote Originally Posted by exchemist View Post
    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Last edited by molecool; March 25th, 2014 at 12:50 AM.
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    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
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    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
    Eh?? NO!

    What I said was that the spinning and orbital motion of the electron, and the motion of electrons in a wire, are BOTH movement. The MOVEMENT in both cases gives rise to a magnetic field. So although they are certainly different forms of motion, both give rise to magnetism in the same way.

    As to whether electron spin and orbital motion are the same thing, no they are not. Both give rise to magnetism, due to motion of an electric charge, but spin is motion like the Earth rotating on its axis once a day, while orbital motion is like the Earth going round the sun once a year.
    Last edited by exchemist; March 25th, 2014 at 05:53 AM. Reason: reworded
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    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
    Eh?? NO!

    What I said was that the spinning and orbital motion of the electron, and the motion of electrons in a wire, are BOTH movement. The MOVEMENT in both cases gives rise to a magnetic field. So although they are certainly different forms of motion, both give rise to magnetism in the same way.

    As to whether electron spin and orbital motion are the same thing, no they are not. Both give rise to magnetism, due to motion of an electric charge, but spin is motion like the Earth rotating on its axis once a day, while orbital motion is like the Earth going round the sun once a year.
    Haha yeah that's what I was trying to say lol sorry for the miscommunication (I was trying to say different movements). Anyway, the electron's intrinsic spin gives rise to its own magnetic field only, while its orbit gives rise to the whole atom's magnetic field is that right? And is the direction of orbit same as the direction of spin?

    As a side question, the electron pairs in an atom often tend to repel each other too, right? (VSEPR theory)

    Once again thanks for your great explanation!
    Last edited by molecool; March 25th, 2014 at 08:49 AM.
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    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
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    Quote Originally Posted by exchemist View Post
    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
    Eh?? NO!

    What I said was that the spinning and orbital motion of the electron, and the motion of electrons in a wire, are BOTH movement. The MOVEMENT in both cases gives rise to a magnetic field. So although they are certainly different forms of motion, both give rise to magnetism in the same way.

    As to whether electron spin and orbital motion are the same thing, no they are not. Both give rise to magnetism, due to motion of an electric charge, but spin is motion like the Earth rotating on its axis once a day, while orbital motion is like the Earth going round the sun once a year.
    Haha yeah that's what I was trying to say lol sorry for the miscommunication (I was trying to say different movements). Anyway, the electron's intrinsic spin gives rise to its own magnetic field only, while its orbit gives rise to the whole atom's magnetic field is that right? I'm a bit confused at this part.

    As a side question, the electron pairs in an atom often tend to repel each other right? (VSEPR theory)
    The overall magnetic moment of the atom is the resultant (i.e. vector sum) of the magnetic moments due to spin and to orbital contributions from each electron. This is what the discussion about spin-orbit coupling was getting into. There are several different possible orientations of each with respect to the others, so you can get a variety of different values, all of which may in principle have slightly different energies.

    By introducing VSEPR, you are really jumping to chemistry. VSEPR is applicable to covalently bonded atoms, in which the symmetry of an atom's environment has been reduced, by the presence of the atoms it is bonding with. All the discussion about the hydrogen atom so far has considered isolated atoms, i.e. atoms in a spherically symmetrical environment. (We are in the Physics section.)

    The phenomenon of "hybridisation", formation of "lone pairs", and all that stuff we chemists use in predicting molecular shapes and reactivity - including VSEPR - represent the lowest energy configurations of electron orbitals in these lower symmetry environments. Obviously, trying to solve Schroedinger's equation for systems with several nuclei, disposed in a geometrical pattern around the atom of interest, is going to lead to a set of solutions (=standing wave, resonant states for the electrons) that looks rather different from that of a simple isolated atom.

    According to my understanding, you won't find "lone pairs" of mutually repelling electrons sticking out of an isolated oxygen atom for example - though you will when it is bonded to other atoms.

    P.S. Symmetry is very important indeed in understanding chemical bonding, as it allows the arrangement and relative energy levels of molecular orbitals to be derived in a qualitative or semi-quantitative way. (If we had to try to solve the wave equation every time, we'd never get anywhere - it's far too difficult.) You can get a taste of it here:
    Molecular symmetry - Wikipedia, the free encyclopedia

    But if we go into this at all it needs to be in the Chemistry section or the Mods will be after us.
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    Quote Originally Posted by exchemist View Post
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    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
    Eh?? NO!

    What I said was that the spinning and orbital motion of the electron, and the motion of electrons in a wire, are BOTH movement. The MOVEMENT in both cases gives rise to a magnetic field. So although they are certainly different forms of motion, both give rise to magnetism in the same way.

    As to whether electron spin and orbital motion are the same thing, no they are not. Both give rise to magnetism, due to motion of an electric charge, but spin is motion like the Earth rotating on its axis once a day, while orbital motion is like the Earth going round the sun once a year.
    Haha yeah that's what I was trying to say lol sorry for the miscommunication (I was trying to say different movements). Anyway, the electron's intrinsic spin gives rise to its own magnetic field only, while its orbit gives rise to the whole atom's magnetic field is that right? I'm a bit confused at this part.

    As a side question, the electron pairs in an atom often tend to repel each other right? (VSEPR theory)
    The overall magnetic moment of the atom is the resultant (i.e. vector sum) of the magnetic moments due to spin and to orbital contributions from each electron. This is what the discussion about spin-orbit coupling was getting into. There are several different possible orientations of each with respect to the others, so you can get a variety of different values, all of which may in principle have slightly different energies.

    By introducing VSEPR, you are really jumping to chemistry. VSEPR is applicable to covalently bonded atoms, in which the symmetry of an atom's environment has been reduced, by the presence of the atoms it is bonding with. All the discussion about the hydrogen atom so far has considered isolated atoms, i.e. atoms in a spherically symmetrical environment. (We are in the Physics section.)

    The phenomenon of "hybridisation", formation of "lone pairs", and all that stuff we chemists use in predicting molecular shapes and reactivity - including VSEPR - represent the lowest energy configurations of electron orbitals in these lower symmetry environments. Obviously, trying to solve Schroedinger's equation for systems with several nuclei, disposed in a geometrical pattern around the atom of interest, is going to lead to a set of solutions (=standing wave, resonant states for the electrons) that looks rather different from that of a simple isolated atom.

    According to my understanding, you won't find "lone pairs" of mutually repelling electrons sticking out of an isolated oxygen atom for example - though you will when it is bonded to other atoms.

    P.S. Symmetry is very important indeed in understanding chemical bonding, as it allows the arrangement and relative energy levels of molecular orbitals to be derived in a qualitative or semi-quantitative way. (If we had to try to solve the wave equation every time, we'd never get anywhere - it's far too difficult.) You can get a taste of it here:
    Molecular symmetry - Wikipedia, the free encyclopedia

    But if we go into this at all it needs to be in the Chemistry section or the Mods will be after us.
    Ah so that means the direction of electron spin is same as the direction of orbit? P.S. Thanks for your time on explaining VSEPR! I'll look into that soon. (Though I'm a bit disappointed that I couldn't fully understand anything just by reading through wikipedia alone.)
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    Quote Originally Posted by molecool View Post
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    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
    Eh?? NO!

    What I said was that the spinning and orbital motion of the electron, and the motion of electrons in a wire, are BOTH movement. The MOVEMENT in both cases gives rise to a magnetic field. So although they are certainly different forms of motion, both give rise to magnetism in the same way.

    As to whether electron spin and orbital motion are the same thing, no they are not. Both give rise to magnetism, due to motion of an electric charge, but spin is motion like the Earth rotating on its axis once a day, while orbital motion is like the Earth going round the sun once a year.
    Haha yeah that's what I was trying to say lol sorry for the miscommunication (I was trying to say different movements). Anyway, the electron's intrinsic spin gives rise to its own magnetic field only, while its orbit gives rise to the whole atom's magnetic field is that right? I'm a bit confused at this part.

    As a side question, the electron pairs in an atom often tend to repel each other right? (VSEPR theory)
    The overall magnetic moment of the atom is the resultant (i.e. vector sum) of the magnetic moments due to spin and to orbital contributions from each electron. This is what the discussion about spin-orbit coupling was getting into. There are several different possible orientations of each with respect to the others, so you can get a variety of different values, all of which may in principle have slightly different energies.

    By introducing VSEPR, you are really jumping to chemistry. VSEPR is applicable to covalently bonded atoms, in which the symmetry of an atom's environment has been reduced, by the presence of the atoms it is bonding with. All the discussion about the hydrogen atom so far has considered isolated atoms, i.e. atoms in a spherically symmetrical environment. (We are in the Physics section.)

    The phenomenon of "hybridisation", formation of "lone pairs", and all that stuff we chemists use in predicting molecular shapes and reactivity - including VSEPR - represent the lowest energy configurations of electron orbitals in these lower symmetry environments. Obviously, trying to solve Schroedinger's equation for systems with several nuclei, disposed in a geometrical pattern around the atom of interest, is going to lead to a set of solutions (=standing wave, resonant states for the electrons) that looks rather different from that of a simple isolated atom.

    According to my understanding, you won't find "lone pairs" of mutually repelling electrons sticking out of an isolated oxygen atom for example - though you will when it is bonded to other atoms.

    P.S. Symmetry is very important indeed in understanding chemical bonding, as it allows the arrangement and relative energy levels of molecular orbitals to be derived in a qualitative or semi-quantitative way. (If we had to try to solve the wave equation every time, we'd never get anywhere - it's far too difficult.) You can get a taste of it here:
    Molecular symmetry - Wikipedia, the free encyclopedia

    But if we go into this at all it needs to be in the Chemistry section or the Mods will be after us.
    Ah so that means the direction of electron spin is same as the direction of orbit? P.S. Thanks for your time on explaining VSEPR! I'll look into that soon. (Though I'm a bit disappointed that I couldn't fully understand anything just by reading through wikipedia alone.)
    Well hold on, what I said was that the total magnetic moment is a vector sum of spin and orbital contributions and that these could take a number of relative orientations. Let's think about this a little. Take sodium, to keep it simple i.e. just one valence shell electron and let's think of the electron in the 3p orbital, where it has one unit of orbital angular momentum and therefore has spin and orbital contributions to the magnetic moment. The spin moment can either align with the orbital moment, adding to it, or can oppose it, subtracting from it. In the first case the direction of spin is the same as the orbital motion. In the second case it is opposite. That's what we mean by the moment pointing in the opposite direction: the direction of spin is reversed, so that what was the North pole becomes the South and vice versa.

    The reference I gave you was to symmetry in chemistry rather than VSEPR. I did this to try to give some insight into the QM way of looking at VSEPR, to link the two topics. VSEPR in itself is a fairly free-standing model of bonding and steric effects, which does not require much understanding of QM.
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    Quote Originally Posted by exchemist View Post
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    But indeed, if the electron is in an orbital with orbital angular momentum (azimuthal quantum number, l, >0, so for example a p, d or f orbital) it will have an orbital magnetic moment as well.
    Sorry for barging in randomly, but does that mean the magnetic field generated by an electric current in a wire is actually just the orbital (not spin) magnetic moment of the electrons flowing in the wire? I mean why doesn't the magnetic field generated by an electric current in a wire have a north pole and south pole?
    Haha. You need to look at it the other way round.

    A magnetic field is generated by any moving electric charge. So the source of the magnetism is the same for a "spinning" or pseudo-"orbiting" electron, and for electrons moving linearly in a wire. In both cases it is the movement that is responsible. (In fact, as we did discuss recently on another thread in this section, magnetism is a consequence of the way relativity affects how electric charge is felt by another charge in relative motion with respect to it. But that's another story).

    The magnetic field is at right angles to the direction of motion of the charge and is conventionally represented by closed loops ("field lines", lines of "magnetic flux", etc). So around a linear wire you get a circular loop, but if you wrap the wire round into a ring or coil, the loops become toroidal like a doughnut. In this configuration the field lines all come out of one end of the ring and re-enter at the other, creating what we call North and South poles. In the case of the electron, in both the spin and orbital cases, it is also moving in a ring, and so will give rise to a toroidal field, with a North and South pole.

    Which is exactly what we mean when we speak of a magnetic dipole.
    Ah, I see. So moving electric charge and electron spin/orbit are two different things. Thanks for clearing up my mind.

    But then again, are electron spin and orbit the same thing? Like, for example, if an electron is spinning left on its own axis, is its orbit around the nucleus going left too? This question's been stuck for quite some time in my mind but for some reason I can't seem to find the answer to it.
    Eh?? NO!

    What I said was that the spinning and orbital motion of the electron, and the motion of electrons in a wire, are BOTH movement. The MOVEMENT in both cases gives rise to a magnetic field. So although they are certainly different forms of motion, both give rise to magnetism in the same way.

    As to whether electron spin and orbital motion are the same thing, no they are not. Both give rise to magnetism, due to motion of an electric charge, but spin is motion like the Earth rotating on its axis once a day, while orbital motion is like the Earth going round the sun once a year.
    Haha yeah that's what I was trying to say lol sorry for the miscommunication (I was trying to say different movements). Anyway, the electron's intrinsic spin gives rise to its own magnetic field only, while its orbit gives rise to the whole atom's magnetic field is that right? I'm a bit confused at this part.

    As a side question, the electron pairs in an atom often tend to repel each other right? (VSEPR theory)
    The overall magnetic moment of the atom is the resultant (i.e. vector sum) of the magnetic moments due to spin and to orbital contributions from each electron. This is what the discussion about spin-orbit coupling was getting into. There are several different possible orientations of each with respect to the others, so you can get a variety of different values, all of which may in principle have slightly different energies.

    By introducing VSEPR, you are really jumping to chemistry. VSEPR is applicable to covalently bonded atoms, in which the symmetry of an atom's environment has been reduced, by the presence of the atoms it is bonding with. All the discussion about the hydrogen atom so far has considered isolated atoms, i.e. atoms in a spherically symmetrical environment. (We are in the Physics section.)

    The phenomenon of "hybridisation", formation of "lone pairs", and all that stuff we chemists use in predicting molecular shapes and reactivity - including VSEPR - represent the lowest energy configurations of electron orbitals in these lower symmetry environments. Obviously, trying to solve Schroedinger's equation for systems with several nuclei, disposed in a geometrical pattern around the atom of interest, is going to lead to a set of solutions (=standing wave, resonant states for the electrons) that looks rather different from that of a simple isolated atom.

    According to my understanding, you won't find "lone pairs" of mutually repelling electrons sticking out of an isolated oxygen atom for example - though you will when it is bonded to other atoms.

    P.S. Symmetry is very important indeed in understanding chemical bonding, as it allows the arrangement and relative energy levels of molecular orbitals to be derived in a qualitative or semi-quantitative way. (If we had to try to solve the wave equation every time, we'd never get anywhere - it's far too difficult.) You can get a taste of it here:
    Molecular symmetry - Wikipedia, the free encyclopedia

    But if we go into this at all it needs to be in the Chemistry section or the Mods will be after us.
    Ah so that means the direction of electron spin is same as the direction of orbit? P.S. Thanks for your time on explaining VSEPR! I'll look into that soon. (Though I'm a bit disappointed that I couldn't fully understand anything just by reading through wikipedia alone.)
    Well hold on, what I said was that the total magnetic moment is a vector sum of spin and orbital contributions and that these could take a number of relative orientations. Let's think about this a little. Take sodium, to keep it simple i.e. just one valence shell electron and let's think of the electron in the 3p orbital, where it has one unit of orbital angular momentum and therefore has spin and orbital contributions to the magnetic moment. The spin moment can either align with the orbital moment, adding to it, or can oppose it, subtracting from it. In the first case the direction of spin is the same as the orbital motion. In the second case it is opposite. That's what we mean by the moment pointing in the opposite direction: the direction of spin is reversed, so that what was the North pole becomes the South and vice versa.

    The reference I gave you was to symmetry in chemistry rather than VSEPR. I did this to try to give some insight into the QM way of looking at VSEPR, to link the two topics. VSEPR in itself is a fairly free-standing model of bonding and steric effects, which does not require much understanding of QM.
    Wait, I don't get it? When an electron is spinning in the opposite direction of its orbit movement the north pole becomes the south? Either way, they're still magnetic right?
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    Quote Originally Posted by molecool View Post
    Wait, I don't get it? When an electron is spinning in the opposite direction of its orbit movement the north pole becomes the south? Either way, they're still magnetic right?
    Please, molecool, learn to edit quoted text. Just keep the relevant bits, rather than quoting the whole preceding thread. Your current habit makes reading nearly iimpossible.
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    Quote Originally Posted by molecool View Post
    [
    Wait, I don't get it? When an electron is spinning in the opposite direction of its orbit movement the north pole becomes the south? Either way, they're still magnetic right?
    Well, yes. If you reverse the direction of current flow in an electromagnet, the North becomes the South and vice versa.

    Same with the direction of spin or orbital angular momentum of an electron. Reverse the direction of the motion and the associated magnetic field will be oriented in the opposite direction.

    That makes sense to you, doesn't it?
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    Quote Originally Posted by exchemist View Post
    Well, yes. If you reverse the direction of current flow in an electromagnet, the North becomes the South and vice versa.

    Same with the direction of spin or orbital angular momentum of an electron. Reverse the direction of the motion and the associated magnetic field will be oriented in the opposite direction.

    That makes sense to you, doesn't it?
    A little. That means a pair of electrons can spin in the opposite direction while orbiting in the same direction like this? http://www.coertvonk.com/wp-content/uploads/electron-orbiting3.png

    2) When an orbital magnetic field's pointing up while the electron's is pointing down, what will happen? (When the orbital movement's direction is opposite to the electron spin's direction)

    3) What direction of orbital movement gives a pointing up magnetic field anyway? (Like, for example an electron spinning left gives rise to a magnetic field that's pointing up)
    Last edited by molecool; March 26th, 2014 at 04:30 AM.
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    Quote Originally Posted by molecool View Post
    Quote Originally Posted by exchemist View Post
    Well, yes. If you reverse the direction of current flow in an electromagnet, the North becomes the South and vice versa.

    Same with the direction of spin or orbital angular momentum of an electron. Reverse the direction of the motion and the associated magnetic field will be oriented in the opposite direction.

    That makes sense to you, doesn't it?
    Oh, ok. That means a pair of electrons can spin in the opposite direction while orbiting in the same direction like this? http://www.coertvonk.com/wp-content/...-orbiting3.png
    Yup. And not only that, but if they occupy the same orbital, they must have opposed spin orientations

    This is because, being fermions, no two electrons are allowed to have exactly the same set of the 4 quantum numbers that define their state in the atom (Pauli Exclusion Principle: Pauli exclusion principle - Wikipedia, the free encyclopedia). These quantum numbers are n, l, m(l), m(s).

    - n tells you which main shell it is in,

    - l tells you what type of orbital it is(s, p, d, f etc) and consequently how much orbital angular momentum it has,

    - m(l) tells you which one, of the multiple sets of each type of orbital, it occupies (e.g. px, py or pz, corresponding to the orientation of the orbital angular momentum), and

    - m(s) tells you the orientation of its spin.

    If for example you were to have a pair of electrons in the same pz orbital, n, l and m(l) would be the same. So the only way to comply with the Exclusion Principle is for m(s) to be different. That means they must have opposed spin in order to do this.

    But if you had unpaired electrons, say, one in the pz and one in the px, m(l) would be different. Then the spin orientation can be as it likes, both up, one up and one down, or both down. Because these 3 possibilities all have slightly different energies, such a state is called a "triplet" state (It tends to show up as 3 closely spaced lines in the spectrum - a triplet of lines).
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    Ok thanks now I'm clear about electron spins. However still I'm a little confused 'bout the orbital magnetic moment. i.e. When an electron spins left it creates a magnetic field pointing up, so does that mean an electron orbiting to the left also creates a magnetic field pointing up?
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    Quote Originally Posted by molecool View Post
    Ok thanks now I'm clear about electron spins. However still I'm a little confused 'bout the orbital magnetic moment. i.e. When an electron spins left it creates a magnetic field pointing up, so does that mean an electron orbiting to the left also creates a magnetic field pointing up?
    Yes it would. In both cases you have a negative charge rotating in the same sense, so the magnetic field associated with this motion will be oriented the same in both cases.
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    Ah now it all makes sense. Still wondering why wikipedia doesn't mention this in the orbital magnetic dipole moment section though...

    Electron magnetic dipole moment - Wikipedia, the free encyclopedia

    I
    guess the magnetic moment of something ultimately depends on the electron spin.

    Thanks!!!
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    Quote Originally Posted by molecool View Post
    Ah now it all makes sense. Still wondering why wikipedia doesn't mention this in the orbital magnetic dipole moment section though...

    Electron magnetic dipole moment - Wikipedia, the free encyclopedia

    I
    guess the magnetic moment of something ultimately depends on the electron spin.

    Thanks!!!
    I expect the author considered it would be obvious to his or her intended readership. It is in any case implicit in the initial formula given in the Wiki article, which you may notice is said to apply to both spin and orbital motion.
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    Oh and one last thing, an electron spinning in a certain direction does not necessarily mean the electron is also orbiting in the same direction right? (Yes I know it's a stupid question)
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    Quote Originally Posted by molecool View Post
    Oh and one last thing, an electron spinning in a certain direction does not necessarily mean the electron is also orbiting in the same direction right? (Yes I know it's a stupid question)
    Indeed, obviously not, from what I was saying previously. The electron spin can EITHER align with, OR be opposed to, the orbital motion. In the sodium 3p 3/2 state the spin and orbit angular momentum are in the same direction, and in the 3p 1/2 state they are in opposite directions. (The latter combination has slightly lower energy.) Of course, an electron in an s orbital has NO angular momentum and NO orbital magnetic moment, so in s states all you have is the magnetic moment due to electron spin.
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    Quote Originally Posted by exchemist View Post
    The electron spin can EITHER align with, OR be opposed to, the orbital motion. In the sodium 3p 3/2 state the spin and orbit angular momentum are in the same direction, and in the 3p 1/2 state they are in opposite directions. (The latter combination has slightly lower energy.) Of course, an electron in an s orbital has NO angular momentum and NO orbital magnetic moment, so in s states all you have is the magnetic moment due to electron spin.
    Are you referring to QM or classical model, exchemist ? Could you explain why that?
    This wiki article (Orbital motion (quantum) - Wikipedia, the free encyclopedia) is a bit confuse, I understood that the value of orbital angular momentum is paradoxically the same in QM:

    "When quantum mechanics refers to the orbital angular momentum of an electron, it is generally referring to the spatial wave equation that represents the electron's motion around the nucleus of an atom. Electrons do not "orbit" the nucleus in the classical sense of angular momentum, however the mathematical representation of L = r × p still leads to the quantum mechanical version of angular momentum."

    but I do not understand how that can be possible, since you said that radius r and speed v are different in QM
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    Quote Originally Posted by exchemist View Post
    The electron spin can EITHER align with, OR be opposed to, the orbital motion. In the sodium 3p 3/2 state the spin and orbit angular momentum are in the same direction, and in the 3p 1/2 state they are in opposite directions..
    Another question: when the spin changes direction (up => down or viceversa) does it do that instantaneously, or it rotates by 180°?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    The electron spin can EITHER align with, OR be opposed to, the orbital motion. In the sodium 3p 3/2 state the spin and orbit angular momentum are in the same direction, and in the 3p 1/2 state they are in opposite directions. (The latter combination has slightly lower energy.) Of course, an electron in an s orbital has NO angular momentum and NO orbital magnetic moment, so in s states all you have is the magnetic moment due to electron spin.
    Are you referring to QM or classical model, exchemist ? Could you explain why that?
    This wiki article (Orbital motion (quantum) - Wikipedia, the free encyclopedia) is a bit confuse, I understood that the value of orbital angular momentum is paradoxically the same in QM:

    "When quantum mechanics refers to the orbital angular momentum of an electron, it is generally referring to the spatial wave equation that represents the electron's motion around the nucleus of an atom. Electrons do not "orbit" the nucleus in the classical sense of angular momentum, however the mathematical representation of L = r × p still leads to the quantum mechanical version of angular momentum."

    but I do not understand how that can be possible, since you said that radius r and speed v are different in QM
    OK. I am referring to the QM model of the atom, as I have done throughout this thread. I have referred to the electron's "orbital motion" rather than calling it an "orbit". I see the extract you quote also speaks of "motion around the nucleus" - and then goes on to stress that one should not think of this as a classical orbiting. So we are saying the same thing.

    When we discussed the presumed variation of the electron's energy between potential and kinetic, and I compared this with harmonic motion, this was in the context of the hydrogen atom where, to keep things manageable, I was really thinking of a way to picture what is going on in the ground state, 1s. In the 1s, remember, the electron has zero angular momentum. So I invited you to picture the electron "swinging", a bit like a pendulum, from a distance out from the nucleus (high potential energy but low KE) to a distance very close to it (low potential, high KE) - essentially a radial motion rather than a circumferential one - but in a way that is not well-defined, in that we cannot know the momentum and position simultaneously (indeterminacy).

    When we come to orbitals in which l>0 (p, d, f…), this radial motion continues, but in addition there is now also a circumferential component of the motion. This is reflected in the fact that in the p, d, f…. orbital there are nodal planes (planes of zero electron density) through the nucleus, so that, if one were to go round the atom in a complete circle, at a given radius, one would encounter alternating phases of +ve and -ve amplitude of (the square root of) probability. In other words, the "waves" are now "circumferential" waves as well as "radial" ones. (These are just the standing waves you get with a resonating sphere - spherical harmonics. See this Wiki article for nice pictures of them:Spherical harmonics - Wikipedia, the free encyclopedia )

    None of this is saying that there is any defined speed of circulation of the electron round the nucleus - the angular "speed" is not "well defined", any more than the radial motion. But this indeterminacy notwithstanding, there is a well-defined angular momentum (as measured by the orbital magnetic moment for example).

    So once more, I return to my theme: picturing what is happening in the QM world requires one to think in terms of waves, rather than particles.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    The electron spin can EITHER align with, OR be opposed to, the orbital motion. In the sodium 3p 3/2 state the spin and orbit angular momentum are in the same direction, and in the 3p 1/2 state they are in opposite directions..
    Another question: when the spin changes direction (up => down or viceversa) does it do that instantaneously, or it rotates by 180°?
    It is best to think of it as being turned it upside down while spinning, rather than being stopped and being made to spin in the opposite direction. This is because the spin is intrinsic to the particle. There is no such thing as an electron that is NOT spinning. In Electron Spin Resonance, electron spins are caused to align with an external magnetic field, which makes the "spin up" and "spin down" states separate in energy and then microwave radiation can induce transitions between the two states. The radiation is interacting with the electrons and tipping them over from spin down to spin up and vice versa. (At least that is the pseudo-classical physical picture of what happens: the QM model does it all with maths as you can imagine.) More about ESR here: Electron Spin Resonance

    What happens during these transitions will be described by the time-dependent form of Schroedinger's equation, which you may recall also accounts for the process by which electrons emit or absorb EM radiation when making transitions between orbitals. So yes, it takes a finite (though very short) time to flip over. No magic involved!
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    Quote Originally Posted by exchemist View Post
    It is best to think of it as being turned it upside down while spinning, rather than being stopped and being made to spin in the opposite direction. This is because the spin is intrinsic to the particle. There is no such thing as an electron that is NOT spinning.!
    So the spin takes ALL directions 0-180° while flipping over to the opposite direction. Why the Stern_gerlach doesn't catch a single electron in di fferent direction? the B field makes them topple one way or the other?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    It is best to think of it as being turned it upside down while spinning, rather than being stopped and being made to spin in the opposite direction. This is because the spin is intrinsic to the particle. There is no such thing as an electron that is NOT spinning.!
    So the spin takes ALL directions 0-180° while flipping over to the opposite direction. Why the Stern_gerlach doesn't catch a single electron in di fferent direction? the B field makes them topple one way or the other?
    Yeah but think about it: it is only "toppling" while in the act of emitting or absorbing radiation. There is no intermediate state that can exist on its own - intermediate positions are only passed through while EM radiation is coupling to the wavefunction and altering the state. So it cannot emit or absorb another photon: that takes the same length of time to take place as the change it is already going through.
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    Quote Originally Posted by exchemist View Post
    There is no intermediate state that can exist on its own - intermediate positions are only passed through
    It's like a coin standing on its side that topples down one way or other and then 'sits' still only at 0 or 180°.
    So if the Stern-Gerlach machine catches it when at , say, 45 or 135°, flattens it either to 0 or 180° which direction is nearer, is this the way it works?
    Now what is odd : they ran H atoms in that machine and they had the same result they got with silver, 50% sipin up/down, right?
    But we know that in 1s state the electron can only spin in one direction (as the spin antiparallel to proton spin is allowed only in interstellar space http://en.wikipedia.org/wiki/Hydrogen_line ) so , why the machine found two different spins?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    There is no intermediate state that can exist on its own - intermediate positions are only passed through
    It's like a coin standing on its side that topples down one way or other and then 'sits' still only at 0 or 180°.
    So if the Stern-Gerlach machine catches it when at , say, 45 or 135°, flattens it either to 0 or 180° which direction is nearer, is this the way it works?
    Now what is odd : they ran H atoms in that machine and they had the same result they got with silver, 50% sipin up/down, right?
    But we know that in 1s state the electron can only spin in one direction (as the spin antiparallel to proton spin is allowed only in interstellar space Hydrogen line - Wikipedia, the free encyclopedia ) so , why the machine found two different spins?
    Do we know this? I'd have thought you would have an equal mixture of the two hyperfine spin states. I could not see anything in the Wiki article to suggest you only get antiparallel spin in interstellar space. Can you quote me your reference for this, please?
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    Quote Originally Posted by exchemist View Post
    Do we know this? I'd have thought you would have an equal mixture of the two hyperfine spin states. I could not see anything in the Wiki article to suggest you only get antiparallel spin in interstellar space. Can you quote me your reference for this, please?
    If the transition is forbidden on Earth , doesn't that mean that in the Stern machine there ought to be only antiparallel spin?
    it is clear why the parallel magnetic dipole moments (ie, antiparallel spins) has lower energy.[2]
    This transition is highly forbidden with an extremely small rate of 2.9×10−15 s−1, and a lifetime of around 10 million (107) years. It is unlikely to be seen in a laboratory on Earth, but is commonly observed in astronomical settings such as hydrogen clouds in ours and other galaxies.

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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    Do we know this? I'd have thought you would have an equal mixture of the two hyperfine spin states. I could not see anything in the Wiki article to suggest you only get antiparallel spin in interstellar space. Can you quote me your reference for this, please?
    If the transition is forbidden on Earth , doesn't that mean that in the Stern machine there ought to be only antiparallel spin?
    it is clear why the parallel magnetic dipole moments (ie, antiparallel spins) has lower energy.[2]
    This transition is highly forbidden with an extremely small rate of 2.9×10−15 s−1, and a lifetime of around 10 million (107) years. It is unlikely to be seen in a laboratory on Earth, but is commonly observed in astronomical settings such as hydrogen clouds in ours and other galaxies.

    Sorry for the delayed reply: I was on holiday in Martinique with my family.

    I think you are confusing spin states with transitions. You can perfectly well have 50% nuclear and electron with spin parallel and 50% opposed, irrespective of the probability of a direct transition between the two states. Don't forget that electrons get excited out of the ground state, and drop back to it again, by numerous "allowed" processes. In some of these processes may fall back with a change of spin orientation compared with what they had last time they were in the ground state.

    So the "forbidden" character of the direct spin flip, i.e. flipping while still in the ground state, is not the only game in town.
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    Quote Originally Posted by exchemist View Post
    electrons get excited out of the ground state, and drop back to it again, by numerous "allowed" processes. In some of these processes may fall back with a change of spin orientation compared with what they had last time they were in the ground state.
    when an electron is in 2p state is excited and can have antiparallel orientation , but when it reverts to ground state fine structure split shows that the exited 3/2 spin loos their orientation and surplus energy, so all of them must be parallel. Why should an electron fall down in the ground state with antiparallel spin?

    Also, wiki says that 1H atoms have +1/2 spin, but we know that both proton and electron have 1/2 spin and orbital spin is 1 , shouldn't the total angular momentum be 2?

    I'll start a new thread on normal force on moving/orbiting body
    Last edited by whizkid; April 8th, 2014 at 04:23 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    electrons get excited out of the ground state, and drop back to it again, by numerous "allowed" processes. In some of these processes may fall back with a change of spin orientation compared with what they had last time they were in the ground state.
    when an electron is in 2p state is excited and can have antiparallel orientation , but when it reverts to ground state fine structure split shows that the exited 3/2 spin loos their orientation and surplus energy, so all of them must be parallel. Why should an electron fall down in the ground state with antiparallel spin?

    Also, wiki says that 1H atoms have +1/2 spin, but we know that both proton and electron have 1/2 spin and orbital spin is 1 , shouldn't the total angular momentum be 2?

    I'll start a new thread on normal force on moving/orbiting body
    Actually, I think I was guilty of introducing a red herring here, or at least not making clear what I was trying to get at. Let me have another go.

    In the 2p excited state, there are 2 options for the orientation of the spin of the electron and its angular momentum: aligned and opposed. These can be called the j=3/2 and j=1/2 sub-states, where j is the vector sum of the angular momentum l and the spin momentum s. The selection rules are that l must change by +/- 1 and that j can change by 0 or +/-1. So transitions to 1s are allowed from both 2p (3/2) and 2p (1/2).

    Now this is about electron spin orientation relative to electron angular momentum, so doesn't directly help us understand about electron spin orientation relative to nuclear spin. However the principle is analogous.

    What we have to consider is electron spin orientation relative to nuclear spin. The proton has a spin i = 1/2 and the electron has spin s = 1/2. We define the vector sum of the spins as f = i + s. So f =1 when the spins are aligned and f =0 when they are opposed, f=1 being of fractionally higher energy than f = 0.

    From what I have read, the electric dipole selection rules are that f can change by 0 or +/1, just like j above. So, if you have a transition from, say 2p (1/2) to 1s, actually you have 2 fractionally different ground sub-states that can be reached; one with f =1 and one with f=0. But you can reach either, so you expect both to be populated equally.

    The catch is that the transition between the 2 hyperfine states in the same subshell (so no change in l) cannot be initiated by the electric dipole mechanism. There is no change in the electron probability cloud, so nothing for the electric vector of radiation to grab hold of. It is instead just a magnetic change and this requires a magnetic dipole transition mechanism. The magnetic dipole selection rules can be different from electric dipole ones responsible for electronic spectra (please don't ask me about them in detail as I do not know anything about them) and, from what I read, this 21cm transition is "magnetic dipole forbidden" - which means it is very improbable, not that it can't take place at all.

    But in any case, regardless of the probability of the direct transition between f=1 and f=0 in the 1s, you have equal population of both hyperfine states.

    I hope this may be clearer.

    P.S. I don't understand what you are saying about total angular momentum being 2. In the ground state the orbital angular momentum is zero, as I have said several times now. All you have is electron spin and nuclear spin. So f = 1 or 0, never 2. It would be different in an excited state like 2p of course.
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    Quote Originally Posted by exchemist View Post
    P.S. I don't understand what you are saying about total angular momentum being 2. In the ground state the orbital angular momentum is zero, as I have said several times now. All you have is electron spin and nuclear spin. So f = 1 or 0, never 2. It would be different in an excited state like 2p of course.
    You were referring, I think, to magnetic moment, I was referring to angular momentum.
    Please correct me if I am wrong, but
    the orbital quantum number is = 0, and the value of L(o) is
    both e and p (as fermions) have intrinsic momentum = 1/2 , so the sum (1+1/2+1/2) makes 2

    Apart from this there are the 3 magnetic momenta
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    P.S. I don't understand what you are saying about total angular momentum being 2. In the ground state the orbital angular momentum is zero, as I have said several times now. All you have is electron spin and nuclear spin. So f = 1 or 0, never 2. It would be different in an excited state like 2p of course.
    You were referring, I think, to magnetic moment, I was referring to angular momentum.
    Please correct me if I am wrong, but
    the orbital quantum number is = 0, and the value of L(o) is
    both e and p (as fermions) have intrinsic momentum = 1/2 , so the sum (1+1/2+1/2) makes 2

    Apart from this there are the 3 magnetic momenta
    No, you have been careless. Check your maths. The orbital angular momentum is proportional to √( l(l + 1)), not √(l +(l + 1)).

    So it is zero in the 1s, as I have repeatedly explained to you. See for example my very first reply to you, right at the start of this thread! If you have not taken this in by now, I start to wonder if we are going to go round in circles.
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    Quote Originally Posted by exchemist View Post
    No, you have been careless.
    Sorry, does that apply only to QM or also to classical Bohr model?
    but still the two intrinsic moments make 1, do not they?
    And don't we have to sum up angular and magnetic momenta?
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    No, you have been careless.
    Sorry, does that apply only to QM or also to classical Bohr model?
    but still the two intrinsic moments make 1, do not they?
    And don't we have to sum up angular and magnetic momenta?
    Again I repeat, everything I have said to you is from QM. The Bohr model doesn't work and is only useful as a simple shorthand for those chemical and physical concepts that do not require QM.

    Again as I have explained to you before, at some length, angular momentum of charged particles is the cause of magnetic moment. In the context of atomic properties they are inseparable and go hand in hand. So you do not add them up.

    "Magnetic momenta" is a garbled phrase, please stop using it. You have angular momentum, the plural of which is angular momenta, and you have magnetic moment, the plural of which is magnetic moments.
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    Quote Originally Posted by exchemist View Post
    everything I have said to you is from QM. The Bohr model doesn't work.....
    as I have explained to you before, at some length, angular momentum of charged particles is the cause of magnetic moment. ... So you do not add them up.
    That's just what is confusing me, exchemist, and makes me go round and round. Probably I am thick or ignorant (or both)but,
    if you say that L is the cause of mu, then I think you are thinking classical and not QM, since the latter states that L is an intrinsic property and that e is not circling/spinning. To be consequential you should say that L and mu are both intrinsic property. If a charged body is not circling cannot cause a magnetic field/moment.
    Moreover I knew that the formula I used comes from classical model:
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    everything I have said to you is from QM. The Bohr model doesn't work.....
    as I have explained to you before, at some length, angular momentum of charged particles is the cause of magnetic moment. ... So you do not add them up.
    That's just what is confusing me, exchemist, and makes me go round and round. Probably I am thick or ignorant (or both)but,
    if you say that L is the cause of mu, then I think you are thinking classical and not QM, since the latter states that L is an intrinsic property and that e is not circling/spinning. To be consequential you should say that L and mu are both intrinsic property. If a charged body is not circling cannot cause a magnetic field/moment.
    Moreover I knew that the formula I used comes from classical model:
    OK. Your problem may be that you have not let go of classical thinking quite enough. Let's take it in stages:

    1) In QM, remember, certain properties are not "well-defined", due to quantum indeterminacy. We discussed this at the start of our discussion, in the context of the "speed" of an electron. Yes? Well, the same goes for "orbital" motion, too. In QM, something can have "orbital" angular momentum, even though it does not move in a well-defined classical "orbit". This is true of an electron in any orbital with l>0. You may recall I explained that the absence of a well-defined speed does not mean the electron is motionless. It is moving, but we cannot track it.

    Similarly, that the absence of a "well-defined" orbital path does not mean it is not - on average - revolving in some way about the nucleus. It is, but we can only know the probability cloud it occupies, not where it is at any instant. Consequently we cannot trace out a nice, neat, deterministic 2D ellipse. So, although on average, it performs "orbital" motion of a sort, it does not follow an "orbit". Do you understand this, or do we need to go over it?

    2) Just as in classical physics, a charge with angular momentum (whether spin or orbital) gives rise to a magnetic moment. Are you OK with this?

    3) Consequently, magnetic moment and angular momentum are intimately related. Do you understand this?

    4) The electron and proton always have a spin magnetic moment, due to their intrinsic spin.

    5) An electron in an orbital in which l>0 also has an orbital magnetic moment. But an electron in an s orbital, in which l=0, has no orbital magnetic moment, as it has no angular momentum. Do you follow?

    If you have difficulty with any of these statements, let me know which ones and why.

    Lastly, I do not see how you can possibly have got the formula √(l(l+1) hbar from classical physics. This expression contains both h bar and the quantum number, l. Neither of these exists in classical physics. If you say you got this from something in classical physics, please provide a reference, as I feel sure there must be a misunderstanding somewhere.
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    Quote Originally Posted by exchemist View Post
    1) It is moving, but we cannot track it.
    ... revolving in some way about the nucleus. It is, but we can only know the probability ". Do you understand this, ?
    2) .. magnetic moment and angular momentum are intimately related... The electron and proton always have a spin magnetic moment, due to their intrinsic spin.
    5) an electron in an s orbital, in which l=0, has no orbital magnetic moment, as it has no angular momentum.
    If you have difficulty with any of these statements, let me know which ones and why.
    Thanks, exchemist , for your patience, I hope we can eventually pin down my doubts:
    - 1 vs 5: even in 1s (ground state) the electron is circling around the nucleus, how can it have no momentum?
    - 2 : a) intrinsic means that it is not spinning, right? we do not know about e, but p is definitely spinning because of the electrostatic attraction by the circling e, why intrinsic?
    : b) what is the physical meaning of intrinsic angular momentum? e can have intrinsic dipole moment as it could easily be a natural magnet, but how can an object that is not spinning have the properties of a spinning body?
    : c) when we exert a torque applying an external magnetic field B (z) do we have to consider the resistance offered by both L and mu, do we have to sum them up, or not?

    I think you could help me a lot if you correct and complete the following scheme of 1H, ground state, and explain how its total angular momentum is 1/2+ ?(Hydrogen atom - Wikipedia, the free encyclopedia )


    L (e) = 1/2...........L(o) = 0 .....L(p) = 1/2
    mu(e)= 1.00165..mu(o)=0 ....mu(p) = ?

    total angular momentum ....+.....+......+..... = 1/2
    Last edited by whizkid; April 10th, 2014 at 03:43 AM.
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    Quote Originally Posted by whizkid View Post
    Quote Originally Posted by exchemist View Post
    1) It is moving, but we cannot track it.
    ... revolving in some way about the nucleus. It is, but we can only know the probability ". Do you understand this, ?
    2) .. magnetic moment and angular momentum are intimately related... The electron and proton always have a spin magnetic moment, due to their intrinsic spin.
    5) an electron in an s orbital, in which l=0, has no orbital magnetic moment, as it has no angular momentum.
    If you have difficulty with any of these statements, let me know which ones and why.
    Thanks, exchemist , for your patience, I hope we can eventually pin down my doubts:
    - 1 vs 5: even in 1s (ground state) the electron is circling around the nucleus, how can it have no momentum?
    - 2 : a) intrinsic means that it is not spinning, right? we do not know about e, but p is definitely spinning because of the electrostatic attraction by the circling e, why intrinsic?
    : b) what is the physical meaning of intrinsic angular momentum? e can have intrinsic dipole moment as it could easily be a natural magnet, but how can an object that is not spinning have the properties of a spinning body?
    : c) when we exert a torque applying an external magnetic field B (z) do we have to consider the resistance offered by both L and mu, do we have to sum them up, or not?

    I think you could help me a lot if you correct and complete the following scheme of 1H, ground state, and explain how its total angular momentum is 1/2+ ?(Hydrogen atom - Wikipedia, the free encyclopedia )


    L (e) = 1/2...........L(o) = 0 .....L(p) = 1/2
    mu(e)= 1.00165..mu(o)=0 ....mu(p) = ?

    total angular momentum ....+.....+......+..... = 1/2
    1 & 5) No, it is NOT "circling" in the ground 1s state. I have told you this several times. Suggest re-reading my first reply to your original post at the the start of the thread, and then read this one again too: http://www.thescienceforum.com/physi...ost536409.html

    You really have to grasp that the electron in an s orbital is not circling: it is moving but has no average revolving component to its motion, hence no angular momentum. Forget Bohr and classical physics - they do cannot explain this, while QM does.

    2) No. Intrinsic certainly doesn't mean not spinning, quite the opposite. Why on earth would you say a thing like that? "Intrinsic" means a property that is unchangeably fundamental to it. (Google definition: belonging naturally; essential.) If an electron exists it is spinning: there is no such thing as an electron with its spin "stopped".

    I won't address the rest of your reply until we have these 2 points cleared up, as I think if we can do this a number of your other queries may evaporate.
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    Quote Originally Posted by exchemist View Post
    1 & 5) No, it is NOT "circling" in the ground 1s state.
    You really have to grasp that the electron in an s orbital is not circling: it is moving but has no average revolving component to its motion, hence no angular momentum..
    Thanks,
    - is there an explanation why this should happen only in 1s? and what is the evidence?or is it just a theory?
    - do the laws of electrostatics apply in 1s? how can the motion of the e be not-circular and respect them?
    Last edited by whizkid; April 10th, 2014 at 08:54 AM.
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    Quote Originally Posted by exchemist View Post
    2) No. Intrinsic certainly doesn't mean not spinning, quite the opposite. Why on earth would you say a thing like that? "Intrinsic" means a property that is unchangeably fundamental to it. (Google definition: belonging naturally; essential.) If an electron exists it is spinning: there is no such thing as an electron with its spin "stopped".
    As well as that, "intrinsic" is also used sometimes to emphasise that an electron "has" spin but is not a little ball spinning on an axis.
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    Quote Originally Posted by Strange View Post
    As well as that, "intrinsic" is also used sometimes to emphasise that an electron "has" spin but is not a little ball spinning on an axis.
    Now you are confusing me, Strange, does it spin or not? we can't have it both ways? or is it like the notorious cat?
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    You still seem to be hung up on the classical notion of spin, in QM "spin" is intrinsic angular momentum, it has no classical counterpart and in no way implies "spin" in a classical sense. It is one of the (IMO) truly badly named concepts as it causes exactly the sort of confusion you are now having. If it was called "wibble" or something else and did not carry the ingrained baggage of "spin" as used in everyday conversation it would cause a lot less head scratching...
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    Quote Originally Posted by whizkid View Post
    Now you are confusing me, Strange, does it spin or not? we can't have it both ways? or is it like the notorious cat?
    It has angular momentum. That is all you need to know.

    The electron is modelled as a point particle and as far as we can tell has no size. So saying that it physically spins doesn't make much physical sense. But it has spin (it just isn't going round and round).
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    Quote Originally Posted by PhDemon View Post
    You still seem to be hung up on the classical notion of spin, in QM "spin" is intrinsic angular momentum,.
    No, I am not, I am prepared to accept any new concept as long as it makes some sense and it is consistent and has a minimal explanation or justification.
    If 'spin/wibble' in QM is something different, is also the notion of angular momentum different or is it 100% identical to L in a spinning wheel? how can 'wibble' generate or be related to magnetic moment? How do you conclude the its value is hbar/2?
    I have read that there is no evidence of the intrinsic spin in e, outside an atom, when it is at rest, in a beam or in a potential well. So how intrinsic is it?
    That is what troubles me
    Last edited by whizkid; April 10th, 2014 at 10:36 AM.
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    I'm sorry but your questions indicate to me that this is something you are confused about and in the way I suggested. Specifically:

    ...does it spin or not? we can't have it both ways? or is it like the notorious cat?
    this implies to me you are still thinking of spin in a classical sense. It has "spin" (i.e. intrinsic angular momentum) but does not spin (in a classical sense).

    As to the rest exchemist has already given you the information and I can't add any more to his excellent answers. You may find this useful http://en.wikipedia.org/wiki/Spin_(physics)

    EDIT to address your edit.

    I have read that there is no evidence of the intrinsic spin in e, outside an atom, when it is at rest, in a beam or in a potential well.
    Where did you read this? It is incorrect, there are problems measuring it due to experimental uncertainties but it is there (that is what INTRINSIC means, it is not a case of it being present in bound electrons and not there in free electrons), the issue is addressed theoretically here: http://sro.sussex.ac.uk/22128/ but this is now getting beyond my knowledge, you may need one of the physics specialists to step up to continue down this line.
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    Quote Originally Posted by PhDemon View Post
    I'm sorry but your questions indicate to me that this is something you are confused about and in the way I suggested.
    this implies to me you are still thinking of spin in a classical sense. It has "spin" (i.e. intrinsic angular momentum) but does not spin (in a classical sense).
    I told you I accept that e has an intrinsic property called 'wibble'. Fair enough. it is like L. OK.
    But please tell how do you calculate wibble? L is mvr, right? to what parameters is it related? how does it react to torque? how does it generate a magnetic field, a magnetic moment , how is this latter related to wibble by what ratio?
    If someone provides all this answer wibble is acceptable, a scientific concept, and not just a name.
    I know exchemist is doing his best, probably it is the theory that is incomplete
    Last edited by whizkid; April 10th, 2014 at 10:58 AM.
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    Quote Originally Posted by whizkid View Post
    But please tell how do you calculate wibble? L is mvr, right? to what parameters is it related?
    It isn't calculated; it is measured. As far as I know it isn't derived from anything else. (Although you know that if a particle is a boson it will have integer spin, and if it is a fermion it will have half integer spin.)

    how does it react to torque?
    I can't answer that.

    how does it generate a magnetic field, a magnetic moment , how is this latter related to wibble by what ratio?
    This starts to get into Feynman's "why questions" territory (1). Charged particles with spin also have a magnetic moment. Does one cause the other? Why is there a factor of 2 from the classical prediction? <shrug> I suspect these are unanswered, and possibly unanswerable.

    (1) Richard Feynman - Magnets (And 'Why' Questions) - YouTube
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