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Thread: Classical lifetime of hydrogen

  1. #1 Classical lifetime of hydrogen 
    Forum Ph.D. william's Avatar
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    Warning: math alert

    This problem was inspired by a question in the Astronomy & Cosmology section which asks if classical physics will eventually replace quantum physics.

    Picture the hydrogen atom as an electron orbiting a proton. As the electron moves, it will radiate away its energy. The power radiated away from a moving charge (for v << c) is given by the Larmor formula which is

    P = mu<sub>0</sub>q<sup>2</sup>a<sup>2</sup>/(6*pi*c) = dE/dt,

    where q = charge,
    a = acceleration (centripetal in this case),
    c = speed of light,
    mu<sub>0</sub> is the permeability of free space, and
    E is energy (all in SI units).

    For your convenience;
    q = e ~ 1.6 * 10<sup>-19</sup> C (C = Coulombs)
    c ~ 3 * 10<sup>8</sup> m/s
    mu<sub>0</sub> ~ 1.3 * 10<sup>-6</sup> kg m/C<sup>2</sup>

    From this, one can calculate the time it would take for the electron to spiral into the proton as it loses its energy, that is, the "classical lifetime" of hydrogen.

    To solve this, assume all the electron's energy is in the form of kinetic energy, the acceleration (a) is centripetal, and make use of the Coulomb force to find a. There is a minor integration involved but you can get the answer to within an order of magnitude without using calculus. Oh, and you'll have to use the
    Bohr radius (a<sub>0</sub> ~ 5 * 10<sup>-11</sup> m), and the
    mass of the electron (m<sub>e</sub> ~ 9 * 10<sup>-31</sup> kg)
    somewhere along the way....



    The challenge then, is for someone to calculate the classical lifetime of hydrogen. (To within an order of magnitude is good enough for me.) It wouldn't hurt to show that v << c through most of the electron's journey for the Larmor formula to apply, but I won't require it. This is a problem I'm sure most of you can do (with or without using calculus)....

    Hope you all find this educational.

    Cheers,
    william


    Addendum:
    By the way...
    you may have guessed that the "classical lifetime" of hydrogen is very small. This kept many physicists awake at night before the advent of quantum physics....


    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  3. #2  
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    A quick comment about this problem;

    Anyone interested in going into physics as a career may encounter problems like this on the "GRE-Physics" exam. This exam one usually takes during their senior year of undergrad. It is a physics "aptitude" exam usually required when applying to grad school.

    Only... the question you'd be asked would be something like;
    According to classical mechanics the atom will decay in a very short time. Roughly how long does it take for the electron to spiral into the nucleus as it emits electromagnetic radiation?
    That's it! No equations, no constants given, no hints. (And the above quote was taken from a GRE practice exam....)

    Not that bad one might say... except there are 100 of these type of questions and only 3 hours to take the exam! Oh yeah, and the exam costs something like $100 USD.



    So... a certain "Borg" comes to mind. It might be a good exercise to try to solve some of the problems I have posed including this one. So far, that certain "Borg" seems to be "all fluff and no stuff."

    N.B. If you're not sure who the "Borg" is, then it's probably not you....

    Cheers,
    william

    P.S. Anyone who will eventually take the physics GRE, feel free to PM me and perhaps I can give some more insight on this exam and possible strategies for taking it.


    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  4. #3  
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    Hmmm, I think I am gonna take a shot at this one. All I have under my belt is Physics B and BC Calculus (high school sophomore) but I think I might be able to do this one.
    I demand that my name may or may not be vroomfondel!
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    Not trying to make a thing of this but; isn't the fact that the electron orbit doesn't decay explained by the internal alternating forces created by the quarks of protons and neutrons. Isn't this also in essence a classical explanation and is Quantum physics not just an observation with no explanation...................... I love particle physics and am just trying to clear this kind of stuff up in my head. You could technically use the Larmor equation on the quark level too and modify it for strong forces add them all together and Bam! Do you happen to know any good equations for modelling large atomic structures like Oxygen and Iron electron orbital functions and such?
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  6. #5  
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    Quote Originally Posted by Beky
    Not trying to make a thing of this but; isn't the fact that the electron orbit doesn't decay explained by the internal alternating forces created by the quarks of protons and neutrons. Isn't this also in essence a classical explanation and is Quantum physics not just an observation with no explanation...................... I love particle physics and am just trying to clear this kind of stuff up in my head. You could technically use the Larmor equation on the quark level too and modify it for strong forces add them all together and Bam! Do you happen to know any good equations for modelling large atomic structures like Oxygen and Iron electron orbital functions and such?
    A "quick and dirty" response to your question is this;
    I don't know why things are the way they are in the quantum world. For instance, the electron in the hydrogen atom is only allowed to be in certain energy states. I liken this to an ATM machine. When I withdraw money from one, I can only withdraw money in quantized amounts ($20 to be exact). I also am not up on the latest theories in quantum and particle physics.

    I personally don't have any experience with any models of the large atomic structures you mention. And I think the atomic theorists are still working out the details....

    So I guess I'm not much help. Sorry....

    Cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  7. #6  
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    A little bit of quick and dirty math yeilds the following formula for me:

    r(t) = r<sub>0</sub>*e^kt

    where k = -12*pi*c*m/(u<sub>0</sub>*q<sup>2</sup>)

    when I set in r<sub>0</sub> = bohr radius and r(t) = radius of a proton, i get t = 3.63*10<sup>-23</sup> seconds.
    I demand that my name may or may not be vroomfondel!
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    Well first off no quicker, nor dirtier than I. Second off at least you didn't give me crap(I mean most people right off would've said if there was variation, the charge wouldn't be a constant. Yeah, baby!). I don't know how to explain any of it myself, or do I . The only thing that gets me is how did they know that Uranium 235 needed a critical mass of 15Kg before it would explode because it needs 32 generations of neutron decay(I swear they did it mathematically and not just through observation)? And what are super computers really for? 10 years ago the standard military desktop used for meteorology had 1Gig RAM I wonder what the engineers where using and are now. Any rate I don't have a computer or $10 000 for a workstation equivalent to a super computer of 30 years ago. I wonder how much paper a scientist would go through in the early 1900s. I'll stop interupting you now.
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  9. #8  
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    permeability of free space units are Newtons/Ampere^2 or Henrys/meter to break it into its components it would be Kg(m/s^2)/C^2

    and the equation gives me watts/second^2???????????
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  10. #9  
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    Quote Originally Posted by Beky
    permeability of free space units are Newtons/Ampere^2 or Henrys/meter to break it into its components it would be Kg(m/s^2)/C^2
    I caught you before you could get away this time beky... :wink:

    Wanna try this again...?

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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    huh
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  12. #11  
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    Quote Originally Posted by Vroomfondel
    A little bit of quick and dirty math yeilds the following formula for me:

    r(t) = r<sub>0</sub>*e^kt

    where k = -12*pi*c*m/(u<sub>0</sub>*q<sup>2</sup>)

    when I set in r<sub>0</sub> = bohr radius and r(t) = radius of a proton, i get t = 3.63*10<sup>-23</sup> seconds.
    I'm still trying to figure out what you did.

    But for now, I'll say that you are a few orders of magnitude off. The time is not quite that fast....

    Consider this;

    P = dE/dt => t = integral(dE/P) with the proper limits of course....


    The non-calculus way would be just

    t ~ E/P using the Bohr radius yada yada, and this gets the right order of magnitude.

    A valiant attempt though. Thanks!

    Cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  13. #12  
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    I know what I did I forgot to convert A to C/s I've never seen it written that way Kg m/C^2
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  14. #13  
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    Quote Originally Posted by Beky
    huh
    [mu<sub>0</sub>] = N/A<sup>2</sup> = (kg m/s<sup>2</sup>)*(s<sup>2</sup>/C<sup>2</sup>) = kg m/C<sup>2</sup> no?

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  15. #14  
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    I'm blonde does that make any difference!
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  16. #15  
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    Quote Originally Posted by Beky
    I know what I did I forgot to conver A to C/s I've never seen it written that way Kg m/C^2
    The reason why I gave the units the way I did was because kg m/C<sup>2</sup> was most natural to this problem. Otherwise you would have to do some conversions yourself if you wanted to check the dimensions (and we all should...).
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  17. #16  
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    Hey William

    The answer is around 10^-11 seconds if i remember correctly.

    Can you ignore the potential energy of the system though? Im too lazy to check
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  18. #17  
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    Quote Originally Posted by river_rat
    Hey William

    The answer is around 10^-11 seconds if i remember correctly.

    Can you ignore the potential energy of the system though? Im too lazy to check
    Hi double-r,
    You are correct as usual. I was wondering where you've been....

    I did ignore the potential energy. It might be worthwhile to check how it affects the outcome if it is considered though....

    Cheers and thanks,
    william
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  19. #18  
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    Hey william

    Ive been pretending to do research lol. Im a bit stuck at the moment and am awaiting enlightenment from the gods of mathematics

    mmm, will take a look at the problem again - i just looked up the other answer from when we solved it many moons ago in undergrad.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    I got to play with this one rearrange it back to like t = Ke m ln ratio of radius
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    can you show where you used the gyroscopic ratio and the bohr magneton. That was part of your original question wasn't it how to use it but the bohr magneton was already in P and then you added the g to give the K. I could be out of wack, I often am.
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  22. #21  
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    Quote Originally Posted by Beky
    can you show where you used the gyroscopic ratio and the bohr magneton. That was part of your original question wasn't it how to use it but the g was already in P and then you added the bohr magneton to give the K. I could be out of wack, I often am.
    Hi Beky,
    Well, h-bar isn't in the problem but I suppose you can manipulate some constants if you really like the Bohr magneton....

    By "g" do you mean q, the electric charge (elementary charge in this case)?

    Help me out here... what do you mean by "K"?

    Cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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    William, how do you handle the sign problem if you ignore the potential energy of the electron, proton system? Without the potential energy, the atom should never decay - it should actually explode.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  24. #23  
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    hmmm... For my solution i didnt ignore the potential energy. I took the sum of the potential and the kenetic energies, differentiated it, and set it equal to the little formula that you gave me that is also equal to the rate of change of energy... but i just noticed that i messed up somewhere so i wil try it out again and see if i get the right answer. If not, i will try it again and ignore potential energy.
    I demand that my name may or may not be vroomfondel!
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  25. #24  
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    Quote Originally Posted by river_rat
    William, how do you handle the sign problem if you ignore the potential energy of the electron, proton system? Without the potential energy, the atom should never decay - it should actually explode.
    I'm not sure what sign problem you're talking about....

    Is it that you are getting a negative time? If that is the case, you may be integrating from 0 to a<sub>0</sub> and not from a<sub>0</sub> to 0....
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  26. #25  
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    Nope - thats not it.

    Its because dE/dt = -P (energy is radiated away) and that negative cancels the negative from actually subing in for v^2 and a in the actual expression for the total energy of the system (which in your case is q^2/(2r) in Gauss units) and taking time derivatives - so r is increasing with time and not decreasing?

    The potential energy term gives you a new total energy of -q^2/(2r) which sorts out the problem.

    Hope i sketched it out enough without giving the game away.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  27. #26  
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    Quote Originally Posted by river_rat
    Nope - thats not it.

    Its because dE/dt = -P (energy is radiated away) and that negative cancels the negative from actually subing in for v^2 and a in the actual expression for the total energy of the system (which in your case is q^2/(2r) in Gauss units) and taking time derivatives - so r is increasing with time and not decreasing?

    The potential energy term gives you a new total energy of -q^2/(2r) which sorts out the problem.

    Hope i sketched it out enough without giving the game away.
    Still thinking about it... but the
    new total energy of -q^2/(2r)
    gives me pause. It's the concept of a negative energy that puzzles me. Kind of like having a negative radius....

    I haven't worked it out yet but thought I'd let you know what I was thinking.

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  28. #27  
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    Bound states have negative total energy. In this case E = 1/2 m v^2 - q^2/r so the kinetic energy needs to be less then the potential energy etc. Similar thing for keplerian motion if i recall.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  29. #28  
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    Quote Originally Posted by river_rat
    Bound states have negative total energy. In this case E = 1/2 m v^2 - q^2/r so the kinetic energy needs to be less then the potential energy etc. Similar thing for keplerian motion if i recall.
    Oops... You're right.
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  30. #29  
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    Hey dont worry - i had to look half this stuff up anyway, could not remember the coloumb force law in gauss units!

    Its the negative thing that made me think though, as if you ignore the potential energy of the system then the electron always has a postive energy and thus ant be bounded etc.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  31. #30  
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    Yup, with a little math, we can see that the total energy is

    E = U/2 (which is also the virial theorem! (for bound orbits...))

    where U = potential energy which is a negative quantity.
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  32. #31  
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    Anyhoot, the purpose of this thread was to point out that the classical lifetime of hydrogen is about 10<sup>-11</sup> s.

    So... for the person ("HiLe") who asked in the Astronomy & Cosmology section if classical physics will eventually replace quantum physics...
    what do you think now HiLe?

    This was loads of fun (I thought...).

    Cheers,
    william
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  33. #32  
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    had forgotten about that, havent done statmech in a while.
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    not q g the gyroscopic ratio(if I'm not mistaken your K was made up of your P, g and something else). And as for the Ke I was talking about the electro static constant and the coulomb force as you were discussing(the potential). Anyway you actually fixed my math problem I was missing the link and you gave it to me thx. I'm sure I'll figure out the rest. A year ago I was trying to figure this out in a purely classical sense with no observation of permeability and h bar and the such, talk about a mess. I guess thats why schools so important! Maybe when I'm 50.
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    Your next assignment William (if you choose to accept it) is to get these people to derive the ultraviolet catastrophy.

    This post will self destruct in 5... 4... 3... 2... 1...

    Now there is a challenge in the making!
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    how about quantum tunneling it's a litle more in tune with this conversation!
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  37. #36  
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    Quote Originally Posted by river_rat
    Your next assignment William (if you choose to accept it) is to get these people to derive the ultraviolet catastrophy.

    This post will self destruct in 5... 4... 3... 2... 1...

    Now there is a challenge in the making!

    That might be a good one!
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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    Ok, i think im just gonna post my math and you guys can tell me where i went wrong. Here it goes:

    E = -kq<sup>2</sup>/r + <sup>1</sup>/<sub>2</sub>mv<sup>2</sup>

    Now, since F = ma = kq<sup>2</sup>/r<sup>2</sup> = mv<sup>2</sup>/r,

    kq<sup>2</sup>/r = mv<sup>2</sup>

    substituting this into our original equation, we get

    E = -mv<sup>2</sup> + <sup>1</sup>/<sub>2</sub>mv<sup>2</sup>, or E = -<sup>1</sup>/<sub>2</sub>mv<sup>2</sup>

    differentiating this with respect to time, we get

    dE/dt = -mv*dv/dt = -mva

    setting this equal to williams formula, we get:

    mu<sub>0</sub>q<sup>2</sup>a<sup>2</sup>/(6*pi*c) = dE/dt = -mva

    which becomes

    mu<sub>0</sub>q<sup>2</sup>a/(6*pi*c) = -mv

    which is the differential equation:

    dv/dt = -(6*pi*c*m/(mu<sub>0</sub>q<sup>2</sup>))*v

    for simplicity, (6*pi*c*m/(mu<sub>0</sub>q<sup>2</sup>)) = K

    the differential equation becomes:

    dv/dt = -Kv

    who's solution is

    v = v<sub>0</sub>e<sup>-Kt</sup>

    substituting this into

    kq<sup>2</sup>/r = mv<sup>2</sup>

    gives us

    r = (kq<sup>2</sup>/mv<sub>0</sub><sup>2</sup>)e<sup>2Kt</sup>

    or

    r = r<sub>0</sub>e<sup>2Kt</sup>

    So where did i mess that up? The electron cant spiral away from the proton last time i checked. I think i messed something up with the potential energy or the electrostatic force.
    I demand that my name may or may not be vroomfondel!
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  39. #38  
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    Remember we are working in 2D polar coordinates so a (the magnitude of the acceleration) is not the same as dv/dt (time derivative of the magnitiude of the velocity) etc. Instead of trying to get an equation for v, try get an differential equation for r. You should get someting of the form

    r^2 dr/dt = constant if i remember what i did last night correctly.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  40. #39  
    Forum Isotope Zelos's Avatar
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    martillo should see this
    I am zelos. Destroyer of planets, exterminator of life, conquerer of worlds. I have come to rule this uiniverse. And there is nothing u pathetic biengs can do to stop me

    On the eighth day Zelos said: 'Let there be darkness,' and the light was never again seen.

    The king of posting
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  41. #40  
    Forum Ph.D. william's Avatar
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    Quote Originally Posted by Zelos
    martillo should see this
    Who is/was martillo?

    cheers
    "... the polhode rolls without slipping on the herpolhode lying in the invariable plane."
    ~Footnote in Goldstein's Mechanics, 3rd ed. p. 202
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  42. #41  
    Forum Junior Vroomfondel's Avatar
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    Martillo was an old person a while back who absolutely refused to believe in relativity. He created his own website, http://www.geocities.com/anewlightinphysics/index.htm, decicated to exploring the consequences of relativity being an incorrect theory. We pretty much disproved everything he claimed, but he would ignore the proof and continue with his own theories.

    Thanks for the tip river rat, I'll try it out.
    I demand that my name may or may not be vroomfondel!
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  43. #42  
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    Ah yes great wise one your ass is so huge and powerfull, there is great respect for ass like yours.

    J/Psi rules! Down with commerciallized education.
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