# Thread: Speed of Gravity, and Light

1. Question: If information about changes in gravitational attraction due to distance travels at the speed of light, then how does this effect light itself?

Say the sun emits a beam of light. We know the sun's gravity is initially effecting that light, pulling it backwards toward the sun, because the beam's trip through space started right at the sun's surface.

So.... when does the light figure out that it has moved away from being right next to the sun?

The information is traveling at the speed of light, but the beam of light is traveling at the speed of light as well, so how would the information ever catch up?

2.

3. I'd say that the light is only indirectly affected by gravity. The mass of the sun warps the surrounding space. Ligth travels then in this affected space and is affected (red shifted). IMO

4. I wonder, then, if it's the warping effect that moves at the speed of light maybe? So maybe there'd only be a problem if the sun were the object moving too fast?

So, if the sun instantly moved to another galaxy, the space around it would remain warped for an amount of time equal to the amount needed to travel that far at C. I guess?

It's confusing, because we know that gravitational attraction is caused by both objects, rather than just the more massive one, at least in Newtonian physics. Einstein's space-time-curvature seems a lot more complicated, however.

5. G'day from the land of ozzzzzz

There are varies schools of thought about what gravity is. Not only what it is, but! how fast it goes.

Some think that it travels at the speed of light.

Some think that it travels at 20 times the speed of light.

A black hole is supposed to hold onto everything. But! gravity escapes as though it did not exist.

If you look at the wave structure of matter, maybe we could understand the process.

or maybe if you have time read some papers

General Relativity and Quantum Cosmology
http://aps.arxiv.org/list/gr-qc/9909

or this one

Solar System tests of some models of modified gravity proposed to explain galactic rotation curves without dark matter
Authors: Lorenzo Iorio, Matteo Luca Ruggiero
(Submitted on 2 Nov 2007 (v1), last revised 25 Jun 2008 (this version, v4))

Abstract: We consider the recently estimated corrections \Delta\dot\varpi to the Newtonian/Einsteinian secular precessions of the longitudes of perihelia \varpi of several planets of the Solar System in order to evaluate whether they are compatible with the predicted precessions due to models of long-range modified gravity put forth to account for certain features of the rotation curves of galaxies without resorting to dark matter. In particular, we consider a logarithmic-type correction and a f(R) inspired power-law modification of the Newtonian gravitational potential. The results obtained by taking the ratio of the apsidal rates for different pairs of planets show that the modifications of the Newtonian potentials examined in this paper are not compatible with the secular extra-precessions of the perihelia of the Solar System's planets estimated by E.V. Pitjeva as solve-for parameters processing almost one century of data with the latest EPM ephemerides.

Quantum Physics Exploring Gravity in the Outer Solar System: The Sagas Project
Authors: P. Wolf, Ch. J. Bordé, A. Clairon, L. Duchayne, A. Landragin, P. Lemonde, G. Santarelli, W. Ertmer, E. Rasel, F.S. Cataliotti, M. Inguscio, G.M. Tino, P. Gill, H. Klein, S. Reynaud, C. Salomon, E. Peik, O. Bertolami, P. Gil, J. Páramos, C. Jentsch, U. Johann, A. Rathke, P. Bouyer, L. Cacciapuoti, D. Izzo, P. De Natale, B. Christophe, P. Touboul, S.G. Turyshev, J.D. Anderson, M.E. Tobar, F. Schmidt-Kaler, J. Vigué, A. Madej, L. Marmet, M-C. Angonin, P. Delva, P. Tourrenc, G. Metris, H. Müller, R. Walsworth, Z.H. Lu, L. Wang, K. Bongs, A. Toncelli, M. Tonelli, H. Dittus, C. Lämmerzahl, G. Galzerano, P. Laporta, J. Laskar, A. Fienga, F. Roques, K. Sengstock
(Submitted on 2 Nov 2007 (v1), last revised 4 Jul 2008 (this version, v4))

Abstract: We summarise the scientific and technological aspects of the SAGAS (Search for Anomalous Gravitation using Atomic Sensors) project, submitted to ESA in June 2007 in response to the Cosmic Vision 2015-2025 call for proposals. The proposed mission aims at flying highly sensitive atomic sensors (optical clock, cold atom accelerometer, optical link) on a Solar System escape trajectory in the 2020 to 2030 time-frame. SAGAS has numerous science objectives in fundamental physics and Solar System science, for example numerous tests of general relativity and the exploration of the Kuiper belt. The combination of highly sensitive atomic sensors and of the laser link well adapted for large distances will allow measurements with unprecedented accuracy and on scales never reached before. We present the proposed mission in some detail, with particular emphasis on the science goals and associated measurements.

http://ourworld.compuserve.com/homep...p5/gravity.htm

Is faster-than-light propagation allowed by the laws of physics?
http://metaresearch.org/cosmology/gravity/LR.asp

The Speed of Gravity - Repeal of the Speed Limit
http://metaresearch.org/cosmology/gr...peed_limit.asp

The Speed of Gravity What the Experiments Say
http://metaresearch.org/cosmology/speed_of_gravity.asp

6. Mass “bends” spacetime. Supposedly, gravitons transmit the effect of gravity at the speed of light to the surrounding spacetime. Light leaving the Sun is traveling through spacetime that is already bent by gravity. Light, like the planets, is moving in a straight line through spacetime that is curved by gravity.

7. Originally Posted by kojax
Question: If information about changes in gravitational attraction due to distance travels at the speed of light, then how does this effect light itself?

Say the sun emits a beam of light. We know the sun's gravity is initially effecting that light, pulling it backwards toward the sun, because the beam's trip through space started right at the sun's surface.

So.... when does the light figure out that it has moved away from being right next to the sun?

The information is traveling at the speed of light, but the beam of light is traveling at the speed of light as well, so how would the information ever catch up?
This requires a very complex solution. So in my own words, I will try to explain this about as well as I can.

As you should know, I am a free thinker so I have my own opinions on a lot of topics.

Gravity was recently discovered to propagate at the velocity of light (c).

I wrote an article on a new cause for the Cosmological Red Shift (CRS) to replace the 'expansion of space' as the cause.

The CRS is caused by an 'intrinsic' force that affects the Electric Charged Field Particles (ECFP) to distribute thenselves in densities according to the 'inverse sqaure law' (ISL) that also applies to the gravitational force.
So the distrutions of these particle densities weakens (widens) with the distance from the sources such as the electrons.
So the weakening of these density distributions is the cause of the weakening forces according to the ISL.
This would/should cause the transfer of momentum to slow down since the ECFP are more widely spaced. ( I know this raises a question of the radio waves velocities being slower) but I believe that could be true although their has beeb no data about the relative velocities of 'c' and the wavelengths.
There is a relstionship though with the 'energy' levels of the wavelengths in relation to 'c'.

So to get back to the question, the propagation of 'c' in relation to gravity is that they would do so equally in accordance with the ISL.

Cosmo

8. It sounds like I'm hearing two versions of what the speed of gravity means:

1) - The effect itself is transmitted by "gravitons", so gravity only affects you if you move into contact with a graviton.

So: Light would feel the effect of leaving the sun's proximity.

or

2) - Changes in the effect are what move at the speed of light, so the beam of light would always experience the same gravitational pull no matter how far it got from the sun.

Any third possibilities? Am I understanding the first two right?

9. Everything in the universe that has mass experiences the gravitational effect of everything else in the universe with mass, at a ratio that is inversely proportional to the distance between them. If that is what you understood, then you are right.

10. Originally Posted by kojax
Question: If information about changes in gravitational attraction due to distance travels at the speed of light, then how does this effect light itself?

Say the sun emits a beam of light. We know the sun's gravity is initially effecting that light, pulling it backwards toward the sun, because the beam's trip through space started right at the sun's surface.

So.... when does the light figure out that it has moved away from being right next to the sun?

The information is traveling at the speed of light, but the beam of light is traveling at the speed of light as well, so how would the information ever catch up?

gravity bends space so then when the light travels it is affected. so it indirectly affected.

11. Originally Posted by kojax
It sounds like I'm hearing two versions of what the speed of gravity means:

1) - The effect itself is transmitted by "gravitons", so gravity only affects you if you move into contact with a graviton.

So: Light would feel the effect of leaving the sun's proximity.

or

2) - Changes in the effect are what move at the speed of light, so the beam of light would always experience the same gravitational pull no matter how far it got from the sun.

Any third possibilities? Am I understanding the first two right?
I do not buy this 'graviton' version of gravity. This reduces gravity to a single 'pulse'?

Gravity is a omnidirectional force like the electric force.
So it is a 'field' radiationg in 'all' directions.

Cosmo

12. Originally Posted by Cosmo
Originally Posted by kojax
It sounds like I'm hearing two versions of what the speed of gravity means:

1) - The effect itself is transmitted by "gravitons", so gravity only affects you if you move into contact with a graviton.

So: Light would feel the effect of leaving the sun's proximity.

or

2) - Changes in the effect are what move at the speed of light, so the beam of light would always experience the same gravitational pull no matter how far it got from the sun.

Any third possibilities? Am I understanding the first two right?
I do not buy this 'graviton' version of gravity. This reduces gravity to a single 'pulse'?

Gravity is a omnidirectional force like the electric force.
So it is a 'field' radiationg in 'all' directions.

Cosmo
And the electromagnetic field is mediated by virtual photons, just like the gravitational field would be mediated by virtual gravitons.

13. Originally Posted by Arch2008
Everything in the universe that has mass experiences the gravitational effect of everything else in the universe with mass, at a ratio that is inversely proportional to the distance between them. If that is what you understood, then you are right.
But the distance at what time?

Light traveling away from the sun is going too fast for the information to catch up, isn't it? How does it become aware that the distance between it and the sun has grown if that information isn't moving fast enough to reach it?

14. Photons do not have mass. As I already posted, the photons move through spacetime that is curved by gravity. They don't have to exchange any "information".

15. Originally Posted by Janus
Originally Posted by Cosmo
Originally Posted by kojax
It sounds like I'm hearing two versions of what the speed of gravity means:

1) - The effect itself is transmitted by "gravitons", so gravity only affects you if you move into contact with a graviton.

So: Light would feel the effect of leaving the sun's proximity.

or

2) - Changes in the effect are what move at the speed of light, so the beam of light would always experience the same gravitational pull no matter how far it got from the sun.

Any third possibilities? Am I understanding the first two right?
I do not buy this 'graviton' version of gravity. This reduces gravity to a single 'pulse'?

Gravity is a omnidirectional force like the electric force.
So it is a 'field' radiationg in 'all' directions.

Cosmo
And the electromagnetic field is mediated by virtual photons, just like the gravitational field would be mediated by virtual gravitons.
There is no such a thing as 'virtual' photons.

My opinion is that photons are concentrated FIELD particles that move through these electric fields as ONE line radiations even though the fields themselves are 3 dimensional.

I originally also reffered to these field particles as 'virtual'.
But after discussions on the internet as one critic replyed, it you are reffering to the fields as real, than the field particles must be real. I agree with that.

So the field particles are REAL particles that I consider to be charged and distribute themselves uniformly throughout these fields by their MUTUAL repulsion to form these field. And these fields are REAL because of their 'action at a distance' in compliance with the inverse square law as to their strength.

Further proof that these photons are real is that we see them. That is the Balmer Series. But these photons do exist at the other energies too.

Cosmo

16. Originally Posted by Arch2008
Photons do not have mass. As I already posted, the photons move through spacetime that is curved by gravity. They don't have to exchange any "information".

I suppose I just don't find the evidence against them having mass to be convincing. Arriving at such a conclusion requires us to assume that any value too small for us to measure is equal to zero, which is a very arrogant assumption to make.

Photons are clearly much less massive than electrons, however, and that's pretty un-massive.

Why would all the other gravitational fields be created by the combination of the two masses present, and then this one special exception exist?

17. If a photon were to have any mass, any mass whatsoever, it would not be able to travel at the speed of light. Photons travel at the speed of light without any more energy added than they exist out of. They travel at C as a direct result of having zero mass.

18. It all hinges on light being exceptional.

Sometimes I've wondered if all our physics could be revised around snails not photons, and the constant speed of snail. Grant snails everything we grant photons, would the math work?

19. Originally Posted by Pong
It all hinges on light being exceptional.

Sometimes I've wondered if all our physics could be revised around snails not photons, and the constant speed of snail. Grant snails everything we grant photons, would the math work?
No, it all hinges on the speed c being exceptional.

20. Originally Posted by Janus
No, it all hinges on the speed c being exceptional.
C being the speed of snail. Whatta ya think? In a world of slow slugs of course.

21. Originally Posted by KALSTER
If a photon were to have any mass, any mass whatsoever, it would not be able to travel at the speed of light. Photons travel at the speed of light without any more energy added than they exist out of. They travel at C as a direct result of having zero mass.
That's a very circular statement. We only know the speed of light by measuring how fast photons move. (Or measuring the time it takes for electrical potentials to be detected in an electrical circuit, but that can be interpreted using photons, and usually isn't measured in a vacuum anyway.)

Even so, how would be notice the difference between the rate an object moves that has a mass of 0.00000000000000000000000000000000000000000000001 picograms and an object that actually had perfect zero mass? Would it move like maybe 1 kilometer per hour slower or something?

I totally forget how plank's constant applies to all this, but I'm pretty sure small enough masses can approach the speed of light, even if they don't quite reach.

22. Alright, here's what I understand to be the formula prohibiting faster than light movement, taken from the Wiki-pedia article on special relativity:

Given an object of invariant mass m traveling at velocity v the energy and momentum are given (and even defined) by

E = \gamma m c^2 \,\!

\vec p = \gamma m \vec v \,\!

where γ (the Lorentz factor) is given by

\gamma = \frac{1}{\sqrt{1 - \beta^2}}

where \beta = \frac{v}{c} is the ratio of the velocity and the speed of light. The term γ occurs frequently in relativity, and comes from the Lorentz transformation equations.
That may not have pasted right, so you should look here:

http://en.wikipedia.org/wiki/Special_relativity

Basically, Gamma approaches infinity as Beta approaches 1 (which means velocity is approaching C). In order to cancel this in the equation :

E = (Gamma) * M * C^2

M has to approach zero as Beta approaches C then. This clearly means that a non-zero mass could approach C so closely as to be indistinguishable from C (maybe only moving 1 kilometer per hour slower, or less) , so long as the mass was very very small.

Basically, we can never reasonably rule out a photon having *some* mass. We can only safely say that, if it has any mass, it's very small.

23. Originally Posted by kojax
Alright, here's what I understand to be the formula prohibiting faster than light movement, taken from the Wiki-pedia article on special relativity:

Given an object of invariant mass m traveling at velocity v the energy and momentum are given (and even defined) by

E = \gamma m c^2 \,\!

\vec p = \gamma m \vec v \,\!

where γ (the Lorentz factor) is given by

\gamma = \frac{1}{\sqrt{1 - \beta^2}}

where \beta = \frac{v}{c} is the ratio of the velocity and the speed of light. The term γ occurs frequently in relativity, and comes from the Lorentz transformation equations.
That may not have pasted right, so you should look here:

http://en.wikipedia.org/wiki/Special_relativity

Basically, Gamma approaches infinity as Beta approaches 1 (which means velocity is approaching C). In order to cancel this in the equation :

E = (Gamma) * M * C^2

M has to approach zero as Beta approaches C then. This clearly means that a non-zero mass could approach C so closely as to be indistinguishable from C (maybe only moving 1 kilometer per hour slower, or less) , so long as the mass was very very small.

Basically, we can never reasonably rule out a photon having *some* mass. We can only safely say that, if it has any mass, it's very small.
The problem is that while a near-zero mass photon could travel at very near c, it is not restricted to traveling at only that speed, and could travel at any speed from zero to near c. Thus, if such were the case, we should have photons in existence that travel at these various speeds and we should have detected them. But we haven't.

The other point to make is that Maxwell's equations require that light in a vacuum travels at c.

24. That's actually a really good point. Light would have the ability to travel at less than C if it had mass.

There's one possibility where it wouldn't though: If traveling at less than C meant it had less than one quanta of energy.

As for Maxwell, the "speed of light" in the sense of his equations doesn't necessarily have to be the speed that real photons of light travel at. It could be a "speed of light" in the sense of the limit set by Einstein, but which photons traveling in a vacuum never quite obtain. I say this because, in electrical systems, the energy is a literal wave with no real particle attributes. (Or rather it's being carried by the electrons in the wire, but in the same sense as how air carries a sound wave. The electrons aren't moving a C themselves.)

25. Originally Posted by kojax
That's actually a really good point. Light would have the ability to travel at less than C if it had mass.

There's one possibility where it wouldn't though: If traveling at less than C meant it had less than one quanta of energy.
"A quantum of energy" is just the the amount of energy a single photon has
and this depends its wavelength. You can look at it as the minimum amount of energy a particular frequency of light can have. But there no such thing a a minimum size a quantum can have. IOW, not all quantum are equal. Thus one quantum of red light is less energetic than one quantum of blue light.
But this energy also relies on the speed of the photon. So you could have a "blue" photon that travels at a slower speed as a "red" photon, and by virtue of of its slower speed have a quantum of energy equal to that of the "red" photon.

As for Maxwell, the "speed of light" in the sense of his equations doesn't necessarily have to be the speed that real photons of light travel at. It could be a "speed of light" in the sense of the limit set by Einstein, but which photons traveling in a vacuum never quite obtain.
The Maxwell equations say that electromagnetic radiation, by its very nature has to travel at a speed of

So no, this cannot be just viewed as an upper limit.

26. Photons of different wavelengths have different energies. A "quantum" is the minimum amount by which the energy of a photon can vary. IE. One photon's energy might be quite a few quanta.

As for Maxwell's equations, you're right that the electromagnetic radiation in a circuit is traveling at the full speed of light, but it's not a photon. EM radiation traveling through a wire is literally a wave being carried by the electrons in the metal. It has no particle nature whatsoever, only a wave nature. (Whereas light traveling through a vacuum can be seen as both a particle and a wave)

27. Originally Posted by kojax
Photons of different wavelengths have different energies. A "quantum" is the minimum amount by which the energy of a photon can vary. IE. One photon's energy might be quite a few quanta.
Wrong. A photon is a quantum. One photon = one quantum of that particular wavelength of light.

As for Maxwell's equations, you're right that the electromagnetic radiation in a circuit is traveling at the full speed of light, but it's not a photon. EM radiation traveling through a wire is literally a wave being carried by the electrons in the metal. It has no particle nature whatsoever, only a wave nature. (Whereas light traveling through a vacuum can be seen as both a particle and a wave)
What in the world are you talking about? You're confusing electromagnetic radiation with electricity. Electromagnetic radiation is just a more inclusive name for radio waves, microwaves, visible light, gamma rays, etc. All of which, according to Maxwell's equations propagates at a the speed determined by the formula already given.(Which by the way is for electromagnetic radiation in a vacuum.)

28. I'm with Janus on this...not only have we always measured light in a vacuum to travel at C and only C, but we can also derive that it would have to go that speed based on our understanding of the fundamental laws of electromagnetism (the maxwell equations).

29. I think sometimes I say stupid stuff just to get those kind of explanations out of people (or more likely it's a fortunate side effect of a foolish practice).

I think I must be confusing photons with that part of light that is supposed to give it a particle-like quality. I thought they were supposed to be the particle item that Newton believed in, and that they had wavelengths in much the same manner as how an electron traveling through space has a wavelength.

If a photon and a quantum are the same thing, then why do we trouble ourselves to have two words?

30. "Quantum" is a name given that was supposed to mean "packet of energy", quanta being the plural. Photon is the name of the particle, that is a quantum of light energy. Or a better explanation: Quantum.

31. And a graviton is a quantum of gravitational radiation energy.

32. Alright, this is from wiki, and it's still vague on the issue:

The energy and momentum of a photon depend only on its frequency ν or, equivalently, its wavelength λ:

E = \hbar\omega = h\nu = \frac{h c}{\lambda}

\mathbf{p} = \hbar\mathbf{k}

and consequently the magnitude of the momentum is

p = \hbar k = \frac{h}{\lambda} = \frac{h\nu}{c}

where \hbar = h/2\pi \! (known as Dirac's constant or Planck's reduced constant); \mathbf{k} is the wave vector (with the wave number k = 2π / λ as its magnitude) and ω = 2πν is the angular frequency. Notice that \mathbf{p} points in the direction of the photon's propagation. The photon also carries spin angular momentum that does not depend on its frequency.[16] The magnitude of its spin is \sqrt{2} \hbar and the component measured along its direction of motion, its helicity, must be \pm\hbar. These two possible helicities correspond to the two possible circular polarization states of the photon (right-handed and left-handed).
It says the energy or momentum of a photon depends on its wavelength. Does that not imply that one photon could have more or less energy than another?

A quantum, however, is a unit of measurement, like an inch or a gram. You can't have an extra energetic quantum any more than you can have an extra long inch or an extra heavy gram.

33. A quantum, however, is a unit of measurement, like an inch or a gram. You can't have an extra energetic quantum any more than you can have an extra long inch or an extra heavy gram.[/quote]

There is no fixed energy value to a quantum. It is not a unit of measurement. If it were you would be able to type "quantum" into google, and one of the first things it would tell you is what the value of one quantum is. I.E. "one quantum equals 10^-x ergs"

34. To All

A MAJOR Discovery:

I was thinking about the Planck Constat as having a real physical substitution for the mathematical expression that appears to be the smallest unit of energy in the universe.

I FOUND IT.

Working on the principal of modifying the deBroglie’s wave formula and Einsteins M/E formula, I decided to use the following components where
h replaces E for the energy source, m sub e as representing matter and l replaced 'c' with Lambda (wavelength) as a single unit of light as the smallest unit of energy.
Since the unknown here is the wavelength, I worked the formula to solve for Lambda.
m sub e = 9.109^-31 = electron mass. So:

Lambda = square root of h divided by m sub e = 6.626^-34/9.109^-31
= root of 7.274^-4 = 2.697^-2 meters

Rearranging the formula for solving the Constant, we have
h = (2.697^-2)^2 x m sub e.

Thus this Lambda has a much lower value than the CMBR wave length of .002 meters that = the temperature of 2.73K.

Weins formula reduces the temperature to a wavelength of one millimeters but at the low end of the radiations, it errs to lower values.
The Rayleigh formula is considered accurate for the low end of the radiation and reduces to v^2T
However, this formula would not be applicable because photons have frequencies of just 'one' as single quanta's.

So my solution for the smallest source of energy in the universe is the 26.97 millimeter wavelength that represents the Planck Constant.

The value (2.73K) than represents the 'equalized' space temperature in accordance with the 2nd Law of Thermodynamics, where heat redistributes itself to create one uniform temperature.

So the PC wavelength has a much lower energy than the CMBR of .002 meters
or 2 millimeters. Shorter wavelengths have higher energies.

Edited for some corrections and modifications from the original post.

Cosmo

35. Originally Posted by Janus
A quantum, however, is a unit of measurement, like an inch or a gram. You can't have an extra energetic quantum any more than you can have an extra long inch or an extra heavy gram.
There is no fixed energy value to a quantum. It is not a unit of measurement. If it were you would be able to type "quantum" into google, and one of the first things it would tell you is what the value of one quantum is. I.E. "one quantum equals 10^-x ergs"[/quote]

Originally Posted by wikipedia

(plural: quanta) is an indivisible entity of a quantity that has the same units as the Planck constant and is related to both energy and momentum of elementary particles of matter (called fermions) and of photons and other bosons.

You'd be surprised how hard it is to get a straight and sensible explanation from most of the online sources out there. So.... you're saying a quantum can have any value out of a certain range of energies, and it's still one quantum?

Behind this, one finds the fundamental notion that a physical property may be "quantized", referred to as "quantization". This means that the magnitude can take on only certain discrete numerical values, rather than any value, at least within a range. There is a related term of quantum number.

So what do they mean by saying the magnitude can only take on discreet values? I had always assumed that a "quantum" was the amount of energy between one value and the next.

I'm not doubting you. I just want to understand this, especially given that my initial impression appears to have been wrong.

36. Originally Posted by kojax

So what do they mean by saying the magnitude can only take on discreet values? I had always assumed that a "quantum" was the amount of energy between one value and the next.

I'm not doubting you. I just want to understand this, especially given that my initial impression appears to have been wrong.
Let's say you have a photon of ultraviolet light. It has a certain energy related to its wavelength. This is the smallest "unit" of energy you can have for this particular wavelength of light. It is also the smallest step you can take between different energy levels of this wavelength. It is one quantum of this light.

Now say you have a photon of infrared light. It also has a certain energy, but it is less than the energy of the ultraviolet photon. It is also the smallest unit of energy you can have for this wavelength. It is one quantum of infrared light.

One quantum of infrared light has less energy than one quantum of ultraviolet light, But each is the smallest discreet value each light can have.

37. One quantum of infrared light has less energy than one quantum of ultraviolet light, But each is the smallest discreet value each light can have.
But the EM spectrum is a continuous spread of frequencies, no? Aren't there then quanta of light with energies corresponding to frequencies all through the spectrum, with the distinction between whether it is an IR or UV quanta only related to if the energy of the quantum corresponds to a frequency that is classified to either or in between? In other words, that quanta can have any energy within certain maxima and minima translating to where it falls in the spectrum?

Edit: On second thought, these photons would be quantized according to the change in the energy level of the electrons that emitted the photons, which can only have certain values. But that is not the only source of photons. Are the other sources similarly quantized? Could one then figure out the origin of the photon by measuring its energy (i.e. if it corresponds to an electron orbital energy level or some other process)

38. Originally Posted by Janus
Originally Posted by kojax

So what do they mean by saying the magnitude can only take on discreet values? I had always assumed that a "quantum" was the amount of energy between one value and the next.

I'm not doubting you. I just want to understand this, especially given that my initial impression appears to have been wrong.
Let's say you have a photon of ultraviolet light. It has a certain energy related to its wavelength. This is the smallest "unit" of energy you can have for this particular wavelength of light. It is also the smallest step you can take between different energy levels of this wavelength. It is one quantum of this light.
Ok, so let me get this straight: A beam of light can have absolutely any wavelength then? There isn't a quantum limit on what discreet values the wavelength can be?

Or is it a matter that, the balance of frequency and intensity must always together add up to certain discreet values?

Or does frequency have nothing to do with it (other than determining what the quantum limits will be)?

Or a better way to ask: Is the quantum limit smaller for UV light, or is it smaller for IR light?

Now say you have a photon of infrared light. It also has a certain energy, but it is less than the energy of the ultraviolet photon. It is also the smallest unit of energy you can have for this wavelength. It is one quantum of infrared light.
So, I'm wondering, does amplitude (ie. the intensity) of the wave vary by the same amount in both cases?

One quantum of infrared light has less energy than one quantum of ultraviolet light, But each is the smallest discreet value each light can have.
So, the energy of a beam of light is able to assume a wider range of values at lower frequencies than it can at higher frequencies? (Assuming the source is just as powerful in both cases)

I mean, if we consider that a higher frequency of light can carry more energy (and I'm not sure it can), then I guess the range of values would be the same, because there's more distance between the maximum amount possible to send, and zero, but there's also wider gaps.

39. Originally Posted by kojax
Ok, so let me get this straight: A beam of light can have absolutely any wavelength then? There isn't a quantum limit on what discreet values the wavelength can be?

Or is it a matter that, the balance of frequency and intensity must always together add up to certain discreet values?

Or does frequency have nothing to do with it (other than determining what the quantum limits will be)?

Or a better way to ask: Is the quantum limit smaller for UV light, or is it smaller for IR light?
Light can have any frequency. The frequency determines how large the "steps" are between allowed energy levels of a given frequency. UV light has larger steps than Infrared.
As an analogy, Ultraviolet light can be considered course sand compared to Infrared being fine sand. You can have equal amounts, off each total mass wise, but you can divide the infrared up into smaller constituent parts.

so, I'm wondering, does amplitude (ie. the intensity) of the wave vary by the same amount in both cases?
If you have x number of photons(quanta) of each, the ultraviolet will be of greater intensity. If you have equal intensity of each, there will be fewer photons(quanta) of ultraviolet

So, the energy of a beam of light is able to assume a wider range of values at lower frequencies than it can at higher frequencies? (Assuming the source is just as powerful in both cases)

I mean, if we consider that a higher frequency of light can carry more energy (and I'm not sure it can), then I guess the range of values would be the same, because there's more distance between the maximum amount possible to send, and zero, but there's also wider gaps.
It's not just a matter of range of values, it is also a matter of how the energy can be delivered.

Another analogy: Say you have two different frequencies of light. The higher frequency is represented by lead bullets and the lower by ping-pong balls.

You have an equal mass(total beam intensity) of both, and you fire them at a concrete wall at equal speed. The total energy imparted to the wall is the same in both cases.
The higher frequency light is delivered in "packets" which are both more compact (shorter wavelenth) and have a larger individual energy.)

You could sit and fire ping-pong balls at the wall all day, and the wall would be none the worse for wear. Bullets, however will chip pieces off the wall, And the same total mass of bullets fired at the wall over the same period of time will cause significant damage.

40. Sorry to take so long replying. I had to wrap my head around this concept. So, basically, the intensity of light is quantized (the amplitude of the wave has to be certain discreet amounts), but its wavelength is not?

It looks like the relationship is linear:

Originally Posted by wiki

....however, as a particle, it can only interact with matter by transferring the amount of energy

E = \frac{hc}{\lambda},

where h is Planck's constant, c is the speed of light, and λ is its wavelength. ....
So the amplitude of the wave is always quantized by the same amount, regardless of wavelength. It's just that the energy depends on both amplitude and wavelength.

I should totally print this whole page of discussion off, and read it when I get bored in class.

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