# Redshifting help?

1. Sep 3, 2009

### Evolver

Okay, I understand the principle of redshifting, but I feel I am missing an element to it's comprehension because as is it doesn't make complete sense to me.

I understand the Doppler effect of sound waves, as in a train passing by; the waves in front are of a higher frequency due to the motion of the train producing them and subsequently lower frequency as the train passes.

Now, the redshifting of light from distant galaxies traveling away at high speeds is where I am confused. I am lead to believe that the galaxies are generating a similar effect in that the light waves are lower frequency behind the moving body and thus appear redshifted (and if the galaxies were heading towards us would appear blueshifted.)

But this makes no sense to me. According to relativity, light in a vacuum travels at a constant speed regardless of the motion of the body that it is emitted from. Therefore, indicating that light is affected by the motion of galaxy is not consistent with this line of thinking.

I do realize it probably has something to do with time dilation, but I am not aware of how. Please someone help me grasp this concept.

Thanks!

2. Sep 3, 2009

### nutgeb

First, you need to distinguish between two kinds of light redshifting, SR Doppler and Cosmological.

SR Doppler redshift occurs as between two inertial frames - think of one as being stationary and the other moving. SR Doppler redshift is the combination of two components (which actually are inseparable). In the case of redshift, the light photons lose energy as a result of the need to accelerate themselves such that they travel at exactly c relative to the observer. This is classical Doppler redshift. You are correct that the speed of light - c- does not vary in any inertial frame. But the energy of a photon does decrease when observed in frame that is receding away from the emitter. The lower the energy of a photon, the longer its wavelength. Longer wavelength = redshift.

The second component of SR Doppler redshift is time dilation. From the perspective of the observer, the emitter is time dilated because it is in motion relative to the observer. Dilated time = slower clock = lower frequency of wavecrests = longer wavelength = redshift. The SR Doppler formula simply multiplies the classical Doppler redshift by the emitter's time dilation.

Cosmological redshift is a different but related phenomenon which occurs in an expanding, gravitational universe. At a simple level, cosmological redshift is exactly proportional to the distance of the emitting object at the time of observeration. Another way to say it is that the cosmological redshift is exactly proportional to the amount by which the scale factor of the universe has expanded in the interval between emission and observation.

But these descriptions of the cosmological redshift do not explain HOW it occurs. It is fairly certain that the 'expansion of space' does not itself directly cause the distance between wavecrests to expand. The 'expansion of space' is inherently incapable of causing an increase in separation between any two things unless those things were already moving apart in the initial condition, and retain some sort of separational momentum (like physical objects such as galaxies have). Over a small time period, all of the wavecrests emitted by a receding emitter have exactly the same fixed wavelength, so the crests cannot be said to be in motion relative to each other in the initial conditions. In addition, the concept of retained separational momentum cannot be meaningfully applied to intangibles such as wavecrests.

SR Doppler redshift does not occur in a cosmological context between galaxies which are receding away from each other exactly at the pace of their local Hubble flow, because no time dilation exists as between fundamental comovers in the FRW metric, which models cosmological expansion.

3. Sep 3, 2009

### sylas

Think about your train example again. The sound waves are also traveling at the same speed with respect to you, no matter whether the train is approaching or receding.

You are also correct about blue shift for an approaching galaxy. Andromeda, for example, is approaching us, and is blueshifted as a result.

Time dilation does mean that the formulae used are a bit different; but you still get a Doppler shift even without any time dilation effects -- as the train example shows.

Cheers -- sylas

PS. This is not a question for which you need to worry about all the details of large scale cosmology. For nearby galaxies, there's no meaningful difference between cosmological redshift and Doppler redshift. The cosmological factors of large scale spacetime become important on very large scales; but not for small redshift galaxies.

Last edited: Sep 3, 2009
4. Sep 3, 2009

### javierR

The speed of light is indeed a constant here, but the doppler effect has to do with observed frequencies of the light, and these can change from observer to observer. I haven't checked the quality of the wikipedia entry, but I'll assume it's alright:
http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

5. Sep 3, 2009

### JesseM

SR Doppler redshift has nothing to do with photons "losing energy as a result of the need to accelerate themselves", it's just time dilation plus the effect of the source moving away from you (or towards you in the case of blueshift), which means that each successive wave peak is emitted at a greater distance from you and thus has longer to travel to reach your eyes than the previous peak. See my post #12 on this thread for a numerical example.

6. Sep 3, 2009

### nutgeb

I agree that each wave has longer to travel.

But are you disagreeing that a redshifted photon has less energy than a non-redshifted one?

7. Sep 3, 2009

### nutgeb

I agree that each wave has longer to travel, which can be correctly described as causing the longer wavelength.

But, of course, a redshifted photon has less energy than a non-redshifted one, and my description is equivalently correct.

Last edited: Sep 3, 2009
8. Sep 3, 2009

### JesseM

How is it equivalent? For one thing the relation between energy and wavelength depends on quantum mechanics, whereas my explanation works for classical EM waves too. And I don't see how you'd actually derive the relativistic Doppler shift from the QM energy/wavelength relation (or from the energy/wavelength relation + time dilation--you appear to have said the energy change has to be considered separately from the time dilation effect)...can you provide an actual derivation of the Doppler shift equation in this way, or a numerical example showing how you can get the correct shift using the energy/wavelength relation? If not I am doubtful that your statement makes any sense at all.

9. Sep 3, 2009

### nutgeb

Of course the energy of light is proportionally reduced, as measured by an observer in relative motion away from the emitter, compared to the energy measured by an observer stationary in the inertial frame of the emitter.

If you disagree with that statement, please state simply, "I disagree."

10. Sep 3, 2009

### JesseM

I do not disagree with that statement on its own. But I see no way to derive the Doppler shift equation from the fact that the energy is lower in the frame where the emitter is moving, and that is what you were saying when you said "In the case of redshift, the light photons lose energy as a result of the need to accelerate themselves such that they travel at exactly c relative to the observer. This is classical Doppler redshift." Unless you can show how to derive the Doppler shift from the different energy in the two frames, then this statement is a complete non sequitur, akin to saying "the length of the emitter is contracted in the frame where the emitter is moving. This is classical Doppler redshift". The first sentence is of course true on its own terms, but it has nothing to do with the explanation for the Doppler shift.

11. Sep 4, 2009

### nutgeb

Jesse, I'm not going to get into an extended argument on this point. I agree with your terminology for describing how the classical redshift applies to light, it's a good description.

The classical Doppler formulas, based on velocity and wavelength, are as follows (where outward relative velocity has a positive sign):

Moving emitter, stationary observer: $$\lambda _{o} = \lambda _{e} (1 + v_{e})$$

Moving observer, stationary emitter: $$\lambda _{o} = \lambda _{e} / (1 - v_{o})$$

12. Sep 4, 2009

### JesseM

OK, and these formulas don't seem to be derivable from energy considerations as you were suggesting--that's the only point I was making.

13. Sep 7, 2009

### nutgeb

[Edit: Upon further consideration, I revised the first sentence as follows: The classical Doppler shift is required by energy conservation considerations, and the redshifted energy of a photon in an observer's frame can be calculated in theory by measuring the emitter's atomic recoil in the observer's frame.]

Consider a hydrogen atom that emits a photon, which subsequently is absorbed by an observer moving away from the atom. The atom will recoil away from the observer. The sum of the atom's recoil energy and the photon's redshifted energy, both as measured in the observer's frame, must equal the energy of the atom's motion (relative to the observer) just before the photon was emitted. Classical Doppler shift of the photon is required in the observer's frame in order to avoid violating the conservation of energy.

The energy of a photon in a frame is equal to the Planck Constant times the photon's frequency as measured in that frame.

Last edited: Sep 7, 2009
14. Sep 7, 2009

### JesseM

First of all, a classical or relativistic derivation of the Doppler shift shouldn't depend on knowing formulas from quantum mechanics. Second, even if we assume we know the relation between energy and frequency from QM, it's not clear this is actually helpful in getting the Doppler equation, which is supposed to deal with the case of an emitter moving at constant velocity, not an emitter that recoils and changes velocity each time it emits a photon. The Doppler equation compares the frequency in two frames, the emitter frame and the observer frame, it's not clear what you want the equivalent of the "emitter frame" to be in the above argument--the rest frame of the atom after it has emitted the photon, or before?

If you want to continue to defend the claim that the Doppler equation can be derived from energy considerations, you need to actually provide a derivation with equations--or at least provide a numerical example--the verbal arguments above are much too handwavey.

15. Sep 7, 2009

### nutgeb

Every emission of every photon involves a recoil. This is not QM in any specialized sense, it's basic physics.
As you know, the Doppler equation compares the redshift to the emitter frame before the effect of recoil is considered. The atom's recoil frame is brought into the analysis specifically as a way to derive the redshift from energy conservation. As you know, it is most straightforward to analyze energy conservation with regard to a complete system rather than by reference to pieces of the system in isolation.
This is simple energy conservation. The momentum ptot of the total atom+photon system remains exactly constant before and after emission. Therefore the momentum of the redshifted photon must be ptot - patom, where the latter is the momentum of the recoiled atom as measured in the observer's frame. {Edit: In order to balance the total system momentum to compare vectors in both the +x and -x directions, the atom's momentum needs to be expressed as a negative number, so the redshifted photon's momentum should be expressed as ptot + patom.} Obviously the recoiled atom's momentum (in the observer's frame) is simply the atom's mass times its [initial velocity + recoil $$\Delta$$ velocity].

Last edited: Sep 7, 2009
16. Sep 7, 2009

### sylas

That is not a derivation with equations. It's still hand wavey. Try giving an actual mathematical derivation using equations. Derivations means you express the assumptions as equations (not as English) and get your result with algebra (not with an English declaration that it is "simple").

17. Sep 7, 2009

### JesseM

I was talking about the quantum relationship between the energy of a photon and its frequency/wavelength, which you seem to be assuming (otherwise how are considerations of energy/momentum supposed to tell us anything about frequencies?)
Typically I think the Doppler equation just assumes the effects of recoil can be treated as negligible because the momentum of the emitted waves are very small compared to the momentum of the emitter. In any case, if you want to think in terms of the frame where the atom is at rest before emitting the photon, then actually show how it works using this frame and the fame of the observer. If you don't want to do a general derivation, just give a specific numerical example where we consider an atom in motion at some specific velocity relative to an observer, consider it emitting a photon of some specific frequency, and then (somehow, because you have in no way explained it) use energy considerations to figure out both the frequency in the frame where the atom was at rest before emission, and the frequency in the observer's frame. If you can't even provide a simple numerical example, then it's obvious you don't have an actual well-defined argument in mind.
Huh? You just said you wanted to consider the emitter's frame "before the effect of recoil is considered". Just tell me, what frame are you using for the "emitter frame" in the standard Doppler equation? Is it the frame where the atom is at rest before emitting the photon, or the frame where it is at rest afterward?
All of this is pretty obvious, but it gives me no idea of how you intend to derive the redshift equation, which relates frequency in the observer's frame to frequency in the emitter's frame. In the above you appear to be talking solely about the momentum of the photon and atom in the observer's frame before and after the emission, with no reference to any other frame.

Please, no more vague verbal arguments. If you have a well-defined idea about how to derive the Doppler equation from energy considerations, it should be exceedingly simple to give a numerical example where you actually use energy considerations to calculate the frequency of the photon in the observer's frame and some other frame (what that other frame is, you still haven't made clear) and show that they match what's predicted by the Doppler equation (classical or relativistic, I'm not sure...if it's not relativistic, does that mean you won't assume the photon moves at c in both frames?) If you don't have any clear idea of how to do this, that's fine, but then please acknowledge that your ideas are tentative and don't confidently assert that the Doppler relation can be derived from energy considerations if you don't have a definite procedure in mind.

Last edited: Sep 7, 2009
18. Sep 7, 2009

### nutgeb

The photon recoil analysis is a way of relating a photon's redshifted energy to the total system energy, and to demonstrate that Doppler shift is required for energy conservation. But beyond that, I don't think it is the easiest path to analyze the workings of classical redshift. I have the recoil analysis written out in mathematical terms but sharing it will not advance this discussion.

Let me return to what I intended in my original post. You have used a lot of words to characterize a simple statement of mine, most of which are twisting my words.

A photon's energy in an observer's frame is trivially derived from its momentum in that frame: E = pc.

Further, it is a basic element of the classical Doppler shift that a photon's momentum is reduced by the factor of (1-v)/c if the observer is moving away from the emitter.

It is absurd to challenge these basic propositions or demand that I provide proof of something else. I never intended to suggest that a photon's redshifted energy shift derives from anything other than the decrease in its momentum due to chasing a moving observer. That's why I was perfectly willing to accept the way you described it.

19. Sep 7, 2009

### JesseM

By "recoil analysis" do you mean a derivation of the redshift starting from considerations of energy and/or momentum? If so, providing it would certainly advance the discussion--it would either show you do have the derivation I've been asking for, or myself and others could point out any flaws we found in the analysis.
Your "simple statement" was: In the case of redshift, the light photons lose energy as a result of the need to accelerate themselves such that they travel at exactly c relative to the observer. This is classical Doppler redshift. Then in a subsequent post you said The classical Doppler shift can be derived from energy considerations alone. Is it "twisting your words" to understand these as claims that the correct value for redshift can be derived from considerations of energy or momentum?

If you do think I am twisting your words, the solution is simple: say something like "no, I do not claim the Doppler equation can be derived just from considerations of the energy/momentum in both frames." On the other hand, if my interpretation of your words was correct and you do claim such a derivation is possible, then all you need to do is provide it (as I said, a numerical example rather than a full derivation would be convincing too). But if you won't do either of those things, you're just being evasive, and probably running afoul of the forum rules about not using the forum to advance original claims that are not part of mainstream physics (or that aren't demonstrably derivable from mainstream physics).
Strictly speaking you can only "derive" that by starting from the relativistic equation $$E^2 = m^2*c^4 + p^2*c^2$$ and then including the fact that a photon has zero rest mass. Didn't you say you wanted to talk in classical terms rather than relativistic ones? In relativistic terms your earlier statement about the atom's momentum just being its mass times its velocity would be incorrect, for example.
What are you talking about? The classical Doppler shift equation says absolutely nothing about a photon's momentum, it deals with frequency of a wave in different frames. In a classical universe, if a wave is sent out at c relative to the emitter, and the emitter is moving away from the observer at v, then the observer will see the frequency reduced by 1/(1 + v/c). In quantum physics we also know that momentum is proportional to frequency according to the formula p = h*frequency/c, so if we carelessly combine the quantum formula with the classical one (though this would be physically dubious since the classical formula assumes the wave has a speed c-v in the observer's frame, while the quantum formula assumes photons move at c in whatever frame the formula is being used) then we could conclude the momentum is also reduced by 1/(1 + v/c), which is different from your equation (1-v)/c (a formula that doesn't even seem to make sense dimensionally since v and c have units of distance/time while 1 is dimensionless).
Your (1-v)/c formula for momentum reduction, which supposedly comes from "classical Doppler shift", is not a basic proposition. And it's not absurd to demand that you provide proof of what you asserted at the beginning about Doppler shift coming from energy considerations which got this whole debate going, something you appeared to continue to assert when you said "The classical Doppler shift can be derived from energy considerations alone." If you wish to back off from this claim, then do so.
The debate was never about a photon's "redshifted energy", it was about the Doppler equation which deals with the frequency of a wave (of photon if you prefer) in both frames. Do you or do you not claim this is derivable from considerations of momentum/energy? If you do not, please state that clearly, and if you do, please provide a derivation or a numerical example.

20. Sep 7, 2009

### nutgeb

This is my last post on this subject. It is impossible to have a rational dialogue with you when you are twisting my words and spraying arguments in all directions. You can post whatever you want on this topic as long as you direct it to the OP or others, and leave me and my words out of it.

For the sake of clarity, the following will clarify and REPLACE ALL prior statements I've made on this subject:

A photon's energy as measured locally in the observer's frame is a direct function of its momentum. From Wikipedia article "Photon":

"In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p. For comparison, the corresponding equation for particles with mass m is: $$E^2 = m^2 c^4 + p^2 c^2$$"

Classical Doppler effect causes an observer moving away from the emitter to measure a photon to have a lower momentum than it had in the emitter's frame. This is true regardless that the simple form of the classical Doppler formula does not include a momentum term.

Energy, momentum, frequency and wavelength are all directly related and go hand-in-hand with respect to redshift. Whether the reduction in momentum is caused fewer wave crests hitting the observer in a given period of time, or by an increase in the wavelengths between received crests, or simply from being received in a moving frame, is not crucial to the OP's question. As a practical matter it amounts to the same thing. Various "hand-waving" descriptions can reasonably describe the outcome.

The recoil of an emitter atom demonstrates that the classical Doppler shift of the photon complies with energy conservation.

*** FINIS ***

Last edited: Sep 7, 2009