# B Why do photons allow Doppler shift

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1. Oct 25, 2017

### Buckethead

If we (a detector) are moving toward a star that emits a single photon (due to its distance) and that photon hits our detector, it will be blue shifted. My question is why. If the color of a photon is a reflection of its energy level and since the speed of the photon is always coming at us at c irrespective of the speed at which we are traveling toward the star, then why does the color of the photon change if we increase our speed?

2. Oct 25, 2017

### Paul Colby

In current theory light is composed of electromagnetic waves that have quantized amplitudes. The wave (of which the photon is the energy (and momentum) absorbed by the detector thus changing the waves state by one amplitude step) is Doppler shifted just like a classical wave would be.

3. Oct 25, 2017

### Buckethead

But why, even if we view the photon as a wave, would it be Doppler shifted if its velocity relative to the detector does not change? It seems to me this is equivalent to being in a plane and talking to someone while the plane is in the air. The velocity of the sound wave does not change relative to you (just like light in my example) and as a result its pitch does not change.

4. Oct 25, 2017

### Staff: Mentor

If you consider the receiver to be at rest, then the source is moving towards the receiver. This does not affect the speed with which the waves approach the receiver, but every successive wave crest has to cover slightly distance than the one immediately before it. Thus, the time between the reception of two consecutive crests is not the time between their emission; it is somewhat less because the second crest travels a shorter distance so arrives just a bit sooner than it otherwise would. This will probably be clearer if you try drawing a spacetime diagram showing the paths of successive wave crests through spacetime.

That explains the classical Doppler effect. There is an additional relativistic correction from time dilation because the emitter is moving relative to the receiver.

5. Oct 25, 2017

### Mister T

The speed is the same but the energy is different. It's the energy, not the speed, that determines the frequency (color).

For massive particles the increase in speed with respect to energy approaches zero as the speed approaches $c$. In other words, as you approach speed $c$ huge increases in energy produce negligible increases in speed. My point is that the relationship between speed and energy is strange compared to the non-relativistic relationship and the photon is a purely relativistic particle.

6. Oct 26, 2017

### Buckethead

I'm specifying a single photon which left the star mellenia ago, when the sensor was at rest and then later accelerated to some modest speed toward the star, so the relative speed between the star and the observer couldn't be a factor
But why is the energy different? If the speed of the photon never changes relative to the observer, then why would its energy change?

7. Oct 26, 2017

### Staff: Mentor

There's no quantum mechanics involved in this problem (aside from statistical effects at the receiver, which are a distraction here) so no photons involved - you have a flash of light travelling from the emitter to the receiver. This flash of light is an electromagnetic wave.

If the receiver is approaching the source, then the crests of the wave will be closer to one another using the frame in which the receiver is at rest than using the frame in which the emitter is at rest. This will become clear if you draw a diagram.

The light is moving at $c$ in all frames, but the wavelength is shorter in the frame in which the receiver is at rest. Therefore the frequency and the energy are greater. This is just another example of the general fact that kinetic energy is always frame-dependent; the only surprising thing is that the energy carried by a flash of light is dependent on the frequency and wavelength, not the speed.

8. Oct 26, 2017

### Buckethead

I'm good with this since the property of light (wave or particle) depends on the experiment you are performing, so wave it is.
I'm not sure how to draw a diagram to reflect this, and I get what you are saying with regard to viewing this as the receiver at rest, but let's look at this another way. Suppose the emitter and receiver are at rest and the star emits a short pulse. Sometime later before the light reaches the receiver, the emitter accelerates toward the receiver then coasts to a constant speed. I expect the receiver will detect no color change. Now reverse the experiment and instead accelerate the receiver then coast to a constant speed just before the pulse reaches the receiver. I expect the receiver will see a blue shift. First, is this correct? If so I find this curious since when the pulse was emitted, the relative speed between the two was 0. And at the time the receiver finished accelerating and was in coast mode, its speed relative to the emitter is no longer relevant since the pulse is in space between the two. In other words, the emitter could explode and vanish so that all that's left is the receiver and the pulse of light somewhere in space. The receiver is moving at c relative to the pulse of light and yet when it arrives it will be blue.

The only difference between the emitter accelerating and the receiver accelerating is the acceleration itself. But if the accelerations occur only during the time the pulse is between the emitter and receiver, and if both are moving at a constant velocity relative to each other when the pulse is still in between, then in both cases the situation between the pulse and the receiver is identical. The relative speed between the pulse and receiver remains at c in both cases and in both cases the speed of the emitter should be irrelevant since what the emitter is doing in both cases is isolated from the pulse already on its way.

9. Oct 26, 2017

### PAllen

Another way to look at this is purely kinematically. 4 momentum is a vector that transforms per the Lorentz transform. If something has a 4 momentum in an emission frame, it will have the Lorentz transform of that 4 momentum in another frame, e.g. the receiver. If you Lorentz transform (E,p), then specialize to that case of E=p for a massless particle (or light), you get the relativistic Doppler formula. So given that light must have energy and momentum, it must undergo relativistic Doppler between frames due to Lorentz invariance. I should say, you get the Doppler formula applied to E. Then the energy of a photon determines its frequency.

Last edited: Oct 26, 2017
10. Oct 26, 2017

### Staff: Mentor

Yes.
If the light is moving to the right in some given frame, and the frequency of the light is $\nu$ as measured by an observer at rest in that frame:
- If the receiver is moving to the left in that frame at the moment of reception then the frequency in the frame in which the receiver is at rest will be $\nu+a$; the receiver will measure a blueshift.
- If the receiver is moving to the right in that frame at the moment of reception then the frequency in the frame in which the receiver is at rest will be $\nu-b$; the receiver will measure a redshift.
What happens to the emitter after the light is emitted is irrelevant; all that matter is that the frequency was $\nu$ in the frame in which the emitter was at rest at the moment of emission. Likewise, what happens to the receiver before the light is received is irrelevant; all that matters is the speed of the receiver at the moment of reception.
We get different results in the emitter-accelerates and the receiver-accelerates cases because after the acceleration:
- in the first one the receiver is at rest in the frame in which the frequency is $\nu$.
- in the second one the receiver is moving to the left in the frame in which the frequency is $\nu$.

Last edited: Oct 26, 2017
11. Oct 26, 2017

### robphy

Along the lines of @PAllen 's comment, here are some energy-momentum diagrams of photons in different frames of reference.
(These were taken from my post to a different question on another site https://physics.stackexchange.com/questions/362125/momentum-conservation-with-photons/363478#363478 )

In the rest frame of the source, light signals of the same frequency are emitted in the forward and backward direction.
(The original question I responded to asked about the velocity of the source after emission.)
The diagram visualizes the conserveration of 4-momentum problem:
$$\tilde P_{fin}+\tilde {k_1}+\tilde {k_2}=\tilde P_{init}$$
in the rest frame of the source,
and in the lab frame [which observes the source moving].

Note, in the lab frame,
the forward light-signal's 4-momentum is increased
(compared to that signal's 4 momentum in the rest frame)
and the backward light-signal's 4-momentum decreased (similarly).
Thus, in the lab frame, the light-signals have different frequencies.

12. Oct 26, 2017

### phyzguy

Yes, this is correct. I'm surprised you find this curious. Suppose I throw a baseball at you at 50 miles per hour. While the baseball is in flight, you accelerate towards it to 200 miles per hour. Is it surprising that the baseball will hit harder than if you stayed stationary? I think not. The analogy is not perfect, since light moves at a constant speed, but it shows that the receiver's speed can matter. As Nugatory has explained, the receiver's speed at the point of absorbing the pulse is what matters, because it causes the wave crests to arrive closer together.

13. Oct 26, 2017

### PAllen

The situation is not symmetric at all. What the emitter does after emission is obviously wholly irrelevant, as is what the receiver does after reception. If the emetter changes speed before emission, that will have an effect. If the receiver changes speed before reception, that will have an effect. Both of these before/after statements are invariant because they events along one world line (timelike).

14. Oct 26, 2017

### jartsa

When you do a small acceleration, everything in the universe changes shape (for you). The change is very small for slow moving objects, not very small for fast moving objects. Lorentz-contraction formula can be used to calculate the change of shapes of all things, except those 'things' that move at the speed of light. If we use 0.99999c as the speed that the light pulse moves, we get almost correct results using the Lorentz-contraction formula.

And the derivation of a general contraction formula is trivial.

Oh yes this was about energy. During your acceleration there is a (gravitational) potential difference between every point of the universe (for you), and when you do a small acceleration, everything in the universe changes energy (for you). The amount of change is the original energy of the object multiplied by the potential between the starting position and end position of the object.

Typically fast moving objects, like light pulses, have a large distance between their starting position and end position.

Last edited: Oct 26, 2017
15. Oct 26, 2017

### Mister T

Ask yourself why you expect that the energy should depend on the speed? In non-relativistic physics we expect the energy to be proportional to the square of the speed, but that is just the non-relativistic approximation. It's flawed and the flaws become more and more apparent the closer you get to speed $c$. At speed $c$ there is no dependence at all. Photons of various energies all have the same speed.

16. Oct 26, 2017

### Wes Tausend

Skinnier, but taller photons. :)

17. Oct 27, 2017

### Buckethead

Thanks everyone for contributing to the answers. I've read everyone's post carefully, (understanding some more than others) and enjoy that it is indeed clearing my head a little. I understand the conclusions and have no reservations that red/blue shift occur under the stated circumstances, but am still struggling with this from a strictly logical point of view. Here is my most concise way of putting it:

Emitter and receiver have zero relative velocity. Emitter strikes out, then both emitter and receiver accelerate in the same direction and cease acceleration some time later, all while the pulse of light is somewhere in between the two. At all times the emitter and receiver have zero relative velocity. The pulse reaches the receiver and is blue shifted. During the entire time, the pulse should also have had a velocity of c relative to both the emitter and receiver. Therefore, since the only players in question are the emitter, the pulse, and the receiver and since their relative velocities never changed there should not have been a blue shift.

Do I see a flaw in my argument? Yes. From a third frame watching this whole thing from the side the acceleration caused the relative velocities between the pulse and the receiver to change hence we have blue shift. So there is a way out, but I still don't understand why, if the velocity between the receiver and the pulse never changed from the perspective of the receiver, why it would see blue shift. Not sure I'll ever really understand this.

18. Oct 27, 2017

### Ibix

Not true in relativity from the receiver's perspective due to the relativity of simultaneity - see Bell's spaceship paradox.

19. Oct 27, 2017

### PAllen

I am not sure what the confusion is, but I'll try another clarification. The only thing that matters for Doppler in SR is the relative velocity between the emitter at the emission event and the receiver at the corresponding reception event. What the emitter or receiver do before or after these events is wholly irrelevant (for that reception event). Simultaneity is irrelevant.The Doppler then results from the Lorentz boost for this relative velocity. The analysis can be done in any frame because the relative velocity described is frame invariant.

20. Oct 27, 2017

### jartsa

I mean we pick some speed close to c as the the speed that the light pulse moves according to the observer before the observer has accelerated. When the observer accelerates some small amount the speed that the light pulse moves according to the observer changes a tiny amount.

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