# Doppler effect in accelerating frame.

1. Oct 21, 2013

### Ravi Mohan

Consider a spaceship and a source, with no relative motion but spatially separated. At t=0, spaceship moves with velocity $v \hat{a}_x$ w.r.t source and at t = 0 source emits a photon.
After photon has been emitted, the speed of spaceship starts increasing in same direction. So what parameters will change according to the person sitting in spaceship (when the speed is changing).
a) Speed of photon
b) Energy of photon
c) Momentum of photon

I think that all above parameters will remain same.

a) Speed of photon is same in all frames.
b) Tho there is doppler shift which will change the frequency, but it depends only on the instantaneous relative velocity (between source and spaceship) when the photon was being emitted.
c) E = pc, so p should remain same too.

Please correct me if I am wrong. Thanks

2. Oct 21, 2013

### yuiop

The Doppler shift depends on the relative velocity between the source (at the time the light is emitted) and the ship (at the time the light is received). If the source is inertial and continues emitting at a constant frequency (as seen in the source rest frame), the frequency of the signal received by the ship will be decreasing continually.

3. Oct 21, 2013

### Ravi Mohan

I think that I didn't frame the questions properly. Sorry for that.
The parameters given are related to a single photon emitted at t=0.

4. Oct 21, 2013

### yuiop

You stated the source and the spaceship are spatially separated at t=0. Therefore the time the spaceship receives the photon is T>0 and and the velocity of the spacship relative to the source is v>0, so the spaceship sees the photon as redshifted relative to the source frequency.

5. Oct 21, 2013

### WannabeNewton

For starters, this is not true. The speed of light is the same in all inertial frames; in accelerating frames light does not move at $c$ in general.

6. Oct 21, 2013

### Ravi Mohan

So what should be the answer. Due to some legal binding I am not allowed to reveal the name of the test in which this question appeared, but rest assured that the test is of high standard. The options were
a) all parameters are unchanged
b) all parameters are changing
c) E and P decrease while v (speed of photon) is constant
d) P and v vary such that E is constant
e) v and E vary such that P is constant

7. Oct 21, 2013

### Staff: Mentor

No, it doesn't. It depends on the relative velocity of the source when the photon was emitted, and the receiver when the photon is received. The source's velocity is v = 0 always, and the receiver's velocity is v > 0 when the photon is received (even though it was 0 when the photon was emitted--at least, I think that was your intention in stating the problem), so, as yuiop said, the photon will be redshifted, so its energy and momentum are changed. (Its speed is unchanged, as you noted.)

8. Oct 21, 2013

### Ravi Mohan

Well to be honest, I don't understand it. When I recall the derivation of Doppler shift, I see that we write equation of wave (say a plane wave) in ($S$) and ($S'$) frame. Then we apply lorentz transformation in the wave equation of primed frame and finally equate the coefficients to get the relation of frequencies. The basic assumptions are
1) There is constant relative velocity.
2) Speed of light is same in both frames.

(And both of them are against the question. I didn't know about the second one.)

Also how do we bring in the conditions you mentioned

9. Oct 21, 2013

### dauto

You can use the formula derived for the object moving at constant speed. All you have to do is use the speed of the object at the moment the photon is received and make sure you measure that speed with respect to the source at the moment the photon was emitted.

10. Oct 21, 2013

### yuiop

This question is a bit unfair on the students as it is a bit ambiguous. With a single photon, this implies a single measurement and when the photon arrives at the spaceship, it will measure the local speed of the photon as c and the momentum and energy both reduced by the redshift factor so this corresponds to answer (c).

We can analyse this further by considering a very long (and elastic) spaceship with equal proper acceleration at the front and back. (This is not Born rigid motion!). Now as the photon passes the observer at the front of the spaceship, the frequency and momentum will be lower than the corresponding measurements made when the photon passed the observer at the rear of the ship. Both observers at front and rear will measure the local velocity of the photon to be c as it passes. This still corresponds to answer (c).

However, the wording suggests they want to know what happens to the photon in transit according to a single accelerating observer. We know that the accelerating observer will consider the non local velocity of the photon to be increasing, initially slow and increasing to c. This is because he knows that observers lower down the ship, measure the photon velocity to be c and he also know that their clocks are running slower than his, so he concludes the light was going slower when it passed the rear observers. He also knows the rear observers measured the energy and momentum to be greater than when it arrives at his location, but when he allows for their slow clocks, he concludes that the energy and momentum of the photon does not change in transit. Constant energy and momentum with increasing velocity of the photon is not one of the options, so presumably that is not the answer they are looking for. The question should be reworded to make it clear they are only interested in the local measurement by the spaceship at the time it receives the photon.

11. Oct 21, 2013

### Staff: Mentor

Strictly speaking, this isn't true for the case you gave, where the ship's motion is described by the Bell congruence (the one that appears in the Bell spaceship paradox, with proper acceleration constant along the ship); it's only true for the Rindler congruence (which is the one that would correspond to Born rigid motion, but which has proper acceleration varying along the ship). For the Bell congruence, the positive expansion means that, even when allowing for the slower clocks of the observers to the rear, the given observer will conclude that the photon lost energy and momentum in transit (i.e., he will see an additional redshift over and above the redshift that he can attribute to the difference in clock rates). However, for a reasonably short spaceship, the difference will be very small.

I definitely agree that the question is not worded well.

12. Oct 21, 2013

### yuiop

I was talking about Born rigid case for the final example, but unfortunately forgot to mention that .. oops. Do you happen to have any equations to hand, for the time dilation in the Bell congruence case at the time the photon passes each successive accelerating observer?

13. Oct 21, 2013

### Staff: Mentor

Not handy, no; I'd have to work it out or look it up.

Also it's worth pointing out that in all these scenarios, there will be a Rindler horizon: a spaceship whose acceleration is greater than c^2/x, where x is the ship's distance from the source at time t = 0 in the source frame (when the photon is emitted), will never see the photon at all. (In the case of the Bell congruence spaceship, it's even possible for the photon to be able to catch the rear end but not the front end, if the numbers are just right.)

14. Oct 22, 2013

### Ravi Mohan

Hmm! I still think that option (a) should be the answer. If we are considering the instantaneous speed of spaceship, when photon reaches it, then how can we say that the observer in space ship should see the energy and momentum of photon changing in time. He will just notice changed energy (to which I agree) but that is what, I think, is not being asked.

But maybe the question was not framed in this sense or I recall it incorrectly or I need more time to understand.

Thanks for your help yuiop, Peter and dauto.

15. Oct 22, 2013

### yuiop

I think the best way to tackle this question is consider a series of experiments. Each time the experiment is carried out, the proper acceleration of the ship remains the same, but the head start of the ship is increased, so that in each successive experiment, the velocity of the ship relative to the source is increased. After carrying out such a set of experiments, we would find that with increased velocity of the ship relative to the source, the observed frequency of the photon is reduced, the energy is lower, the speed is c and consequently the momentum is lower. This corresponds to answer (c).

...which is provably wrong.

Last edited: Oct 22, 2013
16. Oct 22, 2013

### Ravi Mohan

The way I see it, consider this experiment
Say at t=0 (source frame), t'=0 (spaceship frame). Now source emits a single photon (source is designed such that it emits only one photon). We ask the observer at spaceship to note the frequency or energy at different time (t') according to his frame. The observer knows the acceleration wrt source and relative velocity and all the required parameters. So what would we see in observer's notebook
t' = 1, E = E0
t' = 2, E = E0
t' = 3, E = E0 and so on
with E0 corresponding to shifted frequency.

17. Oct 22, 2013

### yuiop

Until the photon arrives, the spaceship observer will not even know that the photon exists, never mind what colour it is. Are you asking us what the spaceship observer imagines the frequency of the photon be when it is on its way to him?

As I mentioned before, you could repeat the experiment several times with an increasing spatial separation at the start of each experiment, or have a set of spaceships starting at different locations and compare results later. If you did that you would find this relationship between the frequency observed fo and the source frequency fs is:

$f_o= f_s \left(\sqrt{1+(at/c)^2} -at/c \right)$

where t is the coordinate time in the source reference frame and a is the constant proper acceleration of the spaceship. In terms of the proper time of the spaceship the relationship is:

$f_o= f_s \left(\sqrt{1+\sinh(at'/c)^2} -\sinh(at'/c) \right)$

Last edited: Oct 22, 2013
18. Oct 22, 2013

### Ravi Mohan

I misunderstood the question. Now I think I understand what the question means.
Just out of curiosity, how do we get the last equation?

EDIT:
I am not asking you to tell what observer imagines. I am asking you to tell what the observer, after applying physical laws, will report the properties as. Consider a ball (instead of photon) approaching the ship. Can't the observer predict the physical properties of the ball like acceleration, energy and momentum of the ball without the ball actually reaching the spaceship? We are not playing the game of Quantum Mechanics that it is not there until no one sees it. Are we?

Again I must clarify that I have understood the question and think that (c) is correct answer.

Last edited: Oct 23, 2013
19. Oct 23, 2013

### yuiop

I used the relativistic Doppler equation f0= fs γ(1-v) and then substituted for γ and v using this set of equations for accelerated motion. (Note that they use T instead of t' for the proper time of the spacship). Hope that helps.

20. Oct 23, 2013

### yuiop

The ball has the benefit that light can transmit information to the observer about the location and speed of the ball before the ball arrives. If the ship has no windows, then the observer is predicting where the ball is based on an assumption that he knows when it was launched. In this case he is not really "reporting" or "observing". For a single photon, we cannot guarantee launch time, or send confirmation of its launch that arrives before the photon arrives, so the observer of the photon is neither reporting or observing, but guessing where it is, until it actually arrives. Of course I am being a little playful here, but the point is that we cannot make sensible deductions about the laws of physics as applied to photons in transit based on a single observation of a photon by a single observer at a single event. Of course we can make predictions based on what we know about the laws of physics, but your statement above amounts to if the spaceship "observer" knows all about the laws of physics as applied to the behaviour of a photon in an accelerating reference frame, what would he tell us about the behaviour of a photon in an accelerating reference frame? The spaceship, the photon and the experiment become superfluous.