Relativistic Transverse Doppler Effect

[Buxton]

Homework Statement

In Minkowski spacetime, two observers, A and B, are moving at uniform speeds u and v, respectively, along different trajectories, each parallel to the y-axis of some inertial frame S. Observer A emits a photon with frequency nu_A that travels in the x-direction in S and is received by observer B with frequency nu_B. Show that the Doppler shift nu_B/nu_A in the photon frequency is independent of whether A and B are traveling in the same direction or opposite directions.

Homework Equations

$$\lambda/\lambda' = \nu_B/\nu_A = \gamma(1-\beta\cos\theta)$$

Aberration formula: $$\cos\theta' = (\cos\theta - \beta)/(1-\beta\cos\theta) = -\beta$$
(for transverse case)

The Attempt at a Solution

The answer is apparently that the Doppler shift is independent of the relative direction of motion. I have tried to transform to the frame S' where B is stationary, finding the velocity of A using the addition of velocities formula - to then get gamma. I have used the abberation formula to insert cos(theta') into the Doppler shift formula above to get $$\nu_B/\nu_A = \gamma(1+\beta^2)$$Plugging in the velocity of the emitter A in frame S' doesn't seem to get the required result.

 

[Buxton]

100 views and no replies?!

 

[TSny]

Hello, Buxton. Welcome to PF!

It's not unusual for it to take a day or so to get a response to a question.

Are you familiar with working with 4-vectors? Especially the energy-momentum 4-vector of a photon and the 4-velocity of a particle? If so, I think you can use them to get the answer without too much calculation.

If you know the 4-velocity of a particle and the energy-momentum 4-vector of a photon, do you know how to use them to get the frequency of the photon as "observed" by the particle?

 

[Buxton]

Hello, Buxton. Welcome to PF!

It's not unusual for it to take a day or so to get a response to a question.

Are you familiar with working with 4-vectors? Especially the energy-momentum 4-vector of a photon and the 4-velocity of a particle? If so, I think you can use them to get the answer without too much calculation.

If you know the 4-velocity of a particle and the energy-momentum 4-vector of a photon, do you know how to use them to get the frequency of the photon as "observed" by the particle?

Hello TSny, Thank you very much for your help. I've had a look into the method you've suggested. Am I right in thinking you get:
$$\nu_B/\nu_A = \gamma_v/\gamma_u$$
This is achieved using the formula for Doppler shift and noting that the photon 4-momentum is constant along the worldline.
This is independent of the directions of the velocities since the velocities are squared in the gamma factors.
I hope that's right!

Just wondering... can you see where my initial method went wrong? I'm sure it's possible to do it in this way but I'm not sure what I am doing wrong.

 

[TSny]

Hello TSny, Thank you very much for your help. I've had a look into the method you've suggested. Am I right in thinking you get:
$$\nu_B/\nu_A = \gamma_v/\gamma_u$$

That's the answer I got, too.

Just wondering... can you see where my initial method went wrong? I'm sure it's possible to do it in this way but I'm not sure what I am doing wrong.

I would need to see more details of your calculation, but I think it should lead to the right answer. But it looks like it might get messy this way.

 

[Buxton]

That's the answer I got, too.



I would need to see more details of your calculation, but I think it should lead to the right answer. But it looks like it might get messy this way.

Was I right in thinking I'd need to use the abberation formula?

 

[vanhees71]

Hint: A plane light wave is given in covariant form as
A^{\mu}=A \epsilon^{\mu}(k) \exp(-\mathrm{i} k \cdot x).
Here x=(c t,\vec{x}) is the space-time four vector and k the wave four vector k=\omega/c,\vec{k}.

Now you can use the Lorentz transformation for a four vector to transform the k vector from one frame of reference to another one, which leads to the formulae you cited. Also note the dispersion relation for electromagnetic waves: k^0=\omega(\vec{k})=|\vec{k}|.