# Frequency shifts in rotating frames

1. May 31, 2015

### Mentz114

After the recent discussion here about the Sagnac effect** I wanted to see if there is any frequency shift if light is sent from one point on a rotating worldline to another a small distance away. It looks like this cannot happen because $\gamma=dt/d\tau$ is the same at every point on the curve.

Thinking of the scenario as an Einstein elevator (a box) circling a point so that the 'roof' is the closest to the center and the 'floor' is furthest, the observer in the elevator can send a light beam from one side to the other with the beam parallel to the roof and see no deflection or frequency change ( first posulate).

In the non-comoving frame ( say at $r=0$) the elevator is rotating and when the light gets the other wall, the wall is receding/approaching so there is a frequency shift. (second postulate).

These postulations seem to contradict each other. The obvious solution is that the elevator observer sees the frequency shift and attributes it to the gravitational analog of the centripetal acceleration. But then every point still has the same 'potential' ( or does it ?).

Any ideas ?

**
A good reference for rotation is http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]

and elevators

Juergen Ehlers and Wolfgang Rindler
Local and Global Light Bending in Einsteins and other Gravitational Theories
http://pubman.mpdl.mpg.de/pubman/item/escidoc:152994:1/component/escidoc:152993/330688.pdf

Last edited by a moderator: May 7, 2017
2. May 31, 2015

### George Jones

Staff Emeritus
Can you give the worldlines explicitly?

3. May 31, 2015

### PAllen

When you say 'nearby', you have to specify which direction, tangential or radial. For tangential, gamma would be the same, but for radial it would not. The radial difference can be modeled as a potential, and the simple model of pretending centripetal acceleration corresponds to pseudo-gravity of the same acceleration works to accurately predict the spectral shift.

4. May 31, 2015

### WannabeNewton

The argument in your first paragraph doesn't make sense because the light signal is traveling between two different worldlines whereas $\frac{dt}{d\tau}$ is only defined along a single curve; furthermore it's not clear at all what the explicit set of observers you are referring to (c.f. George's question above) but this is a very easy calculation for the set of observers at rest in the rotating frame. The congruence has a 4-velocity $u = \gamma \partial_t$ where $\gamma = (1 - \Omega^2 r^2)^{-1/2}$ and $\partial_t$ is the timelike Killing field in the rotating frame. If an observer $u_1$ in the congruence sends a light signal with tangent $k^{\mu}$ to another $u_2$ in the congruence then $-k_{\mu}u^{\mu}_1 = -k_{\mu}u^{\mu}_2 \Rightarrow \omega_1 (1 - \Omega^2 r_1^2) = \omega_2(1 - \Omega^2 r_2^2)$, giving the frequency shift of the light signal between the two rotating observers.

This is just the doppler effect, which manifests itself through a gravitational redshift in the rotating frame due to the centrifugal potential. For light signals between other types of observers you would have to be more specific about what the worldline of the observer actually is.

5. Jun 1, 2015

### jartsa

In the non-comoving frame ( say at $r=0$) the elevator is rotating and when the light gets the other wall, the wall is receding/approaching so there is a frequency shift ... and there was a frequency shift when the light was emitted.

Those two frequency shifts cancel each other.

I think that's an answer ... But the box dweller should aim the light a little bit to the "upwards" direction, in order to cancel the bending effect that "gravity" has on the light beam.

Last edited: Jun 1, 2015
6. Jun 1, 2015

### A.T.

If the Einstein elevator undergoes proper acceleration then a beam sent parallel to the roof is deflected towards the floor. This will also happen here, due to the centripetal acceleration of the box. And you additionally get Coriolis deflection.

As, jartsa said, you have to consider the emission too.

7. Jun 1, 2015

### Mentz114

Of course.

I should have specified that the light is going tangentialy to the worldline. The worldline and metric are ( as in the Dieks paper).

$u^\mu=1/\sqrt{1-{r}^{2} {\omega}^{2}} \partial_t,\ \ u_\mu=-\sqrt{1-{r}^{2} {\omega}^{2}}\ dt + {r}^{2}\,w/\sqrt{1-{r}^{2}\,{w}^{2}}\ d\phi$

${d\tau}^{2}=-{dt}^{2}\left( 1-{r}^{2} {\omega}^{2}\right) +2\ {r}^{2}\omega d\phi\,dt\,+{d\phi}^{2}\,{r}^{2}+{dz}^{2}+{dr}^{2}$

8. Jun 1, 2015

### Mentz114

I've given specs above. The consnesus seems to be that the rotating elevator shifts can be seen by the elevator dweller and explained by the pseudo-gravity also.

9. Jun 1, 2015

### Mentz114

OK, thanks. I hope I've cleared up the ambiguities.

10. Jun 1, 2015

### Mentz114

Do you mean that the shifts cancel out for the round trip ? Do you think the box-dweller sees a deflection and a shift on a one way trip ?

I'm still not clear about my postulates.

11. Jun 1, 2015

### Mentz114

Can you draw some pictures ? I don't think the rotating elevator is exactly the same as the translationaly accelerated case.

12. Jun 1, 2015

### WannabeNewton

I'm afraid I'm still confused. Do you mean instead that the light signal is tangential to the circular path of the rotating observer in the background inertial frame? The light cannot be tangential to the worldline of the observer as the latter is time-like whereas the former is null. Are you therefore talking about the rotating observer emitting a signal to an infinitesimally neighboring rotating observer separated from the first only in the tangential direction? If so, do these two observers rotate with the same angular velocity?

13. Jun 1, 2015

### Mentz114

Yes. The distance is all $d\phi$. The emission point is on the circumference of the ring and so is the absorption/reflection point. The angular velocity $\omega$ is the same.

14. Jun 1, 2015

### WannabeNewton

Ok in that case there is no frequency shift as can be seen in the formula above by setting $r_1 = r_2$. Intuitively this is simply because the accelerations of the observers are radial so the doppler shift will only occur radially.

15. Jun 1, 2015

### Mentz114

That is what I figured. But the observer who is not comoving sees motion and kinetic frequency shift because the reception point on the opposite wall is receding, is it not ?

16. Jun 1, 2015

### Staff: Mentor

Do you mean what the observer who is not comoving detects with his own detector (which is also not comoving)? Or do you mean what the observer who is not comoving computes will be detected by the detector at the reception point that is comoving?

The latter is an invariant, so it must be the same regardless of who is computing it. The non-comoving observer might interpret the result as indicating a shift in one direction on emission (relative to a non-comoving observer) and a compensating shift in the other direction on reception (again relative to a non-comoving observer), so the final received frequency at the comoving detector ends up being the same as the original emitted frequency at the comoving emitter. But the observed frequency by the comoving detector is the same either way.

17. Jun 1, 2015

### Mentz114

No, the non-comoving observer has no detectors. But he can detect that the emitter and detector on opposite sides of the box are in relative motion.

If this is a round-trip then I can see a cancellation, but not for a one-way trip ( which could be spinwards or antispinwards)

No that's wrong, the emitter is also moving. So there is no frequency shift.

Last edited: Jun 1, 2015
18. Jun 1, 2015

### A.T.

Of course it's not exactly the same, for example you don't have Coriolis acceleration in the frame of a linearly accelerated box. But a light beam sent perpendicular to the proper acceleration of the box will be deflected in the frame of the box in both cases.

19. Jun 1, 2015

### WannabeNewton

If by "observer who is not comoving" you mean an inertial observer then yes. This can also be calculated easily as it is just a Lorentz boost. We have $k'^{\hat{\mu}} = \Lambda^{\hat{\mu}}{}{}_{\hat{\nu}}k^{\hat{\nu}}$ so $\omega' = \Lambda^{\hat{0}}{}{}_{\hat{\nu}}k^{\hat{\nu}} = \Lambda^{\hat{0}}{}{}_{\hat{0}}\omega+ \Lambda^{\hat{0}}{}{}_{\hat{\phi}}k^{\hat{\phi}}$. Now $\vec{k}^2 = 0 \Rightarrow k^{\hat{\phi}} = \frac{\omega}{R}$ hence $\omega' = \omega \gamma(1 + \frac{v}{R}) = \omega \gamma(1 + \Omega)$. This only works if the inertial observer is infinitesimally close to the rotating observer so that $\vec{k}^2 = 0$ holds.

EDIT: Also note this is for an inertial observer that is tangential to the rotating observer. The calculation for an inertial observer that is radially separated from the rotating observer during the Lorentz boost is analogous.

Last edited: Jun 1, 2015
20. Jun 1, 2015

### Mentz114

OK. So which (if any) observers will compute a frequency shift across the box ?