Frequency shifts in rotating frames

[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

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

 

[George Jones]

Can you give the worldlines explicitly?

 

[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.

 

[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.

 

[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. (second postulate).

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.

 

[A.T.]

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).

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.

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).

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

 

[Mentz114]

Can you give the worldlines explicitly?

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).

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

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

 

[Mentz114]

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.

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.

 

[Mentz114]

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.

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

 

[Mentz114]

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.

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.

 

[Mentz114]

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.

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

 

[WannabeNewton]

I should have specified that the light is going tangentialy to the worldline.

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?

 

[Mentz114]

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?

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.

 

[WannabeNewton]

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.

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.

 

[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 ?

 

[PeterDonis]

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 ?

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.

 

[Mentz114]

Do you mean what the observer who is not comoving detects with his own detector (which is also not comoving)?

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.

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.

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)

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

 

[A.T.]

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

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.

 

[WannabeNewton]

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 ?

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.

 

[Mentz114]

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.

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

 

[Mentz114]

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.

That's nice. Your final expression looks like $\omega$ times the relativistic Doppler effect.

 

[Mentz114]

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.

Peter, I am really confused about this cancellation.

At the proper time of emission the opposite wall is say $\phi$ angular distance away. When the light reaches the opposite wall it is $\phi+\delta \phi$ from the emission point. It has moved so it must be receding.

So in the inertial frame a frquency shift is computed. There is no cancellation.

If there is a frequency shift computed in the inertial frame it must also be seen in the elevator. This can be ascribed to the 'fall' in the pseudo-gravitational field.

So I'm going with that (until I change mind again)

 

[PeterDonis]

I am really confused about this cancellation.

Remember that you're not computing the behavior of the light relative to an inertial observer; you're computing what the inertial observer would compute regarding the behavior of the light relative to the emitter and the receiver. So you have to take into account the motion of the emitter and receiver as well as the motion of the light itself.

To take the simplest case, suppose the inertial observer is at rest at the center of the circle around which the emitter and receiver are moving. The motion of the emitter and receiver is purely transverse and at the same speed relative to the inertial observer, so the only Doppler effect is transverse. Transverse Doppler will induce a redshift when the light is emitted, but an equal and opposite blueshift when the light is received, for no net shift at all.

Other positions or motions of the inertial observer relative to the emitter and receiver are harder to analyze from scratch, but we can invoke Poincare invariance to argue that the result we just got should be valid for any inertial observer, regardless of their location or motion relative to the emitter and receiver.

 

[Mentz114]

Remember that you're not computing the behavior of the light relative to an inertial observer; you're computing what the inertial observer would compute regarding the behavior of the light relative to the emitter and the receiver. So you have to take into account the motion of the emitter and receiver as well as the motion of the light itself.

Yes, I believe I have done this.

To take the simplest case, suppose the inertial observer is at rest at the center of the circle around which the emitter and receiver are moving. The motion of the emitter and receiver is purely transverse and at the same speed relative to the inertial observer, so the only Doppler effect is transverse. Transverse Doppler will induce a redshift when the light is emitted, but an equal and opposite blueshift when the light is received, for no net shift at all.

The transverse Doppler depends on the angle $\theta$
$\frac{f_o}{f_s} = \frac{ 1 - \frac{ \|\vec{v_o}\|}{\|\vec{c}\|} cos(\theta_{co}) } { 1 - \frac{ \|\vec{v_s}\|}{\|\vec{c}\|} cos(\theta_{cs}) } \sqrt{ \frac{ 1-(v_s/c)^2 }{ 1-(v_o/c)^2 } }$

I think transverse Dopplers don't cancel for a macroscopic change of $\phi$ because $\theta$ has changed from the perspective of the central observer.

The transverse Doppler is a red-herring. I'm looking at what the central observer computes. And there is a recession going on between emitter and receiver in this frame in that frame.

Other positions or motions of the inertial observer relative to the emitter and receiver are harder to analyze from scratch, but we can invoke Poincare invariance to argue that the result we just got should be valid for any inertial observer, regardless of their location or motion relative to the emitter and receiver.

Yes.

 

[PeterDonis]

The transverse Doppler depends on the angle $\theta$

Which, if the inertial observer is in the center of the circle, is the same for both emitter and receiver; both are moving exactly perpendicular to the line connecting them to the inertial observer, so $\theta = \pi / 2$ in both cases.

 

[Mentz114]

Which, if the inertial observer is in the center of the circle, is the same for both emitter and receiver; both are moving exactly perpendicular to the line connecting them to the inertial observer, so $\theta = \pi / 2$ in both cases.

Not sure about that. But what the center observer would measure with their instruments is not at issue.

But the issue is - is the receiver receding from the emission point ? The emitter itself is moving, but if we send a short pulse, after the emitter stops sending the receiver is now moving away from the point where the pulse transmission stopped.

 

[PeterDonis]

Not sure about that.

The definition of $\theta$ is the angle between the direction of motion of the emitter/receiver and the line from the emitter/receiver to whatever spatial point you are using as a reference. If that point is the center of the circle, then $\theta = \pi / 2$ for both emitter and receiver. If you are using some other point as a reference, then $\theta$ might change, true, but there will also be other changes, and that more general case is more complicated than I want to analyze here.

the issue is - is the receiver receding from the emission point ?

No, the issue is, what is the relative velocity of the emitter and the receiver? Or, more precisely, how do the following two inner products compare:

(1) The inner product of the light pulse's 4-momentum with the emitter's 4-velocity;

(2) The inner product of the light pulse's 4-momentum with the receiver's 4-velocity.

These two inner products represent the emitted and received frequencies of the pulse, respectively.

if we send a short pulse, after the emitter stops sending the receiver is now moving away from the point where the pulse transmission stopped.

Relative to an inertial frame, yes, but that's irrelevant. The pulse is not emitted by a point in space; it's emitted by an emitter, which puts out wave fronts with a given frequency relative to the emitter.

 

[PeterDonis]

Transverse Doppler will induce a redshift when the light is emitted, but an equal and opposite blueshift when the light is received,

On thinking this over, I think I got it backwards; I think it should be a blueshift on emission and an equal and opposite redshift on reception. The reason is that the emitter is moving "towards" the emitted pulse (meaning, the angle between the emitter's direction of motion and the direction the pulse is emitted towards is less than 90 degrees), whereas the receiver is moving "away from" the received pulse (meaning, the angle between the receiver's direction of motion and the direction the pulse is received from is more than 90 degrees). The angles are supplementary (one is 180 degrees minus the other), so the shifts are of equal magnitude.

 

[Mentz114]

The definition of $\theta$ is the angle between the direction of motion of the emitter/receiver and the line from the emitter/receiver to whatever spatial point you are using as a reference. If that point is the center of the circle, then $\theta = \pi / 2$ for both emitter and receiver. If you are using some other point as a reference, then $\theta$ might change, true, but there will also be other changes, and that more general case is more complicated than I want to analyze here.
...
Relative to an inertial frame, yes, but that's irrelevant. The pulse is not emitted by a point in space; it's emitted by an emitter, which puts out wave fronts with a given frequency relative to the emitter.

(my emphasis)
The central observer is inertial, the elevator is rotating with the space-time. I don't follow all of that. But I'm still not convinced.

I'll try a proper calculation because I've had some insights from the replies, but no-one has produced any numbers ( including me) so this has been hand-waving for the most part ( exception being WBN )

Thanks to you all.

 

[WannabeNewton]

I think something that would help, at least me anyways, is a finalized, coherent, and detailed description of the problem that you (Mentz) want solved because I'm still not sure what exactly the scenario you have in mind is.

 

[PeterDonis]

The central observer is inertial

Which means that a frame in which this observer is at rest (and the emitter and receiver are moving around a circle) is inertial. That's the frame I'm using.

the elevator is rotating with the space-time.

What does "rotating with the spacetime" mean? The spacetime is not "rotating"; it's flat Minkowski spacetime. Using a rotating chart, so that the emitter and receiver of the light pulse are at rest in the chart and there are fictitious forces present, does not change the spacetime.

 

[jartsa]

Do you mean that the shifts cancel out for the round trip ?

No, I meant there's a frequency shift f, when light is emitted at one side of the box, and frequency shift -f when the light is absorbed at the other side of the box.

Well ok, according to the inertial observer the light actually gives the absorber this energy: frequency of the light during flight * hbar.

It's just that some percentage of the energy of the light becomes kinetic energy of the absorber, according to the inertial observer, but not according to box dwellers.

So careful thinking tells us that there is a frequency shift according to the inertial observer, but not according to box habitants.

 

[PeterDonis]

some percentage of the energy of the light becomes kinetic energy of the absorber, according to the inertial observer, but not according to box dwellers

This raises a good point: if momentum is to be conserved when the light is emitted and absorbed, the motion of the emitter and absorber must change at the emission and absorption events. So they can't both be moving on the same perfect circle, as seen from an inertial frame. My analysis was really of the limiting case where the energy of the light is negligible compared to the mass and kinetic energy of the emitter and absorber.

There is another aspect to be considered: are the emitter and absorber physically connected (other than by the light pulse itself)? How their motion changes on emission/absorption of the light will be different if they are, than if they are not.

Note also that, at emission, some percentage of the energy of the light was taken from the kinetic energy of the emitter. (We are assuming here that the emission process and the absorption processes can be treated as inverses of each other, and that the emitter and absorber have identical masses.) In the center of mass frame of the emitter/absorber system, the two kinetic energy shifts are equal and opposite, so the frequency shifts are also equal and opposite, as you say.

 

[Mentz114]

No, I meant there's a frequency shift f, when light is emitted at one side of the box, and frequency shift -f when the light is absorbed at the other side of the box.

Well ok, according to the inertial observer the light actually gives the absorber this energy: frequency of the light during flight * hbar.

It's just that some percentage of the energy of the light becomes kinetic energy of the absorber, according to the inertial observer, but not according to box dwellers.

So careful thinking tells us that there is a frequency shift according to the inertial observer, but not according to box habitants.

Why bring the energy absorbed by the wall into this. It's not relevant surely.

I don't see how this can be correct. I'm talking about kinematics and Doppler shifts.

"careful thinking" is fine, but how about some calculation ?

 

[Mentz114]

Which means that a frame in which this observer is at rest (and the emitter and receiver are moving around a circle) is inertial. That's the frame I'm using.

yes, me too. The observer at $r=0$.

What does "rotating with the spacetime" mean? The spacetime is not "rotating"; it's flat Minkowski spacetime. Using a rotating chart, so that the emitter and receiver of the light pulse are at rest in the chart and there are fictitious forces present, does not change the spacetime.

I used that phrase as careless shorthand for the 'coordinate basis vectors' are rotating. I know it is a chart not a spacetime. But thanks for pulling me up on that.