Frequency shifts in rotating frames

[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.

 

[PeterDonis]

I used that phrase as careless shorthand for the 'coordinate basis vectors' are rotating.

But if you're using an inertial frame, the coordinate basis vectors are not rotating. So I'm confused about what frame you're using.

 

[Mentz114]

But if you're using an inertial frame, the coordinate basis vectors are not rotating. So I'm confused about what frame you're using.

Peter, I've tried hard to distinguish between the elevator whose worldline is given above, and the center observer whose worldline is the elevator worldline where $r=0$, which gives $u_C^\mu=\partial_t$, which is at rest in the Minkowski chart.

The problem is that if there is no frequency shift, then there can be no deflection. Ehlers&Rindler show that a beam of light traveling across the gravitational field gradient will fall. And so it must have a frequency shift. If we shone a light across the room we're in, it would fall, wouldn't it ?

Maybe that's where I'm going wrong.

 

[jartsa]

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 ?

Well here's one calculation that ignores time dilation, as the speed is quite low:

A laser gun moving at speed 0.1 c shoots forwards a laser pulse whose duration is 1 s.

Observer whose speed is zero says the length of the beam is 0.9 light-seconds. (because the front of the beam is 0.9 c * 1 s away when the rear of the beam leaves the laser gun). Now this observer calculates how long it takes for the pulse to be absorbed by an absorber that is co-moving with the laser gun.

It takes 1 s, because at closing speed 0.9 c, a 0.9 light-seconds long beam disappears into the absorber in 1 s. (I mean, a 0.9 light-seconds long thing dives into another thing at closing speed 0.9 c)

Observer calculated that co-moving observer received the wave crests at equally quick succession as they were created, because the wave crests were created in one second and absorbed in one second.

 

[Mentz114]

Well here's one calculation that ignores time dilation, as the speed is quite low:

A laser gun moving at speed 0.1 c shoots forwards a laser pulse whose duration is 1 s.

Observer whose speed is zero says the length of the beam is 0.9 light-seconds. (because the front of the beam is 0.9 c * 1 s away when the rear of the beam leaves the laser gun). Now this observer calculates how long it takes for the pulse to be absorbed by an absorber that is co-moving with the laser gun.

It takes 1 s, because at closing speed 0.9 c, a 0.9 light-seconds long beam disappears into the absorber in 1 s. (I mean, a 0.9 light-seconds long thing dives into another thing at closing speed 0.9 c)

Observer calculated that co-moving observer received the wave crests at equally quick succession as they were created, because the wave crests were created in one second and absorbed in one second.

Yes, thank you. I drew myself a diagram with spherical wave fronts and they are squeezed on emission and then stretched again on reception. I can't work out the deflection though. It looks as if the inertial frame would see deflection.

 

[A.T.]

he problem is that if there is no frequency shift, then there can be no deflection.

If the beam arrives at the same height in the box frame, then there is not frequency shift, even if the beam is bent. The inertial frame just sees the beam going straight.

 

[harrylin]

[..] The problem is that if there is no frequency shift, then there can be no deflection. Ehlers&Rindler show that a beam of light traveling across the gravitational field gradient will fall. And so it must have a frequency shift. If we shone a light across the room we're in, it would fall, wouldn't it ?

Maybe that's where I'm going wrong.

Perhaps - it depends on what frequency shift you are thinking of. The equivalence principle is about predicted observations; interpretation is a different matter.

Einstein as well as Okun explained that in vacuum in a gravitational field in rest, the number of wave crests must be conserved. A predicted frequency shift due to Doppler effect from acceleration in an accelerating frame without gravitation, must therefore in an equivalent gravitational field in rest be interpreted as due to difference in reference frequencies (clock frequencies).

PS. it was that insight that led to the first prediction of gravitational time dilation.

 

[Mentz114]

If the beam arrives at the same height in the box frame, then there is not frequency shift, even if the beam is bent. The inertial frame just sees the beam going straight.

I now take the view that there is no deflection of frequency shift. See Below.

Perhaps - it depends on what frequency shift you are thinking of. The equivalence principle is about predicted observations; interpretation is a different matter.

Einstein as well as Okun explained that in vacuum in a gravitational field in rest, the number of wave crests must be conserved. A predicted frequency shift due to Doppler effect from acceleration in an accelerating frame without gravitation, must therefore in an equivalent gravitational field in rest be interpreted as due to difference in reference frequencies (clock frequencies).

PS. it was that insight that led to the first prediction of gravitational time dilation.

The Einstein elevator differs from the elevator on a string in a crucial way. If the EE is not in a gravitational field it moves in the same direction as the 'field' gradient ( up-down). The rotating elevator moves sideways in relation to the field. So one would not expect the same phenomena for the transverse light beam., viz deflection and frequency shifts.

 

[A.T.]

it moves in the same direction as the 'field' gradient ( up-down). The rotating elevator moves sideways in relation to the field.

The proper acceleration of the box is relevant here.

So one would not expect the same phenomena for the transverse light beam., viz deflection and frequency shifts.

They will differ quantitatively if the box is large enough. But for small enough boxes they are very similar. In both cases Light rays (send perpendicular to the proper acceleration of the box) will be curved in the frame of the box. But if received at the same height (along the proper acceleration of the box) as emitted there will be no frequency change.