Universal Compass Speedometer Redux: Relativistic Doppler?

[Chris Miller]

Could relativistic Doppler, as in measuring the cosmic microwave background's wavelength from all directions, be used to determine one's velocity and direction in space? With maybe a gyroscopic device initialized to some arbitrary x,y,z grid?

 

[PeterDonis]

Could relativistic Doppler, as in measuring the cosmic microwave background's wavelength from all directions, be used to determine one's velocity and direction in space?

No, but it could be used to determine one's velocity and direction relative to the local frame in which the CMBR is isotropic.

 

[Chris Miller]

Thanks, I understand that there's no universal system of coords. But couldn't one use the arbitrary gyroscopically defined/initialized/remembered frame as a reference.

 

[PeterDonis]

I understand that there's no universal system of coords.

That's true, but it's not the objection I was making.

couldn't one use the arbitrary gyroscopically defined/initialized/remembered frame as a reference.

Yes, you could, but it would be telling you "one's velocity and direction in space". That concept does not make sense. With appropriate setup of the frame, it would be telling you your velocity and direction relative to a frame (or better, a hypothetical observer at your location) in which the CMBR was isotropic.

 

[Chris Miller]

Thanks again for taking the time, Peter. I'm so sloppy/careless with my terminology. By "direction in space" I meant relative to the initial arbitrarily assigned and gyroscopically remembered coordinates. Like if I were to use gyroscopes to fix my current north/south/east/west/up/down, the CMBR could always tell (via relativistic Doppler effect) my velocity and direction relative to this initial CMBR-isotropic frame?

 

[Ibix]

Does a gyroscope actually work to maintain an orientation in this context? Doesn't Thomas precession mean that an inertial gyroscope and an initially co-aligned one that is moved away and returned will not necessarily be co-aligned?

 

[PeterDonis]

By "direction in space" I meant relative to the initial arbitrarily assigned and gyroscopically remembered coordinates.

A better term for such a setup is "frame", since it is not really defining "space" but rather a frame of reference for you to use.

the CMBR could always tell (via relativistic Doppler effect) my velocity and direction relative to this initial CMBR-isotropic frame?

I already said in post #2 what the Doppler shift you observe in the CMBR can be used to tell you: it can tell you your direction and speed relative to the local frame in which the CMBR is isotropic. Or, if "local frame" isn't concrete enough, imagine a hypothetical observer at your location who sees the CMBR as isotropic; the Doppler shift you observe in the CMBR tells you your direction and speed relative to such an observer.

I'm not sure if the above is equivalent to what you are asking, but it's what the CMBR can tell you.

 

[Chris Miller]

Does a gyroscope actually work to maintain an orientation in this context? Doesn't Thomas precession mean that an inertial gyroscope and an initially co-aligned one that is moved away and returned will not necessarily be co-aligned?

Right. And thanks for the pointer. Couldn't corrections/forces applied to the macroscopic gyroscope, relativistic and accelerated, theoretically be measured and adjusted for?

 

[Chris Miller]

Thank you again, Peter. Rereading your posts I think affirms my hypothesis. I find it interesting that the CMBR is not seen as isotropic in all inertial frames. I wonder then if it were measured precisely enough if it'd be perfectly isotropic in any.

 

[PeterDonis]

Couldn't corrections/forces applied to the macroscopic gyroscope, relativistic and accelerated, theoretically be measured and adjusted for?

They could, but that won't fix the issue Ibix is talking about. Thomas precession is not due to "corrections/forces" applied locally to the gyroscope. It's due to a fundamental property of relativistic spacetime (more precisely, of flat relativistic spacetime: in curved spacetime there are additional effects, like de Sitter precession and Lense-Thirring precession, which also affect gyroscopes), which is that, unlike in Newtonian physics, you cannot assume that a gyroscope with no local forces acting on it will always "point in the same direction" globally.

 

[PeterDonis]

I wonder then if it were measured precisely enough if it'd be perfectly isotropic in any.

We have already measured it precisely enough to know that it isn't perfectly isotropic in any frame--it has "wrinkles" which have been mapped in some detail by the COBE and WMAP satellites. But there will always be some frame in which it is "on average" isotropic (i.e., averaging over angular fluctuations in intensity)

 

[Ibix]

I find it interesting that the CMBR is not seen as isotropic in all inertial frames. I wonder then if it were measured precisely enough if it'd be perfectly isotropic in any.

There exists a frame in which the temperature of the CMB is equal in all directions (on average at least). But if you are moving in that frame then you will see the part in front of you blue-shifted and the part behind you red-shifted - i.e. not isotropic. In fact we do see this because the Earth is not quite at rest in the isotropic-CMB frame.

 

[Chris Miller]

So how fast is the Earth moving relative to the isotropic-CMB frame? Would it be possible to then find a truly isotropic-CMB frame?

 

[Chris Miller]

They could, but that won't fix the issue Ibix is talking about. Thomas precession is not due to "corrections/forces" applied locally to the gyroscope. It's due to a fundamental property of relativistic spacetime (more precisely, of flat relativistic spacetime: in curved spacetime there are additional effects, like de Sitter precession and Lense-Thirring precession, which also affect gyroscopes), which is that, unlike in Newtonian physics, you cannot assume that a gyroscope with no local forces acting on it will always "point in the same direction" globally.

Interesting. Then is there no known way to "remember" any orientation?

 

[Ibix]

So how fast is the Earth moving relative to the isotropic-CMB frame? Would it be possible to then find a truly isotropic-CMB frame?

600km/s
Source: http://astronomy.swin.edu.au/cosmos/C/Cosmic+Microwave+Background+Dipole

Interesting. Then is there no known way to "remember" any orientation?

Just leave a gyroscope at home. Or even just a plain old rod. That's your reference.

 

[PeterDonis]

There exists a frame in which the temperature of the CMB is equal in all directions (on average at least). But if you are moving in that frame then you will see the part in front of you blue-shifted and the part behind you red-shifted - i.e. not isotropic.

Is this right? The observed temperature of the CMB is affected by Doppler shift; if the temperature is isotropic on average then there should also be no average Doppler shift, i.e., no dipole moment.

Would it be possible to then find a truly isotropic-CMB frame?

What does this mean? If it means a frame in which the CMB is absolutely isotropic (not just on average), then I already answered that in post #11: no, it isn't possible to find such a frame.

Then is there no known way to "remember" any orientation?

It's not a question of "remembering"; it's a question of the relationship between local "orientation", which is defined by gyroscopes (or some similar method of constructing a local frame of reference), and global "orientation", which is defined by the directions to distant objects like the stars. The point is that this relationship is not in general fixed even for local frames that do not feel any forces (whereas in Newtonian physics it would be).

 

[Ibix]

Is this right? The observed temperature of the CMB is affected by Doppler shift; if the temperature is isotropic on average then there should also be no average Doppler shift, i.e., no dipole moment.

I thought that was what I said. There exists a frame in which the CMB is the same temperature in all directions - the one in which co-moving observers are at rest. If you are not stationary in this frame (i.e. you are moving with respect to a nearby co-moving observer) you see a dipole due to Doppler.

 

[PeterDonis]

If you are not stationary in this frame

Ah, I see, I was confused by the previous version of this: I interpreted "moving in this frame" as "moving such that this frame is your rest frame".

 

[Ibix]

Ah, I see, I was confused by the previous version of this: I interpreted "moving in this frame" as "moving such that this frame is your rest frame".

The joys of natural language. I'll add that one to my rapidly growing list of "ways in which choice of language could accidentally mislead someone about relativity".

 

[PeterDonis]

I'll add that one to my rapidly growing list of "ways in which choice of language could accidentally mislead someone about relativity".