No redshift in a freely falling frame

PeterDonis

Ah, I see. But this derivation only holds for purely radial motion; in fact it only holds when one object is static at a constant height and the other is moving purely radially. The case stevendaryl talks about, as here...

...is more general. You gave a "hint" as to how a more general formula could be derived, but you haven't actually done the derivation. (A fully general formula would need to include the effects of motion, both radial and non-radial, for *both* objects, not just one.)

Also, as you noted in a later post, what you call "v" in your formula is a coordinate velocity, not an actual observed relative velocity.

Finally, since you mentioned Ashby's paper, it's worth noting that your derivation uses Schwarzschild coordinates, but his paper does not. He uses Earth Centered Inertial (ECI) coordinates, which differ from Schwarzschild coordinates in several important respects:

(1) They are isotropic;

(2) The coordinate time dt is scaled to the rate of time flow of observers on the "geoid" (the equipotential surface at "sea level") who are at rest relative to the actual Earth (i.e., rotating with it); however, the simultaneity convention is that of hypothetical inertial observers moving with the Earth's center of mass but *not* rotating with it (which would be the simultaneity convention of Schwarzschild coordinates centered on the Earth);

(3) The "potential" [itex]\Phi_{0}[/itex] in the metric, which is the potential on the "geoid", includes not only the effect of the Earth's rotation, but also includes corrections for the Earth's quadrupole moment, so it differs in two ways from the Schwarzchild potential.

These differences don't affect the general points under discussion, but since you referenced Ashby's paper, I think it's worth pointing out the ways in which his equations and notation differ from those being used in this thread.

GAsahi

1. stevendaryl denied for the longest time that it is correct, resulting into a monumental waste of time

2. the derivation is easy to extend to more complex cases, like the one of a stationary receiver and a rotating transmitter, the methodology is the same.

3. the derivation extends easily to other metrics (as in the Ashby paper). The point is that the methodology is STANDARD and that it produces predictions that are confirmed by experiment (Pound Rebka for radial motion, GPS for more complex motion, etc)

...because the derivation is trivial


Agreed. The point was to show stevendaryl that the applied methodology is STANDARD.

stevendaryl

My mistake, but either way, it's incorrect in the case that I was talking about, namely two observers, at the equator, moving at relative speed v in opposite directions (east for one, west for the other). In that case
[tex]\frac{d \tau_1}{d \tau_2}[/tex] = 1.
[tex]\frac{f_2}{f_1}[/tex] = √((1-v/c)/(1+v/c))

stevendaryl

No, I said that your DERIVATION was wrong. From the very beginning, I said that it was your method that was incorrect, not the result. It happens to give the right answer in one situation, but not in other situations.

PeterDonis

I agree, but it's worth noting that part of what is standard is the adoption of particular coordinates--you used Schwarzschild coordinates, the Ashby paper used ECI coordinates (which seem to be the "standard" for these kinds of computations). That includes adopting a particular simultaneity convention, which is crucial for defining the time differentials and intervals that appear in the equations. I agree it's a "natural" choice of simultaneity convention for the purpose, but it's still a specific choice.

PeterDonis

The general method he used is correct, provided you adopt the particular coordinates and simultaneity convention he adopted. (The particular derivation he gave was for a special case, yes.) If you are saying those coordinates and that simultaneity convention are not the only possible ones, that's true; and in different coordinates you would use a different method for defining "relative clock rates". An observer on a satellite in low Earth orbit, for example, would see Earthbound clocks (clocks at rest on the Earth's surface and rotating with it) to be "running slow" if he used his own local inertial coordinates; but in the ECI frame his clocks would be "running slow" relative to Earthbound clocks (I believe that's right for low enough orbits--the GPS satellite orbits are very high, 4.2 Earth radii IIRC). However, the observed frequency shift for light traveling between the two observers would be independent of which coordinates you adopted.

stevendaryl

I don't see that his derivation works in the case of two observers on the surface of the Earth, one moving east and one moving west, with relative speed v. His method would give a redshift of zero (or at best, of the order of (v/c)²), which is not the correct answer. The correct answer should, in the limit as r_s→0 approach the Doppler shift formula.

That's exactly my point: redshift is coordinate-independent, while "relative clock rate" is coordinate-dependent. The two are only equal in certain circumstances.

GAsahi

First off, the formula [itex]\frac{f_2}{f_1} = \sqrt{((1-v/c)/(1+v/c))}[/itex]
does not apply for circular motion. I know for a fact that the relativistic Doppler is more complicated for accelerated motion. Note the presence of the terms in [itex]\frac{v^2}{c^2}[/itex] here.
More importantly, you do not get redshift as reflected in your formula, you get blueshift for half the circle (decreasing closing distance) and redshift only for the other half (increasing closing distance) so, on average, [itex]\frac{f_1}{f_2}=1[/itex] due to problem symmetry.
Lastly, we were discussing the correctness of my formulas/derivations for radial motion. Failing to prove your point , you quietly moved the goalposts to circular motion. But even then, you failed to acknowledge my repeated references to the Ashby paper that shows how this problem gets solved for the case of circular motion. The solution uses a different metric but the methodology is the same, all the effects (with the notable excption of the Sagnac effect) fall out the metric.

PeterDonis

The correct answer for what? For the observed frequency shift of light sent between the two observers? Or for the relative clock rates of the two observers? You yourself pointed out that the two are not necessarily the same; this is a case where they are different.

A more general formula for the rate of a moving clock relative to coordinate time, generalizing the one that GAsahi derived, is this:

[tex]\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^{2} r}}\sqrt{1 - \frac{v^{2}}{c^{2}}}[/tex]

which is similar to the formula in Ashby's paper, except that I have used Schwarzschild coordinates instead of the ECI coordinates that he used. (Also, he expands out the square roots and throws away higher order terms.) Note also that v is relative to a non-rotating observer, so an observer at rest on the Earth's equator has a v of about 450 m/s eastward.

So the ratio of clock rates for two observers, both at the same radius r but moving at different velocities, is:

[tex]\frac{d\tau_{1}}{d\tau_{2}} = \frac{\sqrt{1 - \frac{v_{1}^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{2}^{2}}{c^{2}}}}[/tex]

where the terms in M cancel since both observers are at the same radius r. This does look different from the Doppler formula, but so what? Why would we expect the two to be the same for this case?

PeterDonis

Missed a couple of items in an earlier post, just wanted to clarify:

Yes.

Yes.

Austin0

Originally Posted by PeterDonis View Post


Perhaps you could clarify a point of confusion.
1)You stated that wrt the launch frame the front and rear clocks would be related by the ratio of gammas.

2)GAsahi declared that incorrect and said they would be related by the Doppler factor.
Indicating that they were not equivalent.

3)You then agreed with him that they were not equivalent in general

4) But were equivalent in this instance

So my question is:
Is the ratio of gammas as measured in the launch frame exactly the same ratio as derived from the Rindler metric ?

Ratio of Velocity gammas =Doppler ratio=Rindler ratio??

Or not?

thanks