No redshift in a freely falling frame

[GAsahi]

I should elaborate on what I mean by that:
What I thought was being proposed was that a way to compute
f₁/f₂
is the following:

The Schwarzschild metric:

d$\tau$² = (1-r/r_s) dt² - 1/(1 -r/r_s) dr² - r² dθ²

So let one observer be at "rest" at r=R. Then we have for that observer:
d$\tau$₁ = √(1-R/r_s) dt

Correct.

Let the other observer be also at r=R, moving at speed v along the θ direction; that is R dθ/dt = v. Then we have:
d$\tau$₂ = √(1-R/r_s - v²) dt

Correct.

Now, my claim is that d$\tau$₁/d$\tau$₂ will NOT give the correct redshift for signals sent from the first observer to the second observer.

Why not? This is the bee under your bonnet that you keep repeating with no formal justification.

 

[PeterDonis]

yes,of course.

Well, I suppose it's true that the paper wasn't hard to find, but despite the "of course", it's helpful to post an explicit link, to be sure there's no question about what you are referring to.

$$\frac{d \tau_1^2}{d \tau_2^2}=\frac{1-r_s/r_1}{1-r_s/r}\frac{1}{1-\frac{v^2}{(1-r_s/r)^2}}$$

Can you point out where in Ashby's paper you are deriving this from?

I can easily do that for the case of circular motion but I will leave that as an exercise for you. Hint: you use the fact that $dr=0$ and you use the full Schwarzschild metric (you do not drop the rotational term in $d \theta$.

There is no "rotational term in $d \theta$" in the Schwarzschild metric. If you are referring to the metric in Ashby's paper, that metric is *not* "the Schwarzschild metric". It uses features of the Schwarzschild metric, but it's not the same thing. Also, there is more than one metric referred to in Ashby's paper; if there are particular equations in Ashby's paper that you are using, it would be helpful to give explicit references.

 

[stevendaryl]

Easy, as already explained for the case of radial motion:

$$\frac{d \tau_1^2}{d \tau_2^2}=\frac{1-r_s/r_1}{1-r_s/r}\frac{1}{1-\frac{v^2}{(1-r_s/r)^2}}$$

The first factor represents the "gravitational redshift" component, the second factor (speed dependent) represents the "Doppler" component.

I thought you were saying that there was no Doppler component! So maybe we are on the same page now. The ratio of clock rates is given by the "gravitational redshift" component. This is NOT equal to the observed redshift, except in the special case in which the sender and receiver are at rest (no Doppler component).

So you cannot compute redshifts by simply taking a ratio of "clock rates", which is what I thought you were claiming.

 

[stevendaryl]

There is no "rotational term in $d \theta$" in the Schwarzschild metric.

Sure there is. In all its gory detail, the Schwarzschild metric is:

d$\tau$² =
(1-2GM/(c²r)) dt²
- 1/(1-2GM/(c²r) dr²/c²
- r²/c² d$\theta$²
- r² sin²($\theta$)/c² d$\phi$²

 

[PeterDonis]

There is no "rotational term in $d \theta$" in the Schwarzschild metric.

I hadn't read the Word document GAsahi posted earlier when I wrote this; having looked at it, I see that by "rotation term" he meant what would usually be called the "angular term". Sorry for the confusion.

 

[GAsahi]

I thought you were saying that there was no Doppler component! So maybe we are on the same page now.

The whole thing started with your claim that my derivation is not correct. The formulas were posted since the beginning of this thread. Are we done now?

 

[GAsahi]

Can you point out where in Ashby's paper you are deriving this from?

I derived it myself, at the beginning of this thread, here.

 

[stevendaryl]

Why not? This is the bee under your bonnet that you keep repeating with no formal justification.

At first, I was not clear whether there was just some miscommunication going on, or there was a serious error of understanding on your part. This shows me that, whatever miscommunication there may have been, you are mistaken about some fundamental facts about GR and SR and Doppler shift.

Try taking the limit as r_s → 0 and v << c. Then the redshift formula should reduce to the nonrelativistic Doppler shift formula.

Your way of doing things would give:
d$\tau$₁ = dt
d$\tau$₂ = √(1-(v/c)²) dt
So the ratio d$\tau$₁/d$\tau$₂ gives
1/√(1-(v/c)²)

That's NOT correct. It should be, instead
√((1-v/c)/(1+v/c))

 

[stevendaryl]

The whole thing started with your claim that my derivation is not correct.

Your derivation is certainly not correct, even though it happens to give the right answer in a specific case. In the case of two observers at the same radius R, moving away from each other at speed v, your derivation gives the wrong answer.

 

[GAsahi]

At first, I was not clear whether there was just some miscommunication going on, or there was a serious error of understanding on your part. This shows me that, whatever miscommunication there may have been, you are mistaken about some fundamental facts about GR and SR and Doppler shift.

Try taking the limit as r_s → 0 and v << c. Then the redshift formula should reduce to the nonrelativistic Doppler shift formula UNLESS one replaces v with its function of the relative speed between source and receiver.

Your way of doing things would give:
d$\tau$₁ = dt
d$\tau$₂ = √(1-(v/c)²) dt
So the ratio d$\tau$₁/d$\tau$₂ gives
1/√(1-(v/c)²)

That's NOT correct. It should be, instead
√((1-v/c)/(1+v/c))

In my formula $\frac{d \tau_1^2}{d \tau_2^2}=\frac{1-r_s/r_1}{1-r_s/r}\frac{1}{1-\frac{v^2}{(1-r_s/r)^2}}$
$v=\frac{dr}{dt}$, so $v$ is NOT the relative speed between source and receiver. This is why the above formula will not reduce to the relativistic Doppler formula.

On the other hand, in the formula $\sqrt{\frac{1-v/c}{1+v/c}}$ v IS the speed of the receiver wrt the source (see my .doc attachment on the subject. Though you are using the symbol "v" in both cases, the meaning is different. We have been over this before.
If you are fixated on finding an "error" in my derivation , you would do best to ask the meaning of the variables first.

 

[stevendaryl]

If you are fixated on finding an "error" in my derivation , you would do best to ask the meaning of the variables first.

I gave you a formula for d$\tau$₁/d$\tau$₂ for the special case in which the first observer is at the constant value of r=R, and constant $\theta$ and $\phi$, the second observer is at the constant value of r=R, but is traveling so that R d$\theta$/dt = v. I thought you AGREED with that formula. You certainly said that you did.

I said: "Now, my claim is that dτ₁/dτ₂ will NOT give the correct redshift for signals sent from the first observer to the second observer."

You replied: "Why not?"

Which to me seemed to indicate that you thought it WOULD give the correct redshift formula. Do you now agree that it doesn't?

 

[GAsahi]

I gave you a formula for d$\tau$₁/d$\tau$₂ for the special case in which the first observer is at the constant value of r=R, and constant $\theta$ and $\phi$, the second observer is at the constant value of r=R, but is traveling so that R d$\theta$/dt = v. I thought you AGREED with that formula. You certainly said that you did.

I said: "Now, my claim is that dτ₁/dτ₂ will NOT give the correct redshift for signals sent from the first observer to the second observer."

You replied: "Why not?"

Which to me seemed to indicate that you thought it WOULD give the correct redshift formula. Do you now agree that it doesn't?

I am done wasting my time.

 

[stevendaryl]

Yes, but when the twins come together, the traveling twin has experienced less elapsed proper time, which of course is *not* a coordinate-dependent statement. So if I define "ratio of two clock rates" in terms of elapsed proper time between some pair of events common to both worldlines, then the ratio is not coordinate-dependent.

Well, that's kind of a funny way to define "rate". By that definition, the rate is undefined for clocks that DON'T get back together eventually.

But anyway, if we don't get hung up on word choice, I think that you agree that
the instantaneous ratio (d$\tau$₁/dt)/(d$\tau$₂/dt) is always defined for any coordinate system, but is coordinate-dependent.

Of course, if the two worldlines don't cross, there won't be a pair of events common to both worldlines. But there may still be a coordinate-independent way to pick out "common" events on both worldlines. For example, if the spacetime has a timelike Killing vector field which is hypersurface orthogonal (as Schwarzschild spacetime does), I can pick two spacelike hypersurfaces orthogonal to the Killing vector field and say that the "common" events on each worldline are the events where the worldlines intersect the two surfaces. This, of course, is a roundabout way of saying "pick the events on each worldline with Schwarzschild coordinate times t1 and t2", but you'll note that I've stated it in a coordinate-independent way. I could do the same thing for a pair of Rindler observers with non-intersecting worldlines.

I would put it a different way, which is that in cases in which there is a special symmetry, there is a "natural" choice for a time coordinate.

Didn't you point out that there are some pairs of observers (such as two observers on Earth's equator but at opposite points) for whom ratio (1) different from ratio (2) even in Schwarzschild coordinates?

Yes, that's why I had my full qualification: If there are coordinates such that (A) the metric components are independent of time, and (B) the sender and receiver are both at rest in that coordinate system, then the ratio of "clock rates" is the same as the redshift for a light signal sent from one to the other. If (A) or (B) fails, then they won't be the same.

It may also be worth noting that Schwarzschild coordinates and Rindler coordinates both have a Killing vector field as "dt", so the two definitions of R1 amount to the same thing in those coordinates.

That was part of my point (Isn't "there is a timelike Killing vector field" and "there is a coordinate system in which the components of the metric are time-independent" the same thing? Is there some circumstance in which one holds and not the other). For specific coordinates and for "stationary" observers in that coordinate system, you can compute a ratio of "clock rates" and get the same answer as the redshift formula, but not in other coordinates or for other observers. For example, in the case of an accelerating rocket with clocks in the front and rear, if you use the inertial coordinates of the "launch" frame, then the ratio of clock rates will not be the same as the redshift.

 

[stevendaryl]

I am done wasting my time.

I can't tell whether that means that you now realize that d$\tau$₁/ d$\tau$₂ is not equal to f₁/f₂, or still think that they're always the same.

 

[GAsahi]

I can't tell whether that means that you now realize that d$\tau$₁/ d$\tau$₂ is not equal to f₁/f₂, or still think that they're always the same.

You need to get your ratios right:

$$\frac{d \tau_1}{d \tau_2}=\frac{f_2}{f_1}$$

 

[PeterDonis]

I derived it myself, at the beginning of this thread, here.

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...

In the case of two observers at the same radius R, moving away from each other at speed v, your derivation gives the wrong answer.

...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" $\Phi_{0}$ 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 Schwarzschild 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]

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.

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)

...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.)

...because the derivation is trivial

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.

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

 

[stevendaryl]

You need to get your ratios right:
$$\frac{d \tau_1}{d \tau_2}=\frac{f_2}{f_1}$$

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
$$\frac{d \tau_1}{d \tau_2}$$ = 1.
$$\frac{f_2}{f_1}$$ = √((1-v/c)/(1+v/c))

 

[stevendaryl]

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

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]

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

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]

I said that it was your method that was incorrect

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]

The general method he used is correct, provided you adopt the particular coordinates and simultaneity convention he adopted.

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.

(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.

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]

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
$$\frac{d \tau_1}{d \tau_2}$$ = 1.
$$\frac{f_2}{f_1}$$ = √((1-v/c)/(1+v/c))

First off, the formula $\frac{f_2}{f_1} = \sqrt{((1-v/c)/(1+v/c))}$
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 $\frac{v^2}{c^2}$ 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, $\frac{f_1}{f_2}=1$ 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]

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.

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:

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^{2} r}}\sqrt{1 - \frac{v^{2}}{c^{2}}}$$

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:

$$\frac{d\tau_{1}}{d\tau_{2}} = \frac{\sqrt{1 - \frac{v_{1}^{2}}{c^{2}}}}{\sqrt{1 - \frac{v_{2}^{2}}{c^{2}}}}$$

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:

I think that you agree that
the instantaneous ratio (d$\tau$₁/dt)/(d$\tau$₂/dt) is always defined for any coordinate system, but is coordinate-dependent.

Yes.

Isn't "there is a timelike Killing vector field" and "there is a coordinate system in which the components of the metric are time-independent" the same thing?

Yes.

 

[Austin0]

Let me do an explicit calculation to prove my point.

In Rindler coordinates (X,T), we have two clocks, one at X = X₁, and one at X₂. The Rindler interval is:

d$\tau$² = X² dT² - dX²

So for clocks at rest in the X,T coordinates, we have:
d$\tau$ = X dT

So the ratio of the rates is: d$\tau$₁/d$\tau$₂ = X₁/X₂

Conclusion: the "higher" clock (with greater X) runs faster.

Now, do the same calculation in the coordinate system (x,t) related to (X,T) through:

x = X cosh(gT)
t = X/c sinh(gT)

So d$\tau$² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d$\tau$ = $\sqrt{1-(v/c)^{2}}$ dt


1)The ratios of the rates in this coordinate system is given by:
d$\tau$₁/d$\tau$₂ = $\sqrt{1-(v_{1}/c)^{2}}$/$\sqrt{1-(v_{2}/c)^{2}}$

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

x = X cosh(gT)
t = X/c sinh(gT)

So d$\tau$² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d$\tau$ = $\sqrt{1-(v/c)^{2}}$ dt


The ratios of the rates in this coordinate system is given by:
d$\tau$₁/d$\tau$₂ = $\sqrt{1-(v_{1}/c)^{2}}$/$\sqrt{1-(v_{2}/c)^{2}}$


2)This part of the derivation is in error. If you did it correctly, you would have gotten that the correct result is $\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-v/c}{1+v/c}}$ where $v$ is the instantaneous speed of the rocket containing the two clocks wrt the launcher frame.

Originally Posted by PeterDonis View Post

It looks to me like these two quantities refer to two different things. The first refers to Schwarzschild spacetime; the second refers to Rindler coordinates on Minkowski spacetime. The answers for those two cases will not be the same, because Schwarzschild spacetime is curved and Minkowski spacetime is flat.

No, both are about flat spacetime. The difference is that
√(1-(v₁/c)²)/√(1-(v₂/c)²) is the ratio of the two clock rates, as measured in the "launch" frame, while √(1-v/c)/√(1+v/c) is the redshift formula for the case in which the "lower" clock sends a signal while at rest, and the signal is received by the "upper" clock when that clock is traveling at speed v. (Since the light signal takes time to propagate, the upper clock will have achieved a nonzero velocity while the light signal is in flight).

3)My point is that the redshift formula is NOT the same as the ratio of clock rates, 4)except in very specific circumstances. Those circumstances actually hold for Rindler coordinates and for Schwarzschild coordinates, but they don't hold for arbitrary coordinates. The conditions for being able to equate "relative clock rates" with "redshift" are: (1) The metric tensor is independent of time, and (2) the sender and receiver are at rest in the coordinate system.

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