No redshift in a freely falling frame

stevendaryl

When they are NOT at rest wrt each other, the redshift formula is NOT the same as the ratio of the clock rates. The redshift differs from this rate by a Doppler correction. They explicitly say that in the Wikipedia article about Pound-Rebka:

Special Relativity predicts a Doppler redshift of :

[itex] f_r=\sqrt{\frac{1-v/c}{1+v/c}}f_e[/itex]

On the other hand, General Relativity predicts a gravitational blueshift of:

[itex]f_r=\sqrt{\frac{1-\dfrac{2GM}{(R+h)c^2}}{1-\dfrac{2GM}{Rc^2}}}f_e[/itex]

The detector at the bottom sees a superposition of the two effects.

GAsahi

The emitter and receiver are NOT at rest wrt each other. The calculations done using Schwarzschild coordinates are confirmed by practice. So, your second claim is false.

GAsahi

You misread the wiki paper: it tells you that the way to measure the gravitational effect is by cancelling it with the appropriate amount of Doppler effect by moving the source wrt the detector at the appropriate speed.

stevendaryl

Because you are making a serious mistake in confusing two different things:
(1) The ratio of clock rates, and (2) the redshift formula. They are not the same, except in special circumstances.

No, they don't. If the emitter and receiver are in motion, then the redshift formula has to be adjusted to include both position-dependent and velocity-dependent effects.

I didn't make a mistake, you did.

You didn't compute d[itex]\tau[/itex] in terms of dt and dr for any coordinate system other than Schwarzschild. My claim is that if you had used a different coordinate system to compute d[itex]\tau[/itex] for the two clocks, and taken the ratio, you would have gotten a different answer than you get for Schwarzschild.

For you to say "if your claims were true the GPS calculations would fail" makes no sense, because what I'm saying AGREES with what you are saying when Schwarzschild coordinates are used to compute d[itex]\tau[/itex]. Since you haven't attempted to compute d[itex]\tau[/itex] for any other coordinate system, the point of disagreement hasn't come up.

Well, it actually has come up, in the Rindler case, but you wisely declined to offer a calculation of d[itex]\tau[/itex] in that case.

stevendaryl

Why do you think that that says something different from what I'm saying? I'm saying that if the two detectors are in motion relative to one another, Doppler shift must be included in the redshift calculation. That's clearly true. They say it right there in the article.

It explicitly says: "The detector at the bottom sees a superposition of the two effects",
where the two effects are position-dependent time dilation, and Doppler shift.

stevendaryl

If the detector and the emitter are not at rest relative to each other (as measured in Schwarzschild coordinates) then the pure position-dependent gravitational time dilation must be corrected by an additional Doppler term. Are you disputing that? That's very bizarre. Think about it: suppose that the receiver and the sender are at the SAME height. Then the redshift is purely due to Doppler.

You CAN'T use the gravitational time dilation to compute redshift without including Doppler, except in the special case in which the sender and receiver are at rest (so that the Doppler effect is zero). You're not seriously disputing that, are you?

GAsahi

Repeating the same error ad nauseaum doesn't make it right. Your so-called "counter-example" has the source and the emitter at rest wrt each other.
You are desperately trying to prove that the method does not apply when the emitter and the detector are moving wrt each other (you changed the goal posts when I showed you that the method works when there is no relative motion). The GPS calculations , as posted by Ashby, disprove your statement.

So, you have a "counter-example" that does not apply and your statements are contradicted by mainstream application of Schwarzschild coordinates to explaining the GPS functionality. You are 0 for 2.

PeterDonis

Hm, ok, I need to go back and read your original posts more carefully. However, I'm not sure GAsahi is talking about flat spacetime (but maybe I need to go back and read his original posts more carefully too).

This is not correct as you state it; the circumstances are not coordinate-dependent. See below.

It would be better if you stated these conditions in coordinate-free terms, which can be done:

(1) The spacetime has a timelike Killing vector field;

(2) The sender and receiver's worldlines are both orbits of the timelike Killing vector field.

That should make it clear that the conditions you are talking about depend on particular properties of the spacetime and the worldlines, but *not* on coordinates; the mathematical description of the conditions looks simpler in Schwarzschild coordinates (or Rindler in flat spacetime), but that doesn't mean it's only "true in" those coordinates.

GAsahi

I showed you how to do the calculations using the Schwarzschild solution for the case of relative motion between source and detector. You do not need any "additional Doppler term", the answer is fully contained in the Schwarzschild solution. You seem to have this bee under your bonnet that you can only use the Schwarzschild solution when the source and the detector are stationary.

stevendaryl

Whether two objects are at rest wrt each other is a COORDINATE-DEPENDENT fact. In Rindler coordinates, two clocks at different values of the X coordinate are at rest relative to one another. In inertial coordinates, they are not at rest relative to one another.

It clearly doesn't. You know that's the case. If the receiver and the sender are at the SAME height, and are moving relative to one another, then there will be a nonzero redshift. The redshift formula in that case is not the same as the position-dependent gravitational time dilation formula. I can't believe you're disputing that.

No, they DON'T. They are in complete agreement. What is true is that the Schwarzschild relative clock rate calculation gives the same answer as the redshift calculation in the case where the sender and receiver are stationary in the Schwarzschild coordinates. If they are NOT stationary in the Schwarzschild coordinates, then there is an additional Doppler effect that must be taken into account. Are you seriously disputing this?

GAsahi

Good, you finally got this right despite multiple previous protestations.


You are either missing the point or you are trying desperately to move the goalposts. If the source and the receiver are moving wrt each other, the effect is WHOLLY described by using the Schwarzschild solution, Doppler AND gravitational effect all rolled in ONE formula, the one formula derived SOLELY using the Schwarzschild solution. You can find that solution posted in this forum. You seem to be disputing that the solution is valid though it is the standard approach to solving such problems (see the references to Neil Ashby).

stevendaryl

You didn't do the case in which the sender and receiver are at the SAME radius r, and have a relative velocity v in the direction perpendicular to the radius. Your method gives the WRONG answer for this case, if you don't include the Doppler effect.

PeterDonis

GAsahi, I can't find a link in this thread to the Ashby paper you are referencing. Do you mean this one?

http://relativity.livingreviews.org/...es/lrr-2003-1/

stevendaryl

No, they're NOT. You are deeply confused about this point. Consider the case in which the sender and the receiver are at the SAME radius r. For example, they are both on the surface of the Earth, on the equator. But they are in relative motion. They are traveling in opposite directions, one traveling east and the other traveling west. One observer sends a signal to the other. Let f₁ be the frequency of the signal as measured by the sender, and let f₂ be the frequency as measured by the receiver.

In this case, the two frequencies will NOT be the same. They will differ by a Doppler shift. How are you proposing to compute that Doppler shift solely using the Schwarzschild metric?

The answer is: you can't. f₁/f₂ is NOT equal to d[itex]\tau[/itex]₁/d[itex]\tau[/itex]₂ in that case.

stevendaryl

He's not, but flat spacetime is a special case of curved spacetime. If the technique works in general, then it should work in flat spacetime, as well.

The claim that I'm making, which is really an indisputable claim, it's pure mathematics, is that the ratio of two clock rates for distant clocks is a coordinate-dependent quantity. This is easily seen to be true in SR: In the twin paradox, during the outward journey, each twin's clock is running slow, as measured in the coordinate system in which the other twin is at rest. The ratio of two clock rates is a coordinate-dependent quantity. It's true in SR, and it doesn't become less true in GR.

My point is that there are two different ratios to compute:

(1) The ratio f₁/f₂ of a light signal sent from one observer to another, where f₁ is the frequency as measured by the sender, and f₂ is the frequency as measured by the receiver.

This quantity is completely independent of coordinates, and you can calculate it using whatever coordinates you like.

(2) The ratio R₁/R₂ of clock rates for the clocks of the two observers.

This quantity is coordinate-dependent. If you use different coordinates, you get a different ratio.

Specifically, R₁ = d[itex]\tau[/itex]/dt = √(g_αβ dx^α/dt dx^β/dt. This rate has different values in different coordinate systems.

What's special about Schwarzschild coordinates (or Rindler coordinates) is that ratio (2) is equal to ratio (1) for those coordinates, but not for other coordinates.

You are right, that if there is a Killing vector field, then we can come up with a corresponding ratio by defining R₁ = d[itex]\tau[/itex]/dt, where dt is the timelike Killing vector, instead of a coordinate. In that case, R₁ is no longer coordinate-dependent.

GAsahi

yes,of course.

PeterDonis

For one definition of "ratio of two clock rates", yes this is true. But there are other possible definitions.

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.

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.

In a sense all these choices of "common events" are arbitrary; but they do match up with particular symmetries of the spacetime, so they're not completely arbitrary. They do have some coordinate-independent physical meaning.

Yes, agreed.

This depends, as above, on how you define "relative clock rates". You note this as well, since you agree that we could use a timelike Killing vector field as the "dt" in R1.

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? Perhaps what you meant to say is that ratio (1) is equal to ratio (2) for observers who are *static* in these coordinates?

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.