Equivalence Principle in muon experiment?

[Ich]

I don't think that's what he did. I think he assumed there were such an f and g, and then wrote down what the time dilation would look like under that assumption.

No, not really. I knew beforehand that the dilation must be the product of a gravitational and a kinematical component by the way it works: A static observer sees something at his position and relays it to the reference observer. Whatever he sees will look even slower for the reference observer due to gravitational dilation only. So if you rewrite the velocity to what it really is in the static observer's frame, the expressions must be separable, no matter how complicated they look in terms of coordinate velocity. But it's still good to do the math and see how everything falls into place. And the math is, as Dale Spam presumed, a coordinate transformation (basically some scaling, as the basis vectors still point in the same directions).

It looks to me like that assumption was based on something like his claim about the time dilation being the dot product of the object's 4-velocity with the appropriate Killing vector; but as I showed in post #61, that only works for static observers (i.e., observers following orbits of the Killing vector field).

Right. But, as my thinking goes, it was based rather on the operational thoughts mentioned above, the geometric formulation came afterwards and a bit too hastily, I fear. But you see the (hopefully!) correct version in my preceding answer to you.

In post #39 (Ich's first post in this thread), he wrote down a formula which looks, superficially, like it's separable into such an f and g, but that formula mixes time dilations relative to two different coordinate charts. The one involving $U$ is relative to the global Schwarzschild chart, but the one involving $v$ is relative to the local inertial frame of a static observer (note that that's how he *defines* $v$).

Sure. But isn't is a basic right for physicists to use the chart they find most useful? The question was never whether in Schwarzschild coordinates the contributions of potential and coordinate velocity are separable. It was whether one can sepatate "GR" and "SR" contributions in a static spacetime, which I understood to mean gravitational and kinematic time dilation, respectively.

 

[atyy]

It was whether one can sepatate "GR" and "SR" contributions in a static spacetime, which I undersood to mean gravitational and kinematic time dilation, respectively.

Could you comment on how this is related to the OP, and the equivalence principle? The OP sets up the question in special relativity, so it is a static spacetime, but the accelerated muon presumably is something like a "non-static spacetime" (yes, I know that's not right, but the point here is not the answer since we all know how to calculate that, it's to what extent the OP's heuristic can be justified)?

 

[Ich]

Could you comment on how this is related to the OP, and the equivalence principle? The OP sets up the question in special relativity, so it is a static spacetime, but the accelerated muon presumably is something like a "non-static spacetime" (yes, I know that's not right, but the point here is not the answer since we all know how to calculate that, it's to what extent the OP's heuristic can be justified)?

Sorry, my entry point in this thread was Dale Spam's #22 , I never read the OP and did not intend to start a parallel discussion. Concerning the OP, the answer is easy: proper accelerations don't matter courtesy of the clock hypothesis. Gravitational fields do matter, however, but there are none if you describe the accelerator in an inertial frame, as he obviously does. You can introduce some by switching to a corotating, and yes, there would be gravitational time dilation, but all it does is to compensate for the now absent (as the particles are at rest in the new frame) kinematic time dilation. Just many ways of expressing the same thing.

 

[PeterDonis]

The product of both is the total time dilation

But this only works because the $g_{tt}$ factors cancel, so you're just left with the reciprocal of $u^t$. It doesn't show that $u^t$ factors into two contributions, which is what you've been claiming.

if you rewrite the velocity to what it really is in the static observer's frame, the expressions must be separable

But as I noted in a previous post, that just amounts to saying that, if you switch definitions of "time dilation" in midstream, so to speak, you can make "time dilation" seem separable. What you're doing is defining "gravitational time dilation" relative to the global Schwarzschild coordinates (i.e., it's 1 for the reference observer at infinity), and "kinematic time dilation" relative to the local static observer. Obviously the product of these two things is separable--it's a product! But all that's saying is that time dilation is separable if you define the two terms in the product in terms of variables in two different frames. It's not the same as saying that time dilation, defined consistently (gravitational and kinematic both defined using variables relative to the same frame) is separable.

Now you could say that you're not defining the two terms in the product in terms of variables in two different frames--you're defining them in terms of two invariants (the norm of a Killing vector and the dot product of two vectors), which just happen to match up to variables in two different frames. Which works fine as long as the motion of the "moving" object doesn't change altitude (which is the simplified case we've been discussing for the latter part of this thread)--but what if it does?

 

[PeterDonis]

isn't is a basic right for physicists to use the chart they find most useful?

Sure, but you can't switch charts in the middle of a formula.

 

[Ich]

The product of both is the total time dilation

But this only works because the $g_{tt}$ factors cancel, so you're just left with the reciprocal of $u^t$. It doesn't show that $u^t$ factors into two contributions, which is what you've been claiming.

Pardon me, but you lost me here. I showed exactly two contributions, one gravitational and one kinematical, the product of which happens to be the total time dilation, the reciprocal of $u^t$ namely. As an answer, you say I didn't show that [the reciprocal of] $u^t$ factors into two contributions.
There seems to be something wrong here.

But as I noted in a previous post, that just amounts to saying that, if you switch definitions of "time dilation" in midstream, so to speak, you can make "time dilation" seem separable.

No. Definitely no. They are separable. It's a two step process, one from the moving body to the static observer, the next from the static observer to the reference observer. Which coordinates I use for each is none of your business, you can't accuse me of switching definitions to make something seem separable. I'm using the appopriate charts to calculate the respective contributions, and show that their product is the total time dilation.

What you're doing is defining "gravitational time dilation" relative to the global Schwarzschild coordinates (i.e., it's 1 for the reference observer at infinity), and "kinematic time dilation" relative to the local static observer. Obviously the product of these two things is separable--it's a product!

Well, but isn't the point that this product is not just some product, but also happens to be the total time dilation? Proving thus that the latter is separable, if one admits that the factors are of gravitational and kinematical origin, respectively?

Now you could say that you're not defining the two terms in the product in terms of variables in two different frames--you're defining them in terms of two invariants (the norm of a Killing vector and the dot product of two vectors), which just happen to match up to variables in two different frames.

Yes, that's actually also what I'm really saying. But it's not sheer coincidence - my starting point is indeed the invariant quantities, which I then try to express in the appropriate frames. So it's by "intelligent design" that they "happen" to match up to variables in the respective frames.

Which works fine as long as the motion of the "moving" object doesn't change altitude (which is the simplified case we've been discussing for the latter part of this thread)--but what if it does?

Yeah, what if? I claimed that the equations also hold if $v \parallel$ is not zero; I also gave the transformations that led me to this result. It shoud be easy to check.

 

[Dale]

I don't think that's what he did. I think he assumed there were such an f and g, and then wrote down what the time dilation would look like under that assumption.

That is a good point. I think that it is reasonable to assume such coordinates exist since there is so much freedom in choosing coordinates, but an explicit chart would be preferable.

 

[PeterDonis]

I showed exactly two contributions, one gravitational and one kinematical, the product of which happens to be the total time dilation, the reciprocal of ##u^t## namely.

I think my question here is a matter of interpretation, so I'll drop it. I agree that, as long as you view the "total time dilation" as a product of two invariants, it's obviously separable because it's a product of two separate things. I think the issue is whether the term "total time dilation" is really the appropriate term for this, since "time dilation" is usually viewed as coordinate-dependent, but invariants are coordinate-independent.

I'm using the appopriate charts to calculate the respective contributions

You can calculate invariants in any chart you like, yes, and there's no need to use the same chart for different invariants, as long as you end up with invariants. But then the expression you end up with, though it is obviously separable when viewed as the product of two invariants, as above, will not necessarily be separable if one insists that it has to be written in terms of quantities defined in a single coordinate chart. Whether that matters is probably also a matter of interpretation.

However, physically, there is certainly something different about these two invariants. One, the kinematic invariant, involves a purely local quantity, the relative velocity of the moving observer and a (momentarily) co-located static observer. The other, the gravitational invariant, involves a non-local measurement; there is no way for a static observer to determine the length of the Killing vector at his location by purely local measurements. He has to receive light signals from a source located at the reference Killing vector and measure their blueshift (or redshift, as the case may be). Part of the intuitive tug that some of us seem to be feeling towards expressing everything in a single chart, the global Schwarzschild chart, may be due to the fact that the gravitational invariant involves a non-local measurement, so if you express it in terms of a chart at all, it can only be a global chart.

I claimed that the equations also hold if $v_{\parallel}$ is not zero; I also gave the transformations that led me to this result.

Can you point to the specific post? For some reason I'm not finding it.

 

[Dale]

Maybe I do. I always try to see things from the perspective of their operational implementation and, correspondingly, try to find a covariant, geometric expression for what is happening. Coordinates come into the game sometimes abstractly as a calculation tool, but more often by their physical meaning.

Coordinates do not generally have a physical meaning, and it doesn't make sense to try to find a covariant geometric expression for a quantity which is not covariant.

But to me, time dilation is the differential aging between such and such events, usually with an operational setup in mind and a geometric representation available. An easy example is the dilation between two observers in relative motion at the same event. Time dilation is the (!inverse! ;) ) dot product of the respective velocity vectors. That's as coordinate-independent as it can get. ... Time dilation is a relation by nature

While that is an interesting covariant quantity, it is simply not time dilation.

When two observers are in relative motion at the same event we do not say that they have a time dilation relation between them. We pick a reference frame and we say that each has a given amount of time dilation wrt that coordinate chart. We unambiguously say that the clock at rest in our chosen frame is undilated and we say that the clock that is moving is time dilated. We do not describe it as a relationship between the two clocks, but a relationship between each clock and the frame. Remove either clock and we still can speak of the other clock's time dilation wrt any given frame. It is inherently frame variant, because in different frames the first clock may be undilated and the second dilated, or in yet another frame they may both be equally time dilated.

If you are talking about a covariant geometric quantity then you are not talking about time dilation. In the rest frame of one of the clocks the covariant quantity is equal to the time dilation of the other clock, but that does not make the covariant quantity the same as the time dilation any more than the invariant mass is the same as the energy.

I accept without proof that there exists a coordinate chart where the gravitational and kinematic time dilations are separable. That does not change the fact that they are not separable in general charts and specifically that they are not separable in the usual coordinate charts that I have examined on Scwharzschild spacetime. Hence, the OP's concern about the GR part and the SR part of time dilation was not something that he should worry about in general since those parts are not always separable.

 

[atyy]

Coordinates do not generally have a physical meaning, and it doesn't make sense to try to find a covariant geometric expression for a quantity which is not covariant.

While that is an interesting covariant quantity, it is simply not time dilation.

When two observers are in relative motion at the same event we do not say that they have a time dilation relation between them. We pick a reference frame and we say that each has a given amount of time dilation wrt that coordinate chart. We unambiguously say that the clock at rest in our chosen frame is undilated and we say that the clock that is moving is time dilated. We do not describe it as a relationship between the two clocks, but a relationship between each clock and the frame. Remove either clock and we still can speak of the other clock's time dilation wrt any given frame. It is inherently frame variant, because in different frames the first clock may be undilated and the second dilated, or in yet another frame they may both be equally time dilated.

How about gravitational red shift of two clocks at slightly different heights in a potential well? That does seem to be a relationship between two clocks, and also seems to be something like a Doppler shift, which at least naively, seems to be "real". I don't understand Ich's construction, but I wonder if something like this lies behind it. In the special relativity twin paradox, we usually don't consider the Doppler shift as time dilation, since according to the time dilation, the other clock runs slow throughout the trip, but the Doppler shift has the observed clocks running slow then running fast (I may have garbled that, reading quickly off http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html: "Terence computes that Stella's clock is really running slow by a factor of about 7 the whole time, but he sees it running fast during the Inbound Leg because each flash has a shorter distance to travel. And Stella computes the same for Terence.").

Edit: I didn't intend everything to be blue after the link, but it's somehow automatically coloured.

 

[Dale]

How about gravitational red shift of two clocks at slightly different heights in a potential well? That does seem to be a relationship between two clocks, and also seems to be something like a Doppler shift, which at least naively, seems to be "real".

I agree. Back in post 68 I even specifically mentioned redshift and differential aging as two standard invariant quantities that are associated with time dilation.

Edit: I didn't intend everything to be blue after the link, but it's somehow automatically coloured.

No worries, I fixed it. I think that the colon confused the parser into thinking that what followed was still part of the URL.

 

[PeterDonis]

I accept without proof that there exists a coordinate chart where the gravitational and kinematic time dilations are separable.

I don't. I only accept that the two invariants Ich defined are separable in the sense that you can obviously define them separately and then multiply them. I don't think there is a single chart in which you can express their product as a separable formula. You have to switch charts to do that.

 

[Ich]

I think my question here is a matter of interpretation, so I'll drop it. I agree that, as long as you view the "total time dilation" as a product of two invariants, it's obviously separable because it's a product of two separate things.

That's good.

I think the issue is whether the term "total time dilation" is really the appropriate term for this, since "time dilation" is usually viewed as coordinate-dependent, but invariants are coordinate-independent.

No Problem. For me, there is an operational definition and a measurement that indicates that u's clock is "ticking slower" as judged by the reference observer. And that is time dilation. If you insist on defining it in terms of a global coordinate chart - well, I think I would disagree, but there's no need to discuss this here, as it is ultimately a matter of taste, not physics.

However, physically, there is certainly something different about these two invariants. One, the kinematic invariant, involves a purely local quantity, the relative velocity of the moving observer and a (momentarily) co-located static observer. The other, the gravitational invariant, involves a non-local measurement; there is no way for a static observer to determine the length of the Killing vector at his location by purely local measurements. He has to receive light signals from a source located at the reference Killing vector and measure their blueshift (or redshift, as the case may be). Part of the intuitive tug that some of us seem to be feeling towards expressing everything in a single chart, the global Schwarzschild chart, may be due to the fact that the gravitational invariant involves a non-local measurement, so if you express it in terms of a chart at all, it can only be a global chart.

Agreed to all.

Can you point to the specific post? For some reason I'm not finding it.

That was #53 . I made the calculation, but didn't show it here, nothing interesting about it.

 

[Ich]

Coordinates do not generally have a physical meaning, and it doesn't make sense to try to find a covariant geometric expression for a quantity which is not covariant.

Right. So if you insist that time dilation is just some coordinate issue, I'm not talking about time dilation. Call it "transverse Doppler effect" or something, it doesn't matter.

When two observers are in relative motion at the same event we do not say that they have a time dilation relation between them. [...]We do not describe it as a relationship between the two clocks, but a relationship between each clock and the frame.

Ok, but: if time dilation were just a coordinate artifact (you did't claim that, I know), we would not bother talking about it. We're talking about it because there is something very physical behind those frames you're using. It is the (imagined) latticework of meter sticks, equipped with many clocks synchronized in a certain and very sensible way with a master clock, which may be thought of being carried by some canonical observer A. Comparing the moving clock B with said clocks showing the coordinate time, we get time dilation.
And this whole process has a very neat geometric description: Synchronizing the frame clocks according to the Einstein convention, you define the directions orthogonal to the four velocity of the observer. Comparing the moving clock with a frame clock thus simply gives you the orthogonally projected length of B's unit time vector onto A's unit time vector. That is, the dot product of their four velocities. The process is reciprocal, as is the dot product.

At least, that's my way of thinking about it. As long as there is no error in the reasoning above, I'm happy with it and the "unnamed" geometric quantity I've been constructing here. No problem if you disagree, but to me, it'll always be my little time dilation. :)

 

[PeterDonis]

Synchronizing the frame clocks according to the Einstein convention, you define the directions orthogonal to the four velocity of the observer. Comparing the moving clock with a frame clock thus simply gives you the orthogonally projected length of B's unit time vector onto A's unit time vector.

I think you mean "parallel" here, not "orthogonal". Projections orthogonal to the observer's 4-velocity would give spatial vectors, not timelike vectors.

Comparing the moving clock with a frame clock thus simply gives you the orthogonally projected length of B's unit time vector onto A's unit time vector. That is, the dot product of their four velocities.

There are some complications here that affect how your interpretation generalizes.

First: the actual dot product involved is between the 4-velocity of a static observer at some finite radius and a moving observer at that same radius. (I think those are the ones you've been calling A and B.) The 4-velocity of a static observer at infinity, who is your "reference" observer, is spatially separated from A and B, and you can't take dot products between spatially separated vectors in curved spacetime.

However, since A happens to be static, there is a well-defined way to "convert" a dot product of B's 4-velocity with A's 4-velocity, to a sort of "effective dot product" of B's 4-velocity with the reference observer's 4-velocity: you just multiply by A's "redshift factor" relative to the reference observer. This is the product of the two invariants that you have been talking about.

But now let's ask a question: how does this computed "time dilation factor" for B compare to the actual redshift of light emitted by B and received by the reference observer? (We know they're the same for A, but A is static.) We'll idealize by imagining, not just a single reference observer, but a whole fleet of them, in a big sphere at infinity, so that no matter where B is in his orbit, he can always emit a flash of light radially outward and have it received by a reference observer. (We'll also ignore the fact that he has to aim the light at an angle to make sure it travels radially outward, because of aberration; he has a supercomputer that can calculate all that.)

Given all that, if B is in a circular orbit at a constant radius (the same radius as A), then I think the redshift of his light, when received by a reference observer, will be given by your "time dilation" factor--the product of A's "redshift factor" with B's time dilation relative to A. However, if B is in an elliptical orbit, so that he has nonzero radial velocity, I think this will not be the case. The radial velocity leads to a Doppler shift that has to be included in the computation of the overall redshift.

(To see this in an extreme case, suppose that observer B is moving radially outward from A at exactly escape velocity--that is, he will eventually reach infinity and just be at rest there. If he emits light as he passes A, this light will have zero redshift when it reaches infinity--we know this because B's energy at infinity is equal to his rest mass, since he's moving at escape velocity, meaning he has zero "redshift factor". A would interpret this as the Doppler blueshift from his outward motion exactly cancelling the gravitational redshift from the change in altitude. But B's "time dilation" relative to A is still given by $\sqrt{1 - v^2}$, so A still interprets B's clocks as "running slower" than his, even though an observer at infinity would say the opposite based on the respective redshifts of the light they emit.)

 

[Dale]

if time dilation were just a coordinate artifact (you did't claim that, I know), we would not bother talking about it.

If only that were true! We talk endlessly about things that are just coordinate artifacts.

We're talking about it because there is something very physical behind those frames you're using. It is the (imagined) latticework of meter sticks, equipped with many clocks synchronized in a certain and very sensible way with a master clock, which may be thought of being carried by some canonical observer

I find this type of thinking somewhat dangerous. It inevitably leads to confusion where people assign physical significance to coordinate artifacts because they believe that the coordinates are "something very physical". This is precisely the notion which leads to the endless discussions of things which are merely coordinate artifacts.

Even where the coordinates are assigned using some physical rods and clocks, any coordinate-dependent quantities remain coordinate dependent quantities and it is important to recognize them as such. They do not become "something very physical" simply because the coordinate-system involves physical rods and clocks.

 

[Ich]

I think you mean "parallel" here, not "orthogonal". Projections orthogonal to the observer's 4-velocity would give spatial vectors, not timelike vectors.

I think an orthogonal projection onto U gives a vector parallel to U.

The radial velocity leads to a Doppler shift that has to be included in the computation of the overall redshift.

True, but I don't try to calculate redshifts. The time dilation factor I'm talking about is without the Doppler factor. An observer at infintiy would not attribute the redshift to time dilation alone. He would do this for A's light, of course, but would rely on A to tell him B's measured time dilation at A.

I think (but that's another quick shot and not really thought through) that, geometrically, the difference is whether you transport B's wave vector or his velocity vector to the observer at infinity. The former gives redshift, the latter the gamma factor. OTOH, this may also be utterly wrong, it's just an idea.

 

[Ich]

I find this type of thinking somewhat dangerous. It inevitably leads to confusion where people assign physical significance to coordinate artifacts because they believe that the coordinates are "something very physical". This is precisely the notion which leads to the endless discussions of things which are merely coordinate artifacts.

To the contrary. It is most important to know clearly what your coordinates are and what they are not. I did not say that coordinates are something very physical, I think you misrepresented my stance there grossly. But "those frames you're using" are not some numbers, but local inertial frames after Einstein. The very paper that introduced the concept of time dilation to physics did so after extensively defining how inertial frames are constructed and what the coordinates mean. Time dilation in that paper was not some coordinate quirk, it was something expressed in rod and clocks - and differential aging.

It is indeed dangerous to think of all coordinates as something physical. I know this kind of discussions very well. But thinking of all coordinates just as numbers is not the solution. You have to know what you're dealing with, what you can immediately see and what not.
There are quite a few quantities that are observer-dependent. If you specify the observer, you have an invariant quantity with physical meaning. Imho, local time dilation belongs to this group just as redshift does. You don't need certain frames to calculate or express them, but they may be useful. That has nothing to do with reading off coordinate values and brainlessly taking them as real intervals.

Even where the coordinates are assigned using some physical rods and clocks, any coordinate-dependent quantities remain coordinate dependent quantities and it is important to recognize them as such. They do not become "something very physical" simply because the coordinate-system involves physical rods and clocks.

Well, to me, the covariant vector product or the operational definition is as physical as it can get. It needs an observer for its definition. And yes, it is physical - not because the coordinate-system involves physical rods and clocks, but because the coordinates you're using are a shorthand notation for the actual (or imaginary) rods and clocks that construct/define the effect.
But let's not go into a discussion of "physical" or "real". I think I stated the case for local time dilation as an observer-dependent effect that is not just a coordinate artifact. If you see it differently, well, I know of no experiment which would decide who is right and who wrong.

 

[atyy]

But let's not go into a discussion of "physical" or "real". I think I stated the case for local time dilation as an observer-dependent effect that is not just a coordinate artifact. If you see it differently, well, I know of no experiment which would decide who is right and who wrong.

But if you define it using Killing vectors, that seems very nonlocal, which would be ok, just not in line with the idea that time dilation is something that can be derived entirely from the equivalence principle.

 

[PeterDonis]

I think an orthogonal projection onto U gives a vector parallel to U.

The term "orthogonal" usually means "perpendicular", i.e., zero dot product. So a projection orthogonal to U of some vector would give the component of the vector that has zero dot product with U. I don't think that's what you mean. Can you give the math explicitly for the kind of projection you are referring to? Or a reference that uses it?

 

[PeterDonis]

The time dilation factor I'm talking about is without the Doppler factor. An observer at infintiy would not attribute the redshift to time dilation alone. He would do this for A's light, of course, but would rely on A to tell him B's measured time dilation at A.

I'm not sure I understand. Are you saying that redshift = time dilation, by your definition, for A only, but not for B? Or are you saying that redshift = time dilation for B too, but that somehow the observer at infinity has to ask A what B's time dilation is (instead of just directly measuring B's redshift)?

 

[PeterDonis]

I think (but that's another quick shot and not really thought through) that, geometrically, the difference is whether you transport B's wave vector or his velocity vector to the observer at infinity. The former gives redshift, the latter the gamma factor.

Do you mean the wave vector of the light B emits? This will be determined by B's 4-velocity (i.e., the dot product of the emitted wave vector with B's 4-velocity is determined; that's the initial condition that determines the emitted frequency of the light). Parallel transporting the emitted wave vector along the null geodesic the light follows from B to the observer at infinity (call him O), and then taking the dot product of the parallel transported wave vector with O's 4-velocity, does indeed give the redshift of B's light as observed by O. (In fact, this general prescription for determining observed redshift works in any spacetime whatever.)

As for the velocity, do you mean B's 4-velocity vector? If so, I'm not sure how you would transport it to infinity to compare with O's 4-velocity. Parallel transporting it along the null geodesic the light follows won't work, because parallel transport preserves dot products, so the "gamma factor" for B calculated this way would end up being equal to the redshift of the light as observed by O. I think parallel transporting along a purely radial spacelike geodesic lying in a surface of constant Schwarzschild coordinate time would give the gamma factor you are referring to (the product of the two invariants you specified), but I haven't done a calculation to verify that. But even if it works, it's important to note that the sense of "transport" being used is different from that used for the wave vector above.

 

[Ich]

The term "orthogonal" usually means "perpendicular", i.e., zero dot product. So a projection orthogonal to U of some vector would give the component of the vector that has zero dot product with U. I don't think that's what you mean. Can you give the math explicitly for the kind of projection you are referring to? Or a reference that uses it?

Wikipedia, here and here. Zero dot product between image and projection direction, not between image and the vector you project onto. I think that's the usual definition.

Do you mean the wave vector of the light B emits? This will be determined by B's 4-velocity (i.e., the dot product of the emitted wave vector with B's 4-velocity is determined; that's the initial condition that determines the emitted frequency of the light). Parallel transporting the emitted wave vector along the null geodesic the light follows from B to the observer at infinity (call him O), and then taking the dot product of the parallel transported wave vector with O's 4-velocity, does indeed give the redshift of B's light as observed by O. (In fact, this general prescription for determining observed redshift works in any spacetime whatever.)

Synge 1960, I think. I came across this result a few times.

I'm not sure I understand. Are you saying that redshift = time dilation, by your definition, for A only, but not for B? Or are you saying that redshift = time dilation for B too, but that somehow the observer at infinity has to ask A what B's time dilation is (instead of just directly measuring B's redshift)?

That's not my definition - at least I'm not aware of it. A has zero two-way redshift all the time, so everything has to be attributed to gravitational redshift = time dilation. Doppler shift, on the other hand, is also two-way.

As for the velocity, do you mean B's 4-velocity vector? If so, I'm not sure how you would transport it to infinity to compare with O's 4-velocity. Parallel transporting it along the null geodesic the light follows won't work, because parallel transport preserves dot products, so the "gamma factor" for B calculated this way would end up being equal to the redshift of the light as observed by O. I think parallel transporting along a purely radial spacelike geodesic lying in a surface of constant Schwarzschild coordinate time would give the gamma factor you are referring to (the product of the two invariants you specified), but I haven't done a calculation to verify that. But even if it works, it's important to note that the sense of "transport" being used is different from that used for the wave vector above.

Must be along the spacelike geodesic. You're right, you can't transport both along the same line.

 

[Ich]

But if you define it using Killing vectors, that seems very nonlocal, which would be ok, just not in line with the idea that time dilation is something that can be derived entirely from the equivalence principle.

Right. In my discussion with Dale Spam, I concentrated on local time dilation only, as we seemed to disagree on that already. Just basic SR.

 

[vanhees71]

I think now you throw out the baby with the bathwater. Not all coordinate times are unphysical. To come back to the original topic with the life-time dilation of the muons, this is a measurable physical effect. The amount muons from the cosmic radiation reaching us is larger than naively expected when not taking into account the relativistic space-time structure.

Of course, here are two times involved, but both times are physical and not mere coordinates without physical meaning. First the life-time, $\tau$ of an unstable particle is defined in its rest frame. This is just a convenient convention, because it uniquely defines an intrinsic parameter of the particle in a uniquely defined reference frame which is indeed preferred by the physical situation to look at this particle. The second time is the eigentime of an observer in an inertial frame (discussing the special-relativistic case here), which is at the same time the coordinate time of this frame. It's a well defined physical time, measurable by a clock at rest relative to the observer.

The life-time of the particle, defined in its rest frame can be evaluated as its proper time. In terms of the coordinate time of the observer it's given by
$$\tau=\int \mathrm{d} t \sqrt{\dot{x}^{\mu} \dot{x}_{\nu}}$$,
which is an invariant.

The most simple case is a muon in uniform motion. Then you have (setting $c=1$ as is natural in relativity and using the west-coast convention, most common in HEP)
$$\dot{x}_{\mu} \dot{x}^{\mu}=1-\dot{x}^2=1-v^2.$$
Thus, the proper time (setting the time origin such that $t=0 \; \Leftrightarrow \; \tau=0$, you get
$$\tau=\sqrt{1-v^2} t.$$
So the measured mean-life time of the muon of the observer is
$$t=\frac{\tau}{\sqrt{1-v^2}}=\gamma \tau,$$
i.e., longer by the Lorentz time-dilation factor.

As stressed already at the beginning of this thread this is a very precisely measured effect of relativistic kinematics. A recent measurement by my "Alma Mater", GSI(=Gesellschaft für Schwerionenforschung=Helmholtz institute of Heavy-Ion Research) in Darmstadt, Germany, made it to the public-science media:

http://phys.org/news/2014-09-ions-relativistic-dilation-precision.html

The research article can be found here:
http://dx.doi.org/10.1103/PhysRevLett.113.120405

 

[PeterDonis]

Zero dot product between image and projection direction, not between image and the vector you project onto. I think that's the usual definition.

Hm, yes, I see. Probably I just haven't read enough formal treatments of this. The terminology still seems weird to me, but it probably makes sense to a mathematician. ;)

A has zero two-way redshift all the time

I don't understand; zero two-way redshift relative to what observer? And what, exactly, do you mean by "two-way redshift"? Do you mean the net redshift of light doing a two-way round trip from A to somewhere else and then back to A? That's not what I've been talking about all this time; I've been talking about the redshift of A's light as observed by O, who is at infinity.

 

[Dale]

I did not say that coordinates are something very physical, I think you misrepresented my stance there grossly.

My apologies. I was not intending to imply that I was representing your stance at all. I was just trying to show what I see as the danger of the idea you expressed because of the common conclusion that many novices reach, starting from that point.

I have endured many very long discussions with other people who could not understand why they were reaching incorrect conclusions because of their focus on coordinates. Their justification was always the thought that you conveyed. That is why I consider it a dangerous idea. Not because you were using it to make any incorrect conclusions, but because of how often I have seen other people do so.

But thinking of all coordinates just as numbers is not the solution.

I disagree completely, in my opinion it is the only theoretically justifiable solution. All coordinates are always just numbers. Even when those numbers are assigned by some very reasonable and well-accepted physically-based convention, the coordinates themselves are still always just numbers. That is simply a fact of the mathematical structures used by relativity. Coordinates are never anything more than a computational convenience. All of the physical content is contained only in invariants.

Consider even the simplest example in the most reasonable of coordinates, the SR inertial frame. A coordinate time difference of 1 s along some worldline conveys no physical information whatsoever, you cannot even know if the worldline could represent a material object. A proper time difference of 1 s, on the other hand, does convey physical information.

 

[vanhees71]

But if you have an inertial reference frame in SR you just need to take an observer at rest in this frame. Then his proper time is the coordinate time of this reference frame, and has thus a physical meaning.

 

[Dale]

But if you have an inertial reference frame in SR you just need to take an observer at rest in this frame. Then his proper time is the coordinate time of this reference frame, and has thus a physical meaning.

Even in that case the proper time and the coordinate time are not the same thing. The proper time is defined only along the observer's worldline where it has physical meaning. The coordinate time has no physical meaning at any point off the observer's worldline.

In particular, there is no physical sense in which the time of any event off the observer's worldline is "the same time" as a given event on the observer's worldline. The simultaneity established by the coordinate system is entirely a matter of convention, not physics.

 

[vanhees71]

But already Einstein in his famous paper in 1905 has provided a physical way to provide the synchronization of the coordinate time via light signals. In GR such a synchronization is possible only locally. For a derivation see Landau and Lifshits vol. 2.

 

[Dale]

But already Einstein in his famous paper in 1905 has provided a physical way to provide the synchronization of the coordinate time via light signals.

But that synchronization remains only a convention. If it were more then simultaneity could not be relative.

 

[Ich]

I don't understand; zero two-way redshift relative to what observer? And what, exactly, do you mean by "two-way redshift"? Do you mean the net redshift of light doing a two-way round trip from A to somewhere else and then back to A? That's not what I've been talking about all this time; I've been talking about the redshift of A's light as observed by O, who is at infinity.

I was just arguing why I treat the redshift from A to O as purely gravitational time dilation, but not the redshift B to O. The difference is that A and O are static, so there's no contribution from a varying distance. B's redshift on the other hand does have such a contribution. I mentioned the two-way (round trip) redshift as a possible operational definition of "no changing distance", to be sure that this sentence is not frame-dependent.

 

[PeterDonis]

I mentioned the two-way (round trip) redshift as a possible operational definition of "no changing distance"

The usual operational definition of "no changing distance" is a constant round-trip travel time for light signals. I think this is equivalent to a zero round-trip redshift (i.e., redshift one way exactly canceled by blueshift the other way), but I haven't done a computation to prove it. You're right that either one is invariant and therefore not frame-dependent.