Equivalence Principle in muon experiment?

[atyy]

In the example given in the Usenet Physics FAQ, the EP applies just fine, because the experiment can be analyzed within a single local inertial frame. The fact that you also get gravitational time dilation between observers whose difference in height is too large for them to both fit in a single local inertial frame is just an additional fact; it doesn't change the analysis of the case where the height difference is not too large.

Yes, the FAQ example of two observers in a spaceship is fine, since that is all "local". But how about the muon going round in a ring? It seems to differ from the spaceship example, because (1) how is going round in a ring "local" and (2) in the muon case one observer is inertial and one is accelerating, whereas the spaceship example has both observers accelerating.

 

[Ich]

If this is your definition of "time dilation", then there is no such thing as gravitational time dilation to begin with.

It's just how you calculate the kinematic time dilation of two observers at the same event. The SR part of time dilation, if you like.
It was my point that this expression definitely has nothing to do with gravitational time dilation.

 

[Ich]

I got the formula from the Wikipedia site I linked to, with c=1 and the radial component of velocity = 0 for simplification, as I mentioned earlier.

Ah, I see. But they use the coordinate velocity dx/dt. The decomposition is in terms of relative velocity to the local static observer. If I'm not mistaken, that is $v=v_{rel} =v_{co}/ \sqrt{1-2U}$. So you have

$\sqrt{1-2U-v_{co}^2}=\sqrt{1-2U-(1-2U)v^2}=\sqrt{1-2U}\sqrt{1-v^2}$

We can keep the radial velocity if you want, but it makes the gravitational and kinematic components even less separable.

It's just more math, but with $v_{r, rel} =v_{r,co}/ (1-2U)$ you still get $\sqrt{1-2U}\sqrt{1-v^2}$.

 

[Dale]

Ah, I see. But they use the coordinate velocity dx/dt. The decomposition is in terms of relative velocity to the local static observer. If I'm not mistaken, that is $v=v_{rel} =v_{co}/ \sqrt{1-2U}$. So you have

$\sqrt{1-2U-v_{co}^2}=\sqrt{1-2U-(1-2U)v^2}=\sqrt{1-2U}\sqrt{1-v^2}$
It's just more math, but with $v_{r, rel} =v_{r,co}/ (1-2U)$ you still get $\sqrt{1-2U}\sqrt{1-v^2}$.

Sure, I agree, but as I said earlier $v_{co}=g(U,v)\ne g(v)$. I.e. different observers at different U will disagree on the value of $v_{co}$.

 

[Ich]

Sure, I agree, but as I said earlier $v_{co}=g(U,v)\ne g(v)$. I.e. different observers at different U will disagree on the value of $v_{co}$.

Sorry, I can't follow. I don't see why a decomposition should use, of all things, a coordinate velocity.
The thing is: $\gamma(U,v)= f(U) \, g(v)$. f is a function of the potential only, and g is a function of the relative velocity only. That is a geometric, frame-independent, unique decomposition of time dilation, exactly what you've been looking for.

Ok, for the geometric formulation you need the Killing vector and the four velocity at the event in question, and also a reference Killing vector. The ratio of the lengths of the Killing vectors gives gravitational time dilation, the product of the normalized Killing vektor with the four velocity gives kinematic time dilation. Is this right?

 

[Dale]

I do understand the two frames will agree on the measurements, but how does that involve the equivalence principle?

I think that the OP's idea for involving the equivalence principle is that the centrifugal acceleration felt by the muon in the storage ring should be equivalent to a gravitational field and produce gravitational time dilation in the muon's frame. The muon's frame is a little bit of an unusual frame since it has the Coriolis acceleration as well as the more normal centrifugal acceleration, but for a lab clock at the center is at rest in the rotating frame, so for that specific scenario you can make a gravitational analogy where the muon is dilated because it is deep in a "gravitational potential" (from the centrifugal force).

At least, that is how I understood the OP.

 

[Dale]

Sorry, I can't follow. I don't see why a decomposition should use, of all things, a coordinate velocity.

Because time dilation is a comparison of proper time to coordinate time, so naturally the velocity should be the coordinate velocity of the same coordinate system for which the coordinate time is being used.

The thing is: $\gamma(U,v)= f(U) \, g(v)$. f is a function of the potential only, and g is a function of the relative velocity only.

Not for any $\gamma(U,v)$ of which I am aware. Again, if you don't like the one I am using can you post the one you are thinking of? There may very well be some, but I have never seen a separable one.

That is a geometric, frame-independent, unique decomposition of time dilation, exactly what you've been looking for.

There is no such thing as frame-independent time dilation.

Ok, for the geometric formulation you need the Killing vector and the four velocity at the event in question, and also a reference Killing vector. The ratio of the lengths of the Killing vectors gives gravitational time dilation, the product of the normalized Killing vektor with the four velocity gives kinematic time dilation. Is this right?

I have never seen anyone do this, so I am not sure. However, it does sound interesting. What do you mean by "reference Killing vector"?

 

[Dale]

So, here is my thought process when calculating time dilation. I start with the line element with the (+,-,-,-) signature: $d\tau^2$. To calculate the time dilation formula I simply divide by $dt^2$ and take the square root: $\sqrt{d\tau^2/dt^2}$.

Generally that quantity will be a function of the coordinates, e.g. (x,y,z), as well as a function of the coordinate velocities, e.g. (dx/dt,dy/dt,dz/dt). The question is if you can separate the function of the coordinates from the function of the coordinate velocities.

 

[PeterDonis]

(1) how is going round in a ring "local"

Well, since you can analyze the experiment in a single inertial frame, the rest frame of the ring (because you can assume flat spacetime, the effect of Earth's gravity is negligible over the period of the muon's orbit in the ring), it is certainly local in that sense.

If you try to analyze it in the instantaneously comoving frame of the muon, then as DaleSpam pointed out, you have to deal with the fact that the muon has a tangential velocity, so it's not quite the same as the case covered in the Usenet Physics FAQ article. (Strictly speaking, in the muon's instantaneous comoving frame, the center of the ring has a tangential velocity. See further comments below.)

(2) in the muon case one observer is inertial and one is accelerating, whereas the spaceship example has both observers accelerating.

Yes, that's because the muon's acceleration is not linear, it's circular. But you could still do a perfectly valid analysis in the muon's instantaneously comoving frame; you would just have to account for the relative velocity of the muon and the center of the ring.

However, you would then have to ask the question, what does such an analysis tell you? Does it tell you the time dilation of the muon that we have been talking about? That is, does it tell you how the muon's elapsed proper time over one orbit differs from the inertial clock's proper time between the same two events in spacetime? (More precisely, between the same two spacelike surfaces?) The answer to that is no, it doesn't. To get that answer, you have to do the analysis in either the rest frame of the ring (inertial), or a frame in which the muon remains at rest over an entire orbit (non-inertial). The latter analysis is the one in which a "gravitational field" would come into play.

In short: you are correct that the muon case is different from the case described in the FAQ; but that doesn't change the answer to the OP's question, which is that the FAQ is correct when it says acceleration does not affect clock rates. The reason a "gravitational field" affects clock rates is not, as the OP implies, because the acceleration it takes to remain at rest in the field varies with position. (See my posts #6 and #12 in this thread.)

 

[Ich]

Because time dilation is a comparison of proper time to coordinate time, so naturally the velocity should be the coordinate velocity of the same coordinate system for which the coordinate time is being used.

Ok, but see my answer to your third reply. In my opinion, the coordinate time is relevant only if is connected with something physically interesting.

Not for any $\gamma(U,v)$ of which I am aware. Again, if you don't like the one I am using can you post the one you are thinking of? There may very well be some, but I have never seen a separable one.

There seem to be misunderstandings. I used exactly the one(s) you used also. I just replaced the coordinate velocity with the relative velocity. The expressions then reduce neatly to the one given by me.

There is no such thing as frame-independent time dilation.

If we're talking about static spacetimes, there are the canonical static observers. Their common simultaneity planes are frame independent objects and define gravitational time dilation. Of course, that's the same as Schwarzschild coordinate time, but I think we call the effect time dilation because these coordinates represent the underlying frame-independent preferred observers. If the coordinates were "just numbers" - as they may well be -, you wouldn't call their relation to proper time "time dilation", as it'd be completely arbitrary.

]I have never seen anyone do this, so I am not sure. However, it does sound interesting. What do you mean by "reference Killing vector"?

I'm making this up as I go along, that's why I ask you to follow and see if it is right.
I'm trying to translate time dilation in static spacetimes into a geometric formulation, to give it a clear definition or meaning.
The first thing is the "potential", $\sqrt{g_{tt}}$ in static coordinates. It is defined not by coordinates, but by the timelike Killing field that exists in every static spacetime. More or less by definition it follows that the time dilation between canonical observers corresponds to the length of the respective Killing vectors (you shift for different intervals everywhere to keep everything "the same"). Now you still need a definition where the potential is 1 (no time dilation). That would be at infinity in the Schwarzschild metric, or at the origin in de Sitter space. It could also be at the surface of a planet or star. That is the "reference Killing vector".
So this is time dilation between canonical observers. The time dilation of a moving object would then naturally be the product of its four-velocity with the respective Killing vector. That should give the ratio of proper time to static coordinate time. (Please check!) The contributions of the scaling of the killing vector (=gravitational time dilation) and the angle between Killing vector and four velocity (=kinematic time dilation) are uniquely separable.

 

[PeterDonis]

In my opinion, the coordinate time is relevant only if is connected with something physically interesting

In Schwarzschild coordinates, it is: the timelike Killing vector is $\partial / \partial t$, i.e., the $t$ coordinate is aligned with the timelike Killing vector. That means that if the reference Killing vector (in your terminology) is at infinity, the coordinate time dilation in Schwarzschild coordinates is exactly what you're calling the time dilation relative to the reference Killing vector.

The time dilation of a moving object would then naturally be the product of its four-velocity with the respective Killing vector.

This can't be quite right as a general statement, because the dot product between two 4-vectors is an invariant, but time dilation is coordinate-dependent. However, per my comments above, if we use Schwarzschild coordinates, we can ignore that complication.

Having said that, let's compute the dot product explicitly. It is $u \cdot k = g_{\mu \nu} u^{\mu} k^{\nu}$, where $u^{\mu}$ is the 4-velocity and $k^{\nu}$ is the Killing vector. As I noted above, the Killing vector is just $\partial / \partial t$, so its components are $(1, 0, 0, 0)$; that means the only term in the dot product is the $t$ term, so we have $u \cdot k = g_{tt} u^t = g_{tt} dt / d\tau = g_{tt} / \left( d\tau / dt \right)$.

In other words, the dot product $u \cdot k$ is not the same as what is usually called the "time dilation", which is $d\tau / dt$. It's not even the same as the reciprocal of the time dilation, because of the factor $g_{tt}$. But it just so happens that, for a static observer, since $d\tau / dt = \sqrt{g_{tt}}$, the dot product becomes $u \cdot k = g_{tt} / \sqrt{g_{tt}} = \sqrt{g_{tt}}$, which happens to be equal to $d\tau / dt$. So for a static observer, the relationship you came up with does happen to hold; but unfortunately it doesn't generalize to a non-static observer.

 

[jartsa]

Let's consider a quite massive long rail, and next to it a clock with low mass. This scenario happens in empty space.

When the clock is at rest relative to the rail, the clock experiences a very small gravitational time dilation.

If the rail starts to move very fast relative to the clock, the clock experiences more gravitational time dilation.

If the clock starts to move very fast relative to the rail, the clock experiences more (gravitational?) time dilation, and kinetic time dilation of course.

Right?

Finally I found an example where multiplying kinetic and gravitational time dilation factors does not work.

 

[Dale]

In my opinion, the coordinate time is relevant only if is connected with something physically interesting.

I agree completely with your opinion, but my conclusion is substantially different. My conclusion is that because time dilation is connected to coordinate time and because coordinate time is not generally something physically interesting, time dilation itself is not generally something physically interesting. IMO, time dilation tells you about the coordinates rather than the physics.

I think that the related quantities which are physically interesting are redshift and differential aging. Both of those are unambiguous and frame invariant.

It seems to me that you are trying to change time dilation into something that it is not. I agree with your reason why time dilation may not meet the criteria as something "physically interesting", but there are many such quantities and I would prefer to just put it in with all the rest of the coordinate dependent quantities and recognize it as such.

There seem to be misunderstandings. I used exactly the one(s) you used also. I just replaced the coordinate velocity with the relative velocity. The expressions then reduce neatly to the one given by me.

And I have already agreed that you can clearly do this. Not only can you do this in static spacetimes but you can do this in any spacetime at any event. You can always generate a set of locally inertial coordinates. The only thing that a static spacetime gains is that it naturally picks out one such set of local inertial coordinates at every event.

I don't see what you are not understanding here. The relative velocity depends on the observer relative to whom the velocity is being compared. Different observers at different U will disagree, so the relative velocity implicitly depends on U.

The first thing is the "potential", $\sqrt{g_{tt}}$ in static coordinates. It is defined not by coordinates, but by the timelike Killing field that exists in every static spacetime. More or less by definition it follows that the time dilation between canonical observers corresponds to the length of the respective Killing vectors (you shift for different intervals everywhere to keep everything "the same"). Now you still need a definition where the potential is 1 (no time dilation). That would be at infinity in the Schwarzschild metric, or at the origin in de Sitter space. It could also be at the surface of a planet or star. That is the "reference Killing vector".

I like this approach. It makes it clear for static spacetimes which local frame to use in the decomposition. However, it still doesn't make it independent of the potential. Although you have clearly identified which local observers to use (the ones following the Killing vector), the fact remains that different observers at different potentials disagree! Furthermore, as you mention, the choice of the reference is arbitrary.

Let me give an example. I am in a lab on the surface of a non-rotating Earth watching GPS satellites pass overhead. We will use my lab as the reference since it is following a Killing vector. This means that in my lab, all time dilation is uniquely attributed to velocity, and for other objects following a Killing vector all time dilation is uniquely attributed to gravitational potential.

But what about the GPS satellites? Because of the Killing field I can uniquely identify one specific observer at any event on the GPS orbit, and I can uniquely determine the kinematic time dilation relative to that observer. However, I disagree with his measurement of time, so I disagree with his measurement of velocity. I attribute some of what he calls velocity to the fact that he is at a different potential from me. As such, it still implicitly is inseparable.

 

[Dale]

Hmm. Ich, I think that I am not being self consistent.

If you find that $\gamma(U,V)=f(U) g(V)$ for some functions f and g, then you say that the gravitational and kinematic time dilation are separate.

What you did amounts to a coordinate transform on the spatial coordinates such that there is such a f and such a g. I don't think that it is one of the standard charts, but there is no reason you cannot use it. They are not separable in the normal charts I know of, but it may be that there is a family of charts where there are.

 

[exmarine]

Wow - I've been away for a while. You guys give me a lot to study. Finally, would someone take a stab at giving a rigorous, complete, unambiguous, thorough, precise, definition of the equivalence principle? Thanks again!

 

[Nugatory]

Wow - I've been away for a while. You guys give me a lot to study. Finally, would someone take a stab at giving a rigorous, complete, unambiguous, thorough, precise, definition of the equivalence principle? Thanks again!

I don't think there is such a definition; the EP is more of a heuristic, a way of thinking about problems. There are precise statements that can be made about the relationship between acceleration and the local effects of gravity, but these will still contain words like "locally" which themselves lack precise definition.

 

[PeterDonis]

What you did amounts to a coordinate transform on the spatial coordinates such that there is such a f and such a g.

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. 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).

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* $v$).

 

[atyy]

Wow - I've been away for a while. You guys give me a lot to study. Finally, would someone take a stab at giving a rigorous, complete, unambiguous, thorough, precise, definition of the equivalence principle? Thanks again!

The best form of the equivalence principle within general relativity is the principle of minimal coupling. This says that general relativity is defined using the Hilbert action, and added matter terms do not contain second or higher-order derivatives of the matter fields, and do not couple to any derivatives of the spacetime metric. This form is general enough for all of classical physics.

The most common form of the equivalence principle is that the general relativistic equations for matter take their special relativistic form, but with the Minkowski metric replaced by the spacetime metric. This form fails for "nonlocal" laws of classical physics, for example, those involving second derivatives. However, there is no conflict with the more general form above, because the "nonlocal" laws of classical physics can be derived from local actions.

A third form, related to the second, common form, is that for any free falling observer (who is tiny enough not to perturb the background spacetime), coordinates on his worldline can be chosen so that the laws of special relativity hold on his worldline (but not away from it), provided he does not probe the curvature of spacetime.

All of the above assumes general relativity. However, there are forms of the equivalence principle that are supposed to guide theories of gravity that are not necessarily general relativity. You can look these up as the weak equivalence principle, the strong equivalence principle, and the Einstein equivalence principle. Newtonian gravity, for example, obeys the weak equivalence principle. Newton was aware of some form of equivalence principle as the idea that gravity can be "locally canceled in free fall". Einstein developed the equivalence principle initially as a heuristic to guess the form of a theory of gravity that would be consistent with special relativity and also reduce to Newtonian gravity in the appropriate limit. Most competing theories to general relativity do not obey the strong equivalence principle, except Nordstrom's theory, which was the first theory of gravity consistent with special relativity, but is ruled out because it is not consistent with the observed perihelion precession.

 

[Ich]

In Schwarzschild coordinates, it is: the timelike Killing vector is $\partial / \partial t$, i.e., the $t$ coordinate is aligned with the timelike Killing vector. That means that if the reference Killing vector (in your terminology) is at infinity, the coordinate time dilation in Schwarzschild coordinates is exactly what you're calling the time dilation relative to the reference Killing vector.

Exactly. That is true for all static spacetimes in static coordinates.

This can't be quite right as a general statement, because the dot product between two 4-vectors is an invariant, but time dilation is coordinate-dependent. However, per my comments above, if we use Schwarzschild coordinates, we can ignore that complication.

But it wasn't a general statement. It was made for static spacetimes with a certain reference, as stated.

In other words, the dot product $u \cdot k$ is not the same as what is usually called the "time dilation"

You're right, of course. The product of two four velocites is their respective gamme factor, not its inverse. Here's the correct calculation, with the reference vector assumed to be of unit length:

1. Gravitational time dilation is the length of the killing vector: $\sqrt{k_{\mu} k^{\mu}}$, that is $\sqrt{g_{tt}}$ in static coordinates.
2. Kinematic time dilation:first normalize the Killing vector to get the four velocity of a static observer: $o^{\nu} = k^{\nu}/\sqrt{ k_{\mu} k^{\mu}}$. The inverse (that's where I went wrong) of the dot product with an object's four velocity is their kinematic dilation factor: $1/o_{\nu}u^{\nu}= \sqrt{ k_{\mu}k^{\mu}}/k_{\nu}u^{\nu}$.

The product of both is the total time dilation: $\frac{k_{\mu} k^{\mu}}{k_{\nu}u^{\nu}} = \frac{g_{tt}}{g_{tt}u^t}=\frac{d\tau}{dt}$

 

[Ich]

IMO, time dilation tells you about the coordinates rather than the physics.

I think that the related quantities which are physically interesting are redshift and differential aging. Both of those are unambiguous and frame invariant.

It seems to me that you are trying to change time dilation into something that it is not.

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.
So when you differentiate strictly between differential aging and time dilation, I may agree with you. 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.
Also, I like to say that down in a gravitatioinal well, time goes slower relative to the outside observer in a well defined way. Only allowing for the notion that whenever you send something down and get it back after a while a different time will have passed for it is kind of...parsimonious. Deliberately not saying all that you could say: that there is a pattern behind it and that this pattern suggests that there is a gravitational diation effect depending on the distance. Sure, the latter is not something you directly observed, but imho it's still physics.

And I have already agreed that you can clearly do this. Not only can you do this in static spacetimes but you can do this in any spacetime at any event. You can always generate a set of locally inertial coordinates. The only thing that a static spacetime gains is that it naturally picks out one such set of local inertial coordinates at every event.

...which is exactly the point here. IIRC, the question was whether in a static spacetime gravitational time dilation can be separated from kinematical time dilation. With a unique notion of "at rest", this is possible.

Furthermore, as you mention, the choice of the reference is arbitrary.

Well, that's fine with me. Time dilation is a relation by nature, and you may choose a reference. It's similar with the potential, btw. Can one define a potential in a static spacetime? Yes. And you may still scale it to make it most useful for the problem at hand.

I attribute some of what he calls velocity to the fact that he is at a different potential from me. As such, it still implicitly is inseparable.

But why? Because coordinates - and thus coordinate velocities - are invariably "contaminated" with the potential?
Humor me and forget about coordinates for the moment. Try the operational definition. You have a satellite, and you have a notion of "at rest". If you'd stop the satellite at this event, all time dilation would be clearly and unmistakably of gravitational origin. Now start it again. Time dilation is different, and the only thing that changed is that there is velocity where before there was none. The position is the same, so the gravitational potential or dilation are still the same. So you have two numbers: the first is definitely gravitational, the second is the combination of gravitational with kinematical time dilation.
The only thing you need to make both components uniquely separable is the assumption that these kind of time dilations is mutiplicative, which can be easily argued from the operational point of view: whatever happens at this position gets dilated again by gravitational redshift for the reference observer.

 

[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.