Equivalence Principle in muon experiment?

[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.