Equivalence Principle in muon experiment?

[exmarine]

Someone noted that the famous muon half-life experiment, supporting time dilation in SRT, seems to violate the equivalence principle of GRT. The very large radial acceleration in the experiment does not appear to cause any additional slowing. The acceleration does not seem to have an equivalent effect on slowing the clock rate as a very strong gravitational field should have.

I went searching for the answer in your FAQ and came upon the "clock hypothesis" - that accelerations do not affect clock rates, only velocities do. And then you address that very question of why or how the equivalence principle can still be valid.

But it seemed to me that your defense was a sort of bait and switch. You describe a pair of observers in an accelerating rocket passing signals fore and aft, and noting the red and blue shifts respectively, just like they would occur in a gravitational field. The signal from the aft sender, for example, appears red-shifted to the front observer. Isn’t the critical word in that sentence "appears", i.e., the aft clock is not really running slower, it only appears to be running slower? It must also appear to run faster to any observers even further aft.

I was under the impression that varying clock rates in SRT and GRT were real. I came to that conclusion by re-reading Einstein’s original paper wherein he describes the situation of two stationary synchronized clocks A and B. Should clock A be moved to clock B’s location, it will have lost some time. Isn’t that unambiguous? Different clock rates due to motion in SRT (and acceleration / gravity in GRT) are real?

Your defense seems to be dealing with situations of clock rate appearances only and not real clock rates. The muon experiment seems to be a situation with real clock rates. And it sure seems like that experiment does violate the equivalence principle, as I think everyone agrees that stronger gravity does slow down clocks, as in the GPS clocks, etc. But apparently the very large acceleration in that experiment does not slow down the muon’s clocks even further than that due to the velocity as per SRT.

I hope someone can clarify this for me. Thanks.

 

[ghwellsjr]

I was unaware that there was any acceleration in the muon experiment. Where do you see it?

 

[jtbell]

I think he's talking about muon storage rings, where the muons travel in a circular path.

 

[exmarine]

In the cyclotron over in Europe? Radial acceleration of 10^18 g? Did I not name it correctly? Sorry.

 

[PeterDonis]

exmarine, which FAQ are you referring to? I don't see anything in the PhysicsForums FAQ that talks about the muon experiment.

 

[PeterDonis]

I think everyone agrees that stronger gravity does slow down clocks, as in the GPS clocks, etc.

"Stronger gravity" is not the same as "larger acceleration". Relative clock rates in a gravity well are based on *position* in the gravity well--how "deep" in the well you are--not on the acceleration it takes to keep you there.

For example, consider two clocks, both at the same altitude above the Earth: one is in an orbiting spacecraft (in a circular orbit at that altitude), the other is in a "hovering" spacecraft (which is firing its rockets in order to maintain constant altitude, but has no tangential velocity at all). These clocks will run at different rates, as can be verified by comparing their elapsed times when the orbiting spacecraft completes an orbit (i.e., between two successive instants when the orbiting spacecraft passes the hovering spacecraft ). But the difference in their rates is solely due to the orbital velocity of the orbiting spacecraft ; since both are at the same altitude in the gravity well, the effects of gravity on both their clocks are the same, even though one is accelerated and the other is in free fall. In other words, the acceleration of the hovering clock has no effect on its rate, just as the clock hypothesis says.

 

[A.T.]

The very large radial acceleration in the experiment does not appear to cause any additional slowing.

What additional slowing? It's the same slowing, just explained differently, based on the reference frame. Let's say you have two clocks:

A : Is at rest in an inertial frame.
B : Is moving around A in uniform circular motion.

- In the inertial rest frame of A, B runs slower because of it's motion (kinetic time dialtion).
- In the non-inertial rest frame of both A & B, B runs slower because it's lower in the centrifugal potential (gravitational time dilation).

 

[Dale]

But it seemed to me that your defense was a sort of bait and switch.

Can you link to the specific post or discussion?

In general, whether you are dealing with gravity due to mass or "gravity" due to acceleration the equivalence principle and the clock hypothesis both hold. The time dilation in both cases is due to the gravitational potential, and the acceleration does not directly matter.

 

[exmarine]

Links:

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#particle_lifetimes

http://www.edu-observatory.org/physics-faq/Relativity/SR/clock.html

 

[exmarine]

"Stronger gravity" is not the same as "larger acceleration". Relative clock rates in a gravity well are based on *position* in the gravity well--how "deep" in the well you are--not on the acceleration it takes to keep you there.

Doesn't it take a "larger acceleration" to hold a particular position or altitude the deeper one goes into a gravity well? And the slower a clock runs?

 

[Dale]

Doesn't it take a "larger acceleration" to hold a particular position or altitude the deeper one goes into a gravity well? And the slower a clock runs?

No. If you hollowed out a spot in the center of the Earth the gravitational acceleration would be 0, but the time dilation would be large due to the gravitational potential.

 

[PeterDonis]

Doesn't it take a "larger acceleration" to hold a particular position or altitude the deeper one goes into a gravity well?

Not if you're in a free-fall orbit. ;) Even if you're in an orbit that's not a free-fall orbit, the acceleration it takes to maintain altitude will vary with your orbital velocity. So there isn't a single, well-defined relationship between acceleration and altitude.

Also, the relationship between acceleration and altitude, even if we restrict to "hovering" observers (i.e., zero orbital velocity), will be different for different gravity wells (i.e., different masses of the source--planet, star, whatever). This means that there is no single, well-defined relationship between acceleration and clock rate, even if we restrict to "hovering" observers (meaning, all observers are at rest relative to each other, so there is no velocity effect on clock rates at all); observers with the same acceleration can have different clock rates if they are in different gravity wells.

And, following on to what DaleSpam said, if you include the interior of the gravitating body, there are multiple altitudes at which you can have the same acceleration, i.e., there can be observers in the same gravity well, at rest relative to each other, with the same acceleration but different clock rates. The acceleration needed to stay at rest at the center of the Earth is zero; but also, for any given acceleration up to 1 g (the acceleration at the surface of the Earth), there will be some point in the Earth's interior where you need that acceleration to stay at rest--and therefore, for any given acceleration up to 1 g, there will be two places in the Earth's gravity well where you need that acceleration to stay at rest (one inside the Earth, and one at the appropriate altitude above it), and the clock rates of observers at those two places will be different, even though they have the same acceleration and are at rest relative to each other.

 

[A.T.]

Doesn't it take a "larger acceleration" to hold a particular position or altitude the deeper one goes into a gravity well?

It's not a bottomless well.

 

[Dale]

http://www.edu-observatory.org/physics-faq/Relativity/SR/clock.html

Are you referring specifically to the argument presented in the section entitled "But what about the Equivalence Principle?" If so, it doesn't seem like a bait and switch at all to me. The equivalence principle is all about this kind of "astronauts on a rocket" type of scenario. It seems to be directly addressing the standard Equivalence Principle scenario in a straightforward manner.

I get that you feel unconvinced, but I am not sure why.

 

[exmarine]

Yes, you are right - not the best choice of words. What I meant / should have said was that the defense of the clock hypothesis relied only on "appearances" rather than "real" clock rates. The clock rates of those muons seemed to be "real". Is that a valid distinction? Would someone comment on that?

The gravity potential at the center of the Earth is an interesting idea. Has anyone come up with a metric for the interior of a thin-walled sphere? It seems to me like it would go back to Minkowski-flat? But then I am a novice at all this and don't really know what I am talking about.

 

[jartsa]

But it seemed to me that your defense was a sort of bait and switch. You describe a pair of observers in an accelerating rocket passing signals fore and aft, and noting the red and blue shifts respectively, just like they would occur in a gravitational field. The signal from the aft sender, for example, appears red-shifted to the front observer. Isn’t the critical word in that sentence "appears", i.e., the aft clock is not really running slower, it only appears to be running slower? It must also appear to run faster to any observers even further aft.

Yes, it only appears to be running slow, but if the observer is fooled, then the equivalence principle is fine.

We can not forbid the observer walking right next to the clock. If he does that, the apparent slowness disappears. But then also there appears an apparent speed up of all the other clocks.

I believe the observer is successfully fooled this way, regarding the tick rates of clocks.

What about apparent time differences between clocks? ... Well, that's a little bit too complicated for me right now.

A watch in the observer's pocket simply experiences various time dilations when the observer moves around. When the observer moves around, apparent time differences between the pocket watch and other clocks are converted to real time differences.

 

[A.T.]

Has anyone come up with a metric for the interior of a thin-walled sphere? It seems to me like it would go back to Minkowski-flat?

Yes, flat. But still gravitational time dilation between inside & outside.

 

[Dale]

Yes, you are right - not the best choice of words. What I meant / should have said was that the defense of the clock hypothesis relied only on "appearances" rather than "real" clock rates.

I also did not see anything there which referred to appearances. The description was entirely about the measured gravitational redshift. That is a direct measurement, so it is as "real" as any measurement.

The question is how different reference frames explain the measured redshift. In the inertial reference frame the redshift is attributed to standard Doppler shift due to acceleration during transmission. In the accelerating reference frame it is attributed to "gravitational" time dilation due to the difference in gravitational potential. Both frames predict the same measurement in different ways. The equivalence principle ensures that both ways are legitimate.

The clock rates of those muons seemed to be "real". Is that a valid distinction? Would someone comment on that?

There is no scientific definition of "real". The concept of "real" is a philosophical concept from the study of the philosophical discipline of ontology, part of metaphysics. It is a valid distinction, but not a scientific one.

What you can say scientifically is that the clock rate is frame-dependent. So if you want to consider it to be real then you must allow that reality is frame-dependent. Many people don't like that, so they prefer to say that only the proper-time, which is frame-invariant, is real. Whatever choice you make regarding the reality is a philosophical one with no scientific consequences. Choose whatever makes you feel more comfortable.

The gravity potential at the center of the Earth is an interesting idea. Has anyone come up with a metric for the interior of a thin-walled sphere? It seems to me like it would go back to Minkowski-flat? But then I am a novice at all this and don't really know what I am talking about.

Yes, it does become flat on the interior. If you have a hollowed out sphereical shell there is no gravitational time dilation between different points in the interior.

 

[vanhees71]

I think this thread goes somewhat away from the original topic concerning time dilation. The original question was about muons (or any unstable particles) in a storage ring. This is a problem that has not too much to do with GR, because gravitational effects can be neglected here. The standard treatment in textbooks is very easy and tested many times. Just recently an experiment at GSI Darmstadt (my "Alma Mater" ;-)) has confirmed this treatment very accurately (using excited atomic states of Lithium ions in the Experiment Storage Ring):

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

The original publication is

B. Botermann et al, PRL 113, 120405 (2014)
http://dx.doi.org/10.1103/PhysRevLett.113.120405

The theory is very simple: It says that the clock carried with an object (in this time the lifetime of a particle or quantum state) is given by the proper time
\tau=\int \mathrm{d} t \sqrt{1-\vec{v}^2/c^2},
where t is the usual coordinate time in an inertial reference frame. This hypothesis has been tested many times and confirmed. The above cited experiment reached an accuracy of \pm 2.3 \cdot 10^{-1} for the measurement of \gamma \sqrt{1-\vec{v}/c^2}, where \gamma is the time-dilation factor. This product is 1 if the "proper-time hypothesis" is correct, and this is indeed the case within the above given accuracy.

Now one can argue a bit about this very simple explanation, because it concerns a quantum phenomenon, namely the life time of an unstable quantum state. The \gamma factor of course for this lifetime, can be derived from perturbative quantum field theory. The calculation of a decay width of an unstable-particle state is pretty easy, and the inverse of this decay width is the lifetime of the particle (\hbar/\Gamma to be precise). The quantum-field theoretical treatment, leading to the corresponding Feynman rules, immediately gives the \gamma factor for the lifetime to be E/m c^2, where E is the energy of the unstable particle and m its rest mass. In terms of the three-velocity this is exactly \gamma=1/\sqrt{1-\vec{v}^2/c^2}. So the QFT treatment gives the same \gamma factor as the above naive kinematical argument, and many experiments, including the above cited very recent one confirms these predictions.

Of course, it's and interesting academic exercise to describe everything in terms of the non-inertial co-moving frame of the particle, but it doesn't help with the understanding of relativity.

 

[A.T.]

Of course, it's and interesting academic exercise to describe everything in terms of the non-inertial co-moving frame of the particle, but it doesn't help with the understanding of relativity.

But it seems to be the core of the confusion about the "additional slowing": The idea that you have to add up effects from different frames of reference.

 

[exmarine]

Thanks for all the responses but I am more confused than before. I go back to a case I thought I understood - the GPS clocks:

For SRT and their velocity of 3.9 km/s I get about 7.3 micro-s/day slower. (1-sqrt(1-beta^2))24x3600. For GRT and their orbital radius of 25,560 km versus our surface radius of about 6,437 km, and the Schwarzschild radius of the earth’s mass of about 0.008864 m, I get about 45.10 micro-s/day faster. (sqrt(1-r_s/r_gps)-sqrt(1-r_s/r_surface))24x3600. Which leaves a net faster rate of about 38 micro-s/day.

Are these calculations correct? I recognize that the muon storage ring is not an identical situation, but it sure seems similar to me. Yet the time dilation there is calculated with only the SRT part?

 

[Dale]

There is no SRT and GRT "part" in general. That is only a special situation in a few metrics, like Schwarzschild, where you can do that. In general, the time dilation is given by $d\tau/dt$. That expression is always the time dilation, regardless of if you are dealing with a metric that can be decomposed into two parts or not.

 

[Vanadium 50]

For a typical muon storage ring, the SR effects are a factor of 30. The GR effects are less than one part per billion: the difference between what you would get if you included or ignored GR effects appears only in the twelfth decimal place (at most - it's a good deal smaller than that). That's well beyond the measurement accuracy of these experiments (by about a million).

 

[Dale]

exmarine, since you may not have seen these calculations before, let me show you how they are done (using units where c=1).

For an inertial frame in flat spacetime:
$d\tau^2 = dt^2 -dx^2 -dy^2 - dz^2$
$d\tau/dt = \sqrt{1-(dx^2/dt^2+dy^2/dt^2+dz^2/dt^2)} = \sqrt{1-v^2}$

For the standard coordinates in Schwarzschild spacetime:
$d\tau^2=(1-R/r)dt^2-(1-R/r)^{-1}dr^2-r^2 d\Omega^2$
for $dr=d\Omega=0$ this gives a gravitational part:
$d\tau/dt=\sqrt{1-R/r}$
and in general it includes a motion related part
$d\tau/dt = \sqrt{(1-R/r) -v^2}$
where $v^2=(1-R/r)^{-1}dr^2/dt^2+r^2 d\Omega^2/dt^2$

Whatever the metric is which is appropriate for a given problem, the time dilation is calculated by $d\tau/dt$. The separation in Schwarzschild coordinates is only an approximation even there, and generally doesn't work. Do not get hung up on it or think that it is a general rule.

 

[jartsa]

For SRT and their velocity of 3.9 km/s I get about 7.3 micro-s/day slower. (1-sqrt(1-beta^2))24x3600. For GRT and their orbital radius of 25,560 km versus our surface radius of about 6,437 km, and the Schwarzschild radius of the earth’s mass of about 0.008864 m, I get about 45.10 micro-s/day faster. (sqrt(1-r_s/r_gps)-sqrt(1-r_s/r_surface))24x3600. Which leaves a net faster rate of about 38 micro-s/day.

Ok. Now let's calculate the rate of a clock that revolves around the Earth at the same altitude but at speed 0.86 c.

Answer is:
12 hours/day slower + 45.10 micro-s/day fasterAnd now we can calculate rate of myon decay in a storage ring on the surface of the earth, when myons have speed 0.86 c.
Answer: 12 hours/day slower + 0 s/day fasterOr how about if we lift the myon storage ring into space, same altitude where the GPS clock was. The velocity of the storage ring is 0.
Answer: 12 hours/day slower + 45.10 micro-s/day faster

 

[A.T.]

I recognize that the muon storage ring is not an identical situation, but it sure seems similar to me. Yet the time dilation there is calculated with only the SRT part?

For the GPS clocks you compute the difference for clocks at different heights, so you have to include the effect of gravitational time dilation.

For the muon storage ring you compute the difference for clocks at the same height, so there is no gravitational time dilation. Unless you go to the rotating frame of the muons, where there is artificial centrifugal gravity.

 

[exmarine]

exmarine, since you may not have seen these calculations before, let me show you how they are done (using units where c=1)...

The separation in Schwarzschild coordinates is only an approximation even there, and generally doesn't work. Do not get hung up on it or think that it is a general rule.

Ah, I see what you mean. Thanks!

PS. Any papers / textbooks you guys could recommend would be appreciated.

 

[exmarine]

There is no SRT and GRT "part" in general. That is only a special situation in a few metrics, like Schwarzschild, where you can do that. In general, the time dilation is given by $d\tau/dt$. That expression is always the time dilation, regardless of if you are dealing with a metric that can be decomposed into two parts or not.

hmmm, doesn't look to me like the Schwarzschild metric can be accurately "decomposed" for the gps clock case either, based on your post #24? So my calculation back in post #20 is incorrect? I thought I had picked that up from some reliable source, but can't remember where now.

 

[Dale]

hmmm, doesn't look to me like the Schwarzschild metric can be accurately "decomposed" for the gps clock case either, based on your post #24? So my calculation back in post #20 is incorrect? I thought I had picked that up from some reliable source, but can't remember where now.

That is correct. The decomposition is very inaccurate. To see why, start with the last formula in the "Time dilation due to gravitation and motion together" section here: http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_gravitation_and_motion_together

Now, using units where c=1, if you do a 0-order Taylor series expansion about U=0 (weak gravity) then you get $\sqrt{1-v^2}$ and if you do a 0-order Taylor series expansion about v=0 (slow objects) then you get $\sqrt{1-2U}$. So, the separation is based on a 0 order Taylor series expansion, which means that it is already highly inaccurate, and furthermore it is based on two different 0 order expansions, which is quite odd to me.

I find the whole process quite sketchy, although for GPS it happens to be not too far off.

 

[jartsa]

That is correct. The decomposition is very inaccurate.

An observer points at a distant non-moving black hole and says: "after million seconds that black hole has evaporated. Then said observer accelerates to velocity 0.86 c. Now when asked the observer says: "that moving black hole will be evaporated after two million seconds"

Is that right or wrong?An alarm clock was put into orbit around that same black hole. The clock was set to go off at the same time that the black hole will have been evaporated. First the observer says "that alarm clock will go off after million seconds" Then after the acceleration of the observer the observer says: "that alarm clock will go off after two million seconds"

 

[Dale]

Now when asked the observer says: "that moving black hole will be evaporated after two million seconds"

Is that right or wrong?

I don't know the metric for a moving evaporating black hole. Do you have a reference?

 

[jartsa]

I don't know the metric for a moving evaporating black hole. Do you have a reference?

It's Minkowski metric, I guess. Either it's Minkowski metric, or it's an error to draw the world line of a black hole on a Minkowski diagram ... or the world line of a neutron star, or the world line of the earth, or the world lines of two non-parallel photons.

Probably it's an error to fill one's Minkowski diagram with many large black holes. But I only had one small.:)

 

[Dale]

It's Minkowski metric, I guess.:)

Reference?

 

[jartsa]

Reference?

No I don't have a reference.

But maybe I can redesign the thought experiment so that the black hole is unchanging.

Put one alarm clock near a black hole, another alarm clock far away from the black hole, set the clocks so that they go off at the same time. Move around at great speed, then note that alarms still go off at the same time. Conclude that kinetic and gravitational time dilations seem to be separate in this case.

 

[Dale]

Put one alarm clock near a black hole, another alarm clock far away from the black hole, set the clocks so that they go off at the same time. Move around at great speed, then note that alarms still go off at the same time. Conclude that kinetic and gravitational time dilations seem to be separate in this case.

I have already demonstrated that conclusion to be false for the Schwarzschild metric.

Frankly, your conclusion seems completely unrelated to the proposed scenario. In order to conclude something you have to set up a scenario where different values of the thing being tested will change the outcome.

 

[jartsa]

Frankly, your conclusion seems completely unrelated to the proposed scenario. In order to conclude something you have to set up a scenario where different values of the thing being tested will change the outcome.

I'm trying to find a scenario where a dummy like me can not easily see what the total time dilation is.

Let's consider a moving clock in a gravity well. Maybe that's a difficult case.

There's a scientist in a gravity well, he sets one alarm clock to go off after an hour, another alarm clock he sets to go off after two hours. Then he puts the first clock into a carousel, which is set to such speed that the clocks will go off at the same time, after two hours.

Distant observers agree with the scientist that the both clocks where running the same amount of time, and that one clock's hour hand proceeded one hour forwards, while the other clock's hour hand proceeded two hours forwards.

The guy in the gravity well says the clock rate of the carousel-clock was halved because of kinetic time dilation. The distant observers say the clock rate of the carousel-clock was halved by kinetic time dilation.

What is going on?? Why is this so simple? Maybe the clock in the gravity well is not measuring time?

 

[Dale]

This problem with your proposed scenarios that I am talking about is more about experimental design than relativity. If you have some experimental quantity whose effects you want to test then you need to design an experiment with some parameter which varies that quantity in a systematic manner. If you have two such effects then you need two such parameters.

So here, you need something about your experimental setup which you can vary to change gravitational time dilation only, call that parameter U, and something which you can vary to change kinematic time dilation only, call that parameter V. Then, you want to determine the total time dilation as a function of U and V, $\gamma(U,V)$. 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.

Since we already know $\gamma(U,V)$ cannot be expressed as $f(U) g(V)$ (except to a very crude 0-order approximation) you simply cannot get such an experimental result without violating GR and/or the Schwarzschild metric.

 

[PeterDonis]

Why is this so simple?

Because you only varied the velocity (you put one clock into a carousel), not the altitude (both clocks are still at the same height). In other words, you constructed a test that only tests for kinematic time dilation, so of course you're only going to find kinematic time dilation. As DaleSpam says, if you have two effects (kinematic and gravitational time dilation in this case), then you have to have two different things to vary in order to test the two effects; you can't test for two different effects if you only vary one thing.

 

[Ich]

Then, you want to determine the total time dilation as a function of U and V, $\gamma(U,V)$. 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.

Since we already know $\gamma(U,V)$ cannot be expressed as $f(U) g(V)$ (except to a very crude 0-order approximation) you simply cannot get such an experimental result without violating GR and/or the Schwarzschild metric.

Ok, here's a point where I seem to be misguided.

In my book, in a static spacetime, the time T of a canonical (=static) observer is exactly coordinate time, up to a factor $$\sqrt{g_{tt}}$$. That difference would count as of gravitational origin $f(U)$. Now this observer observes something moving and find its time $\tau$ to be dilated by a factor $\sqrt{1-v^2}$, if v denotes the velocity in said observers frame. So we have $d\tau/dt = dT/dt d\tau/dT = \sqrt{1-2U}\sqrt{1-v^2}$, that is $\gamma(U,V)=f(U) g(V)$.
To check my line of thought, I calculated your example on Wikipedia. There is one problem with it, what they sell as the Schwarzschild metric is obviously just a usual approximation to it. But their formula for the combined time dilation due to gravitational potential and coordinate velocity seems to be correct - because if you re-write it in terms of observed velocity, you find $d\tau/dt = \sqrt{1-2U}\sqrt{1-v^2}$, with clearly separable gravitational and velocity components. That is, I found this to be the case, which might be an example of wishful thinking. I'd appreciate if you could check the result.

So for me, in a static spacetime, time dilation is a two-step thing: from coordinate to observer, then from observer to object. The first step is gravitational, the second needs SR only.
But I don't exclude the possibility that I just overlooked something important.

 

[Dale]

Ok, here's a point where I seem to be misguided.

In my book, in a static spacetime, the time T of a canonical (=static) observer is exactly coordinate time, up to a factor $$\sqrt{g_{tt}}$$. That difference would count as of gravitational origin $f(U)$. Now this observer observes something moving and find its time $\tau$ to be dilated by a factor $\sqrt{1-v^2}$, if v denotes the velocity in said observers frame. So we have $d\tau/dt = dT/dt d\tau/dT = \sqrt{1-2U}\sqrt{1-v^2}$, that is $\gamma(U,V)=f(U) g(V)$.
To check my line of thought, I calculated your example on Wikipedia. There is one problem with it, what they sell as the Schwarzschild metric is obviously just a usual approximation to it. But their formula for the combined time dilation due to gravitational potential and coordinate velocity seems to be correct - because if you re-write it in terms of observed velocity, you find $d\tau/dt = \sqrt{1-2U}\sqrt{1-v^2}$, with clearly separable gravitational and velocity components. That is, I found this to be the case, which might be an example of wishful thinking. I'd appreciate if you could check the result.

So for me, in a static spacetime, time dilation is a two-step thing: from coordinate to observer, then from observer to object. The first step is gravitational, the second needs SR only.
But I don't exclude the possibility that I just overlooked something important.

Let's keep things simple using c=1, and considering only objects with no radial component of velocity. So the Schwarzschild metric time dilation formula is:
$\gamma(U,V)=\sqrt{1-2U-v^2}$. Now, if $f(U)=\sqrt{1-2U}$ and $g(v)=\sqrt{1-v^2}$ then $f(U)\;g(v) = \sqrt{1-2U-v^2+2Uv^2} \ne \gamma(U,V)$

You can, as you describe, build a local inertial frame around any event on the worldline of an observer down in a gravity well. In that local inertial frame the observer can serve as a local "reference clock" and attribute any measured time dilation at very nearby events entirely to kinematic time dilation. However, that still does not generally lead to a separation between the gravitational and kinematic time dilation since the "reference clock" is already gravitationally time dilated. In other words, you can use this method to construct a valid $f(U)$, but then you are left with $g(U,v)$ since the v is measured wrt a local "reference clock" which is itself a function of U.

 

[Ich]

Let's keep things simple using c=1, and considering only objects with no radial component of velocity. So the Schwarzschild metric time dilation formula is:
$\gamma(U,V)=\sqrt{1-2U-v^2}$. Now, if $f(U)=\sqrt{1-2U}$ and $g(v)=\sqrt{1-v^2}$ then $f(U)\;g(v) = \sqrt{1-2U-v^2+2Uv^2} \ne \gamma(U,V)$

Yes, that was something I disagreed with, too. Where did you get this formula? To me, it looks like a very special kind of approximation (not eliminating the square root, that is).

You can, as you describe, build a local inertial frame around any event on the worldline of an observer down in a gravity well. In that local inertial frame the observer can serve as a local "reference clock" and attribute any measured time dilation at very nearby events entirely to kinematic time dilation. However, that still does not generally lead to a separation between the gravitational and kinematic time dilation since the "reference clock" is already gravitationally time dilated. In other words, you can use this method to construct a valid $f(U)$, but then you are left with $g(U,v)$ since the v is measured wrt a local "reference clock" which is itself a function of U.

I disagree. You have total time dilation $1/\gamma$, which cosists of two factors. The first is $\sqrt{g_{tt}}$, definitely GR.The second is really SR time dilation and nothing else. Of course, in Schwarzschild coordinates or something, you have to cut it out of the "gravitationally contaminated" coordinates - using $f(U)$. But this step is physically nothing else than the measurement of local time dilation. Mathematically, it is the dot product of two four velocities at the same event. Both are clearly nothig else than SR time dilation, with no contribution of the potential at all. We could do this anywhere in every universe at every place. It's jut the introduction of certain coordinates that make g(U,v) a function of U. It isn't, really. It's g(v) only, if v is a proper local velocity.

Did you check my calculations of Wkipedia time dilation?

 

[PeterDonis]

Mathematically, it is the dot product of two four velocities at the same event.

If this is your definition of "time dilation", then there is no such thing as gravitational time dilation to begin with. But then what do you call the fact that, for example, an observer at rest in a gravity well can exchange light signals with an observer at rest far away from the gravity well ("at rest" means they are at rest relative to each other) and verify that his elapsed proper time between two successive round-trip light signals is shorter than the far-away observer's elapsed proper time between those round-trip light signals? The standard name for that is "gravitational time dilation", and it is certainly not just a matter of the dot product of two 4-velocities at the same event.

 

[atyy]

But then what do you call the fact that, for example, an observer at rest in a gravity well can exchange light signals with an observer at rest far away from the gravity well ("at rest" means they are at rest relative to each other) and verify that his elapsed proper time between two successive round-trip light signals is shorter than the far-away observer's elapsed proper time between those round-trip light signals? The standard name for that is "gravitational time dilation", and it is certainly not just a matter of the dot product of two 4-velocities at the same event.

I understand how to do the calculation, but I am still confused what the reply to the OP is. Is it that the equivalence principle doesn't apply, because one integrates over a path in spacetime, which is nonlocal so the equivalence principle doesn't apply? Or is it that the equivalence principle does apply - but in which case how does one separate out the SR and GR components?

 

[jartsa]

Because you only varied the velocity (you put one clock into a carousel), not the altitude

I put radioactive gas in a bottle, in another bottle I put radioactive gas which decays twice as fast as the first gas. Then I heat the second gas until the gases decay at the same rate.

Then I put both bottles in a basket. Then I dip the basket in a gravity well.

I have varied velocity by heating, and altitude by dipping.

Has one gas decayed more than the other after this procedure?

 

[Dale]

Yes, that was something I disagreed with, too. Where did you get this formula?

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

I disagree You have total time dilation $1/\gamma$, which cosists of two factors. The first is $\sqrt{g-{tt}}$, definitely GR.The second is really SR time dilation and nothing else.

That is just the problem. It doesn't consist of two factors. This doesn't take a long involved chain of physical reasoning, just note that $\gamma(U,v) \ne f(U) \, g(v)$. The factors simply don't exist.

If you don't like the Wikipedia expression then post one that you do like. It is possible that it becomes separable in some circumstance that I am not aware of.

 

[Dale]

Or is it that the equivalence principle does apply - but in which case how does one separate out the SR and GR components?

The equivalence principle does apply, but why should that be at all related to separating out SR and GR components? The requirement is that the two different frames agree on the measurements, not that either be able to separate time dilation into different components.

 

[PeterDonis]

I am still confused what the reply to the OP is.

The OP's question was based on a mistaken implication: that because observers at different heights in a gravity well have different proper accelerations, the difference in their clock rates must be because of the different accelerations. That's not correct. An easy way to see that it's not correct is to note that the equivalence principle analysis of a gravitational redshift experiment goes through just fine even if the acceleration is the same throughout the experiment. As the Usenet Physics FAQ entry notes, the conclusion relies on the difference in speeds between the emission and detection of the light beam in the experiment; it does not rely on any difference in accelerations.

Is it that the equivalence principle doesn't apply, because one integrates over a path in spacetime, which is nonlocal so the equivalence principle doesn't apply?

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.

Or is it that the equivalence principle does apply - but in which case how does one separate out the SR and GR components?

In a local inertial frame, such as that used in the Usenet Physics FAQ example, there is no "GR component"; everything is just straight SR. I think the follow-on discussion about cases where the height difference is too large for a single local inertial frame to cover everything is not really germane to the OP's original question. We just like thread derails here. ;)

 

[PeterDonis]

The equivalence principle does apply

More precisely, the equivalence principle does apply if the experiment can fit within a single local inertial frame. If you're trying to compute the total gravitational redshift of light coming from the Sun when observed by us here on Earth, you can't rely on the EP to get the answer; you have to go through the full computation using the Schwarzschild metric.

 

[atyy]

The equivalence principle does apply, but why should that be at all related to separating out SR and GR components? The requirement is that the two different frames agree on the measurements, not that either be able to separate time dilation into different components.

I do understand the two frames will agree on the measurements, but how does that involve the equivalence principle? In the calculations, one makes use of the clock hypothesis, and integration over the spacetime path. I guess the equivalence principle is involved in the form "comma goes to semicolon" if the clock hypothesis can be stated in a local form, without the integration. Is this why Ich defines the time dilation as the dot product of the four velocities at each event?

Also, how does intuitive picture of the two accelerating spaceships apply? There both observers are accelerating, but in the muon case it seems that one observer is inertial and the other is accelerating.

 

[PeterDonis]

I have varied velocity by heating, and altitude by dipping.

No, you haven't varied the altitude; both gases are at the same altitude (even though that altitude changes during the experiment). Varying the altitude would mean different gases would have different altitudes.

Has one gas decayed more than the other after this procedure?

If the decay rates of both gases were equal before dipping the basket, they will be equal after dipping the basket, because you dipped both gases. (The rate as observed from far away outside the gravity well will be lower after the dipping, but it will still be the same for both baskets.)