Equivalence Principle in muon experiment?

[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.)

 

[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.