# Equivalence Principle in muon experiment?

1. Sep 19, 2014

### 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.

2. Sep 19, 2014

### ghwellsjr

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

3. Sep 19, 2014

### Staff: Mentor

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

4. Sep 19, 2014

### exmarine

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

5. Sep 19, 2014

### Staff: Mentor

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

6. Sep 19, 2014

### Staff: Mentor

"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.

7. Sep 19, 2014

### A.T.

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).

8. Sep 19, 2014

### Staff: Mentor

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.

9. Sep 19, 2014

### exmarine

10. Sep 19, 2014

### exmarine

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?

11. Sep 19, 2014

### Staff: Mentor

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.

12. Sep 19, 2014

### Staff: Mentor

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.

Last edited: Sep 25, 2014
13. Sep 19, 2014

### A.T.

It's not a bottomless well.

14. Sep 19, 2014

### Staff: Mentor

Are you refering 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.

15. Sep 20, 2014

### 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.

16. Sep 20, 2014

### jartsa

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.

17. Sep 20, 2014

### A.T.

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

18. Sep 20, 2014

### Staff: Mentor

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.

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.

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.

19. Sep 20, 2014

### 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.

20. Sep 20, 2014

### A.T.

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.