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Why does Bailey experiment not show gravitational time dilation?

  1. Sep 9, 2012 #1
    Question: Why is the Bailey et. al. (1977) muon storage ring experiment not considered a refutation of GR, rather than being a corroboration of SR theory?

    It seems to establish that only velocity is the cause of time dilation, and the very high acceleration (order of 10^18g) plays no part in time dilation.

    In the framework of GR, both velocity and acceleration (or at least its equivalent in instantaneous velocity) should cause time dilation.

    The Bailey experiment seems to validate only velocity time dilation, and disprove gravitational time dilation, in spite of a very high acceleration. (For comparison, GPS satellite time dilation seems to show both velocity and gravitational time dilation. Why not the Bailey experiment?)

    Any insights on this would be greatly appreciated.
     
  2. jcsd
  3. Sep 9, 2012 #2
    For a detailed description of the "clock hypothesis", i.e. that time dilation depends only on the instantaneous velocity not on accelerations, and the connection with the equivalence principle of general relativity, see

    http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html
     
  4. Sep 9, 2012 #3

    mfb

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    Acceleration around the ring? At least according to wikipedia, this is the same as time dilation in SR.
    As the muons came from decays of high-energetic pions, they were not accelerated in their flight direction.
     
  5. Sep 9, 2012 #4

    Dale

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    This is not correct. GR does not predict any time dilation due to acceleration.

    I suspect that you are thinking of the equivalence principle and are misapplying it. The equivalence principle can be used to show that light blueshifts as it falls through a gravitational field. In other words, there is a time dilation due to changes in gravitational potential, not due simply to gravitational acceleration.
    The Bailey experiment was conducted on a horizontal ring, so the gravitational potential throughout the experiment was uniform. GPS, on the other hand, has the space component and the ground component at significantly different gravitational potentials. So the Bailey experiment is simply insensitive to GR effects, while the GPS system is sensitive to them.
     
  6. Sep 9, 2012 #5

    bcrowell

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    Actually I think it is possible to interpret the Bailey experiment in terms of gravitational time dilation. If you go into the frame that rotates along with the muons, the fictitious centrifugal force can be interpreted as a gravitational field. For any particle with a given world-line, you can break the time effect down into three parts: kinematic time dilation, gravitational time dilation, and a Sagnac effect. This is, for example, how the results of Hafele-Keating-style experiments were analyzed. To get a nonvanishing gravitational effect, you'd have to compare the muons with something else that was not at the same radius.

    Addressing the OP's original question, the reason that this isn't a disproof of GR is that GR is locally equivalent to SR. Therefore no local experiment that is consistent with SR can be inconsistent with GR.
     
  7. Sep 9, 2012 #6
    @Dalespam

    Yes, I am trying to apply the equivalence principle. As I understand it, while an acceleration is not directly responsible for the time dilation, the potential it creates is. Also, the acceleration need not be actually gravitational, but can be anything, including centrifugal. As long as there is an acceleration, there is a potential, and larger the acceleration, larger the potential. Is this not correct?

    If I am not mistaken, Einsteins derivation of GR uses this principle where the centrifugal acceleration at the edge of the disk is considered 'equivalent' to a gravitational acceleration. The potential is derived based on that.

    So, why should a large acceleration as in the muon ring not create an equivalent large potential? How is the muon ring different from Einstein's disc?
     
  8. Sep 9, 2012 #7
    @bcrowell

    I think the way you are considering is a good way to look at it. However, we actually are comparing the muons with something that is not at the same potential, i.e. an Earth clock. I assume by saying 'not at the same radius' you meant 'at a different potential'.

    To the local experiment, I agree we cannot differentiate if the observer and the experiment are in the same gravitational field. However, in this case, the muon is being subjected to a very large 'artificial gravitation' and therefore is in a different/increased gravitational field. Should we not be able to distinguish the effects here?
     
  9. Sep 9, 2012 #8

    bcrowell

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    Einstein used that example as one of several at the beginning of his 1916 paper "The foundation of the general theory of relativity." That doesn't mean that his work depends in any crucial way on those examples. He provided those examples as motivation and as heuristics. He didn't "derive" GR from them. His Machian example about the two planets (one rotating and one not) doesn't even turn out to be correct, but that doesn't have anything to do with the validity of GR.

    The potential here would be a function of radius, since the gravitational field is radial.

    In the rotating frame, you have three additive effects: kinematic time dilation (which applies to earth clocks), gravitational time dilation (for clocks at unequal r), and a Sagnac effect. These three effects are certainly not physically distinguishable, since you can make some of them appear or disappear simply by changing to a different set of coordinates. Their sum is guaranteed to be the same regardless of whether you do the calculation in nonrotating or rotating coordinates. This is because proper time is a scalar, so it doesn't change when you change coordinates.
     
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