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I Did Scott Kelly age less than his twin, or more?

  1. Mar 3, 2016 #1
    ND Tyson tweeted that Scott Kelley aged 1/100th of a second less than his twin, Mark, while spending nearly a year in space.

    Isn't this a playing-out of a real "twin paradox" experiment? It seems that we have two identical twins, one of which goes on a (kind of) long space flight while the other stays home. The resolution of the twin paradox, if I understand it correctly, is that the twin who must accelerate to return home breaks the symmetry of the picture and therefore is the twin who ages less. For the Kellys, though, Scott is spending the vast majority of the time in an inertial frame, while Mark is spending all of his time in a non-inertial frame -- he is experiencing a constant acceleration, while Scott's worldline is largely geodesic. Shouldn't this asymmetry mean that Mark is actually the "moving" (younger) twin and Scott is the "stationary" (older) twin?
     
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  3. Mar 3, 2016 #2

    DrClaude

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  4. Mar 3, 2016 #3
    The news story I saw quoted astrophysicist Jeffery Bennett, who wrote a pop-sci book on relativity. Bennett presented your basic t'= Lorentz transformation to find the time difference, stating that the ISS was the moving frame and Earth was the rest frame. That calculation produces a difference of roughly .01 seconds, according to Bennett.
     
  5. Mar 3, 2016 #4
    There is an orbit where the time dilation of the gravitational potential cancels out the time dilation of the velocity exactly. So it depends whether the ISS is in a higher or lower orbit. A naiive Lorentz type calculation is not enough.
     
  6. Mar 3, 2016 #5

    A.T.

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    This argument works only in flat space time, far away from gravity sources.
     
  7. Mar 3, 2016 #6

    bcrowell

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    Right. The calculation is worked out as ch. 2, example 11 in my GR book: http://www.lightandmatter.com/genrel/ . The ISS is in an atmosphere-skimming orbit, so the kinematic effect certainly dominates, and Kelly aged less than his twin.
     
  8. Mar 3, 2016 #7
    I understand that both gravitational and relative-speed time dilation are at play (I have read that they cancel each other at a 3,174 km orbit). What I don't understand is the physical justification for assuming that Earth is the rest frame and ISS is the moving frame. From Scott's perspective, his twin was moving in a complex spiraling path, even as Scott was not only motionless, but following a geodesic. Taking SR literally, it should be equally arguable that Scott was in the rest frame -- or more arguable, given the mostly inertial path that he took, compared to the constantly accelerated path that Mark took. How then do we assume that Earth is the stationary frame?
     
    Last edited: Mar 3, 2016
  9. Mar 3, 2016 #8

    bcrowell

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    Nobody is assuming anything about frames. General relativity doesn't have global frames of reference. If you look at the calculation linked to from #6, you will see no mention of frames.
     
  10. Mar 3, 2016 #9
    I also don't see any calculations in Chapter 2, example 11, which is about muons and SR.

    The astrophysicist quoted in the news story said that Earth was considered the rest frame and ISS was considered the moving frame, and cited the Lorentz transformation only. That's the assumption of frames I'm referring to. Was that an incorrect/naive statement? What's the rigorously correct way to calculate this time difference?
     
  11. Mar 3, 2016 #10

    bcrowell

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    You seem to have been looking at example 12.

    The rigorously correct way to calculate it is given in example 11. It's probably true that you can get quite a good approximation in the case of the ISS simply by using the Lorentz transformation. That's because the kinematic time dilation dominates and the gravitational time dilation is much smaller. You haven't told us what news story you were referring to, so I couldn't comment specifically on what the physicist quoted there actually said.
     
  12. Mar 3, 2016 #11
    Do you mean example 10? Example 11 is called "Large Time Dilation" and is about muons. FYI, example 10 contains a link to example 11, but it doesn't go to example 11.

    I avoided linking to the news story because it's so cheesy. http://www.huffingtonpost.com/entry/neil-degrasse-tyson-scott-kelly_us_56d7342be4b0871f60ed8d0a

    Pardon my density, but I still don't get why ISS or a GPS satellite experiences any gravitational time dilation at all, relative to clocks on the ground. Being in freefall, isn't its spacetime indistinguishable from a flat spacetime? I thought that's what the equivalence principle tells us.
     
  13. Mar 3, 2016 #12

    PAllen

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    There are many different technical statements of the equivalence principle, none correspond to 'popular statements' of it. The feature of the Einstein Equivalence Principle that most closely matches your statement is called "Local Position Invariance" (using a formulation due, originally, I believer to Brans Dicke, popularized by Clifford Will). It says:

    "Locally, physics matches special relativity (flat spacetime) everywhere and everywhen in the universe"

    For example, the Pound-Rebka experiment is considered a test of local position invariance, in that it matches (to well beyond experimental precision) the SR prediction for an accelerated frame in flat spacetime. That is, an accelerated frame (that is 'stationary') in a gravitationa well, is equivalent to a an accelerated frame 'far away from everything'.

    Note the key word 'local'. Comparing an orbiting satellite to an observer on earth is simply not a local comparison. The equivalence principle does not apply at all, pure and simple. Instead, you have no choice to apply general relativity explicitly. I will not repeat what Bcrowell and A.T have already said. Please re-read their comments in light of the total inapplicability of the equivalence principle.
     
  14. Mar 3, 2016 #13

    bcrowell

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    I see. You're looking at the html version, which is out of date. I'm talking about example 11 in the pdf version.

    No. Spacetime curvature is not something you can get rid of by adopting a local free-falling (inrertial) frame of reference. What vanishes in an inertial frame is the gravitational acceleration, not the curvature. And again, there is no such thing as a global frame of reference in GR.
     
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