Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Local inertial frame

  1. Jul 3, 2009 #1
    Dear PF could you advise me
    Whether I understand properly or not:

    In an arbitrary space-time (with an arbitrary curvature) in any sufficiently little region we can go to Local Inertial Frame of Reference - sit into the free falling lift. Being there our experiments are the same as we have in SR.

    But if I have some mysterious device like a voltmeter where instead of volts there are Riemann curvature tensor valus measuered along its scale....

    What this device will show me? Of course it should show me non-zero curvature tensor values???

    So on the one hand I think that I am in SR (in flat space-time) since my experiments in free falling lift are the same as in SR, but when I look on my "curvature-meter" device it shows me that there are non-zero tensor components...

    If I have similar device called "metric-tensor-meter" it should show me the Minkovski values... correct ?
     
  2. jcsd
  3. Jul 3, 2009 #2

    Dale

    Staff: Mentor

    How would you build such a device?
     
  4. Jul 3, 2009 #3

    JesseM

    User Avatar
    Science Advisor

    On p. 400 of Gravitation by Misner/Thorne/Wheeler they discuss ways one might measure the Riemann curvature tensor:
    The section on the gravity gradiometer can be read on a google book search if you skip forward to p. 401 (click 'Contents' at the top of the page to go directly to page 399):

    http://books.google.com/books?id=w4Gigq3tY1kC&printsec=frontcover&source=gbs_v2_summary_r&cad=0
     
  5. Jul 3, 2009 #4

    Dale

    Staff: Mentor

    Such a device would necessarily be large, not at all what the OP is thinking about. I don't think it is possible to build a device like what the OP is describing, something to measure the curvature at a point.
     
  6. Jul 3, 2009 #5

    JesseM

    User Avatar
    Science Advisor

    Not possible in practice, or theoretically impossible even for a device whose parts could be arranged with arbitrary levels of precision, and their movements measured arbitrarily precisely too? (and ignoring quantum physics which places limits on precision) I got the idea that the question in the OP was more theoretical than practical.
     
  7. Jul 4, 2009 #6
    In Spacetime Physics Taylor and Wheeler define such a local inertial frame of reference as a region of spacetime sufficiently small for tidal effects of gravity not to be detectable by whatever measuring devices you have at your disposal. So if I've understood this right, your curvature-meter would show no curvature in such a frame because, by definition, it wouldn't be sensitive enough. And if you found some way of improving its sesitivity to the point where you could detect some curvature in this region of spacetime, then, by that same definition, it would no longer be in an inertial reference frame.
     
  8. Jul 4, 2009 #7
    I think Rasal's reference answers it....although it's a bit difficult to tell what the OP is most interested in...the chacteristic of the spacetime curvature or the device.

    You defined it so via "sufficiently little region"....in a free falling frame. You will see flat space in arbitrarily small observations and if your meters are good they will reflect that.
     
  9. Jul 4, 2009 #8

    JesseM

    User Avatar
    Science Advisor

    I think the definition of the equivalence principle in GR is actually pretty subtle if you want to define it technically as opposed to heuristically, see this thread and this one for some previous discussions. In particular, in the second thread atyy suggests in the opening post that the equivalence principle may only apply to "first order in Taylor series", suggesting that if you have a device which can measure second-order effects that would allow you to tell you were in a curved spacetime, even though these effects should be measurable in an arbitrarily small region. Also, if I'm understanding Taylor and Wheeler's definition above, they're saying that if you have a device that can measure tidal effects within its limits of accuracy (which I think would be equivalent to second order effects in the Taylor series) then by definition you're in a region of spacetime that's not "sufficiently small" for the equivalence principle to hold; if your device can't measure them then it is "sufficiently small", even though in principle it would always be possible to build a more sensitive device that could measure tidal effects in the same region of spacetime.
     
  10. Jul 4, 2009 #9

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    I have to agree with both of the quotes above.

    Bear in mind that a "locally inertial frame" is something that approximates a truly inertial frame over a "small enough" region. "Approximation" does not have a precise definition; you are neglecting the tidal effects as being too small to be of significance, but mathematically those effects are never exactly zero.

    Curvature is defined in terms of derivatives. Mathematically you evaluate a derivative via Δy/Δx as both Δy and Δx shrink to zero. But that method won't work in the real world because both Δy and Δx are corrupted by errors ("noise"), so you have to stop shrinking the value before the noise gets too great. Any curvature measuring device's accuracy is therefore limited by its physical size and it will become too inaccurate if you make it too small, or don't allow it to operate over a long enough time (it's space-time curvature, not just space curvature).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Local inertial frame
  1. Inertial frame (Replies: 7)

Loading...