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Accelerating frame in GRT?

  1. Feb 8, 2014 #1
    I thought that one of the features of GRT was to allow for ACCELERATING reference frames. Yet none of my textbooks seem to show any examples of this. Even the Kerr metric, for a rotating black hole or mass, seems to be "inertial" in the translational directions.

    (1) Is all that correct?

    (2) And what would a metric matrix look like if I were accelerating in say my own x-direction?
     
  2. jcsd
  3. Feb 8, 2014 #2
    (1) No. The Christoffel symbols are not zero for the Kerr metric.
    (2)
    [itex]ds^2 = \left(1+\frac{\alpha x}{c^2}\right)^2 dct^2 - \left(dx^2 + dy^2 + dz^2 \right)[/itex]
    [itex]||g_{\mu \nu}|| = \begin{bmatrix} \left(1+\frac{\alpha x}{c^2}\right)^2 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1 \end{bmatrix}[/itex]
    [itex]\alpha[/itex] is the proper acceleration you undergo, or the number of "g's" you would experience in pilot speak.
    [itex]\alpha =\sqrt{|g_{\mu \nu A^{\mu }A^{\nu}}|}[/itex] where [itex]A^{\mu }[/itex]is the 4-acceleration "of the accelerated observer" for which this line element is appropriate.
     
    Last edited: Feb 8, 2014
  4. Feb 8, 2014 #3
    Thanks, but... Let's go back to a simpler case. They ain't all zero for the Schwarzschild metric either. Does that mean that the mass at the center of it is, or could be non-"inertial" / accelerating? I must be missing something basic.
     
  5. Feb 8, 2014 #4

    Dale

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    The only metric which is inertial is ##ds^2=-c^2 dt^2+dx^2+dy^2+dz^2##. Any metric different from this one (except by sign convention) is non-inertial.
    Yes, a mass at rest in the Schwarzschild coordinates is non-inertial/accelerating. To verify this simply hold an accelerometer.
     
  6. Feb 8, 2014 #5

    stevendaryl

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    The definition of an accelerating (or non-inertial) frame is that a particle at "rest" in that frame has a nonzero proper acceleration. The proper acceleration [itex]A^\mu[/itex] is defined to be:

    [itex]A^\mu = \frac{d}{d \tau} U^\mu + \Gamma^\mu_{\nu \lambda} U^\mu U^\lambda[/itex]

    where [itex]U^\mu[/itex] is the 4-velocity, which in terms of coordinates is [itex]U^\mu = \frac{d}{d \tau} x^\mu[/itex], and where [itex]\Gamma^\mu_{\nu \lambda}[/itex] is the connection coefficients.

    An object at rest in a frame means that we can get frame-centered coordinates so that the 4-velocity components are all zero except for [itex]U^t[/itex], which is constant. So the expression for proper acceleration becomes:

    [itex]A^\mu = \Gamma^\mu_{t t} U^t U^t[/itex]

    So an accelerated frame is one where the connection coefficients [itex]\Gamma^\mu_{t t}[/itex] are nonzero.
     
  7. Feb 8, 2014 #6

    WannabeNewton

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    Take a set of 4 orthonormal basis vectors ##e_{\alpha}## such that ##e_0 = u## is the 4-velocity of the desired observer; one has to transport the ##e_{\alpha}## along this observer's world-line using the appropriate transport law (e.g. Lie transport) so that orthonormality is preserved everywhere along the world-line.

    The ##e_{\alpha}## constitute a local Lorentz frame and the frame is accelerating if ##\nabla_{e_0}e_0 \neq 0##. The frame is rotating if ##\nabla^{\perp}_{e_0}e_i \neq 0## where ##\nabla^{\perp}## is the orthogonal projection of the derivative operator onto the local rest space of ##e_0##.

    I'm not entirely sure what you mean by this but the canonical rest frames of observers who follow orbits of the time translation symmetry, meaning the Copernican frames of those observers who are at rest with respect to the distant stars, are both accelerating and rotating in Kerr space-time.
     
    Last edited: Feb 8, 2014
  8. Feb 8, 2014 #7
    Wow - it will take me a while to absorb everything you folks are saying. Back to the Schwarzschild for a moment. If I am outside the horizon, I cannot tell the difference between my acceleration, say out some radial, or if there is some mass behind me so to speak? I *think* I understand that, equivalence principle and all. What I am trying to understand is the possible state of motion(s) of that mass behind me. Is it, or must it be "inertial", or can it be accelerating, etc.? I guess that information is not in my metric matrix? THANKS!
     
  9. Feb 8, 2014 #8

    WannabeNewton

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    The Schwarzschild solution is sourced by a static perfect fluid with 4-velocity field ##u^{\mu} = (-\xi_{\nu}\xi^{\nu})^{-1/2}\xi^{\mu}## where ##\xi^{\mu}## is the time-like Killing field of the resulting solution. The fluid has a 4-acceleration field given by ##a^{\mu} = \nabla^{\mu} \ln (-\xi_{\nu}\xi^{\nu})^{1/2}##.
     
  10. Feb 8, 2014 #9

    Dale

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    That has nothing to do with whether or not a frame is inertial. However, according to Birkhoff's theorem any possible spherically symmetric distribution of matter will have the same curvature in the vacuum region outside. So from your local metric you cannot determine if you are outside a solid planet, a collapsing ball of dust, or an exploding star (assuming the same total mass for each).
     
  11. Feb 8, 2014 #10

    WannabeNewton

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    Oh true, a static perfect fluid is just one possible source obeying spherical symmetry. We can just as well have a freely falling collapsing dust sphere. Sorry about that!
     
  12. Feb 8, 2014 #11

    jtbell

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    Use the "Quote" button on the post you want to respond to, instead of the "New Reply" button at the bottom of the thread.
     
  13. Feb 9, 2014 #12
    Every mass in this universe is accelerating, in one direction or another. So it seems that there should be some anti-symmetric aspect in the local metric surrounding any mass?

    Thanks for all the responses.
     
  14. Feb 9, 2014 #13

    PeterDonis

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    This sort of "acceleration" is frame-dependent; it's usually called "coordinate acceleration" since it depends on the coordinates you choose. But an object undergoing coordinate acceleration doesn't necessarily feel any acceleration; it might be in free fall, and you've just chosen coordinates that are non-inertial. For example, if you're standing at rest on the Earth's surface and you drop a rock, the rock has nonzero coordinate acceleration with respect to coordinates in which you are at rest; but the rock is in free fall and feels no acceleration.

    The kind of acceleration that is invariant (i.e., independent of the coordinates you choose) is called "proper acceleration"; stevendaryl defined it in an earlier post. Most masses in the universe are *not* accelerating in this sense, because they're in free fall, feeling no acceleration, just like the rock in the example above.

    I'm not sure what you mean by "anti-symmetric aspect in the local metric". The metric tensor is always symmetric.
     
  15. Feb 10, 2014 #14
    Yes, the masses are in free-fall. Maybe I can totally re-phrase my question. It seems that the near-earth metric would be biased slightly by the sun's metric - the earth is accelerating towards the sun. That's what I mean by anti-symmetric - in the direction towards the sun. Could the solar metric be "added" to the earth metric? I realize GRT is very non-linear, but wouldn't the solar metric be approximately constant over here? What would the metric here near the earth look like if I super-imposed an approximately constant solar metric? Is that possible?
     
  16. Feb 10, 2014 #15

    Dale

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    Several off-topic and unproductive posts were removed to avoid derailing the thread.
     
  17. Feb 10, 2014 #16

    Dale

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    You have hilighted the problem. GR is nonlinear, so you cannot simply add the metrics in general. However, in the case you mention the non linear terms are probably small enough to neglect for all but the most sensitive of experiments.

    I am not sure what you are asking. Curvature is tidal gravity. So in principle you could measure the curvature/tidal effects and determine that the metric differed from Schwarzschild.
     
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