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Metric for a free falling observer

  1. Jun 2, 2015 #1
    Is it true that for a free falling observer in a non homogeneuos gravitational field, the metric according to his reference frame is always Minkowski? If it is true, Is it valid only locally?
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  3. Jun 2, 2015 #2


    Staff: Mentor

    Locally, yes, it's true, but only locally.
  4. Jun 2, 2015 #3
    If so i have a problem.
    In Rindler coordinate i am moving in a uniform accelerated reference frame; the same as saying that i am free falling in a uniform gravitational field, but if i am free falling why Rindler' s metric is not Minkowski?
  5. Jun 2, 2015 #4


    Staff: Mentor

    No, it isn't. If you are at rest in Rindler coordinates, you are not in free fall; you are accelerating. If you are free-falling, you are not at rest in Rindler coordinates.
  6. Jun 2, 2015 #5
    I am sure that i am misleading something, i hope you will help me.
    That s my problem: locally you can' t distinguish the effect of acceleration from those of gravity, right?
    If so why i wouldn' t be able to recover Minkowski from Rindler in a neighbourhood of a point?
  7. Jun 2, 2015 #6


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    I think you may be missing what observer falling in a gravitational field corresponds to which observer in Minkowski space.

    The observer who is at a constant X-coordinate in Rindler coordinates is locally equivalent to an observer at a stationary spatial point in a stationary space-time.
    The observer who is freely falling in the curved space-time is the one who is locally equivalent to an inertial observer in Minkowski space. This would be the observer falling with a gravitational acceleration of 9.8 m/s^2 (at the Earth's surface) in Newtonian physics.
  8. Jun 2, 2015 #7


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    You ask:

    The question of what the metric is depends on your coordinate choices.

    For ANY observer, no matter how they are accelerating or not accelerating, you can choose coordinates so that the metric is locally Minkowskii.

    While you can choose such coordinates, there's nothing that says that you HAVE to choose them.

    When you talk about a "frame of reference", in GR, the common understanding is that you are talking about a tangent space to a manifold. The manifold in general is curved, the tangent space is not curved, it is flat. This is usually illustrated with the manifold being a sphere, and the tangent space being a plane tangent to the sphere at some point.

    Possibly you have some different understanding of what a "frame of reference" is than I do, and it is impeding communication. I'm afraid I haven't seen any good definitive textbook discussions of the issue to point you at to try and resolve the issue. My best understanding is that "a frame of reference" is defined a vector space, defined by it's basis vectors, while a manifold is not in and of itself a vector space. Instead, we say that a manifold has at every point on the manifold a tangent space that is a vector space.

    It really all boils down to convention, the convention is that the "reference frame" is a set of othonormal basis vectors in the tangent space. That's just what the term is taken to mean.

    The metric is a different animal altogether - it defines the manifold itself, not the tangent space. When you make a choice of coordinates, you define what is called a "coordinate basis" set of vectors at each point on the manifold which describe the same tangent space as the reference frame does. The issue is that unless you happen to make a very particular coordinate choice, your coordinate basis will not be orthonormal, and hence will not be what is usually meant by "reference frame".
  9. Jun 2, 2015 #8


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    The local equivalence is between a frame at rest relative to a nearby mass and the Rindler coordinates. Both represent non-inertial frames, where a local measurement will show non-zero proper acceleration.

    There is no such equivalence of either of the above to an inertial Minkowski frame, where a local measurement will show zero proper acceleration.
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