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Is the Schwarzschild spacetime spherically symmetric for all observers?

  1. Nov 7, 2014 #1
    In special relativity a sphere in the rest frame for some observer looks like an ellipsoid for an observer with a relative velocity.

    Can we use the same reasoning for the Schwarzschild spacetime? Namely that a spherically symmetric spacetime produced by a spherical mass look ellipsoidal for an observer with a relative velocity compared to that body?
  2. jcsd
  3. Nov 7, 2014 #2


    Staff: Mentor

    No, because Schwarzschild spacetime itself is not a spherical object. See below.

    Are you asking about a spherical object (like an idealized ball or an idealized planet or star) or about a spherically symmetric spacetime? They're two different things.

    For a spherical object, assuming the effects of any other objects can be neglected, yes, as long as you are far enough away from it, the spacetime curvature due to the body's mass can be ignored and you can treat it as just a spherical object in flat spacetime, in which case things work the same as they do in SR.

    For a spherically symmetric spacetime, you can't look at it "from the outside"; you're in it. The spherical symmetry of the spacetime is a global geometric property, which you can measure and verify regardless of your state of motion within that spacetime.
  4. Nov 7, 2014 #3


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    Gold Member

    As Peter noted, symmetries are invariants, that can be stated in terms if e.g. existence of different types killing vector fields. However, typically, only well chosen coordinates manifest all symmetries the sense of the metric expressed in those coordinates showing them. Thus standard inertial coordinates in flat spacetime show homogeneity and isotropy. However, Rindler coordinates (natural for a uniformly accelerating observer), show neither isotropy nor homogeneity in the metric expression. The symmetries haven't disappeared, but they are not manifest in those coordinates. Similarly, a spherically symmetric spacetime booted ultra-relativistically produces the Aichelburg-Sexl metric, which does not show spherical symmetry in coordinate expression:

  5. Nov 7, 2014 #4


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    GR doesn't really have "frames". So we have to do a bit of translation work to interpret the question. It's not clear to me at the moment the best way to do this :(. My current best attempt at an interpretation is "what does the Schwarschild geometry look like in the fermi-normal coordinates associated with a moving observer". Unfortunately this isn't an easy question to answer, while Fermi normal coordinates are conceptually useful, giving the best extended equivalent to the Newtonian idea of a frame that GR has to offer, they are hard to calculate. I would guess that a spherical shell of constant Schwarzschild time would be converted into a non-synchronized ellipsoidal shell in fermi normal coordinates, but I haven't done the calculations to attempt to show this and I'm unlikely to.

    If you consider a small region around some point where the curvature effects are small enough to be neglected, you can do something much simpler than use Fermi Normal coordinates. Instead, you use the idea of a local "frame field" in the flat tangent space. This also is friendly to one's Newtonian intuition, and it's easier to calculate when you are interested in a region close enough to the observer that you don't have to account for curvature effects.
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