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The bowling ball isn’t falling to the earth faster. The higher perceived acceleration is due to the earth falling toward the bowling ball.

  • BB84@mander.xyzOP
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    4 hours ago

    So if I have a spherically symmetric object in GR I can write the Schwarzschild metric that does not depend on the radial mass distribution. But once I add a second spherically symmetric object, the metric now depends on the mass distribution of both objects?

    Your point about linearity is that if GR was linear, I could’ve instead add two Schwarzschild metrics together to get a new metric that depends only on each object’s position and total mass?

    Anyway, assuming we are in a situation with only one source, will the shell theorem still work in GR? Say I put a infinitely light spherical shell close to a black hole. Would it follow the same trajectory as a point particle?

    • Sasha@lemmy.blahaj.zone
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      28 minutes ago

      Yeah, once you add in a second mass to a Schwarzschild spacetime you’ll have a new spacetime that can’t be written as a “sum” of two Schwarzschild spacetimes, depending on the specifics there could be ways to simplify it but I doubt by much.

      If GR was linear, then yeah the sum of two solutions would be another solution just like it is in electromagnetism.

      I’m actually not 100% certain how you’d treat a shell, but I don’t think it’ll necessarily follow the same geodesic as a point like test particle. You’ll have tidal forces to deal with and my intuition tells me that will give a different result, though it could be a negligible difference depending on the scenario.

      Most of my work in just GR was looking at null geodesics so I don’t really have the experience to answer that question conclusively. All that said, from what I recall it’s at least a fair approximation when the gravitational field is approximately uniform, like at some large distance from a star. The corrections to the precession of Mercury’s orbit were calculated with Mercury treated as a point like particle iirc.

      Close to a black hole, almost definitely not. That’s a very curved spacetime and things are going to get difficult, even light can stop following null geodesics because the curvature can be too big compared to the wavelength.

      Edit: One small point, the Schwarzschild solution only applies on the exterior of the spherical mass, internally it’s going to be given by the interior Schwarzschild metric.

      • Sasha@lemmy.blahaj.zone
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        3 hours ago

        On that first point, calculating spacetime metrics is such a horrible task most of the time that I avoided it at all costs. When I was working with novel spacetimes I was literally just writing down metrics and calculating certain features of the mass distribution from that.

        For example I wrote down this way to have a solid disk of rotating spacetime by modifying the Alcubierre warp drive metric, and you can then calculate the radial mass distribution. I did that calculation to show that such a spacetime requires negative mass to exist.