Tap for spoiler

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

  • Sasha@lemmy.blahaj.zone
    link
    fedilink
    English
    arrow-up
    2
    ·
    edit-2
    1 day ago

    Yeah it would fair point, I’ll be honest I haven’t touched Newtonian gravity in a long time now so I’d forgotten that was a thing. You’d still need to do a finite element calculation for the feather though.

    There’s a similar phenomenon in general relativity, but it doesn’t apply when you’ve got multiple sources because it’s non-linear.

    • BB84@mander.xyzOP
      link
      fedilink
      English
      arrow-up
      2
      ·
      edit-2
      2 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
        link
        fedilink
        English
        arrow-up
        1
        ·
        edit-2
        54 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.

        • Sasha@lemmy.blahaj.zone
          link
          fedilink
          English
          arrow-up
          1
          ·
          edit-2
          47 minutes 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.