Comment on your mom falls significantly faster than g
Sasha@lemmy.blahaj.zone 2 weeks agoPossibly?
A bowling ball is more dense than a feather (I assume) and that’s probably going to matter more than just the size. Things get messy when you start considering the actual mass distributions, and honestly the easiest way to do any calculations like that is to just break each object up into tiny point like masses that are all rigidly connected, and then calculate all the forces between all of those points on a computer.
I full expect it just won’t matter as much as the difference in massed.
BB84@mander.xyz 2 weeks ago
For the bowling ball, Newton’s shell theorem applies, right?
Sasha@lemmy.blahaj.zone 2 weeks ago
Yeah it would fair point, I’ll be honest I haven’t touched Newtonian gravity in a long time now so is forgotten that was a thing.
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.xyz 2 weeks 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 total mass?
But even in a situation with one source, does the shell theorem work in GR? Say I put a infinitely light spherical shell around a black hole. Would it follow the same geodesic as a point particle?
Sasha@lemmy.blahaj.zone 2 weeks 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.