Comment on your mom falls significantly faster than g
BB84@mander.xyz 2 weeks agoSo 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.
BB84@mander.xyz 2 weeks ago
Very interesting! How do you study something like this? Is it classical E&M in a curved space time, or do you need to do QED in curved space time?
Also, are there phenomena where this effect is significant? I’m assuming something like lensing is already captured very well by treating light as point particles?
Sasha@lemmy.blahaj.zone 2 weeks ago
I’ve only ever done QFT in curved spacetime, but I don’t see any reason why you couldn’t do EM, it’ll be a vaguely similar process. I never actually dealt with any scenarios where the curvature was that extreme, and QFT in a curved background is kinda bizarre and doesn’t always require one to consider the specific trajectories, though you definitely can especially if you’re doing some quantum teleportation stuff. In my area it’s simpler to ignore QED and to just consider a massless scalar field, this gives you plenty of information about what photons do without worrying about polarisations and electrons.
It’s been a long time since I did any reading on the geometric optics approximation (in the context of GR this is the formal name for light travelling on null geodesics), but for the most part it’s not something you have to consider, even outside of black holes the curvature tends to be pretty tame (that’s why you can comfortably fall into one in sci-fi), so unfortunately I don’t know of any phenomena (in GR) where it’s important. QFT in curved spacetime generally requires you to stay away from large curvatures, otherwise you start entering into the territory of quantum gravity for which there is no accepted theory.
Outside of GR, it breaks down quite regularly, including I believe, for the classic double slit experiment.
Sasha@lemmy.blahaj.zone 2 weeks 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 mass distribution along the radius. I did that calculation to show that such a spacetime requires negative mass to exist.