this post was submitted on 03 Nov 2024
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Tap for spoilerThe bowling ball isn’t falling to the earth faster. The higher perceived acceleration is due to the earth falling toward the bowling ball.

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[–] [email protected] 18 points 1 week ago* (last edited 1 week ago) (11 children)

If anyone's wondering, I used to be a physicist and gravity was essentially my area of study, OP is right assuming an ideal system, and some of the counter arguments I've seen here are bizarre.

If this wasn't true, then gravity would be a constant acceleration all the time and everything would take the same amount of time to fall towards everything else (assuming constant starting distance).

You can introduce all the technicalities you want about how negligible the difference is between a bowling ball and a feather, and while you'd be right (well actually still wrong, this is an idealised case after all, you can still do the calculation and prove it to be true) you'd be missing the more interesting fact that OP has decided to share with you.

If you do the maths correctly, you should get a=G(m+M)/r^2 for the acceleration between the two, if m is the mass of the bowling ball or feather, you can see why increasing it would result in a larger acceleration. From there it's just a little integration to get the flight time. For the argument where the effect of the bowling ball/feather is negligible, that's apparent by making the approximation m+M≈M, but it is an approximation.

I could probably go ahead and work out what the corrections are under GR but I don't want to and they'd be pretty damn tiny.

[–] [email protected] 9 points 1 week ago (9 children)

Physics books always say to assume the objects are points in doing calculations. Does the fact that the ball is thicker then the feather make a difference?

[–] [email protected] 4 points 1 week ago* (last edited 1 week ago) (7 children)

Possibly?

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 masses.

[–] [email protected] 2 points 1 week ago (1 children)

For the bowling ball, Newton’s shell theorem applies, right?

[–] [email protected] 2 points 1 week ago* (last edited 1 week ago) (1 children)

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.

[–] [email protected] 2 points 1 week ago* (last edited 1 week ago) (1 children)

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?

[–] [email protected] 2 points 1 week ago* (last edited 1 week ago) (2 children)

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.

[–] [email protected] 2 points 1 week ago (1 children)

even light can stop following null geodesics because the curvature can be too big compared to the wavelength

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?

[–] [email protected] 2 points 1 week ago* (last edited 1 week 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.

Edit: Another really cool fact about black holes is that even when you've got really large wavelengths, it often doesn't matter because they get blue shifted to smaller wavelengths once you get close to be horizon.

[–] [email protected] 2 points 1 week ago* (last edited 1 week 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.

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