Yes, that Sasha 🍉

Non-binary 🏳️‍⚧️⬛🟪⬜🟨🏳️‍⚧️
They/them

Anarchist/your local idiot with a guitar

If you’re an Aussie

If you eat food

And if you live on Earth

  • 5 Posts
  • 175 Comments
Joined 11 months ago
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Cake day: December 12th, 2023

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  • 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.


  • 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.


  • 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.



  • 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.


  • I actually thought the answer might be never, but a quick back of the envelope calculation suggests you can do this by dropping a ~1kg bowling ball from a height of 10-11m. (Above the surface of the earth ofc)

    This is an extremely rough calculation, I’m basically just looking at how big a bunch of numbers are and pushing all that through some approximate formulae. I could easily be off by a few orders of magnitude and frankly I didn’t take care to check I was even doing any of it correctly.

    10-11m seems wrong, and it probably is. But that’s still 1,000,000,000,000,000,000,000,000 times further than the earth moves in this situation. Which hey, fun What If style fact for you: that’s about the same ratio of 1kg to the mass of the Earth at ~1024kg.

    That makes perfect sense because the approximations I made are linear in mass, so the distance ratio should be given by the mass ratio.



  • 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.



  • Probably proper knife skills. I’ve always been pretty good with a knife, but I’ve been taking my time to really refine the skill as I do a lot of cooking for large groups so speed is extremely useful. I honestly learnt a lot of it indirectly by just watching how chefs use them, but for the theory and all that I started with Lan Lam’s video on knife skills over at the America’s Test Kitchen yt channel.

    I’m about to be going to an event where I’ll be cooking nearly a thousand meals a day for three days, so I’m going to be putting it to the test. The one nice thing is we’ll have a team of volunteers to help with ingredient prep, so it should be okay but daunting none the less.