4 ms·

> ~0% Violation of conservation of momentum. I don't think the chance of violation of the conservation of momentum would be 0%. Unlikely? Yes. Highly unlikel

> ~0% Violation of conservation of momentum. I don't think the chance of violation of the conservation of momentum would be 0%. Unlikely? Yes. Highly unlikely? Most definitely, but not zero. Why? Because we don't yet have a "theory of everything". For all we know, (not saying this is the case, just a thought experiment/hypothetical) the string theorists have it right, and there are extra dimensions, or brane universes (a multiverse) and maybe this effect is in some way related to that (pulling/utilizing an energy from a higher 4th dimension). Granted, it goes against everything we currently know about physics, but so did a lot of things at one point. And that usually happens when we don't know enough about the area we're researching. And when it comes to quantum mechanics, we're only scratching the surface of what we know. There are plenty of things in quantum mechanics that are completely inexplicable. Things like the double slit experiment, quantum entanglement, quantum tunneling, etc.. We know those things occur, but very little or nothing at all as to the "how" and "why". They seem to defy known physics but there they are...

mikeash 11y ago
That squiggly line in front of the zero means "approximately," so really you're in agreement. The chances are not exactly 0, but more like 0.0000001%.
Symmetry 11y ago
It's certainly not impossible, which is why I put the "approximately" symbol in front of the zero. I would be very, very surprised if the TOE broke conservation of momentum but then again I found it very surprising that there's basically just one experiment that breaks reflective symmetry (I forget what it's Noether[1] equivalent conservation law is). [1]https://en.wikipedia.org/wiki/Noether%27s_theorem https://en.wikipedia.org/wiki/Noether%27s_theorem
- mahranch 11y ago
  > which is why I put the "approximately" symbol in front of the zero. Ah, I missed that somehow... Haven't had my coffee.
  - Symmetry 11y ago
    No problem, I'm sure I've responded after similar misreadings many times.
- gjm11 11y ago
  There isn't a Noether conservation law corresponding to reflective symmetry, is there? The situation where Noether's theorem applies is where you have a smooth family of symmetries, with one of them being the identity (i.e., the trivial symmetry where you don't change anything at all), but reflections don't fit that pattern. On the other hand, if the laws of physics are invariant under reflections then they are also invariant under rotations because you can make rotations out of reflections. So you could say that conservation of angular momentum is the corresponding conservation law. But (so far as anyone knows, I think) the laws of physics really are rotation-invariant and conservation of angular momentum really does hold.
  - JadeNB 11y ago
    > On the other hand, if the laws of physics are invariant under reflections then they are also invariant under rotations because you can make rotations out of reflections. So you could say that conservation of angular momentum is the corresponding conservation law. This is an interesting idea, but, I think, too weak. It is true that every rotation is a composition of two reflections, but there's such an important difference between rotations (orientation-preserving) and reflections (orientation-reversing) that I think one shouldn't say that invariance under the former implies anything about invariance under the latter.
- avmich 11y ago
  And some clarifications at http://math.ucr.edu/home/baez/noether.html http://math.ucr.edu/home/baez/noether.html . And I think Noether's theorem covers regular conservation of momentum - that is, equivalence of laws in every point of space - which is actually at question here.

mikeash 11y ago

That squiggly line in front of the zero means "approximately," so really you're in agreement. The chances are not exactly 0, but more like 0.0000001%.

Symmetry 11y ago

It's certainly not impossible, which is why I put the "approximately" symbol in front of the zero. I would be very, very surprised if the TOE broke conservation of momentum but then again I found it very surprising that there's basically just one experiment that breaks reflective symmetry (I forget what it's Noether[1] equivalent conservation law is). [1]https://en.wikipedia.org/wiki/Noether%27s_theorem https://en.wikipedia.org/wiki/Noether%27s_theorem

mahranch 11y ago
> which is why I put the "approximately" symbol in front of the zero. Ah, I missed that somehow... Haven't had my coffee.
- Symmetry 11y ago
  No problem, I'm sure I've responded after similar misreadings many times.
gjm11 11y ago
There isn't a Noether conservation law corresponding to reflective symmetry, is there? The situation where Noether's theorem applies is where you have a smooth family of symmetries, with one of them being the identity (i.e., the trivial symmetry where you don't change anything at all), but reflections don't fit that pattern. On the other hand, if the laws of physics are invariant under reflections then they are also invariant under rotations because you can make rotations out of reflections. So you could say that conservation of angular momentum is the corresponding conservation law. But (so far as anyone knows, I think) the laws of physics really are rotation-invariant and conservation of angular momentum really does hold.
- JadeNB 11y ago
  > On the other hand, if the laws of physics are invariant under reflections then they are also invariant under rotations because you can make rotations out of reflections. So you could say that conservation of angular momentum is the corresponding conservation law. This is an interesting idea, but, I think, too weak. It is true that every rotation is a composition of two reflections, but there's such an important difference between rotations (orientation-preserving) and reflections (orientation-reversing) that I think one shouldn't say that invariance under the former implies anything about invariance under the latter.
avmich 11y ago
And some clarifications at http://math.ucr.edu/home/baez/noether.html http://math.ucr.edu/home/baez/noether.html . And I think Noether's theorem covers regular conservation of momentum - that is, equivalence of laws in every point of space - which is actually at question here.