4 ms·
> They are completely described by only three properties: their mass, angular moment, and electric charge What about linear momentum, if the black hole has som
by Aardwolf 5y ago
> They are completely described by only three properties: their mass, angular moment, and electric charge
What about linear momentum, if the black hole has some velocity through space is that not also a parameter of it?
- delusional 5y agoI don't think linear velocity is necessary to describe a black hole. Basically, a black hole that moves behaves exactly the same as one that stands still.
- Aardwolf 5y agoWhat's different about angular momentum in that case? Given that rotating or not rotating is also similar to moving or standing still
- deleted 5y ago[deleted]
- Koshkin 5y agoThe (important) difference is that a rotating frame of reference is not inertial.
- deleted 5y ago[deleted]
- dark-star 5y agoRotating black holes behave quite differently from static (non-rotating) black holes.
- DiogenesKynikos 5y agoThe laws of physics are invariant under a change in linear velocity. If you take everything in the Universe and add a constant to its velocity, and then if you yourself add that amount to your own velocity, then everything will appear exactly the same. The laws of physics are not invariant under rotation. If you take everything in the Universe and give it a nudge so that it's spinning around a particular axis, things will change. For starters, everything will fly apart, unless you also create a force field that pulls everything towards the axis of rotation (with the force increasing linearly with distance from the axis of rotation). A technical way to describe this is to say that the laws of physics have Poincaré invariance (or more properly, just Special Relativity has this invariance).[1] That's why linear momentum is viewed as a trivial property of a black hole, but angular momentum is not. 1. https://en.wikipedia.org/wiki/Poincar%C3%A9_group https://en.wikipedia.org/wiki/Poincar%C3%A9_group
- deleted 5y ago[deleted]
- Ma8ee 5y agoNo. Simply put, the laws of physics will be described by different equations in a rotating frame of reference than in one that doesn’t rotate, while it is impossible to distinguish two frames of reference that is just moving relatively to each other without any reference to the outside.
- pfortuny 5y agoAFAIK linear momenum depends on the choice of inertial reference frame, so you may assume it is 0.