Skip to main content
13 votes

How can coordinates be meaningless in General Relativity?

The quote that prompted this question is my fault, so perhaps I should answer this. In context I wrote [C]oordinates are meaningless. You can calculate any physically meaningful quantity using any ...
benrg's user avatar
  • 28.1k
13 votes

How can coordinates be meaningless in General Relativity?

Coordinates are not meaningless. But perhaps a better word would be unimportant - in the sense that the physics does not care what coordinate system you use and all measurements that you could make ...
ProfRob's user avatar
  • 135k
8 votes

How to raise and lower indices as a physicist would handle it?

You've almost got it, but there are a few small things you're missing. First, when a physicist says they're "lowering an index", what they mean is that if you start with the equation $A^{\mu\...
knzhou's user avatar
  • 104k
6 votes

How can coordinates be meaningless in General Relativity?

Aidan Beecher asked: "Why isn't it possible to find any choice of coordinates that correspond to a particular observer, like in special relativity, where it is possible to Lorentz transform into ...
Yukterez's user avatar
  • 12.5k
5 votes

Derivation or origin of projection tensor

Take an arbitrary vector $X^a$. Its projection along $u^a$ direction will be $X^bu_b u^a$. Then the component orthogonal to $u^a$ will be simply $X^a-X^bu_bu^a = (\delta^a_b - u_bu^a)X^b={P^a}_bX^b$
KP99's user avatar
  • 1,867
4 votes

How to raise and lower indices as a physicist would handle it?

Your assumption of $g^{\sigma\epsilon}g_\epsilon^{\;\nu} = \delta^{\sigma\nu}$ is not correct. We know that $g^{\mu\sigma}g_{\sigma\nu} = g_{\;\nu}^\mu$ by definition because that is what it means to ...
Vincent Thacker's user avatar
4 votes
Accepted

Rewriting the Euclidean on-shell action of Schwarzschild-AdS

I think you are almost there. First of all, \begin{align} \beta = \frac{4\pi}{2MG/r_h^2 - 2\Lambda r_h/3} \end{align} is easy to verify and it differs from what you wrote because of a trivial typo. ...
Connor Behan's user avatar
  • 8,371
3 votes
Accepted

Does Black Hole Formation Require Only Mass Density or Also Event Horizon Contact?

I think the question is posing a false dichotomy. It is not a case of either/or but both/and. I'm not sure what is meant in the question by the phrase 'non-contact merger' since a merger surely ...
Andrew Steane's user avatar
3 votes
Accepted

McVittie in Schwarzschild coordinates

Let us start with the metric that you have, with setting $c=1$, suppressing the $t$ dependence of the scale factor and condensing the angular variables: $$ ds^2 = -\left(\frac{\mu_-}{\mu_+}\right)^2dt^...
timebeforespace's user avatar
3 votes

Can gravitational attraction be thought of as a cascade of gravitational length contractions?

No, this is not a good way to explain why spacetime curvature leads to a gravitational acceleration. The gravitational acceleration exists only because the falling object is moving in spacetime so it ...
John Rennie's user avatar
2 votes

GW luminosity depends on the 3rd time derivative but quadrupole formula depends on the 2nd time derivative?

This is in fact a pretty subtle issue. Instead of trivializing this question, let's rephrase it with a concrete example: Because one can easily mimic sin cos functions with piece-wise quadratic ...
Shuo's user avatar
  • 305
2 votes
Accepted

What is the appropiate way of writing the first (torsion-free) Bianchi identity?

Recall that the Riemann tensor for any connection is alternating in its last two lower indices. This is just the very general fact that the curvature of a connection $\nabla$ in a vector bundle $(E,\...
peek-a-boo's user avatar
  • 6,840
2 votes

If information can't move outwards the singularity then how does the event horizon "know" it has to expand simultaneously along the whole surface?

If information can't move outwards the singularity then how does the event horizon "know" it has to expand simultaneously along the whole surface? The horizon doesn’t expand simultaneously ...
Dale's user avatar
  • 105k
2 votes

Does Black Hole Formation Require Only Mass Density or Also Event Horizon Contact?

we have no evidence of non-contact black hole mergers A non-contact merger is a contradiction in terms. If there is no contact then it is not a merger. So now we have two sides: one suggests that ...
Dale's user avatar
  • 105k
2 votes

Gravitational redshift and time dilation

If this is the case, when you bring the earth clock up to the same height as the observer, wouldn't this "illusion" of delay go away, and the two clocks should show the same time? It is not ...
KDP's user avatar
  • 7,398
2 votes

If the electrostatic force is so much stronger than the gravitation force, then how do black holes form?

Gravity is unique because: there is only one type of gravitational charge it is always attractive If, for example, you tried to generate a huge electrostatic force by gathering together a huge ...
John Rennie's user avatar
1 vote

Why Lorentz transformations instead of general linear transformations?

General linear transformations correspond to coordinate transformations. Differential geometry is formulated in a way that geometric statements are independent of coordinates. Poincaire ...
Andrew's user avatar
  • 52.1k
1 vote

Gravitational redshift and time dilation

You can watch a clock continuously as you drop it to a lower altitude, wait a while, and then raise it up again. While it's at the lower altitude it will appear redshifted. While you're lowering it, ...
benrg's user avatar
  • 28.1k
1 vote

Derivation or origin of projection tensor

This is very simple linear algebra, not even differential geometry. You first need to understand things at the level of one vector space; for the differential geometry case, you just apply this ...
peek-a-boo's user avatar
  • 6,840
1 vote

Gravitational redshift and time dilation

Time does actually flow slower in the presence of gravitational fields. This has been observed at the quantum scale, when atomic clocks (relying on a quantum oscillator that has a very precisely-known ...
controlgroup's user avatar
  • 1,409
1 vote

If light can't have relative speed, does it get left behind the earth imperceptibly?

Light travels at the same speed in all reference frames. This fact is well supported by experiments (starting with the famous experiment of Michelson and Morley). Furthermore, consequences of this ...
Nathan C's user avatar
  • 468

Only top scored, non community-wiki answers of a minimum length are eligible