Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
A black hole is a region of spacetime from which nothing can escape. More formally, the future light cone of any observer within the black hole is completely contained in the black hole, and the black hole region is not within the past light cone of any observer that goes to spatial infinity in an infinite amount of time.
2
votes
1
answer
98
views
2nd order perturbation of a charged rotating body
What is the 2nd order perturbation to the flat Minkowski metric $\eta_{ab}$ caused by a charged, rotating body? In particular, if we take a metric $\eta_{ab} + h_{ab} $, what $h_{ab}$ satisfies the Ei …
1
vote
Deriving a Schwarzschild radius using relativistic mass
John Rennie, I think we should clarify that when you go from the second metric to the first, you first perform the transformation $dx^2+dy^2 + dz^2=dr^2 + r^2 d\theta^2+r^2\sin^2\theta d\phi^2$, which …
1
vote
0
answers
24
views
How does temperature correspond to number of particles produced?
I'm looking to calculate the number of particles per unit time produced by a Schwarzschild black hole through Hawking radiation. I know that the temperature of the black hole goes as $\kappa/2\pi$, bu …
1
vote
1
answer
429
views
Quadrupole moment of Kerr spacetime
In this paper, the Kerr black hole is described as having quadrupole moment of $q=J^2/M$ (which means $q=a^2M$ using $J=aM$) whereas in this paper it says in the abstract that the limiting case of Ker …
0
votes
1
answer
574
views
Deriving the Schwarzschild metric in the weak-field regime
I am trying to derive the weak-field Schwarzschild metric, but starting from the same form as Schwarzschild:
$ds^2=-(1+2\Phi(r))dt^2+(1-2\Psi(r))dr^2 +r^2 d\Omega^2$
which has $R=-2\partial_r^2 \Phi …
0
votes
0
answers
142
views
What is the physical reasoning for the weak-field approximation to gravity having a curvatur...
The metric for the weak field approximation to gravity is given by
$ds^2 = -(1-\Phi(r))dt^2 + (1+\Psi(r))\left(dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\phi^2\right)$
When $\Phi(r)=\Psi(r)$, e.g. when …