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 | |
Results tagged with general-relativity
Search options not deleted
user 156895
A theory that describes how matter interacts dynamically with the geometry of space and time. It was first published by Einstein in 1915 and is currently used to study the structure and evolution of the universe, as well as having practical applications like GPS.
9
votes
Accepted
Are fictitious forces still necessary in general relativity?
The generalization of Newton's 2nd law to general relativity is given by
$$ m \frac{d^2 x^\mu}{d\tau^2} + m \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = f^\mu \qquad (\sta …
- 711
3
votes
What is $r$ in a metric signature in general relativity? If $v$ and $p$ are the time and spa...
In general relativity, the metric is assumed to be Lorentzian, which means that it has signature $(1,n-1,0)$ or $(n-1,1,0)$ depending on convention. Because the metric does not have any zero eigenvalu …
- 71.4k
13
votes
Accepted
Is 4-momentum a vector or a 1-form?
Note: A $k$-form is a smooth, totally-antisymmetric $(0,k)$-tensor field. A one-form is therefore a covector field, but the momentum of a particle is not a field, so this answer has been revised to us …
- 71.4k
0
votes
Accepted
Is $dJ(V,V)=0$? where $J$ is a 1-form?
If $dx^j(\partial_k)=\delta^j_k$ then it is equal to 0 but this relation maintain for all basis?
It is always true, regardless of whether or not you use the canonical basis for the 1-forms. Observe …
- 71.4k
0
votes
In Relativity theory, is chronological relation an order relation?
In general, no - the antisymmetric property fails if the spacetime admits a closed timelike curve. If you forbid such curves (e.g. via the chronology protection conjecture) then the antisymmetric prop …
- 71.4k
7
votes
Why do we call the Riemann curvature tensor the curvature of spacetime rather than the curva...
My answer would be: I suppose that in fact the Riemann tensor corresponds to the curvature tensor of tangent bundle (associated with the Levi-Civita connection) but it also gives plenty (if not all) …
- 71.4k
2
votes
Is the metric distance invariant under any coordinate transformation?
Is this true for any coordinate transformation?
Yes. Given a curve $\gamma$ through spacetime, the quantity
$$\Delta s := \int_\gamma \sqrt{g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu}$$
is a scalar, a …
- 71.4k
3
votes
Constant determinant of metric tensor of Schwarzschild solution in $(x, y, z, t)$ coordinates?
A Jacobian determinant is associated to a change of coordinates, not a single coordinate system, so talking about the Jacobian of a Cartesian coordinate chart doesn't make sense.
- 71.4k
2
votes
Accepted
Determinant of metric tensor in Cartesian Coordinates constant in vacuum
$\mathrm{det}(g)$ is not coordinate-independent - it is a scalar density of weight $+2$ (or $-2$, depending on your convention) which generically changes across spacetime. In Minkowski space equipped …
- 71.4k
4
votes
Dust solutions in general relativity
Yes, that would be sufficient (but not necessary). The stress-energy tensor of a perfect fluid takes the form
$$T^{\mu\nu} = \left(\rho + \frac{p}{c^2}\right)U^{\mu}U^\nu + p g^{\mu\nu}$$
for a normal …
- 71.4k
2
votes
Difference between upper and lower indices in Einstein notation
If $\mathbf X$ is a $(2,0)$-tensor, then that means that $\mathbf X$ is a map which eats two covectors $\boldsymbol \alpha$ and $\boldsymbol \beta$ and spits out a scalar $\mathbf X(\boldsymbol \alpha …
- 71.4k
3
votes
How is the interior Schwarzschild metric derived?
In this context, "interior" does not mean inside the event horizon of a black hole. Rather, one imagines computing the metric for a spacetime which features the presence of a star, modeled as a spheri …
- 71.4k
1
vote
Accepted
What does it mean that the metric is static?
If $A(\rho)>0$, then $\partial_t$ is a time-like Killing vector field (so the components of the metric are independent of the temporal coordinate $t$) and there are no spatiotemporal cross terms, and …
- 71.4k
1
vote
Accepted
Check the transformation of coordinates between 2 basis for a tensor of rank 2
$$\newcommand{r}[1]{\color{red}{#1}} \newcommand{g}[1]{\color{green}{#1}} \newcommand{b}[1]{\color{blue}{#1}}$$
The expressions
$$S^{\mu^{\prime} v^{\prime}}=\frac{\partial x^{\r{\mu^{\prime}}}}{\part …
- 71.4k
2
votes
Does the singularity travel along the time axis?
If you want to talk about black hole singularities at a level beyond a pop-sci documentary, then some fundamentals of GR need to be firmly established ahead of time.
Let's assume that all matter tra …
- 71.4k