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0 votes
0 answers
63 views

Stationary, static and strictly stationary/static?

Consider spacetimes which are asymptotically flat at null infinity. How to explicitly show that there exists a hyper-surface orthogonal killing vector field $k^{a}$ that is time-like everywhere in ...
John 's user avatar
  • 21
4 votes
2 answers
344 views

Are there types of spacetime that have no symmetries?

We derive the most basic laws of physics from several fundamental symmetries (those from Noether's theorems, gauge symmetries, Lorentz symmetry...). But are there any types of spacetime where no ...
vengaq's user avatar
  • 2,462
2 votes
1 answer
205 views

Are isometries really global symmetries?

On one hand, a spacetime $(M,g)$ with the Killing vector $\xi^\mu$ and $x^\mu(\tau)$ a geodesic, we can construct the quantity $$Q = \xi_\mu \frac{dx^\mu}{d\tau}\tag{4.32}$$ that is constant along the ...
xpsf's user avatar
  • 1,044
0 votes
2 answers
326 views

Is a static spacetime always spherically symmetric?

I'm a bit confused. In this question it is suggested that a static spacetime can be spherically asymmetric. A static spacetime is one for which the metric doesn't change in time. It's irrotational too....
Deschele Schilder's user avatar
1 vote
1 answer
163 views

How do we know that the actual universe has no Killing vector fields?

This article states the following: The infinitude of conserved energies constructed via Noether’s theorem suffers a startling reversal as soon as Special Relativity is superseded by General ...
Doubt's user avatar
  • 519
4 votes
1 answer
391 views

Prove isometry preserving excision is Killing-like?

(If you think thia is e.g. not well expressed you already understand the request for help.) Theorem: Given a manifold $M$ equipped with a metric $g$ and possessing at least one non-trivial isometry $\...
Julian Moore's user avatar
9 votes
2 answers
2k views

Killing tensor in the Kerr metric

It was famously shown by Carter that the Kerr metric possesses a 4th non-obvious constant of the motion, derived from the separability of the Hamiltonian. This constant is related to a Killing tensor. ...
user1887919's user avatar
  • 1,751
0 votes
2 answers
873 views

How to find the isometry group of space-time?

I am given the next pseudometric: $$ ds^2=dt^2-\frac{r^2+a^2 \cos^2 \theta}{r^2+a^2}dr^2-(r^2+a^2)\sin^2 \theta d\phi^2-(r^2+a^2\cos^2 \theta)d\theta^2. $$ How to find the isometry group for such ...
Ivan Lakhtanov's user avatar