What Are the Killing Vectors of the Schwarzschild Metric?

[jfy4]

Hi,

Does anyone know the explicit form the the killing vector fields for the Schwarzschild metric, specifically in the basis $\partial_{\alpha}$?

On that note, Is the Schwarzschild geometry a space of constant curvature? does it admit 10 killing vectors?

 

[LAHLH]

Well Schwarzschild is fully spherically symmetric, so you instantly have the three rotational Killing Vectors: $$R=\partial_{\phi} =-y\partial_x+x\partial_y$$ (this represents a rotation about z-axis), the other two are just rotations about x,y axis and are given as $$S=z\partial_x-x\partial_z$$ and $$T=-z\partial_y+y\partial_z .$$

It is easy to see also by looking at the Schw metric in the usual Schw coords, that $$K=\partial_t$$ is a KV, the metric components are time independent, it is a stationary (and static) spacetime, and we have time translational invariance.

So that is four KV. Maximally symmetric in 4 dimensions, means that the spacetime admits 1/2n(n+1)=10 KV, just like Minkowski, and as you say; for maximally symmetric spacetime we have constant curvature (say the Ricci scalar is the same everywhere in the spacetime).

Usually for Minkowski the other 6 KV represent spatial translational invariance in x,y,z, and boosts in x,y,z. One could translate these into the equivalent conditions in spherical polars I guess. But note, is Schwarzschild symmetric under translations in r? one could express the metric in the usual coords and see if $\partial_r$ satisfied Killing equation etc etc, but just by inspection do we really have radial symmetry? I would argue that things are clearly different at different radii. As you head out to r infty the spacetime looks Minkowskian, as you get closer to the centre of the BH the spacetime is more curved. Clearly we don't have radial symmetry so don't have a maximally symmetric spacetime.

Another way: what happens to the curvature invariants like $$R^{\rho...}R_{\rho...} ?$$ At r=0 this blows up, and we know that we have a genuine singularity, at the horizon, this is finite (the apparent sing here was just a coordinate sing, and disappears in different coords etc)...thus curvature cannot be invariant, we don't have full translational invariance, and the spacetime is not maximally symmetric...

EDIT: actually I may be wrong about constant curvature here, after all Schw is a vacuum solution that means $$R_{\mu\nu}=0$$ suggesting the Ricci scalar is $$R=0$$ everywhere too, suggesting constant curvature, which is perhaps a distinct notion than the fact that certain curvature invariants like $$R^{\mu\rho\sigma\tau}R_{\mu\rho\sigma\tau}$$ change. So maybe it is non-maximally symmetric but also constant curvature? hmm confused...

EDIT 2: If a spacetime was maximally symmetric it would obey $$R_{\rho \sigma \mu \nu}=\frac{R}{n(n-1)} \left(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu}\right) .$$ In 2d finding that R is constant (i.e. zero in the case of Schw) would be sufficient to show a space max sym, since there is only one indepent component of the Riemann tensor, but in 4d there are 20, so knowing that the Ricci scalar is constant, does not prove constant curvature. We have a stronger condition than R=0 however, we have $$R_{\mu\nu}=0$$ everywhere (this is a 4x4 symmetric tensor, so has 10 indepenent components), thus we are left with 10 independent components of curvature. (Just the same number as the number of KV a Maximally symmetric space would have, but now we don't have ten KV so curvature is not constant).

What about the boost KV?

 

[TrickyDicky]

As LAHLH concludes Schwarzschild spacetime doesn't have cosnstant curvature, the easiest way to understand it is to remember the Weyl curvature, part of the Riemannian tensor, so that even if Ricci tensor equals zero everywhere there is still variable Riemannian curvature due to the Weyl components, specifically its 6 nonzero components.

 

[TrickyDicky]

Regarding the KV part of the OP, 10 KV is the maximum number of KV that a 4 dim Riemannian or pseudo-riemannian spacetime can have, doesn't mean it necsarily has the ten KV, I believe the Schwarzschild spacetime has just 4 KV due to its particular symmetry configuration being a static spacetime.Three due to its spherical symmetry and one time like due to being static.

 

[LAHLH]

Regarding the KV part of the OP, 10 KV is the maximum number of KV that a 4 dim Riemannian or pseudo-riemannian spacetime can have, doesn't mean it necsarily has the ten KV, I believe the Schwarzschild spacetime has just 4 KV due to its particular symmetry configuration being a static spacetime.Three due to its spherical symmetry and one time like due to being static.

Yep, I think so, I can't see anymore than 4 anyway. I think DeSitter and Anti De Sitter are the only Maximally symmetric Lorentzian curved spacetimes?

Is there in general anyway to know when you've found them all, KV that is? I guess the constant curvature condition can tell you straight away something isn't max symmetric, but then how to know if in 4d that means 4,5,6,7,8,9?

 

[TrickyDicky]

Is there in general anyway to know when you've found them all, KV that is? I guess the constant curvature condition can tell you straight away something isn't max symmetric, but then how to know if in 4d that means 4,5,6,7,8,9?

Other than solving the killing equations, and taking into account clues like the type of spacetime and curvature condition, or things like energy conservation conditions, I don't know of a sistematic way or formula that tells you when you've found them all.

 

[jfy4]

Wikipedia has 4 of the killing vectors written at the "Schwarzschild coordinates" page, however there is no citation or references... So I feel shady about them.

 

[LAHLH]

They are correct, and the four mentioned above. If in doubt, plug them into Killing's equation

 

[jfy4]

Other than solving the killing equations, and taking into account clues like the type of spacetime and curvature condition, or things like energy conservation conditions, I don't know of a sistematic way or formula that tells you when you've found them all.

Yep, I think so, I can't see anymore than 4 anyway. I think DeSitter and Anti De Sitter are the only Maximally symmetric Lorentzian curved spacetimes?

Is there in general anyway to know when you've found them all, KV that is? I guess the constant curvature condition can tell you straight away something isn't max symmetric, but then how to know if in 4d that means 4,5,6,7,8,9?

I Know of a theorem in Riemannian geometry that says that the max number is

$$1/2(n+1)(n)$$ where n is the dimension of the space, and I also know of a theorem that says that you can't have

$$(1/2)(n+1)n-1$$

killing vectors. I think this is right, ill check when I get home. I am just trying to remember.

 

[jfy4]

They are correct, and the four mentioned above. If in doubt, plug them into Killing's equation

Thanks, a lot guys.

 

[jfy4]

Oh sorry, to bring this back up!

However, do you know of the order that these killing vectors go in. That is, t is clearly the t component, and same with phi, but the ones with trig terms, I am not sure to what components they correspond to.

 

[LAHLH]

Oh sorry, to bring this back up!

However, do you know of the order that these killing vectors go in. That is, t is clearly the t component, and same with phi, but the ones with trig terms, I am not sure to what components they correspond to.

They are the rotations: rotations about the x, y and z axis. The one about the z axis is $\partial_{\phi}$ when translated to spherical polars, otherwise they are not so simple...they are not things like $\partial_{\theta}$ if that's what you mean by 'components they correspond to'

 

[LAHLH]

I Know of a theorem in Riemannian geometry that says that the max number is

$$1/2(n+1)(n)$$ where n is the dimension of the space, and I also know of a theorem that says that you can't have

$$(1/2)(n+1)n-1$$

killing vectors. I think this is right, ill check when I get home. I am just trying to remember.

$$1/2n(n+1)$$ is the number of KV in maximally symmetric spacetime, which leads us to note that there are 10 in 4d space. But the point was in a non-maximally symmetric spacetime how do you know when to stop looking, in Schw how do we know that 4 is it, not 5?

 

[jfy4]

They are the rotations: rotations about the x, y and z axis. The one about the z axis is $\partial_{\phi}$ when translated to spherical polars, otherwise they are not so simple...they are not things like $\partial_{\theta}$ if that's what you mean by 'components they correspond to'

I mean that $k_{t}=\partial_t$ and $k_{\phi}=\partial_{\phi}$, but to what components to the expressions

$$\sin\phi\partial_{\theta}+\cot\theta\cos\phi\partial_{\phi}$$
and
$$\cos\phi\partial_{\theta}-\cot\theta\sin\phi\partial_{\phi}$$

correspond to in the Killing 4-vector?

Also can $\partial_{\phi}$ be interpreted as translation in the $\phi$ direction?

 

[jfy4]

Also to help you narrow down your quest for the number of killing vectors, here is the actual theorem I mentioned from before:

Fubini's Theorem: A Riemannian manifold V_n of dimension $$n\geq2$$ cannot have a maximal group of motions of $$\frac{1}{2}n(n+1)-1$$ parameters.

So it don't have 9!

 

[LAHLH]

I mean that $k_{t}=\partial_t$ and $k_{\phi}=\partial_{\phi}$, but to what components to the expressions

$$\sin\phi\partial_{\theta}+\cot\theta\cos\phi\partial_{\phi}$$
and
$$\cos\phi\partial_{\theta}-\cot\theta\sin\phi\partial_{\phi}$$

correspond to in the Killing 4-vector?

the basis vectors are typically written (t,r,theta,phi) if that's what you mean, so the first KV there is $$(0,0, \sin\phi,\cot\theta\cos\phi)$$

Also can $\partial_{\phi}$ be interpreted as translation in the $\phi$ direction?

Well, I guess you can think of it like that if you like, translation in phi direction/rotation abt z-axis, what's the difference?

 

[LAHLH]

Also to help you narrow down your quest for the number of killing vectors, here is the actual theorem I mentioned from before:

Fubini's Theorem: A Riemannian manifold V_n of dimension $$n\geq2$$ cannot have a maximal group of motions of $$\frac{1}{2}n(n+1)-1$$ parameters.

So it don't have 9!

I actually don't recall this theorem, googling Fubini's theorem seems to bring up a measure theory theorem to do with ordering integrals, and the only GR book I've found it in so far is an Exact Solutions book, that doesn't exact say much about it, do you recall where you saw this?

I mean we know Schw has four KV, and unless I'm mistaken AdS and De Sitter will be the only curved maximally symmetric Lorentzian manifolds, but this theorem of Fubini saying you can't have 9 is something I have never come across...

 

[jfy4]

I actually don't recall this theorem, googling Fubini's theorem seems to bring up a measure theory theorem to do with ordering integrals, and the only GR book I've found it in so far is an Exact Solutions book, that doesn't exact say much about it, do you recall where you saw this?

Oddly enough, I got it from "Exact solutions to Einstein's field equations" from Cambridge monographs on mathematical physics. its on page 104 of the second edition.

 

[jfy4]

the basis vectors are typically written (t,r,theta,phi) if that's what you mean, so the first KV there is $$(0,0, \sin\phi,\cot\theta\cos\phi)$$
Well, I guess you can think of it like that if you like, translation in phi direction/rotation abt z-axis, what's the difference?

I'm looking for the killing vectors that correspond to the infinitesimal generators of translations in the Schwarzschild metric. That's why I'm asking. The $\partial_t$ one is clearly the one for time, But I would like to know the others for $\phi$, $r$, and $\theta$ if they exists.

 

[jfy4]

Actually, from that same book, here is another theorem which seems to answer the "how many" question.

Theorem: If the rank of these linear algebraic equations

$$\mathcal{L}_{\xi}\mathbf{R}=0$$
and
$$\mathcal{L}_{\xi}(\nabla_{a1}...\nabla_{aN}\mathbf{R})=0$$
for $\xi_a$ and $\xi_{a;b}$ is q, where $\mathbf{R}$ is the Riemann tensor and N runs from 1 on, then the maximal group of motions V_n has is $r=\frac{1}{2}n(n+1)-q$ parameters.

V_n is the Riemannian manifold.

 

[bcrowell]

I'm looking for the killing vectors that correspond to the infinitesimal generators of translations in the Schwarzschild metric. That's why I'm asking. The $\partial_t$ one is clearly the one for time, But I would like to know the others for $\phi$, $r$, and $\theta$ if they exists.

Didn't LAHLH's #2 cover this? You have three rotations (which involve the theta and phi coordinates) plus the temporal translation. That's all there is.

You're not going to get any kind of Killing vector involving $\partial_r$, because points at different r coordinates have different intrinsic properties (e.g., different values of the Kretschmann invariant).