Stationary/ Static Conditions Metric?

[binbagsss]

What are the sufficient / necessary conditions for a metric to be stationary / static?

- If the metric components are independent of time in some coordinate system , is this sufficient for stationary?

- I've read for static if a time-like killing vector is orthogonal to a family of hyper-surfaces than it's static and that this condition is equivalent to there being no cross terms containing $dt$. How are these 2 conditions equivalent? I have no idea how to go about showing this?

Many thanks in advance.

 

[Matterwave]

1. Yes, that condition is sufficient as the basis vectors associated with the time coordinate will be a killing field.

2. Well, if there are no cross terms involving $dt$ then certainly the killing field will be orthogonal to the other 3 coordinate basis vectors right? So that part is easy to see. Showing that the other 3 coordinate basis vectors integrate to form a family of hypersurfaces which foliate the manifold requires the Frobenius theorem. Frobenius's theorem basically says that if a set of vector fields X,Y,Z, etc., have lie brackets which are all contained in this set, then the vector fields will define a foliation. Coordinate basis vectors definitely fulfill this requirement since they all have lie bracket 0 with each other.

Actually, some authors (e.g. Schutz) prefers to use the definition of hypersurface as being a subset of the manifold for which one can find a coordinate system on the manifold where the hypersurface is described by setting one of the coordinates to 0. If that is your definition, then property 2 above should be self-evident.

 

[binbagsss]

Coordinate basis vectors definitely fulfill this requirement since they all have lie bracket 0 with each other.

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I haven't heard/looked into lie derivatives, but, from what I've read I think we need to check that $[Lx,Ly]\partial x,\partial y, \partial z=0$, $[Lx,Lz]\partial x,\partial y, \partial z=0$ and $[Ly,Lz]\partial x,\partial y , \partial z=0$, so nine cases. This is made easier for the case of vector fields, as the basis vectors are, since we have the property $Lx (Y) =[X,Y]$.

All commutators for the basis vectors, $[\partial_{i},\partial_{j}] =0$ and hence the lie bracket $[Lx,Ly]$ is.

Are my thoughts on the right track?

Thanks.

 

[Matterwave]

I haven't heard/looked into lie derivatives, but, from what I've read I think we need to check that $[Lx,Ly]\partial x,\partial y, \partial z=0$, $[Lx,Lz]\partial x,\partial y, \partial z=0$ and $[Ly,Lz]\partial x,\partial y , \partial z=0$, so nine cases. This is made easier for the case of vector fields, as the basis vectors are, since we have the property $Lx (Y) =[X,Y]$.

All commutators for the basis vectors, $[\partial_{i},\partial_{j}] =0$ and hence the lie bracket $[Lx,Ly]$ is.

Are my thoughts on the right track?

Thanks.

I'm not sure about your notations, but it is definitely true that coordinate basis vectors have 0 Lie bracket with each other. That is one of the defining qualities of a coordinate basis. A non-coordinate basis will have non-0 Lie brackets between (at least some of) the basis vectors.

 

[binbagsss]

Frobenius's theorem basically says that if a set of vector fields X,Y,Z, etc., have lie brackets ...[/QUOTE]

Just a quick question..this is equivalent to 'have commutators' right,, for the case of vector fields?
Why do you state lie brackets- does Frobenius's theorem generalise to any tensor field?

Thanks.

 

[Matterwave]

Frobenius's theorem basically says that if a set of vector fields X,Y,Z, etc., have lie brackets ...

Just a quick question..this is equivalent to 'have commutators' right,, for the case of vector fields?
Why do you state lie brackets- does Frobenius's theorem generalise to any tensor field?

Thanks.[/QUOTE]

Lie brackets work on vector fields...commutators (terminology originating from QM) work for operators which vector fields do not necessarily have to be. This is a terminology thing, but this is the terminology that I found in most of my sources.

I don't know what the Lie bracket of tensor fields is, only the Lie derivative of tensor fields.

There's also a Frobenius's theorem that applies to a set of one forms (fields) rather than vector fields.

 

[PeterDonis]

If the metric components are independent of time in some coordinate system , is this sufficient for stationary?

Yes, as Matterwave says, provided that the coordinate you're calling "time" is actually timelike. If it isn't, then while the basis vector for that coordinate still defines a Killing vector field, it is not a timelike one. This happens, for example, at and inside the horizon in Schwarzschild spacetime.

How are these 2 conditions equivalent?

Remember that if two vectors $A^a$ and $B^b$ are orthogonal, that means their inner product is zero, i.e., $g_{ab} A^a B^b = 0$. If $g_{ab}$ is of the form $g_{00}, g_{ii}$, i.e., no cross terms between the "0" coordinate (time, assumed timelike per the above) and the "i" coordinates (the space coordinates), then the vector $\partial / \partial t$, given by $A^a = (1, 0, 0, 0)$, must be orthogonal to any vector of the form $B^b = (0, B^1, B^2, B^3)$. And any vector that lies in a surface of constant $t$ must be of the latter form.