# Stationary/ Static Conditions Metric?

1. Mar 14, 2015

### 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?

2. Mar 14, 2015

### 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.

Last edited: Mar 14, 2015
3. Apr 14, 2015

### binbagsss

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.

4. Apr 14, 2015

### Matterwave

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.

5. Apr 19, 2015

### 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.

6. Apr 19, 2015

### Matterwave

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.

7. Apr 19, 2015

### Staff: Mentor

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.

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.