# Static + Hypersurface Orthogonality

1. Jun 28, 2014

### latentcorpse

Static spacetimes can be defined as having no $g_{tx}$ component of the metric.

Alternatively we can say that they are foliated by a bunch of spacelike hypersurfaces to which the Killing vector field $\frac{\partial}{\partial t}$ is orthogonal.

How are these two statements consistent?

$g_{tx}=0 \Rightarrow g(\frac{\partial}{\partial t}, \frac{\partial}{\partial x})=0$ but I always thought this meant there was no distance between timelike and spacelike vectors rather than a statement about them being orthogonal?

Can someone please clear this up for me.

Thanks.

2. Jun 28, 2014

### PAllen

√g(v1,v2) is the definition of dot product. Dot product=0 is the definition of orthogonality, in this case, 4-orthonality.

3. Jun 28, 2014

### latentcorpse

Why do you have a square root? This would correspond to "ds" rather than ds^2 and I though ds was an infinitesimal line element i.e. a measure of distance?

If the metric is just a statement about orthogonality, how is it connected to length? Or is it only a statement about orthogonality when it vanishes? I seem to be getting myself very confused about this haha

4. Jun 28, 2014

### Bill_K

No, for example the dear old Friedmann cosmology has gtx = 0 but is not static. And contrariwise you can take a static spacetime like Schwarzschild and write it in a coordinate system for which gtx ≠ 0

Last edited: Jun 28, 2014
5. Jun 28, 2014

### latentcorpse

So the correct defn is the hypersurface stuff because it is coordinate invariant.

Is it correct to say that if it's static then it's always possible to find a coordinate system with $g_{tx}=0$?

6. Jun 28, 2014

### WannabeNewton

Static means stationary and hypersurface orthogonal. In geometric terms this means there exists a time-like Killing field $\xi$ for the space-time and this Killing field is hypersurface orthogonal i.e. (locally) there exists scalar fields $f,g$ such that $\xi^{\flat} = g df$. It is very easy to show that this implies there exists a coordinate system $\{x^{\mu}\}$ on the space-time such that $\partial_t g_{\mu\nu} = 0$ and $g_{ti} = 0$. You will find this shown in e.g. Wald, see chapters 6 and 7.

7. Dec 8, 2014

### macrobbair

there might be a notational problem in the above, g(V,V)=g^{ab}V_aV_b

8. Dec 9, 2014

### Staff: Mentor

You have the indexes backwards, at least if you are referring to the implicit statement in the OP that $g_{tx} = g \left( \partial / \partial t, \partial / \partial x \right)$. That translates to $g \left( V, V \right) = g_{ab} V^a V^b$, with the metric indexes lower and the vector indexes upper. Also, the two vectors aren't the same, so you should really be writing $g \left( V, W \right) = g_{ab} V^a W^b$.

Then, if $V$ and $W$ are coordinate basis vectors, as $\partial / \partial t$ and $\partial / \partial x$ are, then the components $V^a$ and $W^b$ are either 1 or 0, depending on which basis vector we are looking at. Thus, $\partial / \partial t = (1, 0, 0, 0)$ and $\partial / \partial x = (0, 1, 0, 0)$, so the formula reduces to $g \left( \partial / \partial t, \partial / \partial x \right) = g_{ab} \left( \partial / \partial t \right)^a \left( \partial / \partial x \right)^b = g_{tx}$, as the OP wrote.