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.