Graduate Static Gravitational Field: Why Must ##g_{m0} = 0##?

[Kostik]

TL;DR

Explaining Dirac's assertion ("GTR", Ch. 16) that in a static gravitational field we must have $g_{m0} = 0, m=1,2,3)$.

In Dirac's "General Theory of Relativity", he begins Chap 16, with "Let us consider a static gravitational field and refer it to a static coordinate system. The $g_{\mu\nu}$ are then constant in time, $g_{\mu\nu,0}=0$. Further, we must have $g_{m0} = 0, (m=1,2,3)$."

It's obvious that static $\rightarrow g_{\mu\nu,0}=0$, but why must $g_{m0} = 0$ ?

What I can think of is this. First, think of ordinary 3D space. Suppose there is no curvature in one of the dimensions, say the $x^1$ dimension. Then the metric $ds^2 = g_{mn} dx^m dx^n$ should have no $dx^1 dx^2$ or $dx^1 dx^3$ terms, since translating along the $x^1$ coordinate direction should not alter how $ds^2$ depends upon $x^2$ or $x^3$.

In the same way, there should be no $dx^0 dx^m$ terms in the metric if the curvature of spacetime is static in time.

Alternatively, if I make the change of coordinates $x'^0=x^0+\text{constant}$, and $x'^m=x^m$ ($m=1,2,3$), and if the gravitational field is static, then $g'_{\mu\nu}=g_{\mu\nu}$ (because the time translation cannot alter spacetime intervals). Hence, $g_{\mu\nu} dx'^\mu dx'^\nu = g_{\mu\nu} dx^\mu dx^\nu$, which implies there are no $dx^0 dx^m$ terms in the metric.

Is this the right way to explain Dirac's assertion?

 

[Ibix]

A static field is one in which the timelike Killing field is everywhere orthogonal to the spacelike surfaces. Thus a vector (1,0,0,0), parallel to the Killing field, must be orthogonal to all vectors (0,a,b,c) that lie in the spacelike planes. The only way that happens is if $g_{m0}=0$.

 

[Orodruin]

Contrast this to a stationary spacetime.

 

[PeterDonis]

In Dirac's "General Theory of Relativity", he begins Chap 16, with "Let us consider a static gravitational field and refer it to a static coordinate system. The $g_{\mu\nu}$ are then constant in time, $g_{\mu\nu,0}=0$. Further, we must have $g_{m0} = 0, (m=1,2,3)$."

Note the key phrase: "and refer it to a static coordinate system". His assertion is only true for such a coordinate system.

The modern way of making this point would be to say that in a static spacetime it is always possible to find a coordinate chart with no $g_{m0}$ cross terms. This is because a static spacetime has a timelike Killing vector field that is hypersurface orthogonal. But it is not necessary for any coordinate chart on a static spacetime to have no $g_{m0}$ cross terms. It is only possible. (Whereas, as @Orodruin mentioned, in a spacetime that is stationary but not static, it is not possible to find such a coordinate chart.)

 

[Kostik]

Note the key phrase: "and refer it to a static coordinate system". His assertion is only true for such a coordinate system.

The modern way of making this point would be to say that in a static spacetime it is always possible to find a coordinate chart with no $g_{m0}$ cross terms. This is because a static spacetime has a timelike Killing vector field that is hypersurface orthogonal. But it is not necessary for any coordinate chart on a static spacetime to have no $g_{m0}$ cross terms. It is only possible. (Whereas, as @Orodruin mentioned, in a spacetime that is stationary but not static, it is not possible to find such a coordinate chart.)

Can you explain / clarify what a "static coordinate system" is, in non-modern language (i.e., the way Dirac would have explained it in 1975)?

 

[Orodruin]

One where the components of $g$ do not depend on $x^0$ and surfaces of constant $x^0$ are orthogonal to the $x^0$ direction.