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

• A
• Kostik
This is the coordinate system in which the metric can be written as$$ds^2 = g_{00} (dx^0)^2 + g_{mn} dx^m dx^n$$and the only freedom you have is choosing the coordinate lines for the (spatial) ##x^m##s.

#### Kostik

TL;DR Summary
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?

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• dextercioby
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##.

• cianfa72 and topsquark
Contrast this to a stationary spacetime.

• cianfa72, vanhees71 and topsquark
Kostik said:
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.)

• cianfa72, vanhees71, topsquark and 1 other person
PeterDonis said:
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)?

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.

• PeterDonis, topsquark and vanhees71