What's the underlying frame of the Einstein's Field Equation?

[Pyter]

What would be measured by a radar measurement’s round trip time (multiplied by c/2) would be
$$
\int \sqrt{-\frac{g_{rr}}{g_{tt}}} dr = \int g_{rr} dr.
$$

Did you get that formula from the metric $ds^2 = g_{tt}\, (c\,dt)^2 - g_{rr}\,dr^2$, by setting $ds^2 = 0$ for the light-like path?

 

[Orodruin]

Did you get that formula from the metric $ds^2 = g_{tt}\, (c\,dt)^2 - g_{rr}\,dr^2$, by setting $ds^2 = 0$ for the light-like path?

Yes. What you want is to integrate dt to obtain the global time difference.

Note: It is $ds^2 = g_{tt} dt^2 + g_{rr} dr^2$. The signs are included in the metric components.

 

[Pyter]

Note: It is $ds^2 = g_{tt} dt^2 + g_{rr} dr^2$. The signs are included in the metric components.

Right. But not $c^2$, I guess? Otherwise either the RHS of the line element is not dimensionally correct, or $g_{tt}$ has a dimension different from $g_{rr}$.

 

[Orodruin]

Right. But not $c^2$, I guess? Otherwise either the RHS of the line element is not dimensionally correct, or $g_{tt}$ has a dimension different from $g_{rr}$.

I always use units with $c = 1$. It is just more ... natural.

 

[cianfa72]

Yes. What you want is to integrate dt to obtain the global time difference.

Note: It is $ds^2 = g_{tt} dt^2 + g_{rr} dr^2$. The signs are included in the metric components.

So for the round-trip coordinate time $\Delta t$ of the complete null path $ds=0$ from the spaceship hovering at fixed $(\theta, \phi, r_1)$ to the star surface at $(\theta, \phi, r_0)$ and back we get the value of that integral multiplied by 2.

 

[Orodruin]

Yes

 

[Pyter]

I always use units with $c = 1$. It is just more ... natural.

Of course, I was talking about MKS units.