- #1
- 2,956
- 1,502
Dear all,
I'm having confusion about the standard derivation of Schwarzschild's gravitational time dilation. For concreteness I'll follow the explanation of Schutz' "gravity from the ground up", but other texts argue the same. So let me rephrase Schutz's explanation (I surpress factors of c in it):
"We know that the time part of the spacetime interval gives the rate at which clocks at a fixed place in space run. If a clock is at rest, so that along its word line [itex]dr=d\theta=d\phi = 0 [/itex], then the Schwarzschild spacetime interval tells us that its proper time lapse [itex]d\tau [/itex] associated with a coordinate time lapse of [itex]dt [/itex] is given by
[tex]
(d\tau)^2 = - (ds)^2 = (1-\frac{2GM}{r}) (dt)^2
[/tex]
This determines etc. etc."
My first confusion is this: to my understanding, the spacetime interval between A and B only equals the elapsed proper time if the corresponding observer traveling between A and B is inertial. But the chosen observer here is clearly not, he's standing still ! So I'd say the equality between the spacetime interval and proper time above does not hold.
My second confusion is that if we use the standard rule for calculating the elapsed proper time for an observer traveling along a path [itex]x^{\mu}(\lambda)[/itex],
[tex]
\Delta \tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}} d\lambda
[/tex]
and evaluate for our observer earlier (his t is proper time),
[tex]
\frac{dx^{\mu}}{d\tau} = (1,0,0,0)
[/tex]
then
[tex]
\Delta \tau = \int \sqrt{-g_{00} } dt = \sqrt{-g_{00} } \Delta t
[/tex]
which coincides with the result Schutz is given. And of course I'm thoroughly familiar with it. Because I'm suspicious about this "coincidence", my question to you is: what am I missing?
I'm having confusion about the standard derivation of Schwarzschild's gravitational time dilation. For concreteness I'll follow the explanation of Schutz' "gravity from the ground up", but other texts argue the same. So let me rephrase Schutz's explanation (I surpress factors of c in it):
"We know that the time part of the spacetime interval gives the rate at which clocks at a fixed place in space run. If a clock is at rest, so that along its word line [itex]dr=d\theta=d\phi = 0 [/itex], then the Schwarzschild spacetime interval tells us that its proper time lapse [itex]d\tau [/itex] associated with a coordinate time lapse of [itex]dt [/itex] is given by
[tex]
(d\tau)^2 = - (ds)^2 = (1-\frac{2GM}{r}) (dt)^2
[/tex]
This determines etc. etc."
My first confusion is this: to my understanding, the spacetime interval between A and B only equals the elapsed proper time if the corresponding observer traveling between A and B is inertial. But the chosen observer here is clearly not, he's standing still ! So I'd say the equality between the spacetime interval and proper time above does not hold.
My second confusion is that if we use the standard rule for calculating the elapsed proper time for an observer traveling along a path [itex]x^{\mu}(\lambda)[/itex],
[tex]
\Delta \tau = \int \sqrt{-g_{\mu\nu} \frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}} d\lambda
[/tex]
and evaluate for our observer earlier (his t is proper time),
[tex]
\frac{dx^{\mu}}{d\tau} = (1,0,0,0)
[/tex]
then
[tex]
\Delta \tau = \int \sqrt{-g_{00} } dt = \sqrt{-g_{00} } \Delta t
[/tex]
which coincides with the result Schutz is given. And of course I'm thoroughly familiar with it. Because I'm suspicious about this "coincidence", my question to you is: what am I missing?