- #1
johne1618
- 371
- 0
What is the time measured by an observer who is in free-fall near an object?
The Schwarzschild metric is given by:
[itex] \large c^2 d\tau^2 = (1 - \frac{r_s}{r}) c^2 dt^2 - ( 1 - \frac{r_s}{r})^{-1} dr^2 - r^2(d\theta^2 + \sin^2 \theta d\phi^2) [/itex]
Now for an observer at a fixed position one uses:
[itex]dr = d\theta = d\phi = 0[/itex]
To give an element of proper time as:
[itex] \large d\tau = \sqrt{1 - \frac{r_s}{r}} dt [/itex]
This is a time interval experienced by an observer who is stationary in the object's gravitational field.
But I want the time measured by a free-fall observer so I can't assume that his position is fixed.
Perhaps this is how to do the calculation.
I work out the world line for a radial light beam using:
[itex] d\tau = d\theta = d\phi = 0 [/itex]
Substituting into the above Schwarzschild metric I find the worldline of a lightbeam given by
[itex] \large (1 - \frac{r_s}{r}) \ c \ dt = dr [/itex]
Now a free-fall observer experiences a flat spacetime locally. He must use a time element [itex]d\tau[/itex] such that a light beam's worldline is diagonal so that
[itex] \large c \ d\tau = dr [/itex]
Therefore his time element must be:
[itex] \large d\tau = (1 - \frac{r_s}{r}) \ dt [/itex]
Does this make sense?
The Schwarzschild metric is given by:
[itex] \large c^2 d\tau^2 = (1 - \frac{r_s}{r}) c^2 dt^2 - ( 1 - \frac{r_s}{r})^{-1} dr^2 - r^2(d\theta^2 + \sin^2 \theta d\phi^2) [/itex]
Now for an observer at a fixed position one uses:
[itex]dr = d\theta = d\phi = 0[/itex]
To give an element of proper time as:
[itex] \large d\tau = \sqrt{1 - \frac{r_s}{r}} dt [/itex]
This is a time interval experienced by an observer who is stationary in the object's gravitational field.
But I want the time measured by a free-fall observer so I can't assume that his position is fixed.
Perhaps this is how to do the calculation.
I work out the world line for a radial light beam using:
[itex] d\tau = d\theta = d\phi = 0 [/itex]
Substituting into the above Schwarzschild metric I find the worldline of a lightbeam given by
[itex] \large (1 - \frac{r_s}{r}) \ c \ dt = dr [/itex]
Now a free-fall observer experiences a flat spacetime locally. He must use a time element [itex]d\tau[/itex] such that a light beam's worldline is diagonal so that
[itex] \large c \ d\tau = dr [/itex]
Therefore his time element must be:
[itex] \large d\tau = (1 - \frac{r_s}{r}) \ dt [/itex]
Does this make sense?