- #1

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The Schwarzchild 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 Schwarzchild 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?