What is the time measured by an observer who is in free-fall near an object?(adsbygoogle = window.adsbygoogle || []).push({});

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 afixedposition 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Time measured by free-fall observer near object?

**Physics Forums | Science Articles, Homework Help, Discussion**