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Cyosis
#15
Apr26-10, 02:31 PM
HW Helper
P: 1,495
There is a sort of relationship between velocity and gravitational time dilation, but it is the escape velocity at a given radius that is required. The Newtonian escape velocity is:

LaTeX Code: v_e = \\sqrt\\frac{2GM}{R}}

See http://en.wikipedia.org/wiki/Escape_velocity

This is the velocity attained by a particle initially at rest at infinity (loosely speaking) when it falls to a radius R as its potential energy is converted to kinetic energy. Inserting this velocity into the SR time dilation equation gives:

LaTeX Code: T single-quote/ T = \\sqrt{1-\\frac{v_e^2}{c^2}} = \\sqrt{1-\\frac{2GM}{Rc^2}}

This is the time dilation ratio of a particle hovering at R. For a particle orbiting at R you have to multiply gravitational time dilation at R by the time dilation due the local orbital velocity of the particle.
I don't see how we can relate velocity and gravitational time dilation of an observer moving along the r direction in a gravitational field by using the definition of proper time from special relativity.

If an observer moves along the r direction the Schwarzschild metric reduces to (spherical symmetry):

[tex]
ds^2=(1-\frac{r_s}{r})c^2dt^2-\frac{dr^2}{1-\frac{r_s}{r}}
[/tex]

which would yield:

[tex]
d\tau=\sqrt{\left(1-\frac{r_s}{r}\right)-\left(1-\frac{r_s}{r}\right)^{-1} \left(\frac{v}{c}\right)^2}dt
[/tex]