How does time dilate in a gravitational field having a relative velocity of v with the field?

Question

Consider a Mass on earth. The time dilation on the surface of Earth is

$$T' = T \sqrt{1 - \frac{2GM}{rc^2}}$$

Now if the mass is moving around the earth at velocity of v w.r.t Earth, what will be the time dilation within the mass as seen from the Earth? The object could be in free fall, orbit or in an escape velocity orbit away from the earth

Is it

$$T'' = T' \sqrt{1 - \frac{v^2}{c^2}}\space - (2)$$

or

$$T'' = T \sqrt{1 - \frac{v^2}{c^2}}\space - (3)$$

if answer is (2) equation is

$$ T'' = T\sqrt{(1 - \frac{v^2}{c^2})({1 - \frac{2GM}{rc^2}})}$$

Answer 1

Let's have a go at doing the calculation for the Schwarzschild metric. I'm going to assume the particle is moving radially because otherwise the algebra gets hairy, but I won't make any restriction on the velocity. Because $d\Omega = 0$ the metric simplifies to:

$$ c^2d\tau^2 = c^2 dt^2 (1 - \frac{r_s}{r}) - \frac{dr^2}{1 - r_s/r} $$

When we say the velocity of the particle is $v$ we mean that in our coordinates $dr/dt = v$ so we can substitute $dr = vdt$:

$$ c^2d\tau^2 = c^2 dt^2 (1 - \frac{r_s}{r}) - \frac{v^2 dt^2}{1 - r_s/r} $$

or:

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

If we graph the ratio of $d\tau/dt$ as a function of $v/c$ we get (this is at $r = 4r_s$):

The magenta line shows the expression I've derived above, and the blue line shows what you get by multiplying the Schwarzschild time dilation factor by the Lorentz factor.

The two curves are obviously different apart from the stationary case when $v = 0$. However there's something odd because the time dilation goes to infinity when $v/c = 0.75$, and remember this is at $r = 4r_s$ so we are well outside the event horizon. The reason for this is that for the Schwarzschild observer the speed of light reduces as you approach the event horizon. For a radially travelling light ray the velocity in the Schwarzschild coordinates is:

$$ v_{light} = c \left(1 - \frac{r_s}{r} \right) $$

So at $r = 4r_s$ an object travelling at $0.75c$ is travelling at the same speed as light and therefore the time dilation is infinite.

Response to comment:

I'll do the calculation at the Earth's surface where the curvature is greatest. The Schwarzschild radius of the Earth is a shade under 9 mm, so the ratio $r_s/r$ at the surface is $1.39 \times 10^{-9}$. Using this value of $r_s/r$ the graph looks like:

There is basically no difference between the two calculations, but this is because the gravitational time dilation is negligable so only the Lorentz factor matters. Still, it's reassuring to note that my calculation derived from the Schwarzschild metric does match the Lorentz time dilation :-)

Answer 2

The weak-field static metric is appropriate for the Earth: $$ds^2 = -(1+2\Phi)dt^2 + (1-2\Phi)\underbrace{(dx^2+dy^2+dz^2)}_{dS^2}\text{,}$$ where $\Phi$ is the gravitational potential, so the time dilation is: $$d\tau = dt\sqrt{1+2\Phi-(1-2\Phi)\frac{dS^2}{dt^2}}\approx\left[1+\Phi-\frac{v^2}{2}\right]dt\text{.}$$ See the parametrized post-Newtonian formalism for refinements of the above metric. Also related is this question regarding orbits that 'cancel' the gravitational and velocity-dependent terms.

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$$ T'' = T\sqrt{(1 - \frac{v^2}{c^2})({1 - \frac{2GM}{rc^2}})}$$

This is not correct in general, but in the weak-field approximation, it's correct if terms higher than $\mathcal{O}(v^2,\Phi)$ are neglected. Since by Taylor-Maclaurin, $\sqrt{1+x} = 1+x/2+\mathcal{O}(x^2)$, we have $$\sqrt{(1-v^2)(1+2\Phi)} = 1+\Phi-\frac{v^2}{2}+\ldots\text{,}$$ where $x = 2\Phi - v^2 - 2\Phi v^2$, and the resulting $\Phi v^2$ term is dropped from $x/2$.