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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

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

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  • $\begingroup$ possible duplicate of Time Dilation in Orbits in the Schwarzschild Metric $\endgroup$ – John Rennie Feb 20 '14 at 10:29
  • $\begingroup$ Almost but slightly different question. The Time necessarily should be a function of velocity v in this case. Mass referred here is moving at a velocity v induced by an external unbalanced force. $\endgroup$ – Deepak Nath Feb 20 '14 at 10:37
  • $\begingroup$ @JohnRennie: for one, this question doesn't specify that the object is in orbit/freefall. $\endgroup$ – Stan Liou Feb 20 '14 at 10:39
  • $\begingroup$ @StanLiou: fair enough, I've retracted the close vote. $\endgroup$ – John Rennie Feb 20 '14 at 10:42
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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 $r = 4r_s$):

TimeDilation

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:

TimeDilation2

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 :-)

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  • $\begingroup$ What will be the equation near Earth? $\endgroup$ – Deepak Nath Feb 20 '14 at 12:35
  • $\begingroup$ Then we should not be seeing anything travelling faster than the speed of 0.75c or so around a black hole. Do you know of any experimental observations? $\endgroup$ – Deepak Nath Feb 20 '14 at 15:33
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    $\begingroup$ @DeepakNath: I've updated my answer to show what happens near the Earth. Re your comment above, we haven't directly observed any black holes yet so we haven't seen the effects on the speed of light. It will be a long, long time before our telescopes are that good. For the time being this is a purely theoretical calculation. $\endgroup$ – John Rennie Feb 20 '14 at 16:00
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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.


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

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 $x/2$.

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  • $\begingroup$ $\phi = -GM/rc^2 \space?$ $\endgroup$ – Deepak Nath Feb 20 '14 at 12:17
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    $\begingroup$ @DeepakNath: Yup, for spherically symmetric case (also, I used units of $c=1$). For including the quadrupole moment, see the linked answer. $\endgroup$ – Stan Liou Feb 20 '14 at 12:47
  • $\begingroup$ From my own derivation I am getting the value sqroot[1/2 + phi - v^2/2] as time dilation ratio. There is a difference in the constant -1/2 from yours $\endgroup$ – Deepak Nath Feb 20 '14 at 12:58
  • $\begingroup$ @DeepakNath: I'm not sure which derivation you're referring to, but if it's the last equation in your OP, I've edited in a response. I'm guessing you might need to be more careful with the Maclaurin series for square root. $\endgroup$ – Stan Liou Feb 20 '14 at 13:17

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