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as per to einstein as we go far from the earth the TIME tends to slow down , so it means when I am one metre above the earth's surface , the time has slow down for me as per http://en.wikipedia.org/wiki/Gravitational_time_dilation what would happen if I am at the centre of earth

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marked as duplicate by Ben Crowell, John Rennie, Qmechanic Jun 24 '13 at 17:31

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  • $\begingroup$ what is your hypothesis? how would you describe the gravitational field at the center of earth as compared to that at the surface of the earth? $\endgroup$ – Joe Jun 24 '13 at 15:25
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    $\begingroup$ You would probably die if you were at the center of the earth (4000 C temperature and gigapascal pressures there). $\endgroup$ – Kyle Kanos Jun 24 '13 at 15:25
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    $\begingroup$ You have it the wrong way round. If you compare a clock far from the Earth with a clock near to the Earth it's the clock near to the Earth that runs slower. So by rising 1 metre above the Earth time runs faster not slower. $\endgroup$ – John Rennie Jun 24 '13 at 15:37
  • $\begingroup$ At the center of the earth there is no gravity (at least classically), so there would be no time dilation. $\endgroup$ – jinawee Jun 24 '13 at 15:59
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    $\begingroup$ duplicate of physics.stackexchange.com/q/10089 $\endgroup$ – Ben Crowell Jun 24 '13 at 17:03
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To a good approximation the time measured far from a gravitational field, $t_\infty$, and the time measured within the gravitational field, $t_0$, are related by:

$$ \Delta t_\infty = \Delta t_0 \left( 1 + \frac{2 \Phi}{c^2}\right)^{-1/2} $$

where $\Phi$ is the Newtonian gravitational potential. So for example outside the Earth at some distance $r$ the gravitational potential is:

$$ \Phi = -\frac{GM}{r} $$

and therefore:

$$ \Delta t_\infty = \Delta t_0 \left( 1 - \frac{2 GM}{c^2r}\right)^{-1/2} $$

which is actually the same result we get from the Schwarzschild metric, though you should note that the Schwarzschild $r$ co-ordinate is subtly different from the $r$ co-ordinate used in Newton's law.

Anyhow, inside the Earth at a distance $r$ from the centre the gravitational potential is:

$$ \Phi = -\frac{GM}{2R^3} (3R^2 - r^2) $$

where $R$ is the radius of the Earth (assuming the Earth is a perfect sphere of uniform density). So the gravitational time dilation is given by:

$$ \Delta t_\infty = \Delta t_0 \left( 1 - \frac{GM}{c^2R^3} (3R^2 - r^2)\right)^{-1/2} $$

or at the centre where $r$ is zero:

$$ \Delta t_\infty = \Delta t_0 \left( 1 - \frac{3GM}{c^2R} \right)^{-1/2} $$

So time at the centre runs more slowly than time at the surface. This shouldn't surprise you as the gravitational potential at the centre is obviously higher than at the surface otherwise objects wouldn't fall down mine shafts. Note that it's the potential that matters not the gravitational acceleration. The gravitational acceleration at the centre of the Earth is actually zero i.e. you would be weightless. However the time dilation is not zero.

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  • $\begingroup$ But in classical mechanics , as per to the equation of acceleration due to gravity , the gravity as we go down it decreases en.wikipedia.org/wiki/Gravity_of_Earth#Depth so how the time runs slower $\endgroup$ – gkshindia Jun 25 '13 at 5:54
  • $\begingroup$ Because the gravitational potential is the integral of the acceleration. Starting with the potential at zero at infinity, as we integrate towards the Earth's surface the potential increases (the magnitude of the potential increases - it actually gets more negative). Then as we integrate from the surface towards the centre the potential still increases so it reaches a maximum at the centre. For the potential to decrease the acceleration below the surface would have to be negative i.e. point outwards. $\endgroup$ – John Rennie Jun 25 '13 at 7:02

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