Jonathan Scott said:
The Wikipedia entry on time dilation is WRONG. I've already added a note to the talk page about that just to record it. The clocks ticks slowest at the center.
Jonathan is right about the Wikipedia entry being wrong.
michelcolman said:
But I thought clocks in inertial reference frames (like the free falling apple) ticked at the same speed? The clock at the center of the Earth is definitely in an inertial reference frame.
And the total field of gravity is zero at the center of the earth. The clock doesn't "feel" any gravity, so the comparison with someone in an accellerating rocket isn't possible either.
Strange, just now when I was starting to think I understood a small part of General Relativity, it turns out I still didn't get it.
It is a little confusing. newcomers to General Relativity are taught the equivalence principle in terms of acceleration but it turns out that gravitational time dilation is more closely related to gravitational potential than to to gravitational acceleration. As you descend towards the centre of the Earth the gravitational potential continues to fall and lower potential means slower clocks. Objects fall from high gravitational potential to lower gravitational potential.
michelcolman said:
OK, I must have misunderstood then.
I thought the equivalence principle said that the "feeling of accelleration" due to gravity was wat was causing the time dilation, just like the time dilation in an accellerating rocket. And that free-falling objects would have no time dilation due to gravity since they were not opposing it. But I guess that was wrong, then.
So, am I finally getting this correctly if I state that free-falling clocks in a field of gravity (actually, clocks that are just starting to free-fall from a stationary position, and don't have any speed yet) are ticking at the same speed as stationary clocks that are held in position inside the gravity field?
The two clocks that you describe would be ticking at the same rate until the free falling clock gains some velocity and distance relative to the stationary clock. This is a good example of one clock feeling acceleration (the stationary clock) while the other is not, yet they both experience the same amount of time dilation.
Here is the equation that makes the relationship clear:
t ' = {t_o \over \sqrt{1-2gm/r_o}} * {(1-2gm/r ') \over \sqrt{1-2gm/R '}}
where t_o is the time rate according to the clock in the hand of a stationary observer located at r_o and t' is the time rate of a test clock located at r' that has been released from location R'.
When the observer is located at infinity the equation reduces to:
t ' = t_o * {(1-2gm/r ') \over \sqrt{1-2gm/R '}}
When the test clock has just been released, R' is equal to r' and the equation reduces to:
t ' = t_o * \sqrt{1-2gm/r '}}
which is the same as the equation for a stationary clock located a r'.
If the test clock is dropped from the infinity while the observer remains at infinity, the equation becomes:
t ' = t_o * (1-2gm/r ')
where part of the time dilation of the test clock is due to gravitational potential at its instaneous location and part is due to its falling velocity.
An even more general equation for relative time that takes the motion of both the test clock and observer into account as well as their respective locations in the gravitational field, (presented without proof, but based on a strong hunch) is:
t ' = t_o *{\sqrt{1-2gm/R_o} \over (1-2gm/r_o)} * {(1-2gm/r ') \over \sqrt{1-2gm/R '}}
where R_o is the height that the observer is released from.
All these equations are based on the radial only Schwarzschild metric which is the static vacuum solution for a non rotating gravitational body.
It is not too difficult to convert these equations into a form where the falling velocity of the test clock is explicit if anyone would prefer them presented that way.
In the case of the Earth the interior Schwarzschild solution has to used for the clock at the centre of the Earth and the equation is:
t ' = t_o * \left( 3 /2 * \sqrt{1-2gm/r_E} -1/2 \right)
which is less than the clock rate of a clock at the surface of the Earth:
t ' = t_o * \sqrt{1-2gm/r_E}
according to an observer at infinity, where r_E is the radius of the Earth and a non rotating Earth is assumed.