Posted:

**Feb 28, 2011 9:20 am**You can apply the twin paradox type stuff when there are accelerations present, no problem, but you can't apply it when spacetime is curved as it is in this example. When this happens, you can't escape using GR. Fortunately for this type of question, we just need the metric, and don't need to compute the connection or curvature.

OK, let's have a crack at it:

I don't bring my old relativity books to work, so I had to resort to Google: Post number 5 in this thread gives the Schwarzschild interior metric (in units where c = G = 1):

ds2 = (3/2sqrt(1-2M/R)-1/2sqrt(1-2Mr2/R3)2)dt2 - (1/(1-2Mr2/R3)2))dr2 - r2dθ2 - r2sin2θdφ2

Consider two points - one at the centre of the earth (r=0), and one at the surface θ = π/2 (to make the maths easy). dr = 0, dθ=0, dφ = 0.

So ds12 = (3/2sqrt(1-2M/R)-1/2)2dt2

and ds22 = (sqrt(1-2M/R)-1/2)2dt2

where ds1 is the interval at a point at the earths centre, and ds2 is the interval at a point at the earths surface.

Putting the c's and G's back I get, where 𝜏1 and 𝜏2 are the proper times at the centre and surface respectively

d𝜏1 / d𝜏2 = (3/2sqrt(1-2GM/c2R)-1/2) / (sqrt(1-2GM/c2R)-1/2)

Shoving in the G, M, R numbers from my Kepler thread I get

d𝜏1 / d𝜏2 = 0.999999999653

So the centre clock is ticking slower (but not by much !!!)

OK, let's have a crack at it:

I don't bring my old relativity books to work, so I had to resort to Google: Post number 5 in this thread gives the Schwarzschild interior metric (in units where c = G = 1):

ds2 = (3/2sqrt(1-2M/R)-1/2sqrt(1-2Mr2/R3)2)dt2 - (1/(1-2Mr2/R3)2))dr2 - r2dθ2 - r2sin2θdφ2

Consider two points - one at the centre of the earth (r=0), and one at the surface θ = π/2 (to make the maths easy). dr = 0, dθ=0, dφ = 0.

So ds12 = (3/2sqrt(1-2M/R)-1/2)2dt2

and ds22 = (sqrt(1-2M/R)-1/2)2dt2

where ds1 is the interval at a point at the earths centre, and ds2 is the interval at a point at the earths surface.

Putting the c's and G's back I get, where 𝜏1 and 𝜏2 are the proper times at the centre and surface respectively

d𝜏1 / d𝜏2 = (3/2sqrt(1-2GM/c2R)-1/2) / (sqrt(1-2GM/c2R)-1/2)

Shoving in the G, M, R numbers from my Kepler thread I get

d𝜏1 / d𝜏2 = 0.999999999653

So the centre clock is ticking slower (but not by much !!!)