Gravitational Time Dilation: What is R in the Equation?

[ilikescience94]

Hey, I'm trying to understand gravitational time dilation, but can not find a good explanation for what R equals in the equation:

T=T0/(sqrt(1-2GM/(Rc^2)))

 

[Nugatory]

That formula is for the time dilation in the gravitational field of a spherical object (like a star or a planet) compared with the time measured by a far away observer.

$R$ is just the distance from the center of the object (although there is a subtlety here, which I won't go into until you're happy with the simple answer). So if you wanted to calculate the gravitational time dilation at the surface of the Earth you'd set $R$ equal to the radius of the earth, about 6400 kilometers.

 

[yuiop]

If you had a large ring centred on the centre of a gravitational body and measured the circumference of the ring, then dividing the circumference by 2*pi gives R as defined in the Schwarzschild equation you mentioned. $T_0$ is the time measured by a clock attached to the (non rotating) ring and T is the coordinate time measured by an observer at infinity.

This is trivially true in flat space, but in general relativity, R as defined in this metric is not the same as what you would obtain if you measured the radius of the ring using a tape measure from the centre of the body to the ring. (This is probably the subtlety that Nugatory is alluding to). This is because Euclidean geometry no longer works in curved space.

 

[Nugatory]

(This is probably the subtlety that Nugatory is alluding to)

yes, that's it. It's not a big deal for weak gravitational fields (planets, ordinary stars, ...) but it matters a lot in stronger gravitational fields.

 

[ilikescience94]

That formula is for the time dilation in the gravitational field of a spherical object (like a star or a planet) compared with the time measured by a far away observer.

$R$ is just the distance from the center of the object (although there is a subtlety here, which I won't go into until you're happy with the simple answer). So if you wanted to calculate the gravitational time dilation at the surface of the Earth you'd set $R$ equal to the radius of the earth, about 6400 kilometers.

Thank you, I like this, and how would I find this new R rather than simply the radius?

 

[pervect]

Thank you, I like this, and how would I find this new R rather than simply the radius?

If you had a large ring centred on the centre of a gravitational body and measured the circumference of the ring, then dividing the circumference by 2*pi gives R as defined in the Schwarzschild equation you mentioned.

You might have missed the answer, yuiop gave it already though.

 

[ilikescience94]

You might have missed the answer, yuiop gave it already though.

I am not a smart man.

 

[pervect]

No need to dis yourself for missing a response, it's easy enough to miss forum responses by being in a hurry, or by having them "sneak in" in front of something you've read.