Confused about gravitaional time dilation math

1. Aug 28, 2007

hover

Hey everyone,

I have a question about the math involving gravitational time dilation. I have become very confused over this now and i need someone to clear this up. I know the formula for gravitational time dilation, which is-

t'=t x square root of(1-(2GM/rc^2))

t'= the time of an observer within the gravitational field of a mass
t= the time for someone who is not in a gravitational field
G= gravitational constant
M=mass
c=speed of light
r=???

I am really confused on what r is. I thought it was suppose to be the distance from the center of a mass. Then i looked on wikipedia and it says that r is

"the radial coordinate of the observer (which is analogous to the classical distance from the center of the object, but is actually a Schwarzschild coordinate)"

I have no clue what that is. Then searching this forum i found this thread. Saying that r= 2GM/c^2. This is really confusing me. Can someone please clarify what r is?

Thanks.

2. Aug 28, 2007

George Jones

Staff Emeritus
Suppose 2 (infinitesimally) thin spherical shells are erected around the black hole at r = r1 and r = r2. Suppose further that A1 and A2 are the surface areas of the 2 shells, and that A1 > A2. In terms of the r-coordinate, A1 = 4*pi*r1^2 and A2 = 4*pi*r2^2, as expected.

Now lower a rope straight down from the outer shell to the inner shell. If spacetime were flat, the length d of the rope would be r1 - r2, the difference in the "radii" of the shells. However, because of the curvature of spacetime, it turns out that d > r1 - r2.

3. Aug 28, 2007

hover

Thanks fot the reply but, unfortunately, i have a hard time understanding what you mean.

I just remembered that 2GM/c^2 is the Schwarzschild radius. Which is the how far away the event horizon is from a black hole. If t'=t x square root of(1-(2GM/rc^2)) and i replace r with 2GM/c^2, then i get t'=t x square root of(1-(2GM/2GM)). Which makes t'= t/0. Meaning that r must be distance from the center of a mass. I hope that this is right.

4. Aug 28, 2007

meopemuk

You are right. r is the distance from the center of the body producing the gravitational field. If you are interested in the time dilation on the Earth surface, then r is Earth radius. If you are interested in the redshift of Solar spectral lines, then r is Sun radius.

Eugene.

5. Aug 28, 2007

hover

Thanks for the responces

6. Aug 29, 2007

pervect

Staff Emeritus
For weak fields this is close to being correct, but it's not quite right.

I thought George's remarks were pretty clear, and correct too.

The Schwarzschild coordinate r is defined such that the area of a sphere of radius r (a set of points with a constant r coordinate) is 4 pi r^2.

This defintion for r is *not* quite the same as saying that r is the radial distance. The difference in defintions won't matter if one is applying the formula here on Earth, but it would be important if one were to apply the formula to a black hole or other "strong field" problem.

7. Aug 29, 2007

hover

I never said what he said was wrong. I just have a bad time interpreting stuff. Sorry .

8. Aug 29, 2007

George Jones

Staff Emeritus
This doesn't quite work.

Erect a shell right at the surface of the Sun. Erect another shell that, according the r-coordinate, is 1 metre above this shell. According to a tape measure, the distance between these shells is 1.000002 metres. For accurate measurements, this is significant.

Do the same for a neuton star that has one solar-mass and a radius of 10 kilometres. Now the tape measure distance between shells is about 1.2 metres. For any type of measurement, this is significant.

Also, for a black hole, there is no "distance from the center of the body."

Here's another way to look at things. Run a tape measure around a great circle of the shell, i.e., measure the circumference of the great circle. The r-coordinate is defined as the circumference of a great circle divided by 2*pi.

The Schwarzschild radius does give the location of the event, but not in terms of how far the event horizon is from the black hole. The event horizon is the boundary between the non-black hole region and the black hole region, so, in some sense, the distance from the event horizon to the black hole is zero.

Last edited: Aug 29, 2007
9. Aug 29, 2007

Thrice

I think it automatically follows from the schw. metric that

$dR^2 = (1-\frac{r_G}{r})^{-1}dr^2$

where dR is the infinitesimal radial distance & dr is the coordinate differential. Clearly dR > dr

10. Sep 1, 2007

antipater

dilation formula only valid outside sphere

The dilation formula is only valid outside the sphere. Once you are inside the sphere, only the mass below you counts. The mass above you forms a hollow sphere and you are weightless inside a uniform hollow sphere. Therefore the gravitational redshift is at its maximum at the surface and deminishes going out in space or going towards the center. It is zero at the center of the sphere. And indeed, your are weightless at the center just like someone far out in space. The equivalence principle applies here too! If you can't tell whether your are at the center or far out in space, your clock will tick the same in both cases.
Think about it, we have an entire universe around us. And yet we are weightless (outside the earth) because it is like being inside a gigantic hollow sphere. If it were true that time dilation would increase as you go towards the center, then the clock would tick slowest where we are and all clocks elsewhere in the universe would tick faster (blue shift). But if anything, there is a (hubble) redshift as we go further out into the universe.

11. Sep 1, 2007

pervect

Staff Emeritus
Unfortunately, this is not quite right. If you have a hollow sphere, the newtonian force inside the sphere will be equal to zero. However, the gravitational time dilation does not "diminish" as you go in towards the center - it is the same on the inside of the sphere as on the outside.

In the weak field (such as on Earth), the gravitational time dilation can be approximated by

$$T \approx 1 + \frac{GM}{R c^2}$$

i.e. basically the time dilation is proportional to U, where U is a postive number equal to the negative of the Newtonian potential energy, i.e U = GM/R. (For a unit test mass, the constant of proportionality is 1/c^2).

Because there is no force inside a hollow sphere, U is the same inside and outside the sphere, and the time dilation is also the same.