# GR to the center of the Earth!

1. Mar 23, 2005

### RandallB

I’m interested in how the rate of time compares in different locations of “deep” gravity fields.
With GR we are good with the idea that a clock on the Moon runs faster than a clock on Earth. Here we’re looking only at GR the effects of acceleration/gravity. We can directly measure the acceleration affecting time in those locations in the Measurement of Gravity by simply weighing a known mass - simple.
But, suppose we place clocks and measuring devices at intervals in a drilling through the center of the Earth. We will see the measured weight (force of gravity) go to zero at the center! What does this mean for the gravity field and the impact the rate of time as you approach the center?

I see to alternatives:
1) Gravity goes to Zero the curve of space time turns back to the same shape as deep space, so time at the center of earth actually runs faster than time on the moon's surface!

2) Although the measurable gravitational force has gone to zero, I’ll call it the “local mass density” around the clock continues to increase as it near's the center, thus the maximum reduction in the rate of time will be found at the center along with max curve in space time.

I sure hope it’s #2 because #1 just gets to weird. The questions are:
A) Do the GR formulas account for this and which one does it reflect 1 or 2?

B) Has someone done a test to verify which is correct? A deep mine experiment confirming the lower weight of a mass (smaller gravity), yet time tested as running even slower than on the surface?

Randall B

2. Mar 23, 2005

### pervect

Staff Emeritus
If you look at the expression for gravitational time dilation in the weak field (where one can sensibly talk about gravity as a force), you'll find that gravitational time dilation is not a function of the "gravitational force", but rather a function of the "gravitational potential energy".

There are some formulas at

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/gratim.html

which illustrate this point, however, the formulas as written do not apply to the situation at the center of the Earth.

There are some past physics forum threads on this topic which you can probably find if you're interested, but the main point is that the gravitational time dilation is

$$T = \frac{T_0}{\sqrt{1-U/c^2}}$$

U here is a positive number, which is the negative of the Newtonian gravitational potential energy per unit mass. You can think of U as representing "how deep" one is in a gravity well, i.e it's the amount of energy / unit mass one would need to escape to infinity.

T is a function of U, not the local "gravitational field".

Below the surface of the Earth, with the assumption that the Earth has a constant density (which is probably not exactly true, but close), you can use the formula

$$U = \frac{GM}{R_0} + .5 \frac{g}{R_0} (R_0^2 - R^2)$$

Here g is the gravitational acceleration at the Earth's surface, R_0 is the radius of the Earth, and R is the distance one is from the center of the Earth (R=0 at the center of the Earth). The formula above is valid only for R<R_0, above that value use the formulas in the web reference.

You can see that U continues to increase, reaching a maximum at the center. This is required because the force is proportional to dU/dR, and as the force always points towards the center of the Earth, dU/dR always has the same sign.

The detailed expression above comes from the fact that for a constant density mass, gravity below the surface follows Hooke's law. This can be derived by a straightforwards calculation based on Newtonian gravity.