MeJennifer said:
Looks like I need to become a bit more rigorous and exact in my understanding of GR because I was under the impression that gravitational time dilation is expressed by
both the curvature of the spatial and temporal dimensions.
I certainly do not want to learn the wrong thing.
Could you perhaps explain how this is the case that only the curvature of the temporal dimension contributes to gravitational time dilation?
Let's take the simplest possible example.
Suppose we have a 2 dimensional metric, one space coordinate, one time coordinate, and that we have a time dilation factor g00 that depends on position, but not on time. This gives us a metric of the form
ds^2 = g00(x) dt^2 - dx^2
I suppose I should add that I've assumed that c=1. The rindler metric for the "uniform gravitational field" of an accelreated observer can be put into this format, for example.
Then the function x(t) which extremizes (maximizes, usually) proper time will be the function which extremizes
\int ds = \int \sqrt{g00 - \left( \frac{dx}{dt} \right) ^2} dt
This function must satisfy the Euler-Lagrange equations:
http://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html
http://en.wikipedia.org/wiki/Euler-Lagrange_equation
with
f(t ,x ,xdot) = \sqrt{g00 - xdot^2}
and we must note that g00 is a function of x, and xdot is defined as dx/dt, the coordinate velocity.
Now we will make tons of simplifications - feel free to work out the full details if you can and if you really want to. These simplifications make the problem essentially Newtonian, so we can compare it with the Newtonian answer, and easy to work.
We will assume that g00 is very nearly unity. This is the normal state of affairs, gravitational time dilation on the Earth's surface is only a part per billion, for instance.
We will also assume that xdot is very small. Since we have assumed that c=1, this means that we assume that we are dealing with velocites much smaller than the speed of light, which is also usually the case.
We can then say by using a Taylor series approximation that
f(t,x,xdot) \approx \sqrt{g00} - \frac{1}{2 \sqrt{g00}} xdot^2
The Euler-Lagrange equations tell us that the path that extermizes proper time is then approximately.
\frac{d}{dt} ( \frac{\partial f}{\partial xdot} ) = \frac{\partial f}{\partial x}
With our simplifying assumptions, this leads us to
- \frac{d}{dt} xdot = \frac{1}{2} \frac{\partial g00(x)}{\partial x}
Since (d/dt) xdot is the acceleration, we see that the acceleration is just
<br />
-\frac{1}{2} \frac{\partial g00(x)}{\partial x} <br />
We see that the rate of change of time dilation with respect to position acts just like a "force" as far as the equations of motion go. If time dilation is not a function of position, bodies following geodesics (i.e. bodies extermizing proper time, which means that they satisfy the Euler-Lagrange equations) do not accelerate. When time dilation is present and varying with position, we can interpret the equations of motion to be the same as if we had an actual force present. We call this force "gravity".
