JDoolin said:
(1) I need to explain the coefficients \left(1-\frac{2 m}{r}\right) and \left( \frac{2m}{r-2m}\right) I've already got somewhat of a handle on the first coefficient (see post 12 and 14 in this thread, except I appear to have an extra 2.)
Part 1a. Finding \left(1-\frac{2 m}{r}\right)
I have once again worked my way through
http://www.mathpages.com/rr/s8-09/8-09.htm and want to document the tricky parts.
The positions of the bottom and top of an elevator are described by: x_1 = \frac{1}{2} a t^2 and x_2 = L +\frac{1}{2} a t^2
It should first be noted that this is an approximation for low acceleration. If you want to treat for large accelerations, you need to make a hyperbola instead of a parabola:
This is only the first of many, many, many approximations. But the rest of the approximations will be based on the Maclaurin series expansions:
\begin{align} \displaystyle \mathrm{ Maclaurin\: Series\: Expansions} \\f(x) &\approx \frac{f(0) x^0}{0!} + \frac{f'(0) x^1}{1!} + \frac{f''(0) x^2}{2!} + ... \\\sqrt{1+x} &\approx \frac{1}{0!}+\frac{x/2}{1!}-\frac{x^2/4}{2!}+\frac{3 x^3/8}{3!}+... \\e^x &\approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}+... \end{align}
We imagine a photon of light leaving the bottom of the elevator at the moment when the elevator slows to a stop, before accelerating upward again. When does the path of the photon,
x = c t
meet the path of the top of the elevator?
x = L + \frac{1}{2}a t^2
Setting the two equal and solving for t, we find there are two solutions.
t=\frac{c}{a}\left ( 1 \pm \sqrt{1-\frac{2 a L}{c^2}} \right )
The lower value of t is the one we want. The higher term is nonsense, because it represents the
second time the elevator meets the photon. This would not happen if we used the hyperbolic equation for the position of the elevator, because the hyperbolic equation never reaches and passes the speed of light. The quadratic equation, however, let's the elevator exceed the speed of light, and run into the photon a second time, producing the nonsense solution.
Our interest now turns to the question of how fast a clock at the top of the elevator runs compared to how fast a clock at the bottom of the elevator. This requires us to define two time variables. I don't know how standard this is, but I'd like to call them t_1(n) and t_2(n), being the time that the nth photon leaves the bottom of the elevator, and the time the nth photon arrives at the top of the elevator.
I am noting this n variable because I think if we are to take a derivative of t1 and t2, we should keep in mind that there must be some dependent variable with respect to which t1 and t2 are changing.
We note that
c = \frac{\Delta x}{\Delta t} = \frac{x_2(t_2)-x_1(t_1)}{t_2 -t_1} = \frac{L+\frac{a}{2}(t_2^2-t_1^2)}{t_2 - t_1}
That is, the distance is the position of the top of the elevator at time 2 minus the position of the bottom of the elevator at time 1.
\begin{align*} c ({t_2 - t_1})&= L+\frac{a}{2}(t_2^2-t_1^2)\\ c (dt_2 -dt_1) &= a (t_2 dt_2 -t_1 dt_1)\\ c dt_2 -a t_2 dt_2 &= c dt_1 -a t_1 dt_1 \\ (c -a t_2) dt_2 &= (c -a t_1) dt_1 \\ \frac{dt_2}{dt_1} &=\frac{c - a t_1}{c- a t_2}\\ \frac{dt_2}{dt_1} &=\frac{1 - a t_1/c}{1- a t_2/c} \end{align*}
However, what we are modeling is a gravitational field, and the idea of the equivalence principle is that if you're watching a descending elevator at the very instant, t1=0, that it ceases to decelerate, stops, and begins to accelerate; at the moment when both the top and bottom of the elevator simultaneously stop, this is a perfect model for watching what is happening to someone standing in a gravitational field.
So, mathematically, what this means is we are going to set t1 = 0.
\left ({\frac{d t_2}{d t_1}} \right ) _{t_1=0}= \left (\frac{1-\frac{a}{c} t_1}{1-\frac{a}{c} t_2} \right ) _{t_1 =0} =\frac{1}{1-\left ( 1 - \sqrt{1-\frac{2 a L}{c^2}} \right )} = \frac{1}{\sqrt{1-\frac{2 a L}{c^2}}}
Now think a moment about waves, with their frequencies, and wavelengths, and periods. What do the dt's represent? They are proportional to the periods of waves, and reciprocal to the frequencies of waves. So note that
\frac{\nu_2}{\nu_1} = \frac{\mathrm{d} t_1}{\mathrm{d} t_2}=\sqrt{1 - \frac{2 a L}{c^2}} \approx 1 - \frac{a L}{c^2}
We don't want to assume, however, that the gravitational field a, within the elevator will be uniform. So what is done next is to divide the elevator into little chunks, noting, first of all, that within each chunk, a(r)=\frac{m}{r^2}.
And in total, the value is a product:
\frac{\nu_{top}}{\nu_{bottom}} = \prod_{bottom}^{top}\frac{ \nu_{upper} }{\nu_{lower}}
Now, did you know that the logarithm of the product of several terms is equal to the sum of the logarithms of those same terms?
\begin{align*} \ln \left (\frac{\nu_{top}}{\nu_{bottom}} \right ) &=\ln \left (\prod_{bottom}^{top}\frac{ \nu_{upper} }{\nu_{lower}} \right ) \\ &= \sum \ln \left (\frac{ \nu_{upper} }{\nu_{lower}} \right )\\ &= \sum _a \ln(1-\frac{a L}{c^2}) \\ &=\sum _r \ln(1-\frac{m L}{r^2 c^2}) \end{align*}
Further, we can break this sum down continuously (and make our second Maclaurin Series Expansion) until we have an integral:
\begin{align*} \sum _r \ln(1-\frac{m L_r}{r^2 c^2}) &=\int _{r_1}^{r_2} \ln \left( 1-\frac{m dr}{r^2 c^2} \right) \\ &\approx \int _{r_1}^{r_2} \left (-\frac{m d r }{r^2 c^2} \right )\\ &=-\frac{m}{c^2}\int _{r_1}^{r_2} r^{-2} dr\\ \ln \left ( \frac{\nu_2}{\nu_1}\right ) &=\frac{m}{c^2}\left ( \frac{1}{r_2}- \frac{1}{r_1} \right ) \end{align*}
Taking e to the power of both sides (and using a Taylor Series expansion of ln(1+x)=x
\begin{align*} \left ( \frac{\nu_2}{\nu_1}\right )&= Exp \left ({\frac{m}{c^2}\left ( \frac{1}{r_2}- \frac{1}{r_1} \right )} \right )\\ &\approx 1 + \left ({\frac{m}{c^2}\left ( \frac{1}{r_2}- \frac{1}{r_1} \right )} \right ) \end{align*}
Now, defining dt as the period of time far away from the mass, and d\tau as the period of time at distance r from the mass:
\begin{align*} \left ( \frac{d \tau}{d t}\right )= \left ( \frac{\nu_\infty}{\nu_r}\right ) &= Exp \left ({\frac{m}{c^2}\left ( \frac{1}{\infty}- \frac{1}{r} \right )} \right )\\ &\approx 1 - {\frac{m}{c^2 r}} \end{align*}
And we are interested in finding the square of this quantity:
\begin{align*} \left ( \frac{d \tau}{d t}\right )^2 &\approx \left (1 - {\frac{m}{c^2 r}} \right )^2\\&\approx 1 - \frac{2 m}{c^2 r} \end{align*}
Set c=1, and we have the value given by the Schwarzschild metric:
g_{11} = \left ( \frac{d \tau}{d t}\right )^2 = 1 - \frac{2 m}{ r}
If I counted right, there are six approximations. The use of a parabola instead of a hyperbola. Four Maclaurin Series approximations, and one Taylor series approximations.
