Anamitra said:
Anamitra, you have to be very careful to say something like this. That statement defines not just
a worldline, but an infinite number of worldlines, all of them going in different radial directions. Actually, what you're describing is inertial (nonaccelerating) particles coming out of an explosion. With the schwarzschild metric, something like this equation might be useful to consider if you want to model a supernova explosion, except in that situation, k would not be a constant (because of acceleration).
Your definition for d\tau using this should be :
d\tau \equiv ds|_{d\theta=d\phi, dr=kdt}=\left ( \sqrt{1-\frac{G M}{r(t) c^2}} \right )dt
Contrast that with my definition:
JDoolin said:
First, a definition of their differentials:
\begin{matrix} <br />
dR \equiv ds|_{d\theta=d\phi=dt=0}=\left ( \frac{1}{\sqrt{1-\frac{G M}{r c^2}}} \right )dr\\ <br />
d\tau \equiv ds|_{d\theta=d\phi=dr=0}=\left ( \sqrt{1-\frac{G M}{r c^2}} \right )dt<br />
\\ r d\theta' \equiv ds|_{dt=d\phi=dr=0} =r d\theta\\ r sin(\theta)d\phi' \equiv ds|_{d\theta=dt=dr=0} =r sin(\theta)d\phi \\ \end{matrix}
(Edit: fixed from original post so second line is an integral with respect to dt instead of dr.)
My definition of \tau (dr=0) is different from Anamitra's (dr=k dt). My definition represents the time on stationary clocks in the gravitational field. Anamitra's definition represents the time on clocks that somehow manage to maintain a constant velocity in the gravitational field.
Ben Niehoff said:
My point was that you could have chosen some other path, such as r = kt, and it would give you an entirely different definition of the new coordinate \tau. So it would be ambiguous to say
\tau = \int \sqrt{1-\frac{2m}{r}} \; dt
because no path has been specified. And it would be an outright lie to say
d\tau = \sqrt{1-\frac{2m}{r}} \; dt
because the right-hand side is not a total differential.
I
could have chosen some other path, but I
didn't. Notice my use of "≡" in my definition of d\tau. Under these circumstances, there is no ambiguity, but I can see how you would say that it is ambiguous in general, if you thought I meant r(t) were an arbitrary path.
Finally:
Ben Niehoff said:
Something is wrong with your numerics. I double-checked my antiderivative in Mathematica. See the attached image of the output.
This is interesting. I can't figure out right exactly what the problem is. One possibility is since acosh is not a true function, that my calculator gave me the wrong part of the curve, or I flubbed it, entering it. If you plug in m=1/2, and r=1.1, into your formula do you get a negative value?
In any case Anamitra's equation from post 126 gave the same result as the numerical integration:
{R}{=}{\sqrt{{r}^{2}{-}{2mr}}}{+}{2m}{ln}{[}\sqrt{r}{+}\sqrt{(}{r}{-}{2m}{)}{]}{+}{C}
Edit: more info:
y=\cosh(\theta)=\frac{e^\theta + e^{-\theta}}{2}= \frac{e^{2\theta }+ 1}{2e^\theta}
reduces to a quadratic equation
(e^\theta)^2-2e^\theta y + 1 = 0
with solution
e^\theta=y\pm\sqrt{y^2-1}
So \theta = arccosh(y)=ln(y\pm \sqrt{y^2-1})
if that helps.