- #36
latentcorpse
- 1,444
- 0
betel said:The second term has two be without the 1/2 because you get twice the same contribution from the t-r and r-t.
Then you can combine the two.
You also know, that it is back at r=0 at tau=4. That will be enough.If you take the correct equation for t it will be easier.
So I find [itex]r(\tau)=-\frac{1}{2} \tau^2+2 \tau[/itex]
and then the t equation becomes
[itex]
\frac{d^2t}{d \tau^2} + \frac{1}{(1+r)} \frac{dt}{d \tau} \frac{dr}{d \tau}=0
[/itex]
What do you mean by now I can combine the two? Do you mean substitute the for r into the t equation or do you mean do some further simplification? Presumably further simplification because substitution yields:
[itex]\frac{d^2t}{d \tau^2} + \frac{- \tau + 2}{ -\frac{1}{2} \tau^2 + 2 \tau + 1} \frac{dt}{d \tau}=0[/itex]
which doesn't look very promising!