Schwarzschild spacetime proper time to fall radially inward

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1. Mar 23, 2017

binbagsss

1. The problem statement, all variables and given/known data

Question attached

My method was going to be:

set $r=R$ and solve for $n(R)$
set $r=2GM$ and solve for $n(2GM)$

I was then going to integrate proper time $s$ over these values of $r$:
$\int\limits^{n=cos^{-1}(\frac{4GM}{R}-1)}_{n=cos^{-1}(1)=0} s(n) dn$#

whereas the solution, all you have to do is plug in these values $n(r)$ and $n(2GM)$ into $s=s(n)$, equation (2).

Is it wrong to integrate or would I get the same answer?

Is the reason that you can plug in and don't need to integrate due to the fact that the lagrangian, is invariant under reparameterisation, or does it have nothing to do with this??:

i.e:
$s=\sqrt{L}=\sqrt{g_{uv}\frac{dx^u}{dn}\frac{dx^v}{dn}} dn$, $n$ the parameter

2. Relevant equations

see above

3. The attempt at a solution

see above

2. Mar 23, 2017

PeroK

I think it's simpler than that. You are given a parameterisation for $s$ in terms of $\eta$. So, if you know the start and end values of $\eta$, you can plug these in and get the start and end proper times.

If you were given $ds$, then you would have to integrate that to get $s(\eta)$, but that has already been done for you.

There's no more to it than that.

Re the metric, the fact that the parameterisation represents a geodesic is all you need to know (although even that doesn't really matter). You might want to check that it's a valid geodesic, but it looks like the question is giving you that as well.