HS: Solve Integral for Observer's Lifetime in BH

[Altabeh]

Homework Statement

Hello

The following integral turns out to be a bit nerve-wracking for me as I’ve tried almost any possible way to get the answer but all my efforts got blocked up and fell flat in their face somehow. The integral which may sound familiar is

\tau=\int_{0}^{2m}{\frac{1}{\sqrt{2m/r-1}}}dr,

and shows the lifetime of an observer swallowed up by the Schwarzschild BH from the time he enters the BH through the hypersurface r=2m to the time he falls into the spacetime singularity at r=0 where he is doomed to die dreadfully.

Any help will be highly appreciated.

AB

 

[Mark44]

The integrand can be written as
\frac{\sqrt{r}}{\sqrt{2m - r}}

I would start first with an ordinary substitution such as u = sqrt(2m - r) and see if that got me somewhere.

Also, you integrand is undefined at r = 0 and r = 2m, so you have an improper integral than you'll need to use limits to find. Since there are discontinuities at both endpoints, you'll need to split the integral into two separate integrals, with limits for each.

 

[Altabeh]

The integrand can be written as
\frac{\sqrt{r}}{\sqrt{2m - r}}

I would start first with an ordinary substitution such as u = sqrt(2m - r) and see if that got me somewhere.

Also, you integrand is undefined at r = 0 and r = 2m, so you have an improper integral than you'll need to use limits to find. Since there are discontinuities at both endpoints, you'll need to split the integral into two separate integrals, with limits for each.

Unfortunately the substitution you offer doesn't work because the integrand then turns again into another indocile function that can't be integrated by any method. I mean the modified integrand

-2\sqrt{2m - u^2}

is not reducible to a simple function which would be capable of being written in terms of elementary functions if one integrated it wrt the new variable u! So what now?

 

[Altabeh]

-2\sqrt{2m - u^2}

Can we put u^2/2m=\cos^2(k) and keep going?

 

[Count Iblis]

Why not put 2m/r - 1 = u^2 in the original integral?

 

[gabbagabbahey]

You might begin with the sub u=\frac{2m}{r}, and follow that with an appropriate trig sub.

 

[elect_eng]

and shows the lifetime of an observer swallowed up by the Schwarzschild BH from the time he enters the BH through the hypersurface r=2m to the time he falls into the spacetime singularity at r=0 where he is doomed to die dreadfully.

Any help will be highly appreciated.

AB

I tried a simple substitution x=1/r and seemed to get an answer very quickly using a table for the new integral. It's possible I worked too fast and made a mistake, but you can check it out quickly enough.