Solving the Radius of a Collapsing Star in Schwarzschild Geometry

[latentcorpse]

I have that \left( \frac{dR}{d \tau} \right)^2 = ( 1 - \epsilon)^2 ( \frac{R_{\text{max}}}{R}-1) describes the radius of the surface of a collapsing star in Schwarzschild geometry. I need to show it falls to R=0 in time \tau = \frac{\pi M}{(1-\epsilon)^{3/2}}

So far I have rearranged to get
\int_{R_{\text{max}}}^0 \sqrt{\frac{R}{R_{\text{max}}-R}} dR = \int_0^\tau (1-\epsilon^2)^{1/2} = (1-\epsilon^2)^{1/2} \tau

How do I do that R integral though?

Thanks.

 

[tiny-tim]

hi latentcorpse!

that π in the answer suggests you should go for a trig substitution

 

[latentcorpse]

hi latentcorpse!

that π in the answer suggests you should go for a trig substitution

I forgot to mention that R_{\text{max}}=\frac{2M}{1-\epsilon^2}

Making the substitution R=R_{max} \sin^2{\theta}

I get \int_{\pi/2}^0 2 R_{max} \sin^2{\theta} d \theta = - R_{max} \frac{\pi}{2} = - \frac{M \pi}{(1-\epsilon^2)}

equating to (1-\epsilon^2)^{1/2}\tau

we get \tau=-\frac{\pi M}{(1-\epsilon)^{3/2}}

i.e. an extra minus sign. and it can't take a negative amount of time for a star surface to collapse to r=0, i have messed up a sign. i think the limits on my integrals are sound but could you check please?

thanks.

 

[fzero]

You have to choose the sign when taking the square root to agree with dR/d\tau<0.