Limit , circular orbit ,schwarzschild s-t ,

[binbagsss]

Homework Statement

[/B]
To take the $lim J \to \infty$, what are the two roots of $r_c$ in this case...

So I believe it says' $J$ big enough it had 2 solutions' is basically saying just avoiding the imaginary solutions i.e. $J^4 \geq 12G^2M^2J^2$ (equality for one route obvs).

Homework Equations

see attachment

The Attempt at a Solution

So I believe it says' $J$ big enough it had 2 solutions' is basically saying just avoiding the imaginary solutions i.e. $J^4 \geq 12G^2M^2J^2$ (equality for one route obvs).

If I write it as

$r_c= \frac{J^2 \pm \sqrt{J^4}\sqrt{1-\frac{12G^2M^2}{J^2}}}{2GM}$

then as $J \to \infty$ the last term vanishes and for the + root clearly get $\frac{J^2}{GM}$ , the first given in 3.48.

However for the - root I would get $0$. I have no idea how to get $3GM$ , do I need to do some sort of expansion?

Many thanks

 

[stevendaryl]

Homework Statement

[/B]
To take the $lim J \to \infty$, what are the two roots of $r_c$ in this case...

So I believe it says' $J$ big enough it had 2 solutions' is basically saying just avoiding the imaginary solutions i.e. $J^4 \geq 12G^2M^2J^2$ (equality for one route obvs).

Homework Equations

see attachment
View attachment 234850

The Attempt at a Solution

So I believe it says' $J$ big enough it had 2 solutions' is basically saying just avoiding the imaginary solutions i.e. $J^4 \geq 12G^2M^2J^2$ (equality for one route obvs).

If I write it as

$r_c= \frac{J^2 \pm \sqrt{J^4}\sqrt{1-\frac{12G^2M^2}{J^2}}}{2GM}$

then as $J \to \infty$ the last term vanishes and for the + root clearly get $\frac{J^2}{GM}$ , the first given in 3.48.

However for the - root I would get $0$. I have no idea how to get $3GM$ , do I need to do some sort of expansion?

Many thanks

Yes, you need to expand the square-root, using $\sqrt{1-\epsilon} \approx 1 - \frac{\epsilon}{2} + ...$