Solving A2/15 B2/23 (Part ii) on Page 50 Cambridge Maths

[Tangent87]

Hi, I am doing question A2/15 B2/23 (specifically part (ii)) at the top of page 50 here:

http://www.maths.cam.ac.uk/undergrad/pastpapers/2001/Part_2/list_II.pdf

I have done everything up to and including showing the impact parameter is given by b=h/E, but I am having trouble doing parts (a), (b), (c). To be honest I don't even know where to start, it says to consider equation (*) and so far all I've been able to show is that if $$\frac{dr}{d\lambda}=0$$ then we have the cubic equation for r: $$r^3-b^2r+2Mb^2=0$$. But looking at the three separate cases in parts a,b, and c I'm guessing we want some kind of quadratic where the discriminant is $$b^2-27M$$?

 

[fzero]

It sounds like they want you to know Cardano's method http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method There is a discriminant for roots of the cubic.

 

[Tangent87]

It sounds like they want you to know Cardano's method http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method There is a discriminant for roots of the cubic.

You may well be right, as using Cardano's method we get this as one of the solutions to the cubic:

$$r=(Mb^2)^{1/3}\left[\left(-1+\sqrt{1-\frac{b^2}{27M^2}}\right)^{1/3}+\left(-1-\sqrt{1-\frac{b^2}{27M^2}}\right)^{1/3}\right]$$

So now we can kind of see where the conditions on b² and M² are going to come from. But I'm not sure what we've done physically, i.e.

Why do we set $$dr/d\lambda$$ equal to zero?
What is the GR significance of the solution r we found above?
Do we have to worry about the other two roots to the equation?

I think we need to answer these questions first before we do anything else.

 

[latentcorpse]

I'm not sure how much use this is but could you not use a particle in a potential approach.

You have the equation $( \frac{dr}{ d\tau} )^2 + V(r) = E^2$

You can show that the potential has a maximum at $r=3M$ with $V(r=3M)=\frac{h^2}{27M^2}$

Now the particle has energy $E$ and we know from the definition of impact parameter that $b= | \frac{h}{E} |$ so $E^2=\frac{h^2}{b^2}$

So if the particle has energy greater than the potential then it has enough energy to reach $r=2M$ and be swallowed by the black hole. This means

$E^2 > V(r) \Rightarrow \frac{h^2}{b^2} > \frac{h^2}{27M^2} \Rightarrow b^2 < 27 M^2$

That would answer part b) I think. You can then do something similar for a).

As for c), I would GUESS (but am far from sure) that it would be some sort of orbit around the black hole. Since if b>27M^2 it's reflected and if b<27M^2 it's swallowed so if b=27M^2 wouldn't it make sense for there to be a balance between attraction and repulsion?

 

[fzero]

You may well be right, as using Cardano's method we get this as one of the solutions to the cubic:

$$r=(Mb^2)^{1/3}\left[\left(-1+\sqrt{1-\frac{b^2}{27M^2}}\right)^{1/3}+\left(-1-\sqrt{1-\frac{b^2}{27M^2}}\right)^{1/3}\right]$$

So now we can kind of see where the conditions on b² and M² are going to come from. But I'm not sure what we've done physically, i.e.

Why do we set $$dr/d\lambda$$ equal to zero?
What is the GR significance of the solution r we found above?
Do we have to worry about the other two roots to the equation?

I think we need to answer these questions first before we do anything else.

$$dr/d\lambda$$ tells us whether the photon is moving towards or away from the BH. For the photon approaching from infinity, $$dr/d\lambda<0$$ for large initial $$r$$ (so we would take the negative root of (*) for consistency). Now if $$dr/d\lambda \neq 0$$ for any positive $$r$$, then the photon will continue traveling to $$r=0$$. If there is a root, then the sign of $$dr/d\lambda$$ changes and the photon is moving away from the BH. I suppose we need to compare with $$r_s=2M$$ to see that if the root is physical.

You will need to consider all roots on principle to determine if there is a physical root corresponding to deflection. You should find that for $$b^2=27M^2$$ there are two degenerate real roots, so $$dr/d\lambda$$ will not change sign.

 

[Tangent87]

Thank you all, although I quite like latentcorpse's approach as we then don't need to solve a cubic.