# Conditional evaluation: black hole cross section

I'm trying to calculate the collision cross section of black holes, more precisely, I want the critical impact parameter $$b_{\text{crit}}$$ for a given initial velocity $$v$$ such that, at this speed, any smaller impact parameter would result in the particle crossing the event horizon. I have code which evaluates each step in the reasoning, but I'm having trouble getting it all to work together.

The idea is:

(1) from the impact parameter and initial speed, numerically integrate the equation of motion to get $$r(\phi)$$

(2) Get the distance of closest approach $$R(b,v)$$ as a function of the initial motion parameters.

(3) solve $$R(b,v)=1$$ varying $$b$$, the value of $$b$$ where this equation holds is $$b_{\text{crit}}$$.

The equation of motion is

$$u''+u=\frac{1}{2(\gamma bv)^2}+\frac{3}{2}u^2$$

We have here $$u=1/r$$. And I'm using units where $$c=1$$, $$2GM=1$$, $$r_{\text{Schwarzschild}}=1$$.

To do this I have the code

Func[b_, v_] :=
NDSolve[{D[u[p], {p, 2}] + u[p] == 1/(2* (b*v) ^ 2) + 3/2* u[p]^ 2,
u == 0, u' == 0}, {u[p]}, {p, 0, 28}];

U[p_, b_, v_] := u[p] /. First[Func[b, v]];


So 1/U(b,v) gives me the motion for a given $$b$$ and $$v$$. Then to find the minimum I have

Rmin[bInit_, vInit_] :=
MinValue[{(1/U[p, bInit, vInit]), 1 > p > 0}, p]


Wherein lies the problem: In some cases, the particle is actually captured by the black hole, so $$r$$ will tend to zero and bad things will happen near the singularity. And in some cases the particle gets captured for very large $$\phi$$ (p in the above code). So I need some condition to set the range of p to look at.

One way to solve this would be to evaluate

NSolve[1/U[p, 100, 0.01] == 1, p]


Where I've chosen some values for $$(b,v)$$. If NSolve finds a solution to this, then the particle gets captured, so the answer will be a larger b, for fixed v, if NSolve finds no solutions to this, then the particle won't get captured and $$b_{\text{crit}}$$ is smaller. Maybe some sort of bisection would work but I'm not sure how to do this in Mathematica.

The ultimate objective is to get a function which returns $$b_{\text{crit}}(v)$$

=======

EDIT: Here is a diagram to make things a bit clearer: If ever $$r$$ is less than one, then the particle gets captured. $$\Delta\phi$$ is the deflection from the straight line path that the particle would take if there was no central mass.

• The equation contains only the product b*v. Why investigate separately the effects of b and v? Mar 18 '19 at 21:55
• I'm looking for the collision cross section as seen by a particle which is at speed $v$ at infinity, i.e. $\pi b_{\text{crit}}^2$, this only depends on $b$. Mar 18 '19 at 22:10
• What is the condition of capture? Mar 18 '19 at 23:21
• @tbfr416 We need to check the equation. How is it received? Mar 19 '19 at 12:29
• This equation 9.33 is for particles that start moving from a state of rest when r->Infinity. Therefore, their energy $E=m_0c^2$. In the scattering problem should be $E>m_0c^2$. Mar 19 '19 at 14:02

I think it is a fundamental property of your differential equation that there is a bifurcation in behavior at bv=2. Greater than that, you have no capture. Less than that, you have capture.

In playing with the relationship, it doesn't appear that there is a value for the product b v such that for some p you have Min[1/u[p]]==1. Rewriting your code in terms of a single term bv...

 Func[bv_] := NDSolve[{u''[p] + u[p] == 1/(2 (bv)^2) + 3/2 u[p]^2, u == 0,
u' == 0}, {u[p]}, {p, 0, 28}];


Now look at the curves using Manipulate.

Manipulate[Plot[Evaluate[1/u[p] /. (Func[k] // First // First)], {p, 0, 28},
PlotRange -> {{0, 28}, {0, 5}}],
{k, 1, 10}]


Here's a plot of 1/u[p] with bv=2 And now with bv=1.9999999

Plot[Evaluate[1/u[p] /. (Func[1.999999] // First // First)], {p, 0, 28},
PlotRange -> {{0, 50}, {0, 5}}] The value of bv=2 seems to be special, as when set, we can directly solve the differential equation.

DSolve[{u''[p] + u[p] == 1/8 + 3/2 u[p]^2, u == 0, u' == 0}, u[p], p]

(* u[p] -> 1/2 Tanh[p/(2 Sqrt)]^2 *)


In playing around, no luck with other values.

• That's odd, I'm quite sure the equation is correct, I've put a reference/link in the question comments to the section of the GR textbook which derives it. Mar 19 '19 at 13:33