I have the equation relating the impact parameter $b$ to the distance of closest approach $R$.

$R^3 - b^2R + 1 = 0$

which can be solved in python. I have a given $b$ and have to find $R$. however, in some cases of $b$ there are up to 3 solutions which are real and positive. How to I choose the true answer?

eg for $b = 5$ i have solutions $[-5.01988126, 4.9798787 , 0.04000256]$

NOTE this is in units of c = r_schwarzchild = 1

  • $\begingroup$ The product of all three solutions has to be $-1$, so how can you have more than two positive roots? $\endgroup$ – Peter Shor Sep 28 '19 at 20:46
  • $\begingroup$ youre right. with the 2 positive roots, is the largest one the one i am looking for? $\endgroup$ – user11339690 Sep 28 '19 at 23:04

Peter Schor's comment that one of the three roots of the cubic must be negative is correct. In general, you have to select the root that is physically relevent. You can reject the negative root, and for the Schwarzschild geometry, any root smaller than the photon sphere radius at $1.5 r_s$.

Also, I believe that you have an error in your equation for $R$, since as written, it does not have the correct solution at the critical impact parameter, that allows the closest approach where it is still possible for a photon to escape.

I assume that you derived the equation by solving for $\frac{dr}{dt}=0$ at closest approach, and using normalized radius and impact parameter as follows, where $r_s$ is the Schwarzschild radius.

$$R = \frac{r}{r_s}$$ $$B = \frac{b}{r_s}$$

I believe that the equation for $R$ should be as follows.

$$R^3 - B^2(R-1) = 0$$

Solving this equation for the critical impact parameter, $B=3 \sqrt{3}/2$, results in a double root at $R = 3/2$, and a negative root at $R = -3$. Solving the equation at a slightly larger impact parameter, say $B=3$, gives roots at about 2.227, 1.185, and -3.411. Clearly only the larger radius can be a solution. The solution at $B=5$ is about 4.394.

  • $\begingroup$ yes!!!!! turns out my only mistake was expanding out an equation, silly. You are brilliant thank you so much!!!!!! $\endgroup$ – user11339690 Sep 29 '19 at 7:50

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