George Jones
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nocks said:That would be
\frac{d^2r}{d\lambda^2} = - \frac{L^2(3M-r)}{r^4}
Which gives me the closest radius for a stable orbit of a photon as 3M
Reduce this second-order equation to two first-order equations by setting p = dr/d\lambda, so that dp/dt = d^2r/d\lambda^2. The set of equations that describes the worldline of a photon then is
<br /> \begin{equation*}<br /> \begin{split}<br /> \frac{d \phi}{d \lambda} &= \frac{L}{r^2} \\<br /> \frac{dr}{d\lambda} &= p \\<br /> \frac{dp}{d \lambda} &= \frac{L^2(r - 3M)}{r^4}.\\<br /> \end{split}<br /> \end{equation*}<br />
Assuming that all required values are given, write a few lines of (pseudo)code that uses the simplest, most intuitive method (Euler's method) to solve these equations.