# Light ray paths near schwarzschild blackhole

by nocks
Tags: blackhole, light, paths, schwarzschild
 Sci Advisor P: 2,193 W(r) should be the effective potential of the system.
 P: 24 Could anyone expand on this please? I would appreciate the help. I have the the effective potential in the schwarzschild metric as (L being angular momentum) $$V_{eff} = ( 1 - \frac{r_{s}}{r})(mc^{2} + \frac{L^{2}}{mr^{2}})$$ Would this be enough information to solve for r and $$\Phi$$ so that I could plot the trajectories. Also could I use the same equation for plotting the course of an observer descending into the black hole?
 Emeritus Sci Advisor PF Gold P: 5,598 Here is some python code I wrote that does pretty much what you're talking about. It's meant to be short and easy to understand, so it uses a pretty crude method of doing the numerical integration. If you want to do accurate numerical calculations of geodesics, you'd want to substitute a better integration method. There are various general-purpose subroutines out there, e.g., in the book Numerical Recipes in C. What my code does is to calculate the deflection of a light ray that grazes the sun. Actually, it calculates half he deflection for a ray that grazes the sun, with the mass of the sun scaled up by a factor of 1000 in order to keep the result from being overwhelmed by rounding errors in my el-cheapo integration method. import math # constants, in SI units: G = 6.67e-11 # gravitational constant c = 3.00e8 # speed of light m_kg = 1.99e30 # mass of sun r_m = 6.96e8 # radius of sun # From now on, all calculations are in units of the # radius of the sun. # mass of sun, in units of the radius of the sun: m_sun = (G/c**2)*(m_kg/r_m) m = 1000.*m_sun # Start at point of closest approach. # initial position: t=0 r=1 # closest approach, grazing the sun's surface phi=-math.pi/2 # initial derivatives of coordinates w.r.t. lambda vr = 0 vt = 1 vphi = math.sqrt((1.-2.*m/r)/r**2)*vt # gives ds=0, lightlike l = 0 # affine parameter lambda l_max = 20000. epsilon = 1e-6 # controls how fast lambda varies while l
 P: 24 I actually have the numerical recipes book next to me although I may avoid solving the elliptic integral I mentioned earlier, and just use the approximation for light deflection, i.e. 4GM/bc$$^{2}$$, to get the einstein ring effect, and focus on the trajectory of the observer descending into the black hole.
Emeritus
PF Gold
P: 5,598
 Quote by nocks I actually have the numerical recipes book next to me although I may avoid solving the elliptic integral I mentioned earlier, and just use the approximation for light deflection, i.e. 4GM/bc$$^{2}$$, to get the einstein ring effect, and focus on the trajectory of the observer descending into the black hole.
Keep in mind that 4GM/bc2 is only a weak-field approximation. It won't give you the right answer if you're close to the black hole.
Mentor
P: 6,248
 Quote by George Jones Yes, exactly. The geodesic equation when $\theta = \pi /2$ then is $$\begin{equation*} \begin{split} \frac{d \phi}{d \lambda} &= \frac{L}{r^2} \\ \left( \frac{dr}{d \lambda} \right)^2 &= E^2 - L^2 W \left( r \left( \lambda \right) \right), \\ \end{split} \end{equation*}$$ where $W \left( r \right)$ is function that I'll specify later, and $E$ and $L$ are constants of motion. In its present form, the second equation is a little difficult to implement on a computer since sometimes a positive square root is needed (increasing $r$) and sometimes a negative square root is needed (decreasing $r$) in the same photon orbit. To get aorund this, differentiate the second equation with respect to the affine parameter $\lambda$ taking into account that the $r$ in $W \left( r \right)$ is itself a function of $\lambda$. What do you get for the second equation after this differentiation?
 Quote by nocks Given it's been a while since i've done any differentiation, is it simply: $$\frac{dr}{d \lambda} \right) &= \frac{ E - L}{W \left( r \right)}$$
No, this isn't quite right. Let's start with the left side. What is

$$\frac{d}{d\lambda} \left[ \left( \frac{dr}{d \lambda} \right)^2 \right] ?$$
P: 24
 Quote by George Jones ...
Given

$$\left( \frac{dr}{d \lambda} \right)^2 &= E^2 - V^2(r) \right)$$
and
$$V^2(r) = \left(1 - \frac{2M}{r} \right)\frac{L^2}{r^2}$$

I have $$\frac{d^2r}{d\lambda^2} = -\frac{1}{2}\frac{d}{dr}V^2(r)$$
P: 24
 Quote by nocks $$V^2(r) = \left(1 - \frac{2M}{r} \right)\frac{L^2}{r^2}$$
Guess I should state that

$$\frac{d}{dr}V^2(r) = \frac{2L^2M}{r^4} - \frac{2L^2(\left 1 - \frac{2M}{r} \right)}{r^3}$$
Mentor
P: 6,248
After simplifying the right side of
 Quote by nocks $$\frac{d}{dr}V^2(r) = \frac{2L^2M}{r^4} - \frac{2L^2(\left 1 - \frac{2M}{r} \right)}{r^3}$$
and plugging back into
 Quote by nocks $$\frac{d^2r}{d\lambda^2} = -\frac{1}{2}\frac{d}{dr}V^2(r)$$
what do you get?
 P: 24 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
P: 24
 Quote by nocks 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
Apologies for the lack of basic physics understanding but for a photon on approach what would it's angular momentum be?
 Mentor P: 6,248 We now have a system of three first-order coupled differential equations that determine the worldlines for "photons". In order to solve these equations, we need three initial conditions. Imagine firing a photon from a laser. One possibility for the three initial conditions: the $\left( r , \phi \right)$ position from which the photon is fired and the direction in which the photon is fired. These initial conditions determine $L$. Alternatively, $L$ could be used as one of the initial conditions. Much more on all of this later.
Mentor
P: 6,248
 Quote by nocks 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

$$\begin{equation*} \begin{split} \frac{d \phi}{d \lambda} &= \frac{L}{r^2} \\ \frac{dr}{d\lambda} &= p \\ \frac{dp}{d \lambda} &= \frac{L^2(r - 3M)}{r^4}.\\ \end{split} \end{equation*}$$

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.
 P: 24 Having a bit of trouble with this but I have p as p = $$\frac{L^2(2M-r)}{2r^3}$$
P: 24
wow it's been a while since I looked at this project.
Thought i'd come back to it :)

 Quote by George Jones 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 $$\begin{equation*} \begin{split} \frac{d \phi}{d \lambda} &= \frac{L}{r^2} \\ \frac{dr}{d\lambda} &= p \\ \frac{dp}{d \lambda} &= \frac{L^2(r - 3M)}{r^4}.\\ \end{split} \end{equation*}$$ 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.

A few questions. Since I lack a good maths background, I would appreciate some advice on how I would implement these 3 equations into a numerical solver like Euler's or Runge-Kutta to get back some results that I could plot.
e.g. Do I pre-define L?, how do I know which direction the photon is travelling?

currently trying to use this : http://www.ee.ucl.ac.uk/~mflanaga/java/RungeKutta.html

 Related Discussions Astronomy & Astrophysics 3 Astronomy & Astrophysics 5 Astronomy & Astrophysics 7 Astronomy & Astrophysics 15