Geodesics in Schwarzschild: Reparametrizing the Equations

  • #1
Hunterc2429
3
0
TL;DR Summary
Hi all, I am currently working through Sean Carroll's "An Introduction to General Relativity," in the process, I am trying to numerically integrate and plot test particles using the geodesic equations provided. However, I want to parameterize them such that the azimuthal angle is the independent variable, but I am unsure of my results.
Hi all,

I am working through Sean Carroll's Textbook, particularly Chapter 5 regarding the Schwarzschild Solution. In this chapter, Energy and Angular Momentum are defined as follows:

$$
\begin{align}
E &= (1-\frac{2GM}{r})\frac{dt}{d\lambda} \Rightarrow \frac{dt}{d\lambda} = (1- \frac{2GM}{r})^{-1}E \\
L &= r^2 \frac{d\phi}{d\lambda} \Rightarrow \frac{d\phi}{d\lambda} = \frac{L}{r^2}
\end{align}
$$

Upon substitution into the four-norm ##\epsilon = g_{\mu \nu}\frac{dx^{\mu}}{d\lambda} \frac{dx^{\nu}}{d\lambda}## one can derive that,

$$(\frac{dr}{d\lambda})^2 = E^2 - (1- \frac{2GM}{r})(\frac{L^2}{r^2} + \epsilon)$$

Using the chain rule,

$$
\Rightarrow \frac{d}{d\lambda}(\frac{dr}{d\lambda})^2 = \frac{d}{d\lambda}(E^2 - (1-\frac{2M}{r})(\frac{L^2}{r^2}+\epsilon) ) \\
\Rightarrow 2\frac{dr}{d\lambda}\frac{d^2r}{d\lambda} = -\frac{2L^2}{r^3}\frac{dr}{d\lambda}+\frac{6ML^2}{r^4}\frac{dr}{d\lambda} + \frac{2M\epsilon}{r^2}\frac{dr}{d\lambda}
$$

And so, we have the following system of first-order geodesic equations,

1. ##\frac{dt}{d\lambda} = (1-\frac{2M}{r})^{-1}E##
2. ##\frac{d\phi}{d\lambda} = \frac{L}{r^2}##
3. ##\frac{dr}{d\lambda}=u##
4. ##\frac{du}{d\lambda} = -\frac{L^2}{r^3}+\frac{3ML^2}{r^4}+\frac{M\epsilon}{r^2}##

Then, by the chain rule, I can divide each of these equations by (2.) to find,

1. ##\frac{dt}{d\phi} = \frac{dt}{d\lambda}\frac{d\lambda}{d\phi}= (1-\frac{2M}{r})^{-1}E \cdot \frac{r^2}{L}##
2. ##\frac{dr}{d\phi}=\frac{dr}{d\lambda}\frac{d\lambda}{d\phi} = u \cdot \frac{r^2}{L} ##
3. ##\frac{du}{d\phi} = \frac{du}{d\lambda}\frac{d\lambda}{d\phi} = (-\frac{L^2}{r^3}+\frac{3ML^2}{r^4}+\frac{M\epsilon}{r^2}) \cdot \frac{r^2}{L}##

Now, what I am having trouble with is,

1. Is setting epsilon=0 for null trajectories enough of a modification to the system above to describe null trajectories?
2. How can I find an expression for ##\frac{d\tau}{d\phi}##? It is my understanding that if a particle is timelike then ##(\frac{d\tau}{d\lambda})^2 = -g_{\mu \nu}\frac{dx^{\nu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}=1## so then ##\frac{d\tau}{d\lambda}=1## and ##\frac{d\tau}{d\phi} = \frac{r^2}{L}##

So then, if timelike, ## \epsilon =1 ##

1. ##\frac{dt}{d\phi} = \frac{dt}{d\lambda}\frac{d\lambda}{d\phi}= (1-\frac{2M}{r})^{-1}E \cdot \frac{r^2}{L}##
2. ##\frac{dr}{d\phi}=\frac{dr}{d\lambda}\frac{d\lambda}{d\phi} = u \cdot \frac{r^2}{L} ##
3. ##\frac{du}{d\phi} = \frac{du}{d\lambda}\frac{d\lambda}{d\phi} = (-\frac{L^2}{r^3}+\frac{3ML^2}{r^4}+\frac{M}{r^2}) \cdot \frac{r^2}{L}##
4. ##\frac{d\tau}{d\phi} = \frac{r^2}{L}##

So then, if lightlike, ## \epsilon=0 ##

1. ##\frac{dt}{d\phi} = \frac{dt}{d\lambda}\frac{d\lambda}{d\phi}= (1-\frac{2M}{r})^{-1}E \cdot \frac{r^2}{L}##
2. ##\frac{dr}{d\phi}=\frac{dr}{d\lambda}\frac{d\lambda}{d\phi} = u \cdot \frac{r^2}{L} ##
3. ##\frac{du}{d\phi} = \frac{du}{d\lambda}\frac{d\lambda}{d\phi} = (-\frac{L^2}{r^3}+\frac{3ML^2}{r^4}) \cdot \frac{r^2}{L}##


Any advice would be greatly appreciated!
 
Last edited:
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  • #2
Haven't checked the maths on detail but the steps seems correct to me.

Note that ##d/d\phi## is ill-defined for radial trajectories, so they cannot be modelled in this approach.
 
  • #3
@Ibix Is this because motion is purely in the r-direction for a radial trajectory with no angular displacement? so then ## \frac{d\phi}{d\lambda}=0 ## and we are effectively dividing by zero? How else might I write my system of equations, then? In the timelike case, use affine parameter ## \tau ##; in the lightlike case, use ## \lambda ##? Or would it be better to simply use ## \lambda ## as my affine parameter? I am trying to write code for numerical integration I could simply define the system as the original lambda system before dividing by ## \frac{d\phi}{d\lambda} ## and just now have another equation ##\frac{d\tau}{d\lambda} = \pm \sqrt{E^2 - (1- \frac{2GM}{r})(\frac{L^2}{r^2}+\epsilon)}##

Python:
def geodesic_system(lam, y, GM, c, L, E, epsilon=1):
    tau, t, p, r, phi = y
    
    dt_dlambda = E / (1 - 2 * GM / r)
    dr_dlambda = p
    dp_dlambda = -L**2 / r**3 + 3 * GM * L**2 / r**4 + GM * epsilon / r**2
    dphi_dlambda = L / r**2
    
    
    if epsilon == 1:
        argument = (1 - 2 * GM / r) * dt_dlambda**2 + dr_dlambda**2 / (1 - 2 * GM / r) + r**2 * dphi_dlambda**2
        
        if argument < 0:
            argument = 0  
        dtau_dlambda = np.sqrt(argument)
    else:
        dtau_dlambda = 0  # Proper time does not evolve for null geodesics
    
    return [dtau_dlambda, dt_dlambda, dp_dlambda, dr_dlambda, dphi_dlambda]
 
Last edited:
  • #4
Hunterc2429 said:
Is this because motion is purely in the r-direction for a radial trajectory with no angular displacement?
Yes. You can't parameterise a path by something that doesn't change.
Hunterc2429 said:
How else might I write my system of equations, then?
I'd just use ##\lambda##, and just decide at the output stage if I want to print ##\tau## instead because it's a timelike path.
Hunterc2429 said:
I am trying to write code for numerical integration
It's generally best to use the proper time or the affine parameter because you know it always changes along the path. You could use the Schwarzschild time coordinate if you really wanted, but that rules out analysing some spatial geodesics (those that travel along constant time surfaces), and rules out horizon crossings (but you've done that anyway by picking Schwarzschild coordinates).
 
  • Like
Likes PeterDonis and Hunterc2429
  • #5
@Ibix, Last question for you: how can I get an expression for ##\frac{d\tau}{d\lambda}## that evolves concerning the trajectory? Because of how I have defined it, I am sure that ##\frac{d\tau}{d\lambda} = 1## or 0 for null geodesics. So, then, proper time evolves linearly with lambda. So I have defined #\frac{d\tau}{d\lambda} # in the following way, where ##p = \frac{dr}{d\lambda}##,

$$
(\frac{d\tau}{d\lambda})^2 = \frac{E^2}{(1- \frac{2GM}{r})} - \frac{p^2}{(1-\frac{2GM}{r})}- \frac{L^2}{r^2}
$$
and
$$
\frac{d\tau}{d\lambda} = \sqrt{ \frac{E^2}{(1- \frac{2GM}{r})} - \frac{p^2}{(1-\frac{2GM}{r})}- \frac{L^2}{r^2}}
$$

So that I can show how proper time evolves as the trajectory nears the mass M, also I am curious about trajectories where proper time evolves with the radius so I have chosen the positive root.

[Mentors' note: edited to fix a Latex formatting error]
 
Last edited by a moderator:
  • #6
Generally, Carroll uses ##\lambda## to mean any affine parameter. Picking ##\epsilon=1## fixes the affine parameter to be the proper time, ##\tau##, so ##\tau=\lambda## for timelike paths.
 

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