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Hunterc2429

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- 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!

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: