# Why does $\frac{d\tau}{d\sigma} = L$?

I am given a (3+1)-dimensional spacetime that has the line element

\begin{equation} ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 + r^2 d\phi^2 \end{equation}

Part (a) of the problem asked me to find the explicit the Lagrangian for the variational principal for geodesics in this spacetime in these coordinates. I did that and got

\begin{equation} L\left(\frac{dx^\alpha}{d\sigma},x^\alpha\right) = \sqrt{ -\left(1-\frac{2M}{r}\right)\left(\frac{dt}{d\sigma}\right)^2 + \left(1-\frac{2M}{r}\right)^{-1} \left(\frac{dr}{d\sigma}\right) ^2 + r^2 \left(\frac{d\phi}{d\sigma}\right)^2 } \end{equation}

Here's where I got stuck

Part (b) of the problem says using the results of (a) write out the components of the geodesic equation by computing them from the Lagrangian.

I was told also to use the following relation:

\begin{equation} \frac{d\tau}{d\sigma}=L \end{equation}

I don't understand where $\tau$ is coming from or why $L = \frac{d\tau}{d\sigma}$. Didn't I just show $L$ was equal to the thing I found in Part (a)?

So can someone

(1) explain how $\tau$ entered the problem

(2) explain why $L = \frac{d\tau}{d\sigma}$

• Firstly in the signature that you are using $L = \sqrt{ - g_{\mu\nu}(x) {\dot x}^\mu {\dot x}^\nu }$. Also $d\tau = - g_{\mu\nu}(x) {\dot x}^\mu {\dot x}^\nu$. – Prahar Jan 23 '15 at 23:31
• So why doesn't $L = \sqrt{d\tau}$? – Stan Shunpike Jan 23 '15 at 23:51
• @Prahar well, $\mathrm{d}\tau = -g_{\mu\nu}(x)\mathrm{d}x^{\mu}\,\mathrm{d}x^{\nu}$, you jumped a step (parametrizing $x^{\mu}=x^{\mu}(\sigma)$)... – Alex Nelson Jan 24 '15 at 0:51
• @AlexNelson - Yes. I guess I was to excited to get the first comment :P (oh, internet!) – Prahar Jan 24 '15 at 1:58
• @StanShunpike: I just noticed you have a sign error in your Lagrangian. See Prahar's first comment, he caught it a lot earlier than I did. – Ryan Unger Jan 24 '15 at 1:59

The Lagrangian you wrote is $$L=\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\sigma}\frac{dx^\nu}{d\sigma}}$$ I'm sure you also know that $$d\tau^2=-g_{\mu\nu}dx^\mu dx^\nu$$ Plugging this into the first equation, we get $$L=\sqrt{\frac{d\tau^2}{d\sigma^2}}=\frac{d\tau}{d\sigma}$$ I'm not sure how much more help you want. Do you want to know why you have been told to use this equation?

I've been working through the topology problems in Wald today, so it's nice to do some calculus again :)

You are solving for the worldline of the particle, which is governed by the geodesic equation $$\frac{d^2x^\mu}{d\sigma^2}+\Gamma^\mu_{\;\,\rho\sigma}\frac{dx^\rho}{d\sigma}\frac{dx^\sigma}{d\sigma}=0$$ These are obviously hard differential equations to solve. However, there is a great way to make life easier. (I hope you don't know what I'm talking about, because this will be lengthy.)

Suppose we use the Lagrangian in the first equation to calculate the equations of motion. We end up with the geodesic equation with all the Christoffel symbols written out. However, $$L=\frac{d\tau}{d\sigma}$$ tells us something interesting. If we write the Lagrangian in terms of the proper time instead of $\sigma$, we end up with another equation $$L=1$$ We can use this in place of the hardest equation in the geodesic equation.

Let me show you what I mean. Write $$A(r)=\frac{1}{B(r)}=1-\frac{2M}{r}$$ for notational simplicity. The path of the particle is parameterized by $\sigma$ and I will use the notation $\cdot=d/d\sigma$ for more simplicity. In this notation, we have $$L=\sqrt{A\dot t^2-B\dot r^2-r^2\dot\theta^2-r^2\sin^2\theta \dot\varphi^2}$$ A page or so of scribbles later, we have the equations of motion $$\frac{d}{d\sigma}(A\dot t)=0$$ $$\frac{d}{d\sigma}(B\dot r)+\frac{1}{2}A'\dot t^2-\frac{1}{2}B'\dot r^2-r\dot\theta^2-r\sin^2\theta\dot\varphi^2=0$$ $$\frac{d}{d\sigma}(r^2\dot\theta)-r^2\sin\theta\cos\theta\dot\varphi^2=0$$ $$\frac{d}{d\sigma}(r^2\sin^2\theta\dot\varphi)=0$$ The second equation is hideous. However, if we used $\tau$ as a parameter, then we would also have $L=1$ which is equivalent to $L^2=1$. So we change parameters to $\tau$. We get rid of the second equation. Thus if $\cdot=d/d\tau$ we now have $$\frac{d}{d\tau}(A\dot t)=0$$ $$\frac{d}{d\tau}(r^2\dot\theta)-r^2\sin\theta\cos\theta\dot\varphi^2=0$$ $$\frac{d}{d\tau}(r^2\sin^2\theta\dot\varphi)=0$$ $$A\dot t^2-B\dot r^2-r^2\dot\theta^2-r^2\sin^2\theta \dot\varphi^2=1$$ The last equation only has first derivatives and thus makes this all much easier to solve.

I could go on and solve these, but I think your exercise would lose value if I did so.

• Oh, that's so simple! Thx! Sure, that would be great. Please feel free to elaborate. – Stan Shunpike Jan 24 '15 at 1:25
• @StanShunpike: Done. Let me know if you want me to continue with the problem. – Ryan Unger Jan 24 '15 at 1:56
• Okay, I follow up to "a page or so of scribbles". What operations are you performing on the rewritten L? No need to write it out. – Stan Shunpike Jan 24 '15 at 2:17
• I plugged $L$ into the EL equations $\partial L/\partial x^\mu-d/d\sigma(\partial L/\partial \dot x^\mu)=0$. – Ryan Unger Jan 24 '15 at 2:29
• @ 0celo7 Although a small thing, that way of redefining the metric in terms of $A(r),B(r)$ etc is very clever. I intuitively thought there was something very clumsy about how I was writing it and that was exactly what I was looking for. Is that common practice to use such tricks? So many of these GR calculations are so ugly, one hopes to save space / paper wherever possible. – Stan Shunpike Jan 24 '15 at 7:40