I'm reading a paper and the author uses some "Lagrangian techniques" to go from

ds^2 = -(1-\frac{2M}{r})dt^2 + (1-\frac{2M}{r})^{-1} dr^2 + r^2(d\theta^2 + sin^2\theta d\phi^2)

to

2L = -(1-\frac{2M}{r})\dot{t^2} + (1-\frac{2M}{r})^{-1} \dot{r^2} + r^2(\dot{\theta^2} + sin^2\theta \dot{\phi^2}), where L is the Lagrangian.

 

Can someone explain what the step between these two equations?

 

PS: I don't know how to make the latex code work here.

I'm reading a paper and the author uses some "Lagrangian techniques" to go from

ds^2 = -(1-\frac{2M}{r})dt^2 + (1-\frac{2M}{r})^{-1} dr^2 + r^2(d\theta^2 + sin^2\theta d\phi^2)

to

2L = -(1-\frac{2M}{r})\dot{t^2} + (1-\frac{2M}{r})^{-1} \dot{r^2} + r^2(\dot{\theta^2} + sin^2\theta \dot{\phi^2}), where L is the Lagrangian.

 

Can someone explain what the step between these two equations?

 

PS: I don't know how to make the latex code work here.

We have three ways. You can use the [ math ] tags or the [ latex ] tags. You can also input LaTeX code into a feature in the text editor.

 

[math]ds^2 = -(1-\frac{2M}{r})dt^2 + (1-\frac{2M}{r})^{-1} dr^2 + r^2(d\theta^2 + sin^2\theta d\phi^2)[/math]

 

[math]2L = -(1-\frac{2M}{r})\dot{t^2} + (1-\frac{2M}{r})^{-1} \dot{r^2} + r^2(\dot{\theta^2} + sin^2\theta \dot{\phi^2})[/math]

Since we know that the equation of motion for a test particle is the geodesic equation, the action should yield the geodesic equation. We also know geodesics correspond to paths of shortest length, as can easily be demonstrated by measuring distances on, for example, the surface of a sphere. The path along a great-circle of the sphere is the shortest path connecting two points. Therefore when you minimize the action, it should correspond to the path of shortest length. This implies that the action should be:

 

[math]S=\int ds=\int \sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}} d\lambda[/math]

 

So the Lagrangian is just: [math]L=\sqrt{g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}}[/math].

 

Similarly, you can also show that the Lagrangian [math]L=g_{\mu \nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda}[/math] also yields the geodesic equation when plugged into the Euler-Lagrange equations. This is the Lagrangian that your author used.