# First integral of Lagrangian

• Identity

#### Identity

I am trying to find the equations of motion for a test particle in the schwarzschild metric. However, I cannot find the correct first integral for the Lagrangian.

The Schwarzschild metric is:
$$ds^2 = -\left(1-\frac{2M}{r}\right)\,dt^2+\left(1-\frac{2M}{r}\right)^{-1}\,dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2$$
If we look at the action, and parameterise it using $s=\frac{\tau}{m}$,
$$S=\int\,ds = \int_{\tau_i}^{\tau_f} \sqrt{g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}m^2}\left(\frac{1}{m}\,d\tau\right)$$
If we DEFINE our lagrangian to be $L=\frac{1}{2}m^2g_{ab}\frac{dx^a}{d\lambda}\frac{dx^b}{d\lambda}$, then
$$S = \int_{\tau_i}^{\tau_f} \frac{\sqrt{2L}}{m}\,d\tau$$
Due to reparameterisation invariance, we can set once again set $\tau=s$, and in that case we get $\frac{\sqrt{2L}}{m}=1\Rightarrow L = \frac{m^2}{2}$.

However, in the book I'm reading, "Black Holes, White Dwarfs, and Neutron stars (Shapiro)", they have $L=-\frac{m^2}{2}$. This is correct, since you need it to find the equation of motion for the test particle. However, I can't seem to get the minus sign. How does it come about?

Thanks

With the signature you're using, gμνdxμ/dτ dxν/dτ is negative along a timelike curve, so you want to integrate ds = √-gμνdxμ/dτ dxν/dτ dτ

Thanks Bill_K, but I'm not exactly sure I understand.

By signature I'm guessing you mean
$$ds^2=-\left(1-\frac{2M}{r}\right)\,dt^2+\left(1-\frac{2M}{r}\right)^{-1}\,dr^2+r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2$$
$$d\tau^2=\left(1-\frac{2M}{r}\right)\,dt^2-\left(1-\frac{2M}{r}\right)^{-1}\,dr^2-r^2\,d\theta^2-r^2\sin^2\theta\,d\phi^2$$

But when we take the action, why do we need to integrate over a timelike curve? What's wrong with integrating over a spacelike curve? After all, isn't the definition

$$S = \int\,ds$$

Particles don't move along space-like curves. They must move on time-like curves.

ah of course. cheers