1. Jan 15, 2010

lync4495

Hello,

I'm beginning to learn GR and ran across the following in the beginning of Stephani's book.

In deriving the equations of motion for a force-free particle, we have the Lagrangian:
$$L = \frac{m}{2} g_{\alpha \beta} \dot{x}^\alpha \dot{x}^\beta$$

To construct the equations of motion we need
$$\frac{\partial L}{\partial \dot{x}^\nu} = m g_{\alpha \nu} \dot{x}^\alpha$$

and

$$\frac{\partial L}{\partial x^\nu} = L_{,\nu} = \frac{m}{2} g_{\alpha \beta, \nu} \dot{x}^\alpha \dot{x}^\beta$$

Then from the Lagrange equations we get

$$g_{\alpha \nu }\ddot{x}^\alpha + g_{\alpha \nu, \beta} \dot{x}^\alpha \dot{x}^\beta - \frac{1}{2}g_{\alpha \beta, \nu}\dot{x}^\alpha \dot{x}^\beta =0$$

I don't understand where the second term in the last equation comes from.

2. Jan 15, 2010

George Jones

Staff Emeritus
It comes from the product and chain rules:

$$\frac{d}{d \tau} \left( \frac{\partial L}{\partial \dot{x}^\nu} \right) = m \frac{d}{d \tau} \left( g_{\alpha \nu} \dot{x}^\alpha \right) = m \frac{d}{d \tau} \left( g_{\alpha \nu} \right) \dot{x}^\alpha + m g_{\alpha \nu} \frac{d}{d \tau} \left( \dot{x}^\alpha \right).$$

In the second-last term,

$$\frac{d}{d \tau} = \frac{d x^\beta}{d \tau}\frac{\partial}{\partial x^\beta}$$

is used.

3. Jan 15, 2010

lync4495

Oh ok. Thank you.

I wasn't realizing the metric tensor was a function of time. Now that I think about it, it makes sense that it is.