# Homework Help: Lagrangian density

1. Oct 5, 2005

### Spinny

Here's the problem. For a neutral vector field $$V_{\mu}$$ we have the Lagrangian density

$$\mathcal{L} = -\frac{1}{2}(\partial_{\mu}V_{\nu})(\partial^{\mu}V^{\nu})+\frac{1}{2}(\partial_{\mu}V^{\mu})(\partial_{\nu}V^{\nu})+\frac{1}{2}m^2V_{\mu}V^{\mu}$$

We are then going to use the Euler-Lagrange equations to show that (for $$m\neq 0$$)

$$\partial_{\mu}\partial^{\mu}V^{\nu}+m^2V^{\nu} = 0 \quad;\quad \partial^{\mu}V_{\mu} = 0$$

Now, the Euler-Lagrange equation (as I found in the textbook) is

$$\frac{\partial \mathcal{L}}{\partial \varphi}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\varphi)} = 0$$

My problem here, with what at first glance would appear to be a rather simple problem, is that I'm confused by all the indices! I haven't a lot of experience working like this, in fact this is all new to me, so I don't quite know how or where to start.

I can, however, give a specific example of what I don't understand, just to get things started.

The first part seemed at first easy enough, as the Lagrangian only contains one part with $$V_{\mu}$$ which is

$$\frac{1}{2}m^2V_{\mu}V^{\mu}$$

and from what I've understood $$V_{mu}V^{\mu}$$ is just the square of each of the components of the vector, so that when you derivate it with respect to $$V_{\mu}$$, I thought you'd get something like

$$m^2V^{\mu}$$

but the text says $$m^2 V^{\nu}$$ which brings up two question, first of all, why is it $$\nu$$ and not $$\mu$$, and why is it an upper index, rather than a lower one?

Furthermore I was wondering if anyone could recommend a good book introductory book about tensor algebra and such, preferably one intended for physicists rather than mathematicians. (It doesn't have to be a book on just about tensors, as long as it contains a good introduction to tensors.)

2. Oct 5, 2005

### dextercioby

It matters whether you differentiate the scalar $V_{\mu}V^{\mu}$ wrt the covector $V_{\nu}$ or wrt the vector $V^{\nu}$. That's why the indices must be treated with great care.

Daniel.

3. May 16, 2010

### TerryW

Hi Spinny,

I worked my way through Schaum's Tensor Calculus which I found very good except for quite a few typos.

Regards

TerryW