# Signs in the Field-Theoretic Euler-Lagrange Equation

1. Mar 8, 2014

### Xezlec

So I have this book that considers the problem of a flexible vibrating string, taking $\phi(x,t)$ as the string's displacement from equilibrium. It then writes a Lagrangian density in terms of this $\phi$, takes $\delta \mathcal{S} = 0$, and eventually concludes that $\frac{\partial}{\partial t}(\frac{\partial \mathcal{L}}{\partial \dot{\phi}}) + \frac{\partial}{\partial x}(\frac{\partial \mathcal{L}}{\partial \phi'}) = 0$. Notice that the time-varying and space-varying terms have the same sign.

Two pages later, it considers a scalar field $\phi(x^0,\mathbf{x})$ with a Lagrangian density $\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)$, and concludes that $\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)})=0$. Now, unless I am having some massive brain fart on how covariant and contravariant work, the time-varying and space-varying terms have opposite signs. Right?

What gives? Why are the signs different between these two situations?

2. Mar 8, 2014

### vanhees71

In
$$\partial_{\mu} \frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}$$
the temporal and spatial components have the same sign. How do you come to the conclusion that might be not so?

The same is true for the four-dimensional divergence.
$$\partial_{\mu} A^{\mu}=\frac{\partial A^{\mu}}{\partial x^{\mu}}=\frac{\partial A^0}{\partial t} + \vec{\nabla} \cdot \vec{A},$$
where I've used natural units, $c=1$, and $x^0=t$.

3. Mar 8, 2014

### Xezlec

Thanks. It's clearly been too long since I've done anything with covariant and contravariant vectors. I need to go back and refresh before jumping back into this stuff.