# Why does field Lagrangian depend on four-derivative?

1. Jan 27, 2015

Hi guys, so this is a pretty generic question.

Starting off with the classical Lagrangian in a case where there is no interaction or explicit time dependence, the functional form is

$L=L(x,\dot{x})=L(x,\partial_{t}x)$.

Now when we look at the Lagrangian density in field theory, the functional dependence is

$\mathcal{L}=\mathcal{L}(\varphi,\partial_{\mu}\varphi)$.

And there's my question. Why does the Lagrangian density depend on $\partial_{\mu}\varphi$ and not only $\partial_{t}\varphi$? i mean why the four-derivative?

I have a few ideas but I'm not sure if they are correct. I'm thinking along the lines that since $\partial_{\mu} = \nabla + \partial_{t}$, maybe the $\partial_{\mu}$ just appears to merge the two together and make it more compact as this one term includes possible interactions and also the field velocity?

2. Jan 27, 2015

### HomogenousCow

That's not true.
The field is assumed to be lorentz invariant, hence derivative terms must be built out of scalars involving the whole four-gradient. (e.g. box operator).

3. Jan 27, 2015