# Total derivative in action of the field theory

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1. Nov 13, 2014

### victorvmotti

When applying the least action I see that a term is considered total derivative.

Two points are not clear to me.

We say that first
$$\int \partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x= \int d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)= (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)$$

And then because the variation at the spatial infinity vanishes this terms is equal to zero.

I do not get the calculation from $$\partial_\mu (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi) d^4x=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$ to

$$d(\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)=\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial t}dt+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x}dx+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial y}dy+\frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial z}dz$$
$$\neq \frac {\partial (\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi)}{\partial x^\mu} dtdxdydz$$

Can you expand this to fill the gap for me.

Also, why we require "spatial infinity" here, isn't it also true that $$\delta \phi$$ in the $$\frac {\partial L}{\partial(\partial_\mu \phi)}\delta \phi$$ vanishes at any two endpoints of the path, why we require infinity here?

2. Nov 14, 2014

### ShayanJ

Let's set $S^\mu=\frac{\partial L}{\partial(\partial_\mu \phi)}\delta \phi$. As you can see, it has a free index, so its a four-vector.
Now if we differentiate it, we'll have $\partial_\nu S^\mu$(which is a 2nd rank mixed tensor). But what you want, is the contraction of this, which becomes:
$\partial_\mu S^\mu=\partial_0 S^0+\partial_1 S^1+\partial_2 S^2+\partial_3 S^3$
So:
$\partial_\mu S^\mu d^4 x=\partial_0 S^0 dt dx dy dz+\partial_1 S^1dt dx dy dz+\partial_2 S^2dt dx dy dz+\partial_3 S^3dt dx dy dz$
Given that you can do the integrations in any order, the result you wanted is concluded.

About your second question, its only that we take the spatial end points to be at infinity and the temporal ones, just two moments in time. But this isn't manifestly relativistic so I guess its better to integrate w.r.t. the proper time but I still don't know how to do that, I should work on it!