Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Total derivative in action of the field theory

  1. Nov 13, 2014 #1
    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. jcsd
  3. Nov 14, 2014 #2


    User Avatar
    Gold Member

    Let's set [itex] S^\mu=\frac{\partial L}{\partial(\partial_\mu \phi)}\delta \phi [/itex]. As you can see, it has a free index, so its a four-vector.
    Now if we differentiate it, we'll have [itex] \partial_\nu S^\mu [/itex](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
    \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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Total derivative in action of the field theory
  1. Total derivative (Replies: 6)

  2. Derivatives of fields (Replies: 1)