About the Lagrange equation involving Coulomb interaction

Question

In the book by Prof Nagaosa, the Lagrange density was given by Eq. (1)
    (1)
Then, according to the Euler-Lagrange equation (Eq.(2)), we can do the derivatives of Lagrange density (e.g., Eq. (1)) with respect to .
    (2)
By doing so, the Book by Prof Nagaosa indicates the following result:
    (3)
Why the factor $1/2$ in front of the integral (over r') disappear?

Paul Chern · Accepted Answer

This can be understood from the definition of functional variation, together with v(r-r')==v(r'-r) of Coulomb potential.
First of all, the functional derivative can be found here:
https://en.wikipedia.org/wiki/Functional_derivative
where you can find this:

Clearly, this gives the derivative of the Lagrange equation:

Next, let me define functional J as a function of function L:

If we follow the definition of the functional derivative
by taking 
we should have

(v(|r-r'|) is used; the derivative is exactly a mimic of fig.1 if you read the wiki carefully)
One interesting fact is that the 1/2 comes from the double-counting, and gets omitted by switching of r and r'.

About the Lagrange equation involving Coulomb interaction

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