This sounds like a science history question, but is not. It is about acceptable forms for the Lagrangian density of electromagnetism. There is also a second question on the distinction between total and partial derivatives in the Lagrangian equation.

I am presently studying Julian Schwinger's paper Quantum Electrodynamics. I. A Covariant Formulation [Phys. Rev. 74 (1948) 1439]. In his expression (1.9) for the Lagrangian density, Schwinger gives the expression

$$ {\cal L}_{\text{field}}^{\text{Schwinger}}=-\frac{1}{2}\left(\partial_{\nu}A_{\mu}\right)\left(\partial_{\nu}A_{\mu}\right) $$

"for the term that refers to the electromagnetic field alone". Schwinger does not distinguish between covariant and contravariant 4-vectors; rather, he inserts an imaginary unit in the time-like part, which corresponds to the implicit use of the "East Coast metric" with $g^{\mu\nu}=\mbox{diag}(-1,+1,+1,+1)$. He furthermore works with Heaviside-Lorentz (HL) units. In this unit system the conventional form of the free-field Lagrangian density is

$$ {\cal L}_{\text{field}}^{\text{HL}}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$

These two forms give different contributions to the Euler-Lagrange equations

$$ \frac{\partial{\cal L}}{\partial A_{\beta}}-\partial_{\alpha}\left(\frac{\partial{\cal L}}{\partial\left(\partial_{\alpha}A_{\beta}\right)}\right)=0\quad (1) $$

The Schwinger form gives

$$ \partial_{\alpha}\left(\frac{\partial{\cal L}_{\text{field}}^{\text{Schwinger}}}{\partial\left(\partial_{\alpha}A_{\beta}\right)}\right)=-\partial_{\alpha}\partial_{\alpha}A_{\beta}, $$

whereas the conventional form gives

$$ \partial_{\alpha}\left(\frac{\partial{\cal L}_{\text{field}}^{\text{HL}}}{\partial\left(\partial_{\alpha}A_{\beta}\right)}\right)=-\partial_{\alpha}F^{\alpha\beta}=-\partial_{\alpha}\left(\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha}\right) $$

In the case of Lorentz gauge ($\partial_{\alpha}A^{\alpha}=0$), the two expressions coincide.

Furthermore, the electromagnetic tensor

$$ F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu} $$

is invariant under a gauge transformation

$$ A_{\mu}\left(x\right)\rightarrow A_{\mu}\left(x\right)-\partial_{\mu}\Lambda\left(x\right) $$

and so this is also the case for the conventional form ${\cal L}_{\text{field}}^{\text{HL}}$ of the Lagrangian density. However, it is not the case for the Schwinger form

$$ {\cal L}_{\text{field}}^{\text{Schwinger}}\rightarrow{\cal L}_{\text{field}}^{\text{Schwinger}}+\left(\partial_{\nu}\partial_{\mu}\Lambda\right)\left(\partial_{\nu}A_{\mu}\right)-\frac{1}{2}\left(\partial_{\nu}\partial_{\mu}\Lambda\right)\left(\partial_{\nu}\partial_{\mu}\Lambda\right) $$

Schwinger rewrites the extra terms as

$$ -\partial_{\mu}\left[\left(A_{\nu}+\frac{1}{2}\partial_{\nu}\Lambda\right)\left(\partial_{\mu}\partial_{\nu}\Lambda\right)\right] +\left(A_{\nu}+\frac{1}{2}\partial_{\nu}\Lambda\right)\left(\partial_{\nu}\partial^2_{\mu}\Lambda\right)\quad(2) $$

and writes: ".. the first term has no effect on the equations of motion. Hence gauge invariance is restricted to the group of generating functions that obey"

$$ \partial^2_{\nu}\Lambda=0. $$

This is incidentally the condition on gauge functions $\Lambda$ to stay within Lorentz gauge.

Let me now come to my questions:

  1. Is the Schwinger form an acceptable alternative form of the Lagrangian density for the free electromagnetic field ? Or, why did Schwinger choose this term ?
  2. Schwinger claims that the first term of $(2)$ does not contribute to the equations of motion. We know that for a point particle adding the total time derivative of a function $f(\mathbf{r},t)$ to the Lagrangian does not change the equation of motion

$$ \boldsymbol{\nabla}L-\frac{d}{dt}\left(\frac{\partial L}{\partial\mathbf{v}}\right)=0. $$

This is because the function argument $\mathbf{r}$ refers to particle trajectory and is thereby time-dependent such that we get

$$ \frac{d}{dt}f(\mathbf{r},t)=\frac{\partial}{\partial t}f(\mathbf{r},t) + \mathbf{v}\cdot \boldsymbol{\nabla}f(\mathbf{r},t);\quad \mathbf{v}=\frac{d\mathbf{r}}{dt} $$

Total derivatives do not seem to appear in the Euler-Lagrange equation for continuous systems $(1)$, or ? In the second edition of Classical Mechanics by H. Goldstein (12-23) we find the alternative form

$$ \frac{d}{dx_{\nu}}\left(\frac{\partial\cal{L}}{\partial\eta_{\rho,\nu}}\right)-\frac{\partial\cal{L}}{\partial\eta_{\rho}}=0;\quad \eta_{\rho,\nu}\equiv\frac{d\eta_{\rho}}{dx_{\nu}}\quad(3) $$

So, which form of the Euler-Lagrange equations -- (1) or (3) -- is correct ? Alternatively, why does the first term of (2) not contribute to the equations of motion, as claimed by Schwinger ?

I would be particularly grateful if answers also provide appropriate references.


1 Answer 1

  1. This choice of the Lagragian is to make possible the quantization of the EM field. If you use the conventional writing, you cannot quantize the field as the field conjugate to $A^0$ is $F^{00}=0$. It is shortly introduced in Quantum Field Theory of Mandl & Shaw, p75.

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