16
$\begingroup$

I have seen numerous 'derivations' of the Maxwell Lagrangian,

$$\mathcal{L} ~=~ -\frac{1}{4}F_{\mu \nu}F^{\mu \nu},$$

but every one has sneakily inserted a factor of $-1/4$ without explaining why. The Euler-Lagrange equations are the same no matter what constant we put in front of the contraction of the field strength tensors, so why the factor of $-1/4$?

$\endgroup$
  • $\begingroup$ Not being an expert, but don't you have that $I_{\mu\nu}I^{\mu\nu}=-4$ in Minkowski space? Then $-1/4$ would be just a natural re-normalization. $\endgroup$ – yo' Nov 21 '13 at 12:39
  • 2
    $\begingroup$ $I_{\mu\nu}I^{\mu\nu} = 4$ (in general, the dimension of space-time). $\endgroup$ – JamalS Nov 21 '13 at 12:41
  • $\begingroup$ Damn, of course $(-2)^2=4$. Still, IMHO it explains the factor $1/4$, and the minus might have some physical explanation. After all, the classical Lagrangian is $L=T-U$, and $U$ is the potential (with minus sign). $\endgroup$ – yo' Nov 21 '13 at 12:48
  • 3
    $\begingroup$ if you omit the $\frac 14$, you'd have to put in an interaction term $4j^\mu A_\mu$; also, I think it can be motivated in the language of differential forms where (hopefully ;)) $F=\sum_{\mu\lt\nu}F_{\mu\nu}dx^\mu\wedge dx^\nu=\frac 12\sum_{\mu,\nu}F_{\mu\nu}dx^\mu\otimes dx^\nu$; the negative sign probably makes sense once you add additional terms to your Lagrangian $\endgroup$ – Christoph Nov 21 '13 at 12:53
24
$\begingroup$

Comments to the question:

  1. First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like.

  2. As Frederic Brünner mentions a normalization of the $J^{\mu}A_{\mu}$ source term with a normalization constant $\pm N$ goes hand in hand with a $-\frac{N}{4}$ normalization of the $F_{\mu\nu}F^{\mu\nu}$ term. Here the signature of the Minkowski metric is $(\mp,\pm,\pm,\pm)$.

  3. Recall that the fundamental variables of the Lagrangian formulation are the $4$-gauge potential $A_{\mu}$. Here $A_{0}$ is a non-dynamical Lagrange multiplier. The dynamical variables of the theory are $A_1$, $A_2$, and $A_3$. The $$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\underbrace{\frac{1}{2} \sum_{i=1}^3\dot{A}_i\dot{A}_i}_{\text{kinetic term}}+\ldots$$ is just the standard $+\frac{1}{2}$ normalization of a kinetic term in field theory. In particular note that the kinetic term is positive definite in order not to break unitarity.

$\endgroup$
  • 1
    $\begingroup$ Note that this would be equivalent to the Lagrangian $F_{ab}F^{ab} + 4 j_{a}A^{a}$. But since we normally get that second term by promoting a derivative to a gauge-covariant derivative, this would mean scaling our charged matter lagrangians by a factor of four, which is a bit goofy when we can instead just scale the maxwell term. $\endgroup$ – Jerry Schirmer Nov 21 '13 at 18:39
  • 6
    $\begingroup$ I think number 3 here is very much the heart of the matter. Of course, it is a matter of scaling fields, since sometimes people will use $-\frac{1}{4g^2}F_{\mu\nu}F^{\mu\nu}$. Also, the factor $\frac{1}{4}$ becomes $\frac{1}{2}$ in non-Abelian theories, due to the normalization of the generators. $\endgroup$ – lionelbrits Nov 21 '13 at 18:59
  • $\begingroup$ can you explain why the positive definite kinetic term is needed for unitarity? $\endgroup$ – tonydo Jun 9 '14 at 12:42
  • $\begingroup$ Comments: 1. The statement should first of all be seen in the light of various traditional sign conventions. 2. One may speculate that the framework of quantum theory, probability & unitarity are more general/fundamental than the Lagrangian and Hamiltonian formalism and the notion of kinetic energy. $\endgroup$ – Qmechanic Jun 9 '14 at 14:55
15
$\begingroup$

The factor is there so that once you add a source term, i.e. $J^\mu A_\mu, $ you get the correct equations of motion, namely Maxwell's equations:

$\partial_\nu F^{\mu\nu}=J^\mu.$

Furthermore, this convention produces the usual $1/2$ in front of the kinetic term of the gauge fields.

$\endgroup$
  • 1
    $\begingroup$ You could absorb the factor into the source term as well, or to the definition of current. Qmechanic's answer is indeed the correct answer. $\endgroup$ – lionelbrits Nov 21 '13 at 18:55
3
$\begingroup$

The interesting point is that with factor -1/4 $$ \mathcal L =-\frac 1 4 F_{\mu\nu}F^{\mu\nu}= \frac 1 2 (\mathbf E^2- \mathbf B^2)\;, $$ From this we roughly have $$\mathcal H =\frac 1 2 (\mathbf E^2+ \mathbf B^2)\;, $$ which consistence with the energy of the EM field$^{[1]}$ $$U=\frac 1 2 \int \big( \epsilon_0 \mathbf E^2 + \frac 1 {\mu_0}\mathbf B^2 \big) d\tau \;.$$

References

[1] D.J. Griffiths, Introduction to electrodynamics Fourth Edition (See back materials)

$\endgroup$
  • $\begingroup$ This is pretty much QMechanic's point $\endgroup$ – JamalS Mar 10 '17 at 11:58
  • $\begingroup$ I think no. My point is on the energy of EM field not about to fix a kinetic part of Lagrangian. $\endgroup$ – Saksith Jaksri Mar 10 '17 at 12:09
  • 1
    $\begingroup$ The normalisation of the kinetic term relates to the normalization of that Hamiltonian. $\endgroup$ – JamalS Mar 10 '17 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.