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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 $-1/4$?

Frederic Brünner
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JamalS
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  • 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. – yo' Nov 21 '13 at 12:39
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    $I_{\mu\nu}I^{\mu\nu} = 4$ (in general, the dimension of space-time). – JamalS Nov 21 '13 at 12:41
  • 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). – yo' Nov 21 '13 at 12:48
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    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 – Christoph Nov 21 '13 at 12:53

3 Answers3

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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.

Qmechanic
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    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. – Zo the Relativist Nov 21 '13 at 18:39
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    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. – lionelbrits Nov 21 '13 at 18:59
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    can you explain why the positive definite kinetic term is needed for unitarity? – dan-ros Jun 09 '14 at 12:42
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    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. – Qmechanic Jun 09 '14 at 14:55
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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.

Frederic Brünner
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    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. – lionelbrits Nov 21 '13 at 18:55
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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)

Saksith Jaksri
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