I can see that $a_{0}$ is not an independent field and Gauss law is a constraint on the theory arising from field equations. But, I don't get the geometrical picture.

Let $A$ be the space of all field configurations and let $G$ be the group of gauge transformations. Then, the actual configuration space $C=A/G$ where we consider $A$ modulo the action of gauge transformations. So, first of all, why don't we do our variational calculus to obtain field equations on configuration space? It has always been taught to me that I should consider variation of "paths" in configuration space only while applying Hamilton's principle. Instead, we first obtain field equations regarding $A$ as our configuration space and then "project" them perpendicular to gauge orbits using Gauss law. Are these two methods equivalent?

Moreover, I don't understand how Gauss law really projects my fields orthogonal to gauge orbits. According to Manton's book (see the relevant pages scanned below) on topological solitons, he starts with a variation of the fields $\phi$ and $a$ in a small interval of time. Then, he claims that a "naive contribution to the kinetic energy" would be some expression involving these variations (see below) which is non-zero for infinitesimal gauge transformations also. But, when I look at the expression of kinetic energy (which is $T=\int(\frac{1}{2}e_{i}e_{i}+\frac{1}{2}\bar{D_{0}\phi}D_{0}\phi)d^{d}x$) in terms of fields, then I find that the expression is gauge invariant. I don't see why the particular expression

$\frac{1}{2}\int\frac{1}{(\delta t)^{2}}(\delta\vec{a}.\delta\vec{a}+\bar{\delta\phi}\delta\phi)d^{d}x$

be the change in kinetic energy.

In short, I want to understand this geometric insight into Gauss law whose details are attached below.

Also, I wonder if this Gauss law has a relation to the Gauss law I learned while doing high school electrodynamics. Any information or resource on this would be very helpful.

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  • $\begingroup$ just curious, what textbook is this from? $\endgroup$
    – DJBunk
    Commented Sep 19, 2012 at 17:09
  • $\begingroup$ @DJBunk It is from "Topological Solitons" by Manton and Sutcliffe. $\endgroup$
    – user4235
    Commented Sep 21, 2012 at 15:51

1 Answer 1


There is a simple reason why we can consider variations on the whole $A$ rather than the quotient $C=A/G$ and the reason is following: all configurations that are $G$-equivalent have the same value of the action $S$. That's what we mean by the statement that the theory has the symmetry $G$. So the variation of the action $S$ in the directions that are equivalent to the action of a gauge transformation in $G$ vanish automatically, by the gauge invariance of the action! The variation of the action $\delta S$ is therefore a combination of the variations $\delta a_\mu$ only of those kinds that are independent of the directions along $G$.

That's also why the equations of motion that we derive from $\delta S=0$ don't determine the evolution of the fields such as $A_\mu$ unambiguously out of the initial conditions: the equations of motion only constrain $F_{\mu\nu}$ and they always allow us to change $A_\mu$ in the future, by a gauge transformation. This ambiguity arises from the "flat directions of the action".

You could be studying $\delta S =0$ on the quotient $C=A/G$ only but it would be cumbersome and $\delta S$ would be physically and literally the same thing as it is on $A$. It's the very point of introducing degrees of freedom which include one redundant one (because of the gauge symmetry) to simplify the picture. In fact, the equations of motion from $\delta S= 0$ on $A$ are manifestly Lorentz-covariant etc. If you were trying to parameterize the space $C=A/G$ by some fields, you would probably have to impose some Lorentz-breaking or otherwise unnatural conditions, e.g. $A_0=0$, and the whole formalism would lose the manifest Lorentz symmetry even though the actual phenomena, when looked at properly, would still obey the laws of relativity.

So yes, the methods on $A$ and on $C=A/G$ are equivalent, and it's the calculus on $A$ that is the smarter one. If the formalism using $C=A/G$ were more convenient from all points of view, we would never talk about the gauge symmetry because it would be a totally counterproductive concept! Be sure that it is a very useful concept.

I can't make sense out of the second part of the question. When we discuss infinitesimal variations of quantities such as energy, they should be linear combinations of the infinitesimal variations of the fields, like $dU=\vec E\cdot d\vec D$. But your expression is "doubly infinitesimal", it is bilinear in $\delta X$ where $X$ is something, and terms this small can be neglected in the usual infinitesimal calculus in which $\delta a$ is sent to zero because they're of higher order.

The energy (and stress-energy tensor) is gauge-invariant in electrodynamics, too, so its (linear, first-order) variation induced by gauge transformations is equal to zero. Talking about some second-order "variations" seems completely misguided to me.

Also, less seriously, I feel uneasy about your usage of the word "orthogonal". In general, one doesn't have an inner product (needed to determine orthogonality) on the full configuration space and it's really not needed for most questions of this sort. The directions in $A$ away from a slice that may be used as representatives of $C=A/G$ come in two types: those that are pure gauge transformations, in the direction of the $G$ "fibers", and those that aren't. Most of them "aren't" but any combination of those that are and those that aren't "isn't" again and no combination is really "fundamentally different" than others. So it is a bit meaningless to look for "orthogonal" directions to the directions along $G$.

Finally, you ask whether "a" Gauss law is related to the standard Gauss law taught at school but you haven't really explained what you mean by "a" Gauss law. There is only one Gauss' law. It's the equation of motion obtained by varying the action with respect to $A_0$, and it gives us something like ${\rm div}\,\vec D = \rho$. It's always the same law – which may be generalized to more complicated theories than electrodynamics, e.g. Yang-Mills theory. This equation ${\rm div}\,\vec D =\rho$ is interesting because it doesn't contain any time derivatives. So it really doesn't dictate the time evolution of anything: it already constrains the initial state. One may prove that if the equation holds at $t=0$, it will hold at any time: the time-derivative of the Gauss' law may be derived as a spatial derivative of other Maxwell's equations involving $\vec D$.

This non-dynamical = constraint character of the Gauss' law (the fact it doesn't contain time derivative) is related to the fact that this equation of motion is derived from the variation of a field that may be interpreted as a completely redundant one in a particular convention how to fix the gauge symmetry. That's why we can identify it with the statement that the states related by gauge transformations are treated as equivalent states by the theory. This is particularly clear in the quantum theory where we may a priori have states not annihilated by ${\rm div}\,\vec D - \rho$ but only states that are annihilated by this operator, i.e. states that respect the equivalence of the states related by gauge transformations, may be considered physical.

  • $\begingroup$ Thanks for your answer. The reason for doing calculus in $A$ seems quite clear now. The expression for kinetic energy contains $(\delta t)^{2}$ in the denominator which is also a second order term. So, the full expression for KE shouldn't be sent to zero in the usual infinitesimal calculus. I have edited my question to clarify what I am asking about the geometrical insight into Gauss law. In a sense, I want to understand what kind of orthogonality the author is referring to and how does his new fields ensure that. Moreover, can you please expand on the last paragraph of your answer? Thanks. $\endgroup$
    – user4235
    Commented Sep 19, 2012 at 9:31
  • $\begingroup$ You may be searching too much in this. He just assumes that there is an inner product on the phase space which allows you to determine the angles of short vectors with the same beginning. An orthogonal vector to g is surely linearly independent of g, a convenient basis. In reality and in general, there is no inner product on geometry, one can't measure the angles, and one doesn't really need the "orthogonal" condition, just "independent". $\endgroup$ Commented Sep 19, 2012 at 9:52
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    $\begingroup$ Dear Lakshya, $A_0=0$ isn't necessarily the most useful gauge choice in all situations but it is always legitimate, representing all possible equivalence classes at least once. It's not hard to see why: you may construct the gauge transformation that gets you from any configuration of the gauge field to one with $A_0=0$. Just find $\lambda$ such that $\partial/\partial t(\lambda)=A_0$, the initial one, with the right coefficients, and if you apply this gauge transformation, the new $A_0$ will be zero (cancellation). $\endgroup$ Commented Sep 19, 2012 at 12:20
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    $\begingroup$ One may always find $\lambda$ obeying this equation $\partial / \partial t(\lambda)=A_0$ for a given field $A_0$: the right $\lambda$ is just the integral of $A_0$ over $t$ from any chosen point $t_0$ to the point $t$ which is the coordinate of the point $(t,x,y,z)$ whose value of $\lambda$ you want to calculate. $\endgroup$ Commented Sep 19, 2012 at 12:21
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    $\begingroup$ It is equally easy in the Yang-Mills case. The transformation is $\ A'_\mu = G A_\mu G^{-1} - \frac{1}{g} (\partial_\mu G)G^{-1}$. Again, you will be able to find a $G$ that makes $A'_0=0$. All "lines" in spacetime with fixed $(x,y,z)$ are still independent of each other and on each of them, you're just "integrating" the $A_0$ matrix in a way to get an element of the gauge group. The required gauge transformation $G$ is the path-ordered exponential of the integral of the matrix $A_0$ along the path of increasing $t$, taken from any point $t_0$. $\endgroup$ Commented Sep 19, 2012 at 13:29

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