# Why can we neglect surface terms in Field Theory?

My teacher says that the integral $$\int _{-\infty }^{\infty }\frac{\partial J^{\mu }}{\partial x^{\mu}}d^4x$$ that we met in QFT can always be neglected since $$\int _{-\infty }^{\infty }\frac{\partial J^{\mu }}{\partial x^{\mu} }d^4x=\int_{-\infty }^{\infty } J^{\mu } n_{\mu } \, ds=0.$$ The second term in the above equation equals zero since $J^{\mu }$ goes to zero at the infinity. I don't quite understand this.

Suppose $J^{\mu }$ is some quantity in space-time, the following figure illustrates the case:

In the picture, I drew a time sequence of the quantity $J^{\mu }$, and the shaded volume should be the domain of the integral. The volume expand to infinity in this integral case.

When we want the integral $\int _{-\infty }^{\infty }\frac{\partial J^i}{\partial x^{i}}d^3x$ over the space volume, the result is also zero due to the reason that $$\int _{-\infty }^{\infty }\frac{\partial J^i}{\partial i}d^3x=\int_{-\infty }^{\infty } J^i n_i \, ds=0.$$

My questions are as follows:

1. I don't see the reason why $J^{\mu }$ is zero at infinity. I only find that $\int _{-\infty }^{\infty }\frac{\partial J^i}{\partial x^{i}}d^3x$ and $\int _{-\infty }^{\infty }\frac{\partial J^{\mu }}{\partial x^{\mu} }d^4x$ go to zero because the amount of the quantities that enter the region are the same as the ones that are leaving the region, but not because the quantity itself goes to zero.

2. We can use the packet idea to construct quantities that are zero at infinity for example:

But even if we do this, the reason that the integral goes to zero is still that in question 1, but not what my teacher says that it goes to zero. Zero seems to be a "superficial" reason.

1. If I adopt my teacher's view, I have an intuition that it may violate the rule that information cannot spread faster than the speed of light. Can the surface condition at faraway influence the phenomena at a local area instantly?
• physics.stackexchange.com/q/57313 Oct 8, 2016 at 7:00
• Your teacher is assuming that he is dealing with field configurations which at infinity tend to vacuum values sufficiently rapidly. This assumption is not completely trivial, but quite often this is done without even mentioning it. Oct 8, 2016 at 7:35

I would like to add that in some cases, even if a term is a total derivative -- so a surface term, by Stokes's theorem -- we can't neglect it in QFT! This will also expand a little on Valter Moretti's point that we need to use some sort of boundary conditions.

For this purpose, let us consider $$S_\theta = \int_M \operatorname{tr} F \wedge F.$$ Here $M$ is 4-dimensional spacetime, $F$ is the gauge field strength for a non-Abelian gauge theory, $F = dA + A \wedge A$, so its components are matrices, that's why we have the trace. You may be unfamiliar with this $\wedge$ and $d$. They're called the "wedge product" and "exterior derivative". In index notation, they correspond to taking the anti-symmetric product; and to taking the derivative, then the totally anti-symmetric part, so: $A \wedge A$ corresponds to $A_\mu A_\nu - A_\nu A_\mu$ and $dA$ to $\partial_\mu A_\nu - \partial_\nu A_\nu$. Note that $A \wedge A \neq 0$ because each component of $A$ is a matrix. You can do all of this with Levi-Civitas, e.g., $$S_\theta = \int d^4 x \, \operatorname{tr} \epsilon_{\mu\nu\rho\sigma} F^{\mu\nu} F^{\rho\sigma} \tag{1}$$ but that leads to writing more indices than I want to.

Anyway, it turns out that $$\operatorname{tr} F\wedge F = d \operatorname{tr} (dA\wedge A + \frac{2}{3} A\wedge A \wedge A)$$ (to show this: $d^2 = 0$, the product rule applies to $d$, and $\operatorname{tr} A^{\wedge 4} = 0$ because of the cyclic property of the trace.) So we have that by Stokes's theorem $$S_\theta = \int_{S^3} \operatorname {tr} (dA\wedge A + \frac{2}{3} A \wedge A \wedge A)$$ where $S^3$ is a 3-sphere at infinity, the "boundary of spacetime". As boundary condition, we should use that $F = 0$ at infinity. Then $$S_\theta = -\frac{1}{3} \int_{S^3} \operatorname A\wedge A \wedge A.$$ Now, if $F = 0$, it is possible to find a gauge transformation such that $A = 0$. So is $S_\theta = 0$? No!

But how? It's topological. $A$ defines a function $S^3 \to G$ where $G$ is the gauge group. Such a function belongs to $\pi_3$, which stands for something called the "third homotopy group", which is a generalization of the concept of winding number. You know how if draw a closed curve -- topologically a circle -- in the plane without crossing through the origin, the number of times it winds around the origin cannot be changed by bending stretching the curve smoothly? The third homotopy group is like that, except it's about how 3-spheres wrap around the target space. (I don't know how to visualize it either!)

If $G = SU(n)$ with $n\ge 2$, then $\pi_3$ is just like the winding number: it's an integer. What this means is that unless the "winding number" of $A$ is $0$, a gauge transformation can't make $A = 0$ everywhere on $S^3$.

So we have shown that $S_\theta \neq 0$ in general, even when we use the boundary condition $F =0$ at infinity. In fact $S_\theta = \theta n$ where $n$ is the "winding number" and $\theta$ is some constant. But why should we care? Because it is a surface term, and variations are taking with the boundary condition that the variation vanish on the boundary, $S_\theta$ makes no contribution to the Euler-Lagrange field equations. But this is quantum field theory, and I really want the path integral \begin{align} \mathcal Z &= \int D[A, \psi, \ldots]\, \exp(-i(S_0[A, \psi, \ldots] + S_\theta)) \\ & = \sum_n \int D[A, \psi, \ldots]\, \exp(-i S_0[A, \psi, \ldots]) e^{-in\theta} \end{align} where on the second line, each integral is only over $A$ such that the winding number is $n$. So $S_\theta$ determines interference terms between different paths. This means it can -- and does -- have physical effects in quantum field theory. For example, from (1) you can see that because there is a Levi-Civita, $S_\theta$ violates parity symmetry.

I think these topological arguments with winding numbers and other similar concepts are really cool. So does the Nobel committee, because this year's prize was about things like this. If you want to learn more, some keywords to look for are instantons, topological quantum field theory (TQFT), theta terms, strong $CP$ problem, axions...

Without some particular hypotheses on $J^\mu$ the statement is simply false. There are so many elementary examples...Take $$J^\mu = x^\mu$$ in Minkowskian coordinates in Minkowski spacetime, for instance.

It is sufficient that $\partial_\mu J^\mu \geq 0$ with $\partial_\mu J^\mu >0$ in a region with non-vanishing four-volume to make false the statement. I guess your teacher is thinking of some very specific cases.

Perhaps your teacher is referring to currents generated in some bounded region $B$ of the spacetime. In this case if the current is associated to some hyperbolic equation, as those of QFT, the support of $J$ is confined in the causal cone $J(B)$ emanated from $B$ in the past and in the future of $B$.

Choosing a sufficiently large cylinder parallel to the temporal axis of a Minkowskian reference frame and completely including this double cone, you may neglect the contribution of the surface integral over the lateral surface of the cylinder.

However I cannot see any compelling reason to prove that the integration over the (spacelike) bases $\Sigma_{\pm T}$ vanishes for $T \to +\infty$. There is no reason to think it. $J$ vanishes in the remote future/past of $B$ but we cannot say that it vanishes so rapidly to make vanishing the mentioned integrals over $\Sigma_{\pm T}$.

To corroborate this idea against the supposed proof of the statement, think of a conserved current, i.e., $\partial_\mu J^\mu =0$ associated to some causal quantum field. In this case $\int \partial_\mu J^\mu d^4x =0$ as wanted, but the reason is not the one you are adducing.

In this case each integral over the mentioned bases $\Sigma_{\pm T}$ of the cylinder does not vanish for $T \to +\infty$.

What it is true is that the sum of them is zero, because they are evaluated referring to opposite timelike normal unit vectors.

However, each integral separately equals (up to sign) the conserved charge $$Q = \int_{\Sigma_t} J^0 d^3x \neq 0$$ associated with the current.

As an example, consider $J_\mu = \phi \partial_\mu \psi - \psi \partial_\mu \phi$ with $\partial_\mu \partial^\mu \phi + f\phi = \partial_\mu \partial^\mu \psi + f \psi=0$ (for every fixed function $f$). It is easy to chose initial conditions of $\psi$ and $\phi$ over $\Sigma_0$ to obtain $Q \neq 0$. In this case, however, $\phi(t,x)$ and $\psi(t,x)$ vanish as $t\to +\infty$ for every fixed $x$.

My conclusion is that, even in QFT, the statement is generally false: It is better to assume $\partial_\mu J^\mu =0$ from scratch and these currents are not directly associated to (free) quantum fields.

• Sorry, I do not understand this paragraph:Perhaps your teacher is referring to currents generated in some bounded region BB of the spacetime. In this case if the current is associated to some hyperbolic equation, as those of QFT, the support of JJ is confined in the causal cone J(B)J(B) emanating from BB in the past and in the future of BB. Oct 8, 2016 at 8:15
• As my second graph showed, if the quantity is like that, is my understanding of the integral to be zero right? Would you please give me any references? Oct 8, 2016 at 8:18
• Think of two solutions $\psi$ and $\phi$ of Klein Gordon equation. Suppose that at time $t=0$ these functions and their temporal derivatives vanish outside a spatial ball $B$. The solutions in spacetime are therefore confined in the region of the events connecting the events in $B$ through timelike or lightlike segments. Oct 8, 2016 at 8:20
• If all things are happening in this region, and I see that the boundary integration is zero. But, how can the boundary influence a local scene. Which is equivalent to say that can we use the Gaussian Law regardless of the time sequence? I mean what if the information of the local change has not been transferred to the boundary, can we still use the Gaussian Law? Oct 8, 2016 at 8:37
• Sorry I do not understand you jargon (and perhaps you do not understand mine :)) Oct 8, 2016 at 8:48

It is common to denote spacetime integrals with $\int d^4 x$ and space integrals with $\int d^3 x$. Integrating by parts, $$\int d^4 x 1\partial_0 J^0=\int d^3 x \int dx^0 1\partial_0 J^0=\int d^3 x\left( \left[J^0\right]^{t=\infty}_{t=-\infty} -\int dx^0 J^0\partial_0 1\right).$$ Since $\partial_0 1 = 0$, we can simplify to $$\int d^4 x \partial_0 J^0 = \int d^3 x \left[J^0\right]^{t=\infty}_{t=-\infty}.$$ If $\lim_{t\to\infty}J^0=\lim_{t\to -\infty}J^0=0$ for all $x\in\mathbb{R}^3$, $\left[J^0\right]^{t=\infty}_{t=-\infty}=0$ so $\int d^4 x \partial_0 J^0 = 0$. We can repeat this argument or $\int d^4 x\partial_1 J^1$, provided this time we define $d^4 x = dx^1 d^3 x$ instead of $d^4 x = dx^0 d^3 x$.

Summing over all $\mu$, the proof that $\int d^4 x \partial_\mu J^\mu = 0$ requires only that $J^\mu\to 0$ "at infinity" in any "direction" in spacetime. For some solutions of some equations of motion, we can prove this explicitly (e.g. due to a Gaussian suppression). In general, the calculus of variations defines equivalence classes on integrals so that integrals are equivalent if they differ by a term of the form $\int d^4 x \partial_\mu J^\mu$. In particular, this is how functional derivatives $\frac{\delta S}{\delta\phi}$ are defined in the axiom $\frac{\delta S}{\delta\phi}=0$ for all dynamic fields $\phi$; $\delta S$ is equivalent in the above sense to $\int d^4 x \sum_\phi \delta\phi \frac{\delta S}{\delta\phi}$.

• If $\partial_\mu J^\mu =0$ you have simple cases where integrals at $t = \pm \infty$ do not vanish separately. Think of the current $J_\mu = \phi \partial_\mu \psi - \psi \partial_\mu \phi$ with $\partial_\mu \partial^\mu \phi= \partial_\mu \partial^\mu \psi=0$. Oct 8, 2016 at 8:10
• "If $\lim_{t\to\infty}J^0=\lim_{t\to -\infty}J^0=0$ for all $x\in\mathbb{R}^3$, $\left[J^0\right]^{t=\infty}_{t=-\infty}=0$ so $\int d^4 x \partial_0 J^0 = 0$" This is a generally dangerous because wrong statement. Oct 8, 2016 at 8:14
• When you are taking the limit of t, don't you need to worry about whether the information of the local event can reach there or not? Would you please explain it a little bit that can we use Gaussian Law without taking care of the causality? Oct 8, 2016 at 8:41
• You're both right that there are conditions in which $\int d^4 x\partial_\mu J^\mu \neq 0$, but the point is that the stationary action principle axiom is $\frac{\delta S}{\delta\phi}=0$, which implies $\delta S$ is a boundary term, "equivalent" to 0 in the sense I defined, though in general not actually equal to 0.
– J.G.
Oct 8, 2016 at 8:47
• OK, but the point of the OP was to prove that the boundary term is $0$ not that it is equivalent to $0$ Oct 8, 2016 at 8:50

The starting point for a field theory is typically an action principle. In order for a variational problem to be well-posed, it is important to impose pertinent boundary conditions (BCs), e.g., Dirichlet BCs. Such BCs may explain why boundary terms can be neglected in the divergence theorem, cf. OP's teacher.

However, there are many field theories where boundary terms are important, or even play a dominant role, such as, e.g., TFTs, or, say, the GHY boundary term in GR. In such situations, it would be inconsistent to throw away boundary terms.