# Is the Green function of electromagnetism a scalar or a tensor?

When I check classical electromagnetism books Maxwell equations

$$\begin{equation} \Box A^\nu (x)=\frac{4\pi}{c}j^\nu (x) \end{equation}$$ can be solved using a scalar Green function $$G(x,x')$$

$$\begin{equation} A^\nu (x)=\int G(x,x')j^\nu (x')d^4x' \end{equation}$$ where the Green function satisfies

$$\begin{equation} \Box G(x,x')=\frac{4\pi}{c}\delta^4(x-x') \end{equation}$$

Examples of this are Jackson, eq. 6.48 on sec. 6.5. Also, on "The classical theory of fields" by Landau, on eqs. 62.9 and 62.10 he uses the scalar green function as well.

This immediately feels strange, since the 4-potential $$A^\nu(x)$$ could, in theory, have different boundary conditions for each component and a scalar Green function simply doesn't have enough degrees of freedom to accommodate that. Evenmore, in the context of quantum field theory, the photon propagator (which is essentially the Green function) is a tensor $$\Pi_{\mu\nu}$$ so I'm confused about the nature of the Green function in classical electromagnetism: Is the scalar Green function $$G(x,x')$$ the most general Green function or in a general case we need a tensorial Green function $$G_{\mu\nu}(x,x')$$?

Note: This question is explicitly about classical electromagnetism, I'm using the quantum field theory propagator as an example to show my confusion but the question applies to the classical theory.

Here's the gist of it:

1. If your field lives in a vector space $$V$$, then the propagator is a map $$V\to V$$, i.e., it lives in $$V\otimes V^*$$. In more down-to-earth terms, if your field has a certain index $$i$$, its propagator has a pair of such indices: $$\psi^i\quad\Longrightarrow\quad G^i{}_j$$ The reason is that, by definition, $$G$$ measures the difference between $$\psi^i(t)$$ and $$\psi^i(t+\mathrm dt)$$, i.e., $$\psi^i(t+\mathrm dt)=G^i{}_j\psi^j(t)$$.

2. In electromagnetism, $$A$$ is a vector, so the propagator is a rank-2 tensor. Strictly speaking it has an upper vector index and a lower vector index, although the metric allows us to write them both as, say, lower indices. So $$G_{\mu\nu}$$.

3. Note that, by Poincaré invariance, the propagator is necessarily of the form $$G_{\mu\nu}(x,x')=G_{\mu\nu}(x-x')$$, with $$G_{\mu\nu}(x)=x_\mu x_\nu f(x^2)+\eta_{\mu\nu}g(x^2)$$ for some functions $$f,g$$. The reason is that the vector indices in $$G_{\mu\nu}$$ must be provided by the tensors in the theory, in this case the only two of which are $$x^\mu$$ and the metric. If there is any other tensorial object, such as a background field or a non-trivial boundary condition, then those contribute to $$G_{\mu\nu}$$ too, and the structure above is to be updated to take that into account.

4. Finally, the current is conserved, which means that the $$f$$-term is irrelevant. (It represents the pure gauge part, and is not fixed by the equations of motion). In other words, in momentum space $$G\sim p_\mu p_\nu f(p^2)+\eta_{\mu\nu}g(p^2)$$ and $$p_\mu j^\mu=0$$. Therefore, you can drop the $$f$$-term, and you are essentially left with a single scalar function $$g(x^2)$$, which can also be referred to as the propagator.

5. All in all, the propagated solution is $$A^\mu(x')=\int G^{\mu\nu}(x-x')j_\nu(x)\mathrm dx=\int g((x-x')^2)j^\mu(x)\mathrm dx$$ which takes the form of a scalar propagator, but only because the tensorial structure is just $$\eta^{\mu\nu}$$, which can be contracted with $$j$$ directly.

So yes: if you have Poincaré-breaking elements, such as fixed fields or boundary conditions which select a preferred direction or point in space, the propagator can no longer be taken to be a scalar, but must take the form of a rank-2 tensor field.

Note that the above discussion is actually quite generic. For example, in the case of a spinor field, the propagator has a pair of spinor indices, and the index structure (in the unbroken Poincaré situation) can be reduced to scalar functions as well. Indeed, one can write $$G^\alpha{}_\beta(x)=\delta^{\alpha}_\beta f(x^2)+(\gamma^\mu)^\alpha{}_\beta\, x_\mu\, g(x^2)+\cdots$$ where $$f,g,\dots$$ are scalar functions. Here "$$\cdots$$" denotes terms with more gamma matrices: $$\gamma^\mu\gamma^\nu$$, $$\gamma^\mu\gamma^\nu\gamma^\rho$$, etc., all the way up to the dimension of the spinor representation ($$\sim 2^d$$). Here, $$\alpha,\beta$$ are spinor indices, and these must be provided by the invariant tensors of the theory. A basis for these is given by the gamma matrices (the number of which depends on the number of spacetime dimensions; four in four dimensions). Note that higher-order terms are redundant, due to the Clifford relation $$\gamma_{(\mu}\gamma_{\nu)}=\eta_{\mu\nu}$$ (so that e.g., $$x^\mu x^\nu \gamma_\mu \gamma_\nu=x^2$$, etc.), so the expression above is actually complete (except in the parity-breaking case, where there is also a $$\gamma^5$$ term).

• Note: the propagator can (and usually does) also depend on $\Theta(x^0)$. This doesn't change the essence of the argument. Oct 27 '20 at 19:04
• Thanks! So basically the reason why some books use the scalar Green function is because they are solving the problem in free space where there's Poincaré symmetry. But if you are working in curved space or if there are boundary conditions or external fields, then you need to stick with the actual tensorial Green function. Is this correct? Oct 27 '20 at 19:07
• sorry then, even though your answer translates to what I say, it is kind of misleading.I mean the important thing that op should understand is, propagator of QED is gauge dependent, because its not observable, and thus does not have a unique universal structure. it has infinitely many forms, which all give the same physics and certain books, papers choose certain forms suited best to their purpose. I mean what you say is correct but, lacks emphasis of gauge dependence which is the key and I think the trouble OP has to grasp. Oct 27 '20 at 21:12
• I mean if i did not know the gauge fixing etc. I would really misunderstand your comment especially point 4. because it almost sounds like only correct gauge is feynman gauge. my criticism is not on physics but on pedagogy. Oct 27 '20 at 21:16
• I mean you can see the misunderstanding of OP in his responses, he said " So basically the reason why some books use the scalar Green function is because they are solving the problem in free space where there's Poincaré symmetry." so he thinks the system they work on makes the greens function scalar. But it is misconception, the correct answer is even though every book on free maxwell theory work on exactly same system some use landau gauge so have tensor GF some use Feynman gauge and have scalar GF. Oct 27 '20 at 21:22

The Electromagnetic free-space green function has two indices $$G_{\mu\nu}(x,{x}')$$. It is a bilocal tensor. When one writes $$A_\mu({\bf x}) = \int G_{\mu\nu}({x},{x}')J_\nu({x'}) \sqrt{g} d^4x,$$ one sees that the $$\mu$$ index in $$G_{\mu\nu}({x},{ x}')$$ transforms as a covariant tensor at the point $$x$$, while the $$\nu$$ index transforms as a covariant tensor at the point $$x'$$.

In flat space one can write a causal Green function as a Fourier transform. $$G_{\mu\nu}({x},{x}')= \int \frac{d^4 k}{(2\pi)^4} \frac 1{{k^2} -(\omega+i\epsilon)^2} \left(\delta_{\mu\nu}- \frac{k_\mu k_\nu}{k^2}\right)$$ Without the $$\left(\delta_{\mu\nu}- \frac{k_\mu k_\nu}{k^2}\right)$$ this would be the scalar Green function.

• Thanks for the answer! My question then is: Why do most books use a scalar Green function? Is it that in flat space the bilocal tensor reduces to a scalar? Oct 27 '20 at 18:31
• No. I think you are misreading the books. Can you give a citation to one that uses a scalar? I'll add some extra to my answer that might help. Oct 27 '20 at 18:32
• I'll edit my question to add citations! thanks Oct 27 '20 at 18:38

Great question. In full generality, the EM Green's function is indeed a tensor $$G^\mu_{\ \ \nu}(x; x')$$ whose Lorentz indices are not necessarily proportional to the 4x4 identity operator. AFT's answer gives the most elegant and general explanation of why we can get away with only considering a scalar function when the boundary conditions are Lorentz-invariant, but a quicker and dirtier explanation comes directly from the first couple sentences in Jackson section 12.11:

The general covariant formulation of Maxwell's equations in terms of potentials is given by $$\square A^\beta - \partial^\beta \partial_\alpha A^\alpha = \frac{4 \pi}{c} J^\beta,$$

or equivalently

$$\left( \square \delta^\beta_\alpha - \partial^\beta \partial_\alpha \right) A^\alpha = \frac{4 \pi}{c} J^\beta.$$

In a generic gauge, the Green's function is intrinsically a Lorentz tensor because of the mixed partial derivative second term. But if we choose to work in Lorenz gauge $$\partial_\alpha A^\alpha = 0$$, then the equation of motion simplifies to D'Alembert's wave equation: $$\square A^\beta = \frac{4 \pi}{c} J^\beta,$$ which is just a set of four uncoupled PDEs of the same form, so in this gauge the Green's function is essentially a scalar (times the identity matrix on the Lorentz indices). So the fact that we can use the scalar version is a nice property of Lorenz gauge but does not hold in a generic gauge.

Index $$\nu$$ has nothing to do with the equation itself. The equation does not know whether $$A$$ and $$j$$ are scalars, 4-vectors on spinors. The equation itself is scalar type, so its Green function can be scalar only.

Another argument: let say $$A^\nu$$ is a solution for $$j^\nu$$. Then $$C^\mu_\nu A^\nu$$ must be solution for $$C^\mu_\nu j^\nu$$. If $$G^\mu_\nu$$ is the general green function then we have (I skip integrals to shorten the notations and imply same indices summation) $$a)\quad A^\mu = G^\mu_\nu j^\nu \qquad b)\quad C^\mu_\lambda A^\lambda = G^\mu_\lambda C^\lambda_\nu j^\nu$$ what immediately gives us for any matrix $$C$$ $$C^\mu_\lambda G^\lambda_\nu = G^\mu_\lambda C^\lambda_\nu$$ i.e. $$G^\mu_\nu \propto \delta^\mu_\nu$$.

Its a tensor $$G_{\mu\nu}(x-y)=\langle \Omega|TA_\mu(x)A_\nu(y)|\Omega\rangle=\int \frac{d^4k}{(2\pi)^4}\frac{ie^{-ik(x-y)}}{k^2+i0_+}\bigg(g^{\mu\nu}+\xi\frac{k^\mu k^\nu}{k^2+i0_+}\bigg)$$ here $$\xi$$ is some parameter that depends on your gauge choice, and $$g^{\mu\nu}$$ is metric of space-time. What does it mean is following.

$$S=-\frac{1}{4}\int d^4xF_{\mu\nu}F^{\mu\nu}=-\frac{1}{2}\int d^4x\int d^4yA^\mu(x)G^{-1}_{\mu\nu}(x-y)A^\nu(y)$$ so this clearly shows the definition of green's function.

It is a scalar or a second rank unit tensor, which amounts to the same. The boundary condition argument does not hold because physically a boundary condition can only be realised by additional charge-current distributions, that is, the boundaries are sources of field themselves.

• It is quite boring to downvote without an argument. It would be more interesting if you could prove Lorentz, Landau, Jackson and Jefimenko wrong. Oct 27 '20 at 23:31