# Green's function for the inhomogenous Klein-Gordon equation

I'm trying to solve the massive Klein-Gordon equation in good old Minkowski space-time: $$(\square + m^2) \phi = \rho(t,\mathbf{x})$$ where $\square = \partial_{\mu} \partial^{\mu} = \partial_{t}^2 - \nabla^2$. So one can use a Green's function approach to find the fundamental solution of the form $$(\square + m^2) \mathscr{G}_{m} = \delta(x^{\mu} - x'^{\mu})$$ where $\mathscr{G}_{m}$ is the familar Klein-Gordon propagator. One then obtains the solution $\phi$ in position space, given as the familiar solution $$\phi(x^{\mu}) = \int d^{4} x' \mathscr{G}_{m}(x^{\mu},x'^{\mu}) \rho(t',\mathbf{x}') \,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ I was perfectly happy with this until I needed to to implement an actual $\rho$ and perform the integrals. My best bet so far has been to use the Bessel function representation I found (here: http://functions.wolfram.com/Bessel-TypeFunctions/BesselJ/31/02/) (which I've assumed has a generalisation) of the form: $$\mathscr{G}_{m}(t,t',\mathbf{x},\mathbf{x}') = \frac {\theta(t-t')} {2 \pi} \delta\Big( (t-t')^2 - |\mathbf{x} - \mathbf{x'}|^2 \Big) - \frac {m} {2 \pi} \theta(t-t' - |\mathbf{x} - \mathbf{x}'|) \frac {J_{1}(m \sqrt{(t-t')^2 - |\mathbf{x} - \mathbf{x}'|^2)}} {m\sqrt{(t-t')^2 - |\mathbf{x} - \mathbf{x}'|^2)}}$$ while this is a nice closed form representation, I am still having real difficulty evaluating the integral $(\star)$. I've looked for quite some time in various places for explicit examples of computing the integral, and have so far come up with very little. Mathematica (my computational program of choice) really disdains these Heaviside functions in the integrals, and offers little guidance. The only case I can do so far is $m \mapsto 0$.

Question: Using the representation of $\mathscr{G}_{m}$ given (or another nicer one), how can one actually go about calculating $(\star)$? Has anyone got a reference wherein some explicit example is calculated where $\rho$ goes beyond a simple $\delta$-function? Even something like $\rho = \rho(r,\theta)$ or $\rho = \rho(r)$ would be of great help.

Thanks!

• Please post the specific $\rho(x,t)$, you have to treat case by case. – JamalS Jun 1 '14 at 7:40
• I'm looking for a generic strategy here, which may be wishful thinking I suppose. The specifics associated with $\rho$ are not all that important (since I have many cases I would look to play around with eventually). I want to know if the representation I have there is nice for computations and how one would go about them. Ideally, I was hoping someone might know a reference wherein some explicit examples are computed. Even something like $\rho = a r^2 + b r + c$ would be helpful, since I can look at this and hope to generalise. As it stands I really don't know how to approach it. – Arthur Suvorov Jun 1 '14 at 8:53
• If the Heaviside functions are the problem, can you try to solve the Problem without them and implement a distinction of cases by hand? The Heavisides just seem to implement causality... – Neuneck Jun 1 '14 at 9:01
• Thanks, @Neuneck, this is definitely worth a shot. With regards to the integrations though, I'm struggling to even evaluate $$\int d^{4} x' \frac {J_{1}(m \sqrt{(t-t')^2 - |\mathbf{x} - \mathbf{x}'|^2}} {m \sqrt{(t-t')^{2} - |\mathbf{x} - \mathbf{x}'|^2}}$$ (that is when $\rho$ is constant). I found this (fh-jena.de/~rsh/Forschung/Stoer/besint.pdf) table of integrals of Bessel functions but this one doesn't appear on there. Is there even a reason to suppose the integral exists in elementary form? I suppose not. – Arthur Suvorov Jun 2 '14 at 1:28
• Do you have a reference of how the green's function is derived ? I mean the formula from wolfram – an offer can't refuse Sep 7 '16 at 10:33

I'm aware this is an old question and may be considered somewhat obsolete, but let me answer it nevertheless - if only for the sake of completeness.

The position space representation of the Klein-Gordon Green function (propagator) clearly looks intimidating. The trick is to do the calculation in momentum space, where the propagator is just a rational function. Before actually doing this let me point out that in the massless case, $m = 0$, and for a static source, $\rho = \rho (\mathbf{x})$, one is actually solving the Poisson equation. If the source is radially symmetric, $\rho = \rho(r)$, as suggested in the question, the solution is the Coulomb potential, $\phi = \phi(r) \sim 1/r$. Allowing for non-vanishing mass, one obtains a Yukawa potential, $\phi \sim \exp(-mr)/r$.

This can be shown explicitly in terms of Fourier integrals. First, transform the field,

$$\phi(k) = \int d^4 x \, e^{i k\cdot x} \, \phi (x) \; ,$$

and similarly the the source, $\rho \to \rho(k)$. The momentum space solution of the KG equation is then

$$\phi(k) = \frac{\rho(k)}{k^2 - m^2 + i \epsilon} \; ,$$

with $k^2 = k_0^2 - \mathbf{k}^2$ and the causal $i\epsilon$-prescription. (Change appropriately for retarded/advanced solutions.) Then transform back to position space,

$$\phi (x) = \int \frac{d^4 k}{(2\pi)^4} \, e^{-i k\cdot x} \, \frac{\rho(k)}{k^2 - m^2 + i \epsilon} \; . \quad (**)$$

As an example, take a Gaussian source, $\rho(r) = \rho_0 \exp(-\alpha r^2)$, with 'normalisation' $\rho_0 = (\alpha/\pi)^{3/2}$. Its Fourier transfrom is $\rho(k) = 2\pi i \, \delta(k^0) \exp (-k^2/4\alpha)$. The $k^0$-integral in $(**)$ is hence trivial, and the $d^3 k$ integration can be done with the usual residue technique writing $\mathbf{k}^2 + m^2 \equiv (\kappa+im)(\kappa-im)$. The result is

$$\phi(r) = e^{m^2/4\alpha} \, \frac{e^{-mr}}{4\pi r} \; .$$

In the point source limit ($\alpha \to \infty$) we reobtain the standard Yukawa potential.

For time dependent sources there will be energy transfer (no $\delta(k^0)$), and the integral $(**)$ will normally be harder. Typically, one can do the $k^0$-integration via residues and the remaining one(s) using stationary phase as e.g. in Ch. 2.1 of Peskin and Schroeder.

• Thanks for the taking the time to write this up. I have seen these approaches before, but was indeed worried that these would not work with time and angle dependent terms. I'll grab Peskin and Schroeder from the library and have a go; thank you for the chapter reference. Surely though the solutions in Fourier and position space must be equivalent under a transformation, so does this then imply that we have some closed form representations for the Bessel integrals? Interesting! May be an interesting research avenue in its own right =) – Arthur Suvorov Jun 2 '15 at 23:19