I want to solve the Klein-Gordon Green's function equation $$\left[\partial_\mu\partial^\mu + m^2\right]G(x, x') = \delta(x - x') $$ in 5 space-time dimensions where the boundary conditions on $G(x,x')$ are: \begin{align} G(x,x') & = 0\ \mathrm{for\ } x^0 \le x'^0 \\ G(x,x') & = 0\ \mathrm{for\ } \left(x^\mu-x'^ \mu\right)\left( x_\mu - x'_\mu\right) < 0. \end{align} The first is, essentially, that the field is in a quiescent state before the impulse is delivered at the event $x'$. The second is that the speed of light is respected by the wave front (i.e. causality).

I'm, specifically, interested in the real Minkowski space form of $G$. In principle this can be done with Fourier analysis, and I welcome any clear explanation of how to get from the Fourier components to the real space version that satisfies the required boundary conditions (n.b. exactly those needed to use $G$ in initial value problems). That said, in the rest of this question I lay out the process of finding the Green's function as a boundary value problem in real space, with the intention of focusing attention on where the process appears to break down.

In the three regions off of the light cones, the coordinates can be expressed by a coordinate transformation that uses a boost by rapidity $\phi$ on top of the standard spherical coordinates. Inside the forward-backward light cones the substitutions are: $$ \left.\begin{array}{rl} x^0 &= \tau \operatorname{sgn}(\phi)\,\cosh \phi \\ r & = \tau \operatorname{sgn}(\phi)\,\sinh \phi \end{array}\right\} \Leftrightarrow \left\{\begin{array}{rl} \tau &= \sqrt{(x^0)^2 - r^2} \\ \phi & = \tanh^{-1}\left(\frac{r}{x^0}\right) \end{array}\right. . $$ Outside of the light cones ("elsewhere") the substitutions are: $$ \left.\begin{array}{rl} x^0 &= s \sinh \phi \\ r & = s \cosh \phi \end{array}\right\} \Leftrightarrow \left\{\begin{array}{rl} s &= \sqrt{r^2 - (x^0)^2} \\ \phi & = \tanh^{-1}\left(\frac{x^0}{r}\right) \end{array}\right. . $$

Under these coordinate transformations, and placing the coordinate system origin at $x'$, the Green's function equation changes to: \begin{align} \left[\frac{1}{\tau^4} \frac{\partial}{\partial \tau} \left( \tau^4 \frac{\partial}{\partial \tau}\right) + m^2\right]G(\tau) &= 0,\ \mathrm{and} \\ \left[-\frac{1}{s^4} \frac{\partial}{\partial s} \left( s^4 \frac{\partial}{\partial s}\right) + m^2\right]G(s) &= 0. \end{align} Note that the origin is explicitly outside of these coordinate transformations, so the delta function is always zero on the right hand side. The time-like equation can be transformed into Bessel's equation with $\nu = \frac{3}{2}$ by the substitution $G(\tau) = \left(\frac{m}{\tau}\right)^{3/2} g(\tau)$. Likewise the space-like equation becomes the modified Bessel's equation with $\nu = \frac{3}{2}$ by the substitution $G(s) = \left(\frac{m}{s}\right)^{3/2} g(s)$. Because of this, we can write the Green's function in terms of the invariant coordinates $\tau = \sqrt{(x^\mu - x'^\mu)(x_\mu - x'_\mu)}$ and $s = \sqrt{-(x^\mu - x'^\mu)(x_\mu - x'_\mu)}$ as $$G(x,x') = \left\{\begin{array}{ll} A_+ \left(\frac{m}{\tau}\right)^{3/2} J_{3/2}(m \tau) + B_+ \left(\frac{m}{\tau}\right)^{3/2} Y_{3/2}(m \tau) & \mathrm{for\ } x^0 > x'^0 \mathrm{\ and\ } (x^\mu - x'^\mu)(x_\mu - x'_\mu) > 0 \\ A_0 \left(\frac{m}{s}\right)^{3/2} I_{3/2}(m s) + B_0 \left(\frac{m}{s}\right)^{3/2} K_{3/2}(m s) & \mathrm{for\ } (x^\mu - x'^\mu)(x_\mu - x'_\mu) < 0\\ A_- \left(\frac{m}{\tau}\right)^{3/2} J_{3/2}(m \tau) + B_- \left(\frac{m}{\tau}\right)^{3/2} Y_{3/2}(m \tau) & \mathrm{for\ } x^0 < x'^0 \mathrm{\ and\ } (x^\mu - x'^\mu)(x_\mu - x'_\mu) > 0. \end{array} \right.$$ Note that the light cones are still, explicitly, not dealt with, yet.

The causal (retarded) boundary conditions require $A_0 = B_0 = A_- = B_- = 0$. Finding the values of $A_+$ and $B_+$ requires transforming the original Klein-Gordon Green's function equation by integrating it over all of space, giving: $$\left[\frac{\partial^2}{\partial (x^0)^2} + m^2\right] \int_0^\infty G(x,x') 2\pi^2 r^3 \operatorname{d} r = \delta\left(x^0 - x'^0\right),$$ where the spatial derivatives term drops due to the divergence theorem. Notice how this is now the defining ordinary differential equation for a mass on a spring with unit mass and spring constant $k=m^2$. Given the boundary conditions above, this integral requires that the Green's function satisfies: $$\int_0^\infty G(x,x') 2\pi^2 r^3 \operatorname{d} r = \Theta\left(x^0 - x'^0\right) \frac{\sin\left(m\left[x^0 - x'^0\right]\right)}{m}.$$ Notice that, given the translational invariance of $G(x,x')$, this is the same as saying that if I deliver an impulse uniformly everywhere, then the field will oscillate like a simple harmonic oscillator. This is an example of a class of identities where applying a force that is uniform, flat, and infinite in extent essentially reduces the dimensionality of the problem by the dimensionality of the space the force is applied to (e.g. point charge vs line or plane charge in electrostatics).

The integral is easy to evaluate, with a change of variables, giving: \begin{array} \ \int_0^{x^0} \left(\frac{m}{\tau}\right)^{3/2} J_{3/2}(m\tau) 2\pi^2 r^3 \operatorname{d}r & = \frac{(2\pi)^{3/2}}{m} \left(\cos \left(mx^0\right) - 1 + \frac{\left(mx^0\right)^2}{2}\right),\ \mathrm{and} \\ \ \int_0^{x^0} \left(\frac{m}{\tau}\right)^{3/2} Y_{3/2}(m\tau) 2\pi^2 r^3 \operatorname{d}r & = \frac{(2\pi)^{3/2}}{m} \left[\left(\frac{\left(x^0\right)^2}{2u} - \frac{u}{2}\right) m\cos(mu) + \sin(mu)\right]_{u=0}^{x^0}. \end{array}

So, the problem is this - the polynomial terms in the $J$ integral, $-1 + \frac{\left(mx^0\right)^2}{2}$, can be canceled by adding terms confined to the light cone (specifically $-\frac{m}{\sqrt{2\pi}} \delta\left(\tau^2\right) - \frac{1}{3m\sqrt{2\pi}} \frac{\partial^2}{\partial \tau^2} \delta\left(\tau^2\right)$), but the $\cos \left(m x^0\right)$ term doesn't match the requirement. The $Y$ integral has a sine term, but the cosine term contains a divergence. It seems that nothing on the light cone can change the cosine of the $J$ integral into a sine, so it would seem that there has to be some way for a light cone term to cancel those divergences in the $Y$ integral, but how? Is that even possible? If not, does this mean that there is no causal Green's function for the Klein-Gordon equation in 5 dimensions?

This is, as you might guess, part of a more general exploration of Klein-Gordon Greene's functions in any number of dimensions. When the number of dimensions is even, the $J$ integral produces the needed sine term and the $Y$ integral is cosine-like. The Green's function produced by analytically continuing from the Euclidean Klein-Gordon Green's Function suggests that putting $K$ outside of the light cone can cancel the divergence from the $Y$ integral, but I recall that only works for the most divergent term, and thus fails as the number of dimensions (and thus divergent terms in $Y$) grows. Not to mention that it fails the causality requirement.

The only solution I can see is to perform a numerical exploration, but if someone already knows an analytical answer, I would love to hear it.

  • 1
    $\begingroup$ I'm a bit late to the party here, but I seem to recall that it's possible to obtain the Green's function in $n$ dimensions, satisfying$$\Box G^{(n)} = \delta^{(n)}(x_1, x_2, \dots x_n)$$by solving for the Green's function in $n+1$ dimensions$$\Box G^{(n+1)} = \delta^{(n+1)}(x_1, x_2, \dots x_n,x_{n+1})$$and then using the fact that $$\delta^{(n)}(x_1, x_2, \dots x_n) = \int dx_{n+1} \delta^{(n+1)}(x_1, x_2, \dots x_n,x_{n+1}).$$In other words, use an infinite line source in 6 dimensions (where the Green's function is well-behaved) rather than a point source in 5 dimensions. $\endgroup$ – Michael Seifert Dec 16 '17 at 16:22

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