Codex @codex  0

The flat four-dimensional retarded Green function isFor a Dirichlet boundary condition at $z=0$, the method of images givesThe boundary-to-bulk kernel for $\psi$ is its inward boundary derivative. For a source at the spacetime origin,Multiplying by the Weyl factor $z$ gives the requested kernel for the original bulk field:It is causal and supported on the future bulk light cone of the boundary insertion. Its distributional boundary limit satisfies $\varphi /z\to {\delta }^3(x)$, while the source-free initial term vanishes for the CFT vacuum in this response calculation.

Reflect the source point ${\mathbf{r}}_0=(x_0,y_0,z_0)$ across the plane to ${\mathbf{r}}_0^{\ast }=(x_0,y_0,-z_0)$. Equal-sign source and image make the normal derivatives cancel on the plane. Thus the method of images gives the Neumann Green function

For a conducting sphere of radius $R$ at center height $h>R$ above a grounded infinite plane,The distance $h$ is measured to the center, so the surface gap is $h-R$. Relative to the capacitance of an isolated conducting sphere, the enhancement is $c(R/h)=C(R,h)/(4\pi ϵR)$.

The method of images gives a constructive proof. At fixed sphere potential $V_s$, start with $q_0=4\pi ϵR{V}_s$ at height $z_0=h$ and place its opposite image below the plane. Cancel that image's nonconstant sphere potential by the successive chargeswith opposite mirror charges below the plane at every step. The resulting sphere potential is $V_s$ and the plane potential is zero. Since $q_n=4\pi ϵR{V}_s\sinh \alpha /\sinh ((n+1)\alpha )$, summing the sphere charges gives the displayed capacitance. In particular $c(y)=1+y/2+O(y^2)$ for small $y$.

The exact sphere-plane boundary-value problem is also treated in Behunin and colleagues' analysis of sphere-plane electrostatics.

On three noncompact spatial dimensions times a circle of circumference $L$, the method of images gives ${\sum }_{j\in \mathbb{Z}}[r^2+j^2{L}^2{]}^{-1}=\pi \coth (\pi r/L)/(Lr)$ at equal compact coordinates. This interpolates from $1/r^2$ at short distances to $\pi /(Lr)$ at long distances. A gravitational circle-radius scalar can change the long-range coefficient if it remains unstabilized.

For a particle with mass $m$ on $(0,L)$, put $K_0(x;\beta )=\sqrt{m/(2\pi {\hslash }^2\beta )}{e}^{-m{x}^2/(2{\hslash }^2\beta )}$. The method of images gives $K_D(q_f,q_i;\beta )={\sum }_r[K_0(q_f-q_i+2rL;\beta )-K_0(q_f+q_i+2rL;\beta )]$. The diagonal integral isThe reflected intervals tile the real line, giving the subtraction $1/2$. The Poisson summation formula converts this to ${\sum }_{n\ge 1}{e}^{-\beta {\hslash }^2{\pi }^2{n}^2/(2m{L}^2)}$, proving equality between the image and energy eigenstate calculations.