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The continuum Gaussian free field on a planar domain $D$ is a centered Gaussian process indexed by finite-energy measures or test functions, with covariance given by the Green function of the Laplacian on $D$.

For an orthonormal basis $(e_n)$ of $H_0^1(D)$, the zero-Dirichlet Green function of the Laplacian is the weak kernel represented byConsequently, whenever the terms are defined and the right side is finite,

For radius $R$, constant mass density $\rho$ and $M=4\pi \rho R^3/3$, normalize the Newtonian gravitational potential to zero at infinity. ThenBy the spherical shell theorem, each spherical shell contributes the potential of a point mass outside it and a constant inside it. Summing inner and outer shells gives $\Phi (r)=-4\pi G\rho [r^2/3+(R^2-r^2)/2]$ inside. Equivalently, angular integration of the free-space Green function of the Laplacian gives ${\int }_{-1}^1(r^2+r^{\prime 2}-2r{r}^{\prime }\mu {)}^{-1/2}d\mu =[r+r^{\prime }-|r-r^{\prime }|]/(r{r}^{\prime })$. The resulting radial integral has the same value.

The potential and its first radial derivative are continuous at $R$. The interior gravitational acceleration is $-\nabla \Phi =-(GM/R^3)\mathbf{r}$, and ${\nabla }^2\Phi =4\pi G\rho$. At the centre $\Phi (0)=-3GM/(2R)$ while its gradient vanishes.

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

Here the sphere's “volume” is its intrinsic $(N-1)$-dimensional area, not the volume of the enclosed ball. Write ${\Omega }_{N-1}$ for the area of the unit sphere. The Gaussian integral givesFor completeness the one-dimensional integral is positive and its square is $2\pi {\int }_0^{\infty }\rho e^{-{\rho }^2}d\rho =\pi$. In hyperspherical coordinates, the same integral isThe substitution is $u={\rho }^2$, and the last integral defines the Gamma function. Equating the two evaluations and scaling lengths by $r$ proves

Take $D$ to mean spacetime dimension, so the number of spatial dimensions is $d=D-1$. This convention is needed when comparing five-dimensional spacetime with four-dimensional spacetime. For a point charge $q$, Gauss's law in $d$ spatial dimensions givesUsing $E_r=-{\varphi }^{\prime }(r)$ and choosing $\varphi (\infty )=0$ for $d>2$ givesThe Green function of the Laplacian with $-{\nabla }^2{G}_d={\delta }^{(d)}$ is consequently $G_d(r)=1/[(d-2){\Omega }_{d-1}{r}^{d-2}]$. Define the gravitational Poisson coupling ${\kappa }_D$ by ${\nabla }^2\Psi ={\kappa }_{D}M{\delta }^{(d)}$, and use acceleration $\mathbf{g}=-\nabla \Psi$. The same flux calculation givesThis explicitly fixes the normalization of the gravitational coupling: in four dimensions ${\kappa }_4=4\pi G_4$, so ${\Psi }_4=-G_{4}M/r$. If a higher-dimensional Newton constant is instead defined by the force law $|\mathbf{g}|=G_{D}M/r^{D-2}$, then ${\kappa }_D={\Omega }_{D-2}{G}_D$ in the displayed expression. In all conventions the force scales as $r^{-(D-2)}$ and the potential as $r^{-(D-3)}$. If “$D$ dimensions” denotes spatial dimensions instead, replace $d$ by $D$ and the potential exponent is $D-2$.

For $d=2$, integration gives $\varphi =-q\log (r/r_0)/(2\pi {\epsilon }_3)$ and $\Psi ={\kappa }_{3}M\log (r/r_0)/(2\pi )$ for the Poisson-normalized gravitational model; these logarithmic potentials have no zero-at-infinity convention. For $d=1$, a fundamental solution of the negative Laplacian is $G_1(x)=-|x|/2$, giving linear potentials up to an additive homogeneous solution.

Now compactify the fourth spatial coordinate $y$ on a circle of circumference $L=2\pi R$, retaining three noncompact coordinates with radial distance $r$. In uncompactified five-dimensional spacetime, $G_4(r,y)=1/[4{\pi }^2(r^2+y^2)]$. A periodic point source is therefore represented by its images:For $r/L\to \infty$, write the sum as $r^{-2}{\sum }_n[1+(nL/r{)}^2{]}^{-1}$ on the source slice $y=0$. The spacing $L/r$ tends to zero, so its Riemann sum approaches the integrable function's integral:Thus the compactified Green function is asymptotic to $1/(4\pi Lr)$, already proving the four-dimensional inverse-distance law.

An exact expression makes the crossover transparent. The Fourier transform of $(r^2+y^2{)}^{-1}$ is $\pi e^{-r|k|}/r$, obtained by closing the integration contour around $y=ir$ or $-ir$. Applying the Poisson summation formula then gives, at $y=0$,The last equality sums a geometric series. Hence the point-source potential with one compact spatial dimension isAt short distances $r\ll L$, $\coth (\pi r/L)\sim L/(\pi r)$ and these reproduce the five-dimensional $1/r^2$ potentials. At long distances $r\gg L$, the hyperbolic factor is $1+O(e^{-2\pi r/L})$, giving the four-dimensional $1/r$ potentials with effective Poisson couplings ${\epsilon }_4=L{\epsilon }_5$ and ${\kappa }_4={\kappa }_5/L$. Equivalently only the massless Kaluza-Klein mode survives at long distance, while the massive modes contribute exponentially suppressed terms.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

For the Dirac delta distribution source $\rho (\mathbf{r})=M\delta (\mathbf{r})$, the stated Green function of the Laplacian givesThis is the Newtonian potential of a point mass; its negative gradient gives the inverse-square gravitational acceleration $-M\hat{\mathbf{r}}/r^2$.

The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$. Substituting $B=\nabla \times A$ and using the Coulomb gauge $\nabla \cdot A=0$ givesThe free-space Green function of the Laplacian therefore givesFor the thin wire current stated in the question this becomes

The zero-Dirichlet Green function of the Laplacian $G_D(x,y)$ is symmetric, vanishes at the boundary in the appropriate sense, is harmonic in $x$ away from $y$, and satisfiesas a distribution. Equivalently, for suitable $\varphi$,

Integral identity and representation. Define $L={\nabla }^2+k^2$ and take $n$ to be the outward unit normal. The product rule givesThe $k^2\varphi \psi$ terms cancel. Integrating and applying the divergence theorem proves the Helmholtz equation version of Green's second identity:

Choose the Dirichlet Green function with the sign conventionwhere $\delta$ is the three-dimensional Dirac delta. Substitute $\psi =G$, use $L\varphi =f$, and evaluate the Dirac delta integral. The boundary term containing $G$ vanishes and $\varphi =g$ on $S$, leavingHenceThe choice of sign in the Green function equation matters: it fixes both the source normalization and these signs. For a bounded region this construction presumes that the chosen $k^2$ is not a Dirichlet eigenvalue of $-{\nabla }^2$; at such an eigenvalue a Green inverse for arbitrary data does not exist.

Infinite space. Put $r=|x-x^{\prime }|$. Away from the source, spherical symmetry reduces the Helmholtz equation toFor $k\ne 0$ its general radial solution isIntegrate $LG=\delta$ over a ball of radius $\epsilon$ about the source. The divergence theorem yieldsSince $G=A/r+Bk+O(r)$, the volume term is $O({\epsilon }^2)$ and the flux tends to $-4\pi A$. Therefore $A=-1/(4\pi )$, while $B$ remains free:The $B$ term extends regularly through the source and solves the homogeneous Helmholtz equation, so it cannot change the unit Dirac delta normalization. A condition at infinity is needed to select it. For example, with time dependence $e^{-i\omega t}$, the outgoing Green function for the three-dimensional Helmholtz equation is $-e^{ikr}/(4\pi r)$, corresponding to $B=-i/(4\pi )$.

Zero wavenumber. For fixed finite $B$, $\sin (kr)/r\to 0$ as $k\to 0$, andThis is the decaying Green function of the Laplacian. Directly at $k=0$, the general radial solution is $A/r+C$; source normalization fixes $A=-1/(4\pi )$, and decay at infinity removes $C$. If the free parameter depends on $k$ as $B(k)\sim C/k$, the limit instead includes that arbitrary homogeneous constant. Thus the usual Laplace limit holds for a fixed parameter, or after a suitable condition at infinity has been selected.

Write $q=|r-r^{\prime }|$. Away from the source, the radial Laplacian gives ${\nabla }^2(-1/(4\pi q))=q^{-2}{\partial }_q[q^2/(4\pi q^2)]=0$. Through a small source-centred sphere, the outward flux of its gradient isThis identifies a unit Dirac delta at the source. More explicitly, for a smooth compactly supported test function $\psi$, apply Green's second identity outside the radius-$\epsilon$ ball. The inner outward normal of that punctured domain is $-e_q$. The term $H{\partial }_n\psi$ has integral $O(\epsilon )$, while $-\psi {\partial }_{n}H$ tends to $\psi (r^{\prime })$. Thus $\int H{\nabla }^2\psi dV=\psi (r^{\prime })$, proving the distributional identityWhenever the integral converges, convolution with this Green function of the Laplacian solves the Poisson equation for Newtonian gravity:The harmonic term represents the homogeneous freedom and is fixed by boundary conditions. For an isolated finite mass distribution, regularity away from the prescribed sources and the normalization $\Phi \to 0$ at infinity select $h=0$.

For a point mass at the origin, the delta integral gives the Newtonian potential of a point mass,

Write $r=|\mathbf{r}|$, $s=|{\mathbf{r}}^{\prime }|$. For a spherical source, the angular integral needed for the Green function of the Laplacian isThis is the angular form of the spherical shell theorem. Substituting the source into the Green function integral therefore yieldsThe elementary integrals areThe terms proportional to $e^{-\beta r}/\beta$ cancel, leavingThe value at the origin is the continuous limit. At large radius $\varphi \sim -1/({\beta }^{2}r)$, consistent with the integrated source $\int \rho dV=-4\pi /{\beta }^2$ and the factor $1/(4\pi r)$ in the far-field potential.

For a direct check, set $u=r\varphi$. Then $u^{\prime \prime }=e^{-\beta r}$, so ${\nabla }^2\varphi =u^{\prime \prime }/r=e^{-\beta r}/r=-\rho$ for $r>0$. Boundedness of $\varphi$ at zero cancels the singular $1/r$ terms and excludes an extra point source. The original density is itself singular but locally integrable; the equation at the origin is understood in the sense of distributions. This is the Poisson potential of an exponential-over-radius source with amplitude $A=-1$.

Translation and rotation invariance allow a radial Green function $G(R)$, with $R=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Outside the source, $G^{\prime \prime }+2G^{\prime }/R=0$, so $G=A/R+B$. Decay at infinity gives $B=0$.

Integrate the defining equation over a small ball centred on the source. The divergence theorem fixes its flux:ConsequentlyThis is the Green function of the Laplacian with the stated negative-delta sign. A homogeneous $1/R$ addition would change the delta normalization, so the singular coefficient is fixed, not an arbitrary integration constant.

In three dimensions the Poisson kernel for the upper half-space isIt gives the harmonic function $\varphi (x,y,z)={\int }_{{\mathbb{R}}^2}{P}_z(x-\xi ,y-\eta )f(\xi ,\eta )d\xi d\eta$ for compatible boundary data. The method of images constructs the zero-boundary Dirichlet Green function by subtracting the reflected-source Green function of the Laplacian from the source kernel. On the boundary plane the outward unit normal is $-e_z$, and $-{\partial }_{n}G=P_z$, giving the integral formula through Green's second identity.

The kernel is positive and has integral one: in polar coordinates this is $z{\int }_0^{\infty }r(r^2+z^2{)}^{-3/2}dr=1$. Its mass outside any fixed disk tends to zero as $z\downarrow 0$. It is therefore an approximate identity, recovering bounded boundary data at continuity points. At a jump across a smooth curve, the vertical limit gives the average of the two adjacent values; assigning a different value only on that curve does not change the harmonic extension.

Let $\kappa >0$ be an inverse length. In three spatial dimensions, seek the Green's function that decays at infinity and satisfieswhere ${\nabla }^2$ is the Laplacian and ${\delta }^{(3)}$ the three-dimensional Dirac delta function. The point source and the differential operator have spherical symmetry. For $r>0$, the radial Laplacian givesWriting $u(r)=r{G}_{\kappa }(r)$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$. Decay at infinity removes the growing exponential function, leaving $G_{\kappa }(r)=C{e}^{-\kappa r}/r$.

The Dirac delta function fixes the coefficient. Integrating over a ball $B_{ϵ}$ and using the divergence theorem givesThe first term tends to $4\pi C$, while the second is $O({ϵ}^2)$. Hence $C=1/(4\pi )$ andThis solves the source equation in the sense of distributions, including at the origin. Merely verifying the equation for $r>0$ would leave the normalization undetermined. The limit $\kappa \to 0$ at fixed $r>0$ is $1/(4\pi r)$, the Green function of the Laplacian for the convention $-{\nabla }^2$.

For an attractive scalar toy interaction, absorb constants into $A>0$ and setHere $A$ has units of energy times length, and negative $F_r$ points inward. At equal $A$, comparison with the unscreened model $V_0=-A/r$ givesThus the potential energy and force have the same exponential scale but different suppression factors. At $x=1$, these ratios are approximately $0.368$ and $0.736$; at $x=3$, they are $0.0498$ and $0.199$. Finite range means rapid decay, not an abrupt cutoff.

For finite Borel measures $\rho ,\nu$ of finite Green energy, the zero-boundary Gaussian free field $h$ on $D$ is characterized bywhere $G_D$ is the zero-Dirichlet Green function of the Laplacian.