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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.

Let $A_s$ denote the surface area of an n-sphere for the unit sphere $S^s$. In this notation $s$ is the intrinsic dimension of the sphere. Cartesian factorization of the Gaussian integral gives $I_N={\pi }^{N/2}$. In hyperspherical coordinates,The substitution $t={\rho }^2$ supplies the defining Gamma function integral. Scaling the sphere radius multiplies its surface volume by $r^{N-1}$, henceThis is the volume of the sphere itself, not of its enclosed ball.

Take $D$ to denote spacetime dimension and $d=D-1$ spatial dimensions. In uncompactified flat space, Gauss's law for a charge $q$ and permittivity ${\epsilon }_d$ gives $E_r{A}_{d-1}{r}^{d-1}=q/{\epsilon }_d$. Since $E_r=-d\phi /dr$, the electric potential for $d>2$, normalized to vanish at infinity, isThus the four-dimensional electric potential falls as $1/r$, and the five-dimensional one as $1/r^2$. If a convention counts only spatial dimensions as $D$, replace $d$ by $D$ in the first formula. For $d=2$, the integral instead gives $\phi =-q\log (r/r_0)/(2\pi {\epsilon }_2)$; for $d=1$, it gives $\phi =-q|x|/(2{\epsilon }_1)$ up to a constant. Neither case has a finite zero at infinity.

For a Newtonian gravitational potential ${\Phi }_g$, define the flux-normalized coupling ${\kappa }_d$ by ${\nabla }^2{\Phi }_g={\kappa }_{d}m{\delta }^{(d)}(\mathbf{x})$ and acceleration $-\nabla {\Phi }_g$. Exactly the same flux integration gives the attractive potentialIn $d=3$, ${\kappa }_3=4\pi G_4$ reproduces $-G_{4}m/r$. If $G_D$ is defined by the $D$-dimensional Einstein-Hilbert action $S=(16\pi G_D{)}^{-1}\int \sqrt{-g}R$, the weak static Einstein field equations give ${\kappa }_d=8\pi G_D(D-3)/(D-2)$ for $D>3$. ThereforeIndeed, $R_{00}\simeq {\nabla }^2{\Phi }_g$ and $T_{00}-g_{00}T/(D-2)=(D-3)\rho /(D-2)$ in signature $(-+\cdots +)$. A Poisson-defined Newtonian theory in two spatial dimensions has a logarithmic potential, but pure three-dimensional Einstein gravity has no analogous local Newtonian point-mass force; the relativistic normalization must not be extrapolated to that exceptional dimension.

For an unwarped product with a fixed extra-dimensional metric, normalize the Einstein-Hilbert action by $\frac{1}{2}{M}_{\ast }^{n+2}\int d^{4+n}x\sqrt{-g}R$. Integrating the part containing the four-dimensional curvature over the internal space gives $\frac{1}{2}{M}_{\ast }^{n+2}\mathrm{vol}(S_R^n)\int d^{4}x\sqrt{-g_4}{R}_4$. Thus, using the reduced Planck mass,This is a volume relation for compactification; a curved sphere additionally requires a mechanism supporting and stabilizing its background. It is not by itself a proof that a sphere times flat spacetime solves a vacuum gravitational theory.

With $M_{\ast }=10^3\mathrm{GeV}$, $M_{\mathrm{Pl}}=2.435\times 10^{18}\mathrm{GeV}$ and $1{\mathrm{GeV}}^{-1}=1.97327\times 10^{-16}\mathrm{m}$, the spherical-volume factors $A_1=2\pi$, $A_2=4\pi$, $A_6=16{\pi }^3/15$ giveThe first estimate is astronomical; the second is roughly a tenth of a millimetre. If both Planck scales are instead defined by $S=M_D^{D-2}\int \sqrt{-g}R/(16\pi )$, the same algebraic volume relation uses the unreduced $M_{\mathrm{planck}}=1.221\times 10^{19}\mathrm{GeV}$. Holding that convention's $M_{\ast }=1\mathrm{TeV}$ gives radii larger by $(8\pi {)}^{1/n}$: approximately $4.69\times 10^{12}\mathrm{m}$, $6.80\times 10^{-4}\mathrm{m}$, $2.54\times 10^{-14}\mathrm{m}$. These are different conventions for what is held fixed at one TeV, not different scaling laws.

Finally, compactify the fifth coordinate on a circle of circumference $L=2\pi R$. For a source and observer at the same compact coordinate, the method of images turns a five-dimensional static potential $C/{\rho }^2$ intoOne can derive this image sum without assuming a summation formula. The Fourier transform of $1/(r^2+y^2)$ is $\pi e^{-r|k|}/r$; Poisson summation then gives $\pi [1+2{\sum }_{\ell \ge 1}{e}^{-2\pi \ell r/L}]/(Lr)$, which is the displayed hyperbolic cotangent. ThereforeThe short-distance limit is $C/r^2$; the long-distance limit is the four-dimensional $1/r$ potential. These exponential corrections are the nonzero Kaluza-Klein modes, with masses $2\pi |\ell |/L$. For electrostatics $C=q/(4{\pi }^2{\epsilon }_4)$, so the effective permittivity is ${\epsilon }_3=L{\epsilon }_4$. For a Poisson-normalized gravitational potential, $C=-{\kappa }_{4}m/(4{\pi }^2)$ gives ${\kappa }_3^{\mathrm{eff}}={\kappa }_4/L$. A massless scalar associated with the circle radius also contributes in an unstabilized gravitational compactification; recovering pure four-dimensional Einstein gravity with $G_4=G_5/L$ requires its stabilization or removal. The $1/r$ crossover itself is independent of this tensor-versus-scalar normalization issue.

Let $V_d$ be the volume of an n-ball of unit radius in ${\mathbb{R}}^d$. A radial shell has volume $d{V}_d{r}^{d-1}dr$, by differentiating $V_d{r}^d$. Integrating shells, or using hyperspherical coordinates, therefore reduces a radial integrand over a Euclidean ball to the displayed one-dimensional integral. The constant $d{V}_d$ is the surface area of an n-sphere of unit radius, with sphere dimension $d-1$.