 Solution

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.