What Is the Gravitational Constant in Higher Dimensions?

In higher-dimensional gravity, Newton’s constant depends on how many spatial dimensions gravity can spread through. It is usually written as $G_D$, where $D$ is the total number of spacetime dimensions. If extra dimensions are compact, the constant measured at large distances is an effective four-dimensional one, related to $G_D$ by the size of those extra dimensions.

The key point is that the higher-dimensional constant is not simply the familiar four-dimensional $G$ with a different numerical value. Its units, force law, and relation to the observed constant all depend on the geometry and on whether the extra dimensions are accessible at the distance being measured.

Units of Newton’s constant

Let $d=D-1$ be the number of spatial dimensions. In the regime where gravity spreads through all $d$ dimensions, the force between two point masses scales as

$$F \propto \frac{G_D m_1m_2}{r^{d-1}}.$$

The proportionality factor can depend on the convention used to define $G_D$; the distance dependence and units do not. Since force has SI units of mass times length divided by time squared, dimensional analysis gives

$$[G_D] = \frac{\mathrm{length}^{d}}{\mathrm{mass}\,\mathrm{time}^{2}} = \mathrm{m}^{d}\,\mathrm{kg}^{-1}\,\mathrm{s}^{-2}.$$

For ordinary three-dimensional space, $d=3$, this is the familiar unit $\mathrm{m^3\,kg^{-1}\,s^{-2}}$. With four spatial dimensions, it would instead be $\mathrm{m^4\,kg^{-1}\,s^{-2}}$; with six spatial dimensions, $\mathrm{m^6\,kg^{-1}\,s^{-2}}$.

In particle physics, natural units often set $c=\hbar=1$. In those units, the same coupling has dimensions of length to the power $D-2$, or mass to the power $-(D-2)$. This is another way to express the dimension dependence, not a different physical law.

Why the force law changes

Gauss’s law explains the changed distance dependence. In $d$ spatial dimensions, the area of a sphere grows in proportion to $r^{d-1}$. If gravitational flux spreads uniformly through all those dimensions, its strength per unit area—and thus the force—falls as $1/r^{d-1}$.

For example, in three spatial dimensions the force falls as $1/r^2$. In four spatial dimensions it falls as $1/r^3$. The gravitational potential is proportional to $1/r^{d-2}$ when $d>2$; the two-dimensional spatial case is an exception, where the potential grows logarithmically with distance.

The formula describes an uncompactified regime, or distances small enough to resolve compact dimensions. If extra dimensions are compact, the force law can change with distance: at short distances gravity may spread through the extra dimensions, while at large distances it can behave approximately as though only the familiar three spatial dimensions were available.

How compactification gives the observed constant

Suppose spacetime has $D=4+n$ dimensions: three large spatial dimensions, one time dimension, and $n$ compact spatial dimensions. For a simple, unwarped geometry, let $V_n$ be the proper volume of the compact dimensions. At distances much larger than their size, the effective four-dimensional Newton constant is

$$G_4 = \frac{G_D}{V_n},$$

or equivalently $G_D=G_4V_n$. This relation assumes the same normalization convention for the higher- and lower-dimensional Einstein–Hilbert actions. It also assumes a simple compactification with no warping or additional fields changing the effective gravitational coupling. More complicated geometries can modify the relation.

The units provide a useful check. $V_n$ has units of length to the power $n$, while $G_D$ has units of length to the power $3+n$ divided by mass and time squared. Dividing by $V_n$ leaves $G_4$ with the usual units of $\mathrm{m^3\,kg^{-1}\,s^{-2}}$.

Worked example: two compact dimensions

Take two extra spatial dimensions compactified on a flat square torus, with each circular direction having radius $R$. Its area is

$$V_2=(2\pi R)^2.$$

The relation between the couplings is therefore

$$G_4=\frac{G_6}{(2\pi R)^2}, \qquad G_6=G_4(2\pi R)^2.$$

For an illustrative radius of $R=1\,\mathrm{\mu m}=10^{-6}\,\mathrm{m}$, the compact volume is approximately $3.95\times10^{-11}\,\mathrm{m^2}$. Using $G_4\approx6.6743\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ gives

$$G_6\approx 2.63\times10^{-21}\,\mathrm{m^5\,kg^{-1}\,s^{-2}}.$$

This is a hypothetical example showing the conversion, not a measurement of a six-dimensional gravitational constant. The result depends on the assumed compactification radius and geometry.

Common mistakes

• Using $D$ for the number of spatial dimensions. Here $D$ counts spacetime dimensions; the number of spatial dimensions is $d=D-1$.
• Assuming $G_D$ has the same units as $G_4$. Its length dimension changes with the number of spatial dimensions.
• Treating $G_4=G_D/V_n$ as universal. It applies to a simple compactification; warping, geometry, or fields that vary across the extra dimensions can change the effective relation.
• Expecting the higher-dimensional force law at every distance. Compact dimensions affect gravity only at scales where their size matters. At much larger distances, measurements can recover the ordinary inverse-square force law.