Probably another question without an answer! ;-)

In most books/papers I saw on General Relativity, everybody writes $\kappa = 8 \pi G_D$ in the right part of Einstein's equation, even for spacetimes of dimensionality $D \ne 4$ (the minus sign is just a matter of convention): \begin{equation}\tag{1} G_{\mu \nu} + \Lambda_D \, g_{\mu \nu} = -\, 8 \pi G_D \, T_{\mu \nu}. \end{equation} I understand that we could always define $G_D$ so we can write $8 \pi$ in front of it. But it's not "natural", since $\Omega_3 = 4 \pi$ is the solid angle in flat space of dimension 3. In general: \begin{align}\tag{2} \Omega_{2 n} &= \frac{1}{(n - 1)!} \, 2 \pi^n, \qquad &\Omega_{2 n + 1} &= \frac{2^{2 n} \, n!}{(2 n)!} \, 2 \pi^n. \end{align} For examples: $\Omega_1 = 2$, $\Omega_2 = 2 \pi$, $\Omega_3 = 4 \pi$, $\Omega_4 = 2 \pi^2$. So it would be more natural to write this, in Einstein's equation (I know, it looks awefull, but it has more sense!): \begin{equation}\tag{3} G_{\mu \nu} + \Lambda_D \, g_{\mu \nu} = -\, 2 \, \Omega_{D - 1} \, G_D \, T_{\mu \nu}. \end{equation} Notice that Newton's constant has a dimension that depends on $D$ (since $G_{\mu \nu} \sim \mathrm{L}^{-2}$ and $T_{\mu \nu} \sim \mathrm{L}^{-D}$ in cartesian-style coordinates): \begin{equation}\tag{4} G_D \sim \mathrm{L}^{D - 2}. \end{equation} That constant has a conformal weight! (it's also true for the fine structure constant by the way: $\alpha \sim \mathrm{L}^{D - 4}$, if you write Maxwell equation in $D$ dimensional spacetime).

So my questions are these:

Why use $G_D = \kappa / 8 \pi$ instead of $G_D = \kappa / 2 \, \Omega_{D-1}$? (it obviously doesn't give the same constant when $D \ne 4$!). $\kappa = 8 \pi G_D$ may be simpler than $\kappa = 2 \, \Omega_{D - 1} \, G_D$, but it doesn't make physical sense!

What is the meaning of the factor $2$ in $\kappa = 2 \, \Omega_{D - 1} \, G_D$ (for any $D$)? Is it related to the graviton's spin? (it is not given by $2 s + 1$ or $s (s + 1)$). In the special case of $D = 4$, what is the interpretation of that factor in $8 \pi G_4 = 2 \cdot 4 \pi G_4$?

In - very - hypothetical spacetimes with many timelike dimensions, what should be the natural expression for $\kappa$? For example, for $D = 4$ and metric signature $\eta = (1, 1, -1, -1)$ (two timelike dimensions, two spacelike dimensions), should we write $\kappa = 2 \cdot 2 \pi G_4$ instead of the usual $\kappa = 8 \pi G_4$?

I'm pretty sure nobody would have an answer to these "idiotic" questions. I'm asking them anyway, in case someone else is crazy enough to scratch his/her head like me! ;-)

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    $\begingroup$ I don't see any real physics here. Which prefactor we assign to $G_D$ is entirely conventional, and the factors involving $\pi$ are simply due to mismatched conventions between Newtonian physics and GR, cf. physics.stackexchange.com/q/71636/50583. The belief that any of the other choices for $G_D$ is objectively "more natural" does not strike me as useful, especially since people are still free to set (by choice of units) whatever multiple of $G$ to unity they like. There is no deep physics in the choices for dimensionful constants, cf. also arxiv.org/abs/1412.2040. $\endgroup$
    – ACuriousMind
    Commented Dec 28, 2018 at 15:46
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    $\begingroup$ @ACuriousMind, I don't agree. In flat space, the elliptic equation part of Poisson's equation do make the $4 \pi$ apparent, in front of $G$. That solid angle is important. In curved space, that $4 \pi$ can be defined from the tangent space, as a limit of a very small region in curved space (the solid angle doesn't depend on the size of the region considered). In $D > 4$, the solid angle is different than in $D = 4$ and should be taken into account, in the elliptic part of Poisson's equation (weak field case). $\endgroup$
    – Cham
    Commented Dec 28, 2018 at 15:50
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    $\begingroup$ I don't want to pick sides here, just wanted to point to a footnote in Ortin's Gravity and Strings, that seems to share the discomfort for this convention. In the second edition of the book, in page 73, at the end of section 3.1.1, footnote 6: "This is an unfortunate convention in the literature in which the factor $4\pi$, which is appropriate for rationalized units in four dimensions, is indiscriminately used in all dimensions.". This book also discusses the implications for Newton's constant in arbitrary dimensions, if only with the books own conventions which you may or may not dislike. $\endgroup$
    – secavara
    Commented Dec 28, 2018 at 16:02
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    $\begingroup$ @ACuriousMind, an example that shows the relative importance of the solid angle. In Schwarzschild coordinates, we have this metric component: $g_{00} = 1 - \frac{2G M}{r}$. Absorbing the solid angle in $G$ would give $g_{00} = 1 - \frac{2 G' M}{4 \pi r}$, which is horrible (and $4 \pi r$ is not even an area)! The $4 \pi$ factor in Einstein's equation is a "chauvinistic" reference to our 3 space dimensions. $\endgroup$
    – Cham
    Commented Dec 28, 2018 at 16:04
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    $\begingroup$ @ACuriousMind, well, I agree that it's a matter of "convention", and I personnaly still use $\kappa = 8 \pi G_D$ (instead of $\kappa = 2 \, \Omega_{D - 1} \, G_D$). But that factor really has something to say about the physics. We may miss something by not taking care of the solid angle. I'm sure (my guts are telling me!) that the extra-factor of $2$ has something to say, but I don't know what (is it related to spin?). $\endgroup$
    – Cham
    Commented Dec 28, 2018 at 16:12

1 Answer 1


$$ G_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi G \, T_{\mu \nu}. $$

This answer is, admittedly, copied from the USENET group sci.physics.research and from an answer 23 years ago from Daryl McCullough regarding that factor of "$2$" in $8 \pi G = 2 \cdot 4 \pi G$

Jan 20, 2000, 3:00:00 AM

[email protected] (John Baez) says...

Yeah, I think Ted Bunn explained it. The $4 \pi$ comes from the area of the sphere, just like electromagnetism. The extra factor of $2$ comes from the fact that $h_{00}$ is $-2$ times the Newtonian gravitational potential $\Phi$.

I thought that the source of this $2$ is the square-root appearing in the proper time formula:

$$ \mathrm{d}T = \sqrt{g_{\mu \nu} \, \mathrm{d}x^{\mu} \, \mathrm{d}x^{\nu}}$$

In weak, time-independent gravitational fields, $g_{\mu \nu}$ takes the form

$$ \begin{align} g_{0 0} &= 1 + h \\ g_{1 1} &= -1 \\ g_{2 2} &= -1 \\ g_{3 3} &= -1 \\ g_{\mu \nu} &= 0 \qquad \text{for } \mu \ne \nu \\ \end{align}$$

So the formula for proper time becomes (using $\left(\frac{v}{c}\right)^2 = \left(\frac{dx^1}{dx^0}\right)^2 + \left(\frac{dx^2}{dx^0}\right)^2 + \left(\frac{dx^3}{dx^0}\right)^2$)

$$\mathrm{d}T = \sqrt{1 + h - \left(\frac{v}{c}\right)^2} \ \mathrm{d}t$$

For $A$ and $v$ small, we can expand the square-root to get

$$\mathrm{d}T = \left(1 + \tfrac12 h - \tfrac12 \left(\frac{v}{c}\right)^2 \right) \ \mathrm{d}t$$

The equations of motion for a test particle are obtained by maximizing the integral $\int \left(1 + \tfrac12 h - \tfrac12 \left(\frac{v}{c}\right)^2\right) \ \mathrm{d}t$. Since the constant $1$ is irrelevant for the equations of motion, this is the same as maximizing the integral $\int D \, \mathrm{d}t$, where

$$ D = \tfrac12 h - \tfrac12 \left(\frac{v}{c}\right)^2 $$

But, in this limit (small $A$ and small $v$), Newtonian mechanics can be used, so we should also be able to obtain the equations of motion by minimizing the integral $\int L \, \mathrm{d}t$, where

$$ L = \tfrac12 m v^2 - m \Phi $$

(where $\Phi$ is the Newtonian gravitational potential).

This suggests that $D$ and $L$ are linearly related. The only way for this to be true is if

$$ h = 2 \frac{\Phi}{c^2} $$

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    $\begingroup$ In other words, you say that the "2" in $\kappa = 2 \cdot 4 \pi G$ comes from the "2" in the metric $ds^2$, which is a way of saying that the metric tensor is a rank 2 tensor (since there's $\mathbf{d}x^{\mu} \otimes \mathbf{d}x^{\nu}$ in it). This is an implicit way of saying that the gravitational field has spin $s = 2$. $\endgroup$
    – Cham
    Commented Jul 30, 2023 at 11:55
  • $\begingroup$ Yes, that is what I believe McCullough is saying in his sci.physics.research post (I linked it above). In another neophyte question of mine, I am trying to make the case that Planck units (time, length, mass, charge) should be "rationalized" to satisfy: $$\begin{align} c&=1\\ \hbar&=1\\ 4\pi G&=1\\ \epsilon_0&=1\\ \end{align}$$ This makes the elementary charge $e=\sqrt{4\pi\alpha}=$0.302822 and I think it should be $4\pi G$ that is normalized rather than $8\pi G$ to normalize both the speed of propagation and the characteristic impedance of EM and gravitational radiation. $\endgroup$ Commented Jul 30, 2023 at 15:22

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