# Baez & Bunn: The Meaning of Einstein's (field) Equation

In this online tutorial, aimed at pea-brains like me, the authors restate Einstein's equation

$$\mathbf{G}_{\mu\nu} = \frac{8 \pi G}{c^4} \mathbf{T}_{\mu\nu}$$

in words:

Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the $x$ direction at that point, plus the pressure in the $y$ direction, plus the pressure in the $z$ direction.

and, defining units so that $8 \pi G=1$ and $c=1$ ("reduced" Planck units), they state it as

$$\frac{\ddot{V}}{V} \Bigg|_{t=0} = -\tfrac12 (\rho + P_x + P_y + P_z)$$

Now I would like to express this Baez/Bunn version of Einstein's equation without a system of natural units, using the constants $G$ and $c$.

I think it's this:

\begin{align} \frac{\ddot{V}}{V} \Bigg|_{t=0} &= -\tfrac12 \frac{8 \pi G}{c^2} \big(\rho_\text{g} c^2 + P_x + P_y + P_z \big) \\ &= \frac{8 \pi G}{c^4} \big(-\tfrac12 \rho_\text{g} c^4 - \tfrac12 c^2(P_x + P_y + P_z) \big) \\ \end{align}

where $\rho_\text{g} c^2$ is the energy density and $\rho_\text{g}$ is the mass density. Now I did this only by fiddling with dimensions and trying to make the equation dimensionally consistent.

To get the scaling constant to $\frac{8 \pi G}{c^4}$, does that mean that the dimension of the left and right side of the equation is the same as the dimension of the elements of the Einstein tensor $\mathbf{G}_{\mu\nu}$ ? This would make $\rho_\text{g} c^4$ to have the same dimension as the stress-energy tensor. Or do we need to multiply $\frac{\ddot{V}}{V}$ by something more to make it dimensionally consistent with $\mathbf{G}_{\mu\nu}$?

Lastly, if I got the above scaled right, then

$$\frac{\ddot{V}}{V} \Bigg|_{t=0} = -\frac{4 \pi G}{c^2} \big(\rho_\text{g} c^2 + P_x + P_y + P_z \big)$$

and it seems that the most natural rationalized (in the manner of Lorentz-Heaviside) Planck units would still be those that set

\begin{align} c &= 1 \\ \hbar &= 1 \\ 4 \pi G &= 1 \\ \epsilon_0 &= 1 \\ \end{align}

That would non-dimensionalize the GEM equations

\begin{align} \nabla \cdot \mathbf{E}_\text{g} &= -\rho_\text{g} \\ \nabla \cdot \mathbf{B}_\text{g} &= 0 \\ \nabla \times \mathbf{E}_\text{g} &= -\frac{\partial \mathbf{B}_\text{g} } {\partial t} \\ \nabla \times \mathbf{B}_\text{g} &= -\mathbf{J}_\text{g} + \frac{\partial \mathbf{E}_\text{g}} {\partial t} \\ \end{align}

as well as the EM counterparts:

\begin{align} \nabla \cdot \mathbf{E} &= \rho \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}} {\partial t} \\ \nabla \times \mathbf{B} &= \mathbf{J} + \frac{\partial \mathbf{E}} {\partial t} \\ \end{align}

In both cases of GEM radiation and EM radiation, using these L-H rationalized Planck units would normalize the speed of propagation in free space to 1 and the characteristic impedance of free space to 1.

My understanding that the $2$ that multiplies $4 \pi G$ that gets us to $8 \pi G$ comes from the square-root appearing in the proper time formula:

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

or

$$(\mathrm{d} t)^2 = g_{\mu \nu} \ \mathrm{d} x^\mu \ \mathrm{d} x^\nu$$

where, in weak, time-independent gravitational fields

$$g_{\mu \nu}=\begin{cases} 1+h \qquad & \text{if } \mu=\nu=0 \\ -1 \qquad & \text{if } 1 \le \mu=\nu \le 3 \\ 0 \qquad & \text{if } \mu\ne\nu \\ \end{cases}$$

dunno what $h$ is. I'm getting this from this very old sci.physics.research thread that happened also to have Baez and Bunn participating. But I think Daryl McCullough had connected the final dots for explaining where the additional $2$ comes from in the $8\pi G$.

Anyway, it seems to me that normalizing $4 \pi G = 1$ and $c=1$ would change the Baez/Bunn statement from using the word "proportional" to "equal" and might make the statement go as

Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is equal to its volume times: the energy density at the center of the ball, plus the pressure in the $x$ direction at that point, plus the pressure in the $y$ direction, plus the pressure in the $z$ direction.

or

$$-\frac{\ddot{V}}{V} \Bigg|_{t=0} = \rho_\text{g} + P_x + P_y + P_z$$

Isn't that the simplest way to express it?

• G is not the metric tensor. It is called the Einstein tensor and is equal to R - 1/2 g R where R is the Ricci tensor, g the metric tensor and R the Ricci scalar. – astronat Jul 16 '17 at 21:09
• okay, i was trying to identify a symbol in that proper time formula. so the $g_{\mu\nu}$ in the proper time equation is not one of the elements of $G_{\mu\nu}$? i guessed that. – robert bristow-johnson Jul 17 '17 at 5:21
• it's not my post, but one from Daryl McCullough in this 17-year old post, and later he says for flat spacetime, $g_{0,0}=1+h$, $g_{1,1}=g_{2,2}=g_{3,3}=-1$, all others zero. i didn't know where these came from, but i guessed some 4x4 matrix. – robert bristow-johnson Jul 17 '17 at 5:30