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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?

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    $\begingroup$ 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. $\endgroup$ – astronat Jul 16 '17 at 21:09
  • $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Jul 17 '17 at 5:21
  • $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Jul 17 '17 at 5:30
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What the tutorial authors say may be true or not. I looked it up, and don't plan on wasting my time with their argument that the equation encompasses all of general relativity, and that tricky assumption he states that it applies in all inertial frames. Even if true it is not what I'd recommend on how to learn GR. A lot of non-simple non-straightforward arguments.

What is definitely true is that aside from the unit questions you bring up (which I won't address) is that his first equation is the same as your last. Nothing gained in your manipulations. However, it is true that the equation he writes is true, and can be useful for some intuition.

The equation is indeed one specific result of GR. It does not state for instance anything about gravitational waves in vacuum, except that the relative volume change in change rate is zero. That's true for waves in vacuum. But it does not say anything about what the volume does. We know how from the Einstein field equations. What they do is they deform that volume. His equation is due to a nonzero stress energy tensor, and affects the Ricci tensor. But there's more, the Weyl tensor describes spacetime in empty spacetime, where the Ricci and stress energy Tensors are zero. That includes many different types of spacetimes, labeled by the number of different Weyl eigenvectors. See the Weyl tensor and what it means at https://en.m.wikipedia.org/wiki/Weyl_tensor. In fact, when the tutorial introduces gravitational waves they say there is distortion, but you can't see how they got that. You just have to believe the words. Does that make you understand anything, or closer to nothing?

So, maybe it may general relativity for pea brains (which does not look like you are if you could write those other equations, and BTW unit translations confuse everyone), or it might just involve a lot of word arguments that are not all straight forward, but I'd recommend that if you want to learn general relativity you pick up some of the math, and follow the process. If you know differential equations, or what they mean, you can learn 4D curved geometry enough to follow the real ideas and effects when they are explained.

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