# Do the units of $G/c^4$ make intuitive sense?

In classical Newtonian mechanics, gravity is explained by:

$$F=G \frac {Mm} {r^2}$$

Where $F$ is the force due to gravitation, $M$ and $m$ are the masses of the two bodies in question, $r$ is the distance between them, and $G$ is the Universal Gravitation Constant as defined by Newton. In SI units, $G$ is measured in $N\frac {m^2}{kg^2}$.

However, Einstein later discovered that $m$ (at least at rest) was equal to $E/c^2$, where $E$ is energy (in this case, the confined energy inherent to rest mass) and $c$ is the speed of light. Plugging this into the Newtonian Gravitation Equation (completely unorthodox, I know) yields $$F=G* \frac {\frac E {c^2} \frac e {c^2}} {r^2}$$$$F= \frac G{c^4} \frac {Ee}{r^2}$$$$F=\bar G \frac {Ee}{r^2}$$ The new quantity $G/c^4=\bar G$ has SI units of $N \frac {m^2} {J^2}$ which after a little math reduces to $m/J$ as $Nm=J$. Thanks to General Relativity, this reduction feels perfectly intuitive - meters of distortion per joule of energy.

However, General Relativity replaced Newtonian Gravity, and I don't actually have any real intuitions for General Relativity, so I have two questions: Is "meters of distortion per joule of energy" a reasonable way of thinking about the General Relativity concept of gravity, and does my derivation of $\bar G$ make any sense?

This probably isn't the best way of thinking about things, for several reasons:

• In GR, the curvature of spacetime is not determined by the amount of matter/energy present (in Joules), but by the density of the energy (in Joules per $\text{m}^3$.)

• In fact, spacetime can be distorted in GR in a region not containing any matter or energy at all, as has been shown 4.9 times by LIGO (to date). Thinking of distortion as inherently produced by nearby energy is liable to lead you astray. (And if you try to think of the distortion on Earth as being related to the mass/energy of the black holes in a LIGO event, you would conclude that the distortion should be gargantuan.)

• "Meters of distortion" is not the right way to think about the distortion of spacetime. For example, in a gravitational wave event, if we have two objects a distance $x$ apart, then the distance between them changes by an amount $\Delta x$ that's proportional to $x$. The quantity $\Delta x/x$ is called the gravitational strain, and it's dimensionless.

To me, your interpretation of "meters of distortion per joule of energy" sounds a bit like you're taking the rubber-sheet analogy too seriously. It's a fine picture for thinking about what's going on in a quantitative sense, but it fails pretty rapidly if you try to go beyond that.

• Regarding your second bullet point, how has LIGO demonstrated something 4.9 times? Oct 27, 2017 at 21:24
• @Asher: Sorry, it's an in-joke among LIGO folks. An event labelled LVT151012 was detected in October 2015 at something like a 90% confidence level. This wasn't high enough to qualify as an "official" detection like the other four gravitational wave events. The joke is therefore that they detected 90% of a gravitational wave. (Also, I guess it's officially 5.9 events as of last Monday.) Oct 27, 2017 at 21:49
• Ah! That makes sense, and now I know to start using "confidence level" instead of "honorable mention" when explaining physics like this to my 2.3rd child. Oct 27, 2017 at 21:54
• actually density is not as determinant as you think of curvature. A galactic black hole has barely the average density of water, while a small black hole might have neutron star densities Mar 22, 2019 at 6:11

Congratulations! You have independently (partially) rediscovered Einstein's theory of general relativity.

The action of general relativity (with cosmological constant $$\Lambda$$) is $$S = \frac{c^4}{8\pi G}\int \varepsilon_{abcd}e^a e^b (d\omega^{cd} + \eta_{lm}\omega^{cl}\omega^{md} + \frac{\Lambda}{6}e^c e^d).$$

Your $$\frac{G}{c^4}$$ is only $$8\pi$$ away from the (inversed) coefficient in front of Einstein's action. The differential forms $$e$$ and $$\omega$$ in the action are just some minute technicalities.

Similarly, the same coefficient is moved to the right hand side of Einstein's field equation $$R_{\mu\nu}-\frac {1}{2}g_{\mu\nu}R+g_{\mu\nu}\Lambda=\frac {8\pi G}{c^4}T_{\mu\nu},$$ where $$T_{\mu\nu}$$ is the energy-momentum tensor, which echoes your observation.

Absolutely, this is a great way to remember and think about G. For example, in GR the Schwarzschild radius of a black hole is: $$R_{Schwarzchild}=2\frac{G}{c^2}M \quad (=3 \ km \ for \ our \ Sun)$$ Many physicists remember $\frac{G}{c^2}=1.5 \ km \ per \ solar \ mass$, and this is what they use to do astrophysics calculations. Particle physics theorists remember $\frac{G}{c^4}=10^{-39} \ Fermi \ per \ GeV$ Actually nobody writes the c's until they have to in their final answers to get the dimensions correct. Almost never does someone have to look up and use $G=6.67 \ 10^{-11} \ \frac{m^3}{kg \ sec^2}$.

• It's nice to see that I got something "right" all by myself. As to my second question, would my derivation of "$\bar G$" be useful? I mean, did Einstein use it as a stepping stone, or is it taught to students like this before they get to the tensors, or what? Oct 26, 2017 at 13:00
• In none of these situations, though, does $G/c^n$ have the interpretation of "meters of distortion per Joule of energy." Oct 27, 2017 at 20:46

$$\frac{G}{c^4}$$ is the inverse upper limit of the classical gravitational force. That is it's true interpretation.

The intuitive interpretation of the units of $$G/c^4$$ is “Riemannian curvature per energy density”, as follows from the Einstein field equations. Riemannian curvature is measured in units of inverse-meters-squared, and energy density is measured in units of Joules per cubic meter. The ratio of these is meters per Joule.

• How did you get ''Riemannian curvature per energy density'' from? G/c^4 is well-known as the Planck force, or an upper limit of the classical gravitational force and has no units that correspond to an ''energy density.'' Apr 10, 2019 at 3:28
• The Planck force is $c^4/G$, not $G/c^4$. $G/c^4$ does not have the units of a force; its units are meters per joule. It does have the units of an inverse force, but that is not a useful way to look at it. My answer is self-explanatory and I don’t know what I could add to it to help you understand it. Apr 10, 2019 at 3:50
• its an inverse force, I thought I need not mention that. What do you mean does not have units of force? Apr 10, 2019 at 7:46
• It doesn't matter if it is written c^4/G or as G/c^4, being inverse of each other, it still wouldn't have units of energy density. Apr 10, 2019 at 7:47
• I never said it did. I said it has the units of Riemannian curvature per energy density, because it does, as my answer explained. Apr 10, 2019 at 14:33