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I understand that Einstein's gravitational constant is a sum of multiple constants and has a value of $8\pi G/c^4$, but what exactly does it represent?

Chris
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Jungwoon Song
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    Constants in physics theories are necessary to convert mathematical results to the units of the particular measurement in order to compare data with theory. – anna v Jun 13 '22 at 04:10
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    Actually the constants in physics are scale dependent, which means that on higher energy scales the strength of the interaction measured by the coupling can get stronger/weaker. A particular value is unimportant, as it depends on manmade constructions (metric system, cgs system), but it must be there to give a general scale to the interactions. – Kregnach Jun 13 '22 at 07:26

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Dale's answer ("just a conversion factor between stress-energy and curvature in SI units") is correct for the numerics, but I beg to disagree on a dimensional level.
When constants like $c$, $\hbar$, or $G$ are employed, they tell us something about the entity of the phenomena that they describe.
Higher powers of $\hbar$ for example mean higher orders in QFT's loop perturbation theory, meaning that you'll have to do very precise quantum measurements to detect stuff with a high power if $\hbar$. In the same way, higher powers of $c$ mean that a higher velocity will be required to "probe" the phenomenon because all velocities (in relativistic theories) are always "with respect to $c$". The same goes for $G$: when it's present it means we care about the gravitational side of the problem we're considering (and the higher the power, the more we care).
To come back to your question, the $8\pi$ is there only on a numerical basis, but the $G/c^4$ is telling us that we are analyzing a gravitational problem and that we need a lot of energy to see it.
The dimensional analysis is the only reason why, sometimes, non-natural units can be useful even to high-energy theoretical physicists.

Mauro Giliberti
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  • @A.V.S. not at all: $c$ is a "big" constant (meaning that velocities are smaller than $c$) so a negative power is natural, in the same way, $\hbar$ is a "small" constant (angular momentum are multiples of $\hbar$) and positive powers are natural. Higher powers (negative for $c$, positive for $\hbar$) mean a higher order in perturbation theory, or higher (relativistic/quantum) precision if you want. – Mauro Giliberti Jun 13 '22 at 10:46
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    @A.V.S. for this reason, I don't like $E=m c^2$ very much, I think that $m=Ec^{-2}$ is much more meaningful when you want to see what energy will be required "probe" the phenomenon of relativistic mass: to see the mass, you need a lot of energy. – Mauro Giliberti Jun 13 '22 at 10:50
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It is just a conversion factor between stress-energy and curvature in SI units. Usually when doing General Relativity we prefer to use natural units where it is equal to 1. SI units are just not very convenient for GR.

The Einstein gravitational constant does not tell you about the physics, it tells you about your units. It can be set to any value you like by a choice of units.

Dale
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  • I have downvoted your answer because I think the answer - „just a conversion factor“ is too superficial. – JanG Jun 13 '22 at 16:54
  • Do what you feel you need to do. The answer is no more superficial than the question. – Dale Jun 13 '22 at 17:33
  • I beg your pardon. I should not use the word superficial in this context. It is opinion speach and too personal. What I wanted to say is that to my knowledge your second sentence is not right. – JanG Jun 13 '22 at 18:28
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If you write Einstein field equation as $$T_{\mu\nu}=R_{\mu\nu}\cdot \kappa^{-1}, $$ you can interpret $\kappa^{-1}$ in terms of maximal possible force in the nature, see https://physics.stackexchange.com/a/707944/281096, or as equal to Plank force.

A physical quantity is a product of number and unit. The first is pure mathematics, the second pure physics. For example, if you want to get the length of a physical object you have to chose some other physical object as unit (yardstick) and measure with it the object. The number tells you then how many units it has. A physical equation can be always made dimensionless (numbers only), but in order to get physical meaning one has express the physical quantity in the form "number times unit". It seems that for all units it is enough to use some combination of the fundamental constants, which have clear physical interpretation.

JanG
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