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Is $G$ just a value to fix the units in the equation? Something like a proportionality constant or a coupling constant?:

$$F=\frac{G m_{1} m_{2}}{r^{2}}$$

Does it have any physical meaning or physical origin?

What does it represent in General Relativity theory?

Markoul11
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3 Answers3

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Is G just a value to fix the units in the equation? Something like a proportionality constant or a coupling constant?

As you say, it is just a value to fix the units in the equation. There is no physical content to its value other than to describe the size of the unit of mass compared to the units of length and time.

What does it represent in General Relativity theory?

Simply a unit conversion. Usually we simply get rid of it by using units such that $c=G=1$. These are so commonly used that they are called "natural units". These include Planck units and Geometrized units among others.

Does it have any physical meaning or physical origin?

No, or rather no physical meaning beyond the choice of units. The physically meaningful constants are dimensionless. To get physically measurable differences requires a change to one of the dimensionless constants.

Dale
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(G) first appeared in Newton's equation and is needed to predict the magnitude of the gravitational force between two known masses. (It's value will depend on the system of units you are working with.) Its value was first determined with the Cavendish apparatus: a spherical mass on each end of a thin rod, suspended on a fine wire, and brought near to two other spherical masses. I'm guessing that it is still the least accurately known of the various physical constants.

R.W. Bird
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This answer is about the physical origin of gravity.

It comes about due to the requirement that the universe should have scaling symmetry and the principle of conservation of energy.

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If the universe were to rescale, then all physical constants and quantities with length dimensions would vary as in the table below, where $H$ is a scaling constant.

\begin{array}{c|c|c} {quantity} & {length-dimension} & {change}\\ \hline length & 1 & e^{Ht}\\ mass & 0 & constant\\ time & 0 & constant\\ h & 2 & e^{2Ht}\\ c & 1 & e^{Ht}\\ G & 3 & e^{3Ht}\\ Area & 2 & e^{2Ht}\\ \end{array}

Then imagine a mass $m$ of energy $mc^2$, it's energy changes to $mc^2e^{2Ht}$.

That's without gravity, however with gravity the total energy due to the mass is

$$mc^2 - \frac{GMm}{R}$$

Where $M$and $R$ represent the mass and radius of the universe up to the Hubble radius and small numerical constants are omitted for simplicity.

Then energy can be conserved as the universe rescales if

$$(mc^2 - \frac{GMm}{R})e^{2Ht} = 0$$

and $$G=\frac{Rc^2}{M}$$

That gives a reason for the value of $G$ and an explanation of the flatness problem.

The equivalent from General Relativity is $$G=\frac{3H^2}{8\pi \rho}$$

i.e the universe is at critical density.

See also this link

John Hunter
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  • Most intriguing answer. I never thought about it in this way. Thank you. – Markoul11 Dec 07 '21 at 09:46
  • @ Markoul11 you're welcome – John Hunter Dec 07 '21 at 09:46
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    @Markoul11 Please keep in mind that this is, to my knowledge, a very non-mainstream view, and a proprosed solution to the flatness problem I never read before. I would not know why the energy of any mass $m$ should include the potential energy of the whole, but only the observable universe. Also, maybe not unimportant, "omitting small numerical constants" lead to omitting a factor of 2 in the final answer... if you follow the "reasoning" correctly, and assume that $M_{Hubble}=4/3\pi R_{Hubble}^3$, the density of the universe would need to be at twice the critical density (which it is not). – Koschi Dec 07 '21 at 10:42
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    @Koschi Interesting argument. So what is your take on this question? I am eager to see your answer. Thank you. – Markoul11 Dec 07 '21 at 10:52
  • @ Koschi the energy mentioned is the total energy due to the existence of the mass $m$. Integrating in shells leads to a factor 1/2 in the last equation instead of 3/8, that's quite close given that Newtonian gravity was used instead of GR. – John Hunter Dec 07 '21 at 11:11
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    If "Newtonian gravity was used" it's circular to say this explains the physical origin of gravity. Also, anyone unsatisfied with the idea that Newton's constant was handed down from the heavens is also going to think this about Hubble's constant. – Connor Behan Dec 07 '21 at 12:27
  • @ Connor Behan at the moment physicists are experimentally measuring two constants, Hubble's constant $H$ and $G$. With the approach of the answer that can be reduced to one, with $G$ being explained in terms of $H$. It leads to the conclusion that the matter density would be measured to be between 0.25 and 0.33 (although really 1.0) https://physics.stackexchange.com/questions/620794/cosmology-an-expansion-of-all-length-scales – John Hunter Dec 07 '21 at 12:34
  • what about the possibility of $G$ being a coupling constant, similar to electrostatics? How this will translate is GR spacetime geometry? – Markoul11 Dec 07 '21 at 20:10
  • @ Markoul11 You might need an expert in GR or particle physics to comment. Apparently there is no reason within GR for the value of $G$, (or $\kappa$), it's got experimentally so that GR matches Newtonian gravity in the weak field limit. The precision of the coupling constant in particle physics depends (according to this website, https://en.wikipedia.org/wiki/Gravitational_coupling_constant) on the values of $G$ and the mass of the electron $m_e$. All the best. – John Hunter Dec 07 '21 at 20:29