TransWikia.com

Is the gravitational constant $G$ a fundamental universal constant?

Physics Asked by Farhâd on July 3, 2021

Is the gravitational constant $G$ a fundamental universal constant like Planck constant $h$ and the speed of light $c$?

7 Answers

Velocity of light $c$, Elementary charge $e$, Mass of the electron $m_e$, Mass of the proton $m_p$, Avogadro constant, $N$, Planck's constant $h$, Universal gravitational constant $G$ and the Boltzmann's constant $k$ are all considered as the fundamental constants in Astrophysics and many other fields.

If any of these values would've to change, there would be a great contradiction differentiating our measured values with that of observed & predicted ones.

  • But, there are cases where $G$ is currently accepted as a variable with some standard deviation 0.003 which is too small. Hence, we use $6.67times10^{-11}Nm^2kg^{-2}$ for doing most of our homeworks. The thing is, It's still fundamental...!

So far, investigations have found no evidence of variation of fundamental "constants." So to the best of our current ability to observe, the fundamental constants really are constant.

Answered by Waffle's Crazy Peanut on July 3, 2021

It is probably constant - at least we have no evidence of any change.

"Is it fundamental?" is the big question of theoretical physics. Nobody has yet managed to derive it in terms of more fundamental constant - but a lot of people have tried.

Answered by Martin Beckett on July 3, 2021

For all intents and purposes, it is a fundamental constant. No one has been able to prove that it isn't fundamental, and within our error in measurement, it's definitely a constant. Like @Crazy Buddy says, $c$ (speed of light), $h$ (Planck's constant), $k_{B}$ (Boltzmann's constant) are all considered to be fundamental constants of the universe. You could have a look at this wiki page.

I think it's important also to realize that the values that they have are only valid within a particular unit convention for measurements. For example, $G = 6.67 times 10^{-11} m^{3} kg^{-1} s^{-2}$ but this value will obviously change if you measure it in say centimeter-gram-second (cgs units). You could also set $G = 1$ (which they do in Planck units), except the rest of your units will have to change accordingly to keep the dimensions correct.

I hope the last part wasn't confusing.

Answered by Kitchi on July 3, 2021

There is an alternative to General Relativity known as Brans-Dicke theory that treats the constant $G$ as having a value derivable from a scalar field $phi$ with its own dynamics. The coupling of $phi$ to other matter is defined by a variable $omega$ in the theory, that was assumed to be of order unity. IN the limit where $omega rightarrowinfty$, Brans-Dicke theory becomes General Relativity. Current experiments and observations tell us that if Brans-Dicke theory describes the universe, $omega > 40,000$. Other theories with a varying $G$ would face similar constraints.

Answered by Jerry Schirmer on July 3, 2021

Real "fundamental" constants should be dimensionless, i.e. numbers that don't depend on units. The existence of $c$ is simply due to the Lorentzian nature of spacetime; it's value is only a matter of choice of unit. The existence of $hbar$ is simply due to the path integral or canonical commutation relations, whose value is again a matter of choice of unit. Similar for Boltzmann constant etc.

On the other hand, the fine structure constant $alphasimeq 1/137$ is dimensionless, so this quantity actually means something other than choice of unit. But the number of the quantity is still not that "fundamental" (we will discuss whether the quantity itself is fundamental in the next paragraph) because the number can change by running renormalization flow - i.e. it changes if you define it on different energy scales. So it's the quantity, rather than the number, that has some actual physical meaning.

In the Standard Model of particle physics there are a bunch of such dimensionless quantities. Are these quantities "fundamental"? People tend to believe NO, because Kenneth Wilson let us realize that quantum field theories like the Standard Model are just low energy effective theories that has some high energy cutoff (just like nuclear physics is effective theory of Standard Model); dimensionless quantities in an effective theory should be depend on those in the higher level theory (just like the dimensionless Reynolds number that tell about the behavior of a fluid depends on the molecular constituent of the fluid). String theorists etc are trying to find a theory that has a least number of dimensionless quantities. Some people think an ultimate theory of everything, if exists, should best has no such quantities at all but only numbers that has math significance (like $1, 2, pi$, or some number with certain analytical, algebraic or topological significance).

In terms of the gravitational constant itself, people generally believe Einstein's General Relativity is an effective theory whose cutoff is about (or lower than) the Planck scale ($sim 10^{19} GeV$, our temporary experimental reach is $sim 10^4 GeV$ in the LHC), above which it needs to be replaced by a theory of quantum gravity. But the quantity $G$ might still be there (just like it was from Newton, but still there after Einstein), we are not sure.

Answered by Jing-Yuan Chen on July 3, 2021

Define the Planck length as

$$L_P=sqrt{dfrac{Ghbar}{c^3}}approx 1.6cdot 10^{-35};m$$

or the Planck area as

$$A_P=L_P^2approx 2.6cdot 10^{-70}; m^2$$

Many theories of quantum gravity argue that something weird must happen at that ludicrous tiny lenght/area that any current theory cannot fully describe. Some people, like Padnabhan and others, are beginning to realize and propose that, maybe, what it is fundamental is the Planck area (or length). You will see often Newton's gravitational law written as:

$$F_N=Gdfrac{Mm}{R^2}=dfrac{L_p^2 c^3}{hbar}dfrac{Mm}{R^2}$$

or

$$F_N=dfrac{L_p^2}{hbar c}dfrac{EE'}{R^2}$$

in some recent research papers. About if it is $G$ or $L_p$ or $L_p^2$ fundamental, there are suspicions from black hole physics that it is the Planck length squared or the Planck area what it is more fundamental, it is just the analogue of phase space "area" ($hbar$ or $hbar cdot 2pi=h$). However, people have good criticism on the fundamental role of Planck area (proposed by Bekenstein as some kind of Bohr-Sommerfeld quantization rule for black holes).

At last, what are the fundamental constants depend on our conventions. If could be wonderful if we could define the speed of light as the velocity that runs one Planck length in one Planck time, but that is trivial. Today, we keep c as fundamental in natural units, AND that allows us to define the meter exactly.

There are different systems of "fundamental natural units": Stoney's, Schrödinger's, Pavsic's dilational natural system, Planck's and some other minor variants (called geometrical or reduced sometimes).

Curiously, there are a curious relationship concerning the "fundamental" role of G (or Planck length/area) and the apparent dimensionality (D=4) of our spacetime. The argument, as far as I know, is due to John D. Barrow (perhaps someone else guessed it, but he is the one to have published it first, to my limited knowledge). John D. Barrow last contribution also hints towards a "maximum tension", also highlighted in theories like (super)string theory/M-theory. In "Maximum Tension: with and without a cosmological constant" John D. Barrow and G. W. Gibbons find the maximum force

$$F_M=dfrac{c^4}{4G}=dfrac{hbar c}{4L_P^2}$$

Following the maximum tension paper, we see that in N-space (not spacetime, but it can be generalized carefully) the quantity

$$Xi (G,hbar, c, e)=hbar^{2-N}e^{N-1}G^{(3-N)/2}c^{N-4}$$ is dimensionless and the higher-dimensional generalization of the fine structure constant alpha $alpha$ (case N=3). Moreover, note that ONLY and ONLY if N=3 is G excluded!!!!!!!! And $hbar$ is excluded if N=2!!!!!! or c if N=4. Nota, also, that $G=M^{-1}L^NT^{-2}$, $e^2=ML^NT^{-2}$, $c=LT^{-1}$ and $hbar=ML^2T^{-1}$, so $c,hbar$ are independent of dimension with these definitions. Why $G$ does not "play" a role when $N=3$? Maybe that and the rewritten G in terms of planck length or planck area is the answer you are searching for, but nobody knows for sure yet the best or ultimate answer to your riddle.

Answered by riemannium on July 3, 2021

If the 3+1 dimensional version of Einstein's field equations are derived from a higher-dimensional version (n+1 dimensional for n > 3), such as they would be in any Kaluza-Klein theory, including super-gravity or string theory, then the coupling κ must arise, in some fashion, from the coupling κ' for the higher-dimensional theory; presumably a composite expression κ = f(κ') involving κ' and integral expressions (or even integral invariants) over the extra dimensions.

So, the going consensus - at least as far as there is a consensus that one of these more fundamental theories holds true - is that neither κ nor G is fundamental.

An additional consequence of this, that people almost universally fail to note, is that this then also means that the Planck units are not fundamental either! Instead, the corresponding "n+1-Planck" units (namely, those derived from c, h and κ', instead of from c, h and κ) would be fundamental. And they can be quite different!

Likewise, a consequence of this (almost equally neglected) is that any discussion that is premised on the fundamentality of the Planck units, must therefore be taken only with a grain of salt. Instead, the appropriate discussion would (in the cases cited above) be that which takes the n+1-Planck units as being the ones which are fundamental. And, again, they can be quite different in size!

The units for the respective coefficients are [h] = ML²/T, [c] = L/T, [κ] = T/M and [κ'] = TLⁿ⁻³/M. For 1+1 dimensional space-time, n = 1, there would be no Planck units at all since, in that case, [hκ'] = 1. Otherwise, the corresponding Planck length would be, up to constant factor, ⁽ⁿ⁻¹⁾√(hκ'). For an 11+1 dimensional theory, for instance, that would be the 10th root of hκ'. On dimensional grounds, an expression for κ would be one of the form κ = f(κ') = (⁽ⁿ⁻¹⁾√(hκ')²) f₀/h, where f₀ is a function only of dimensionless ratios. Either this has a dependency on Planck's constant h, or has a factor f₀ which is a non-trivial function of dimensionless ratios, at least one of which involves h.

Answered by NinjaDarth on July 3, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP