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Why is it that gravity is the weakest of the 4 fundamental forces? I know that from experimental data, we can see that it is the weakest, cf. e.g. this Phys.SE post, but is there any way to prove that it is? And if so, is it intuitive? Is it due to the inverse square law?

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    $\begingroup$ Note that the electrostatic force also obeys an inverse square law. This is known as the "hierarchy problem," and is one of the motivations for string theories with extra spatial dimensions. Perhaps that will steer you towards the right part of the literature. $\endgroup$ – rob Nov 23 '18 at 4:41
  • $\begingroup$ It is actually not that easy to define "weakest" and to compare the "strength" of fundamentally different forces. Think about, why you think that gravity is the weakest force. Is it because two 1-kilogram masses seperated by a centimeter pull/push less strongly than two 1-Coulomb charges seperated by a centimeter? Well, that is just because we have defined one unit of mass as 1 kg and one unit of charge as 1 C. What if we had defined one unit of charge as something a million times smaller? It is suddenly not that clear - it is hard to compare forces that do not share a common parameter. $\endgroup$ – Steeven Nov 23 '18 at 9:47
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To my knowledge there is no proof. There are only the experimental evidence, it is the way our universe is built. It is not the inverse square law's fault : the coulomb's law is also an inverse square law, and if G had been bigger the classification wouldn't be the same.

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  • $\begingroup$ Comparing the value of G with the value of $1/4\pi \epsilon_0$ is meaningless since these constants do have different dimensions. $\endgroup$ – GiorgioP Nov 23 '18 at 9:35
  • $\begingroup$ I'm sorry if I have been unclear, I only wanted to show that it was not the inverse square law's fault from a mathematical point of view. I will edit to better reflect it. $\endgroup$ – Q.Reindeerson Nov 23 '18 at 16:29
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$\def\ns#1#2{#1_{\mathrm{#2}}} \def\qy#1#2{#1\,\mathrm{#2}} \def\10#1#2{#1\cdot10^{#2}}$ I strongly counter the use of term "force" in the present context. It's only meaningful for gravity and electromagnetism, but totally devoid of meaning in the other cases.

Let me explain. Gravity and electromagnetism belong - as to their birth and development - to classical physics. Those fields of physics describe a lot of phenomena of macroscopic scale, whose explanation was initially given using ideas of Newtonian mechanics. First of all that of force. You may measure gravitational, electric, magnetic forces between bodies and give a law of force: gravitational (Newton) electric (Cavendish) magnetic (Ampère). Nothing like that is possible for the so-called nuclear, strong, weak "forces". From whichever point of view, these are not at all forces. But more about this afterwards.

The main reason for difference is that gravity and e.m. forces are long range ones, a technical term describing their dependence on distance. In both cases (Newton, Coulomb) it is $1/r^2$. The other "forces" act only at very short distances and decay exponentially (I will be more precise in the following).

Does this mean that a comparison between strengths of gravitational and electric forces is easy? Not at all. Surely it is meaningless to compare the respective constants. Though both force laws share the same mathematical forms $${G m_1 m_2 \over r^2} \qquad {k\,q_1 q_2 \over r^2}$$ a direct comparison of constants $$G = \qy{\10{6.67}{-11}}{m^3 s^{-2} kg^{-1}}\qquad k = \qy{\10{8.98}9}{N\,m^2 C^{-2}}$$ is unphysical, in the first place because their values depend on units system used. In other words, $G$ and $k$ have different physical dimensions.

You may often find the comparison done using some charged particle, e.g. electron or proton. In both cases you read that gravitational force is largely smaller than electrostatic, for equal distances. For electrons we have $${\ns F{gr} \over \ns F{el}} = \10{2.4}{-43}$$ and for protons $${\ns F{gr} \over \ns F{el}} = \10{8.4}{-37}.$$ This is OK to show the as far as systems of 2 or few more electrons, protons, or other similar particles are concerned gravitational force is likely to be absolutely negligible wrt to electrical one (or other interactions, but it isn't trivial at this point). Actually it was a hard task to show that subatomic particles really feel gravitational force. As far as I can remember, the first direct proof was obtained through neutron interferometry (Staudenmann et al., 1980).

But in macroscopic experiments things are rather different. Force ratios are much less unfavourable for gravity - otherwise Cavendish' experiment would have been impossible. This is because for macroscopic bodies the ratio $q/m$ is not so large as it is for particles. E.g. it is $\qy{\10{9.4}7}{C/kg}$ for a proton, whereas it's largely out of question to give a charge of $\qy1C$ to a body of mass $\qy1{kg}$.


The above disposes of the only real forces in macroscopic world. But when it comes to microscopic (quantum) world the concept of force totally disappears. Just from the very beginning QM never spoke of forces. Even in the simplest and historically first application of Schrödinger equation, the hydrogen atom, electron-proton binding is described in terms of potential energy, not of force. This could be seen a secondary change of viewpoint - after all potential energy already belongs to Newtonian physics.

In fact the first attempts to understand the new "forces" were conducted introducing a "nuclear force" between nucleons (proton-proton, proton-neutron, neutron-neutron). The experimental fact that these forces were of very short range (order $\qy1{fm}=\qy{10^{-15}}m$) explained why they produce no macroscopic effects.

But soon QFT come into play. Yukawa idea (1935) was that nuclear force was mediated by a massive particle he named meson. A "force" mediated by a massive particle has a range linked to the mediator's mass. Yukawa introduced a (Yukawa) potential $$V(r) \propto {e^{-kr} \over r}$$ where $$k={m\,c \over \hbar}$$ if $m$ is mediator's mass. The range of such potential is $1/k$ and equating it to $\qy1{fm}$ a value $$m = {\hbar\,k \over c} = \qy{200}{MeV}/c^2$$ results for meson's mass.

Although I continued to use the word "force" in QFT this idea doesn't exist and even that of a potential energy is a by-product. The basic idea is an interaction term added to free-field lagrangian. The potential energy applies only to a limited subset of situations, of scarce interest: the very low energy interaction between two particles (in present case, two nucleons).

I won't recall ensuing developments which brought us still farther from forces and potentials: QCD, gauge theories. All this leads me to state that present status of fundamental interactions has no place for the concept of force, and the very word should be ruled out, as an inevitable source of confusion for laypersons. The only correct term IMHO is interaction.


I still have to write about "weak force" and its supposed weakness. This was never seen as a force in classical sense (as nuclear force was at the beginning and still is at a phenomenological level in nuclear physics). There are no particles held together or acted on in some other sense by a weak force. It only makes itself felt in some decays - first of all in historical sequence nuclear $\beta$ decay, then neutron decay, muon decay, pion decay, and so on.

But before of all that, when only nuclear $\beta$ decay was known, neutrino hypothesis was born to explain continuous spectrum of electrons emitted and the spin puzzle (Pauli 1930). Fermi (1933) coined the first QFT model of $\beta$ decay, as a 4-line vertex (interaction). The original process was $$n \to p + e^- + \nu$$ later replaced by $$d \to u + e^- + \bar \nu_e \tag1$$ and in electroweak unification (Glashow, Salam, Weinberg, late '60s) by $$d \to u + W^- \to u + e^- + \bar \nu_e.\tag2$$

The reason why Fermi 4-field interaction works at low energies is the heavy mass of $W$ boson, about $\qy{80}{GeV}/c^2$. The $W$ propagator in eq. (2) has a $M_W^2+q^2$ denominator ($q$ momentum transfer). If $q^2\ll M_W^2$ it's almost constant and allows reducing (2) to (1), with a constant factor absorbed in the coupling constant.

Fermi theory predicts a decay rate increasing as the square of excess energy. So only at small energies it is correct to see weak interaction as really weak - in fact it would increase without bounds for increasing energy. Such increase is bounded in electroweak theory because the $W$ propagator at high $q$ begins to decrease.

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It seems relevant in this context to mention the weak gravity conjecture (WGC) by Arkani-Hamed, Motl, Nicolis & Vafa (AHMNV).

One of AHMNV's arguments is that black holes (which satisfy an inequality of the form $M \geq |Q_i|\forall i$ in Planck units), should be able to completely evaporate without a remnant in order to save unitarity, see Fig. 2 on p. 6 in AHMNV. Here $Q_i$ denotes a charge of the $i$'th type of force/interaction. This requires "elementary particles" to obey the opposite inequality $M \leq |Q_i|\forall i$ in Planck units, i.e. gravity should be weaker than the $i$'th interaction.

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It is true that gravity is the weakest force and Nuclear Force is the strongest. Ordered from strongest to weakest, the forces are...

1. the strong nuclear force, 2. the electromagnetic force, 3. the weak nuclear force, and 4. gravity.

But If we consider about There ranges , Gravity has Longest effective range ,and that of nuclear force is smallest. Each force dies off as the two objects experiencing the force become more separated. The rate at which the forces die off is different for each force. The strong and weak nuclear forces are very short ranged, meaning that outside of the tiny nuclei of atoms, these forces quickly drop to zero. The tiny size of the nuclei of atoms is a direct result of the extreme short range of the nuclear forces.

The low value of Gravitational constant might be due to the low coupling factor between matter and the gravitational field.Every force works differently and they have different constants. When you put a large amount of charge with a small opposite charge , the force will be strong. But if you put a small amount of mass , you will need very much mass to attract effectively

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    $\begingroup$ This is not really an answer to the question. $\endgroup$ – StephenG Nov 23 '18 at 7:41
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I think that the only meaningful way to state something about gravity being the weakest of the fundamental forces is to fully qualify such a statement by adding that it refers to the corresponding interactions between any pairs of the known elementary particles. Missing that, it is quite obvious that the gravitational force between two neutral planets dominates over their electrostatic interaction.

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Gravity is not always the weakest "force". In some circumstances (if there are lots of matter/energy content), it may be the strongest force of all, able to crush an entire neutron star against the internal nuclear forces and thus producing a black hole after the gravitational collapse.

The "strenght" of a force depends on the energy scale involved. On the human scale, gravity is weak compared to all other interactions because if it was otherwise, we wouldn't be there to measure it! If gravity was stronger, life wouldn't be possible in our universe. Gravity had to be the weakest so there could be observers at some scale. Our own existence implies that gravity is a "weak" phenomenon at our scale, and that life is dominated by the other interactions (electromagnetism, in our case).

The usual forces couplings are defined with some dimensionless parameter. The fine structure constant of electrodynamics is a good example : \begin{equation}\tag{1} \alpha = \frac{e^2}{4 \pi \varepsilon_0 \hbar c} \approx \frac{1}{137}. \end{equation} In the case of gravity, the coupling parameter is $G$, which has dimensions of a squared length (in natural units) : \begin{equation}\tag{2} G \equiv \frac{\hbar \, G_{\text{N}}}{c^3} \equiv \ell_{\text{Planck}}^2 \approx 2.61 \times 10^{-70} \, \text{m}^2. \end{equation} There's a lot to say about all this, but I'm lacking time to elaborate, and I think this answer is a good start to understand why gravity is "weak" in the usual human scale.

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  • $\begingroup$ Exactly. Gravity is so brutally powerful, it can compress matter into incomprehensible densities. It's so powerful, it can create a 10km diameter-sized nucleus (neutron star). To call it a weak force is quite insane, really. $\endgroup$ – White Prime Nov 24 '18 at 1:41
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OP is not asking the right question. It's "not even wrong"! since comparing a dimension-full interaction (gravity) with a dimensionless interaction (standard model interactions) is an ultimate sin if no circumstance is provided.

Given that the "charge" of the gravitational force is mass (energy tensor), the correct question is: why are the masses of the elementary particles so small compared with the Planck scale? As of yet, mortal physicists are still scratching their heads and fretting about this nasty "naturalness/hierarchy/fine-tuning problem". The Nobel laureate Frank Wilczek actually devoted a whole book about it: The Lightness of Being.

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