# Why is gravity weak at the quantum level?

Why is gravity stronger than other forces at the macroscopic level, yet weaker than other forces at the quantum level? Is there an explanation?

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Possible duplicate: physics.stackexchange.com/q/4243/2451 –  Qmechanic Apr 24 '12 at 10:04

The gravitational constant at the quantum level leads to a very much smaller force than the forces the elementary particles see in their vicinity, in order of strength:

weak, electromagnetic, strong

The weak and the strong are short range forces, their effect disappears when the sizes grow larger than a nuclear radius, order of a fermi. They cannot build up into one strong component that can appear macroscopically.

The electromagnetic force is a long range one like the gravitational force, and stronger, BUT . It has two opposite charges that attract, same charges will repel. This means that mass agglomerates will be mainly neutral, assuming equal positive and negative charges were created at the Big Bang.

Gravity, in contrast is only attractive and can and does build up to the forces we see controlling the space around us, the galaxies and clusters. It is the one that survives at long distances, because of its 1/r collective potential and its attractive only character, so it cannot be masked as the electromagnetic one can be and is.

As an aside, in space the electromagnetic force can be quite evident as a state of matter called plasma which carries magnetic fields and creates storms in space starting from sun explosions. Still the collective effects of massive bodies give gravity the lead macroscopically.

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This may be too simple an answer, depending on whether you're asking for a deeper underlying cause or the reason for the manifestation, but...

Gravity is strong at larger scales because a) it is a long-range force, falling off with the square of distance (as compared to the strong and weak nuclear forces, which are very short range, on the scale of nuclei), and b) all matter has a positive gravitational pull (pardon the imprecise phrasing) on all other matter, in contrast with the electromagnetic force, which has both positive and negative charges that allow for cancellation of electromagnetic forces on larger scales. Since gravity is always attractive, it just keeps adding up.

As to why gravity is weak as an elemental force as compared with the other fundamental forces, perhaps someone more knowledgeable than I will be able to present a reason. To the best of my knowledge, however, there is no "reason." The universe just is that way. Mostly, when physicists say something is fundamental, they mean they don't know why, it just is. You can invoke the Anthropic principle to serve as an explanation, but I find that dissatisfying without evidence that a wide array of possibilities exists, e.g., in some form of multiverse.

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There is a nice argument explaining why gravity must be a little weaker than other forces, but there's no reason why it must be so much weaker than other forces. You might best think of this as a consequence of the power-running of gravity, and the log-running of other couplings. Near the Planck scale, gravity is only a little weaker than the other forces, it's a factor of 10 or so, and it is only at low energies that the other interactions swamp it.

Evey black hole needs to be allowed to decay completely into elementary particles, so that the black hole S-matrix makes sense. If you form a black hole from incoming states, then let the black hole classically decay, it must link up to an out-state S-matrix. There should not be an enormous number of stabilized massive elementary black hole states.

Large black holes with mass M and charge Q neither attract or repel when Q=M (in appropriate Planck units). This is an exact solution to the Einstein/Maxwell system, which you can derive by reexpressing the extremal Reissner Nordstrom solution in isotropic (x,y,z,t) coordinates.

$$ds^2 = - (1-{q\over r})^2 dt^2 + {dr^2\over (1-{q\over r})^2} + r^2 d\Omega^2$$

To find isotropic coordinates, you want a function $u(r)$ such that the inverse function $r(u)$ has the property that

$${ r'^2 \over (1-{q\over r})^2} = {r^2\over u^2}$$

So that the transformed spatial metric is

$$f(u)(du^2 + u^2 d\Omega^2) = f(u)( dx^2 + dy^2 + dz^2)$$

The differential equation gives

$${dr\over r-q} = {du\over u}$$

Which is solved (with correct boundary conditions) by $u= r-q$ (this is remarkable-- the r coordinate for Reissner Nordstrom is also the isotropic coordinate up to a shifting to make the horizon collapse to a point at the origin).

So the metric in isotropic form is

$$ds^2 = - {1\over{(1+ {q\over u}})^2} dt^2 + (1+ {q\over u})^2 (dx^2+dy^2+dz^2)$$

Which you rewrite in a more conceptual form as

$$ds^2 = - {1\over (1 + \phi)^2} dt^2 + (1+\phi)^2(dx^2 + dy^2 +dz^2)$$

Where $\phi$ is the fundamental solution to Laplace's equation. From this, and the locality property of the Einstein Maxwell system, it's reduction at weak fields to Laplace's equation, plus the superposition property, one can immediately see that the solution for two static extremal black holes sitting next to each other at isotropic position 0 and a must be:

$$\phi(x) = {q_1\over |x|} + {q_2\over |x-a|}$$ $$F_{i0} = \partial_i \phi$$ $$g_{\mu\nu} = -{dt^2\over (1+\phi)^2} + (1+\phi)^2(dx^2+dy^2+dz^2)$$

And this is a famous mysterious exact solution in Hawking and Ellis. The above manipulations and the physical argument justify it's form and existence. Two such black holes just sit next to each other, the gravitational attraction is balanced by the electrostatic repulsion.

### Consequence for charged particle spectrum

If the lightest charge particle has $Q<M$, then there isn't enough energy in the extremal or near-extremal classical black hole to decay into these, so it is constipated--- it can't get rid of its charge, and you have a tower of stable states.

If a large number of $Q<M$ particles are thrown together slowly, they will produce a black hole, which then will be large and thermal, and will slowly decay by Hawking radiating photons or gravitons toward extremality. The resulting extremal black hole will not be able to decay at all, so that the process is one-way and seems to violate unitarity.

This physical argument is the level of this principle, but it states that for all charges the lightest particle must have $M\le Q$, the gravitational attraction must be the same or weaker than the electrostatic repulsion. Unfortunately, this is very weak compared to what is observed: which is $M<<<<Q$!

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Ron, you need to copyedit this answer... –  Mitchell Porter Apr 24 '12 at 10:33
Could you give a reference to the existence of a static solution of Einstein's equation that contains two extremal Nordstrom holes? The claim "Large black holes with mass M and charge Q neither attract or repel when Q=M" is of non-obvious truth value to me. Or is this some string theory result? I'll admit to not being a string theory jock. –  Jerry Schirmer Apr 24 '12 at 13:27
@JerrySchirmer: It's not just two static extremal black holes, but any number, including a continuous density of such black holes. It's in the "exact solutions" chapter of Hawking and Ellis, and it's the mysterious looking solution to the Einstein/Maxwell system where the metric is parametrized by an arbitrary solution $\phi(x)$ to Laplace's equation. If you use the solution $\phi=\sum_i {a_i \over |x-c_i|}$ you get the static extremal black hole solution. It's also obvious in string theory, as these are the static branes, but it's more obvious classically. –  Ron Maimon Apr 24 '12 at 15:34
@MitchellPorter: Oops! I had bald less-than signs in the text that were eating chunks of text, as the parser mistook them for html tags. I expanded the answer and copyedited--- thanks. –  Ron Maimon Apr 24 '12 at 15:56