Why is it that the gravitational force acts on large sized objects while the strong and weak nuclear forces act at subatomic levels only? What is that stops each other to enter each others domain?


As you can probably read on Wikipedia, there are roughly-speaking four fundamental forces:

  • Electroweak Force: This force can be both repulsive and attractive. The electromagnetic force is long-ranged, while the weak force has short range.
  • Strong Force: This force is attractive, but has a short range.
  • Gravity: This is an attractive, long-range force.

‘long range’ means a potential of $\frac{1}{r}$, that is, reciprocal in the distance (force proportional to $r^{-2}$), ‘short range’ means an exponentially decaying potential (force proportional to $e^{-r}$).

It is easy to see that the weak and strong force cannot act on large objects: Large objects are usually also far away from each other (else you would have to describe them at a smaller level) and these forces are short-ranged. This short range is usually attributed to the mass of the particles mediating the force.

The electromagnetic force, on the other hand, is a long-ranged force - in theory. However, since atoms themselves are usually neutrally charged, you won’t find large charged bodies in practice.

And this is the reason that the only force acting on large scales is gravity, as it is both attractive (meaning that there aren't ‘gravity-neutral’ objects) and long-ranged.

However, since gravity is so weak (compared to the other forces on atomic scales), it can be safely ignored on small scales.

Note that I assumed a classical understanding of gravity restricted to the four spacetime dimensions rather than thinking of some higher dimensions in which gravity acts as strongly as the other forces.

  • $\begingroup$ exponentially decaying only means decay with a factor $e^{-r/r_0}$ for some value of $r_0>0$. In general, this is not proportional to $e^{-r}$. $\endgroup$ – Arnold Neumaier Nov 16 '12 at 10:17
  • $\begingroup$ You should be able to set $r_0 \equiv 1$ by a suitable choice of coordinates. $\endgroup$ – Claudius Nov 16 '12 at 12:38
  • $\begingroup$ Thanks Cladius for the explanation. In e−r/r0, what does r0 represent here? Is there any explanation, that is theoretical sans the empirical methods? $\endgroup$ – sk1 Nov 16 '12 at 14:25
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    $\begingroup$ Think of $r_0$ in $e^{-r/r_0}$ as the ‘average’ length scale on which the force acts. For the strong force, this would be the diametre of an atomic nucleus. $\endgroup$ – Claudius Nov 16 '12 at 17:45
  • $\begingroup$ But different short range forces require different values of $r_0$, so one cannot choose cooordinates to make all these factors vanish. $\endgroup$ – Arnold Neumaier Nov 18 '12 at 12:22

There are four basic forces in Nature.

The gravitational force exerted upon an object is proportional to its mass, hence very large for very massive objects. (They need not be large; a black hole can be small and very massive.) It is long range, leading to a very slow $1/r^2$ decay of gravitation with respect to distance. For example, the gravitational force exerted by the center of a galaxy is noticeable even across a galaxy.

The electromagnetic force exerted upon an object is proportional to its charge. These can be of opposite sign, and hence tend to cancel in larger objects. But when they do not cancel they have a macroscopically visible effect as they are long range, hence remain measurable at microscopically large distances. This makes this force the most useful one in modern life.

The weak and strong nuclear forces are proportional to corresponding [non-electromagnetic] charges, which can be of opposite sign and hence tend to cancel in larger objects. But even when they do not cancel, they are very short range, which means that their strength decays exponentially with distance. Hence they are not noticeable at macroscopic distances.

  • $\begingroup$ Thanks Sir. Can you please explain the gravitational forces' effect also w.r.t. distance, theoretically? $\endgroup$ – sk1 Nov 16 '12 at 14:31
  • $\begingroup$ @sk1: done in my expanded first paragraph. $\endgroup$ – Arnold Neumaier Nov 18 '12 at 12:26

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