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A force/field which depends inversely on the square of the distance of the source from the point of interest (like electric field depends on $1/r²$, where $r$ is the distance between the the source charge and the test charge) is said to be 'long range'. Why?

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  • $\begingroup$ What fields don't depend on the inverse square law? $\endgroup$
    – user140606
    Commented Feb 17, 2017 at 17:12
  • $\begingroup$ @Countto10 I just want to know why they are called long range ... $\endgroup$ Commented Feb 17, 2017 at 17:17
  • $\begingroup$ If you $\frac{1}{r^2}$ was said to be short-range, then what would you call $\frac{1}{r^6}$? very-very-short-range? $\endgroup$
    – Yashas
    Commented Feb 17, 2017 at 17:22
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    $\begingroup$ I believe "long range force" just means "a force that happens over a distance". In other words, you don't need direct contact. $\endgroup$
    – Steeven
    Commented Feb 17, 2017 at 17:25
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    $\begingroup$ @Yahas Samaga Just curious. $\endgroup$ Commented Feb 17, 2017 at 17:27

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Actually there is a well accepted sense of what long range means. You do have to be careful with different contexts.

In Quantum Field Theory (QFT) and classical FT the field related to a classical force that goes as $1/r^2$ can be radiative, with its amplitude going as 1/r in the far field (i.e., long range). It means that the amplitude of the field that causes that force can propagate and be 'felt' at great distances. The perfect example is electromagnetic forces, and the electromagnetic field. We are able to detect the electromagnetic field emitted billions of light years away. In General Relativity a propagating gravitational field (manifested as spacetime curvature and equivalent to a gravitational force in the weak field limit), also has its amplitude go as 1/r far enough away, and we can detect it also far away. We detected gravitational waves from black holes a billion or so light years away.

In contrast, nuclear forces, or more basically the strong and weak force, have a field that goes down much faster than 1/r, and are almost undetectable outside the nucleus. A mile away you cannot detect it (but an explosion in an A bomb in a very local area can cause heat and radiation (electromagnetic, as well as particles) to go out for miles. Or, from the Sun, across the solar system. Still, those are NOT nuclear or stron or weak waves, rather electromagnetic and particles that were shot out.).

The basic reason from FT or QFT and equivalent theory is that long range forces or fields are carried by zero mass particles, such as the photon or graviton. Massive force fields like the strong and weak FIELDS decay proportionally to exp(-$kmr$)/r, i.e., negative exponentially in r as well for m not equal to zero. This general behavior can be deduced for QFT and is a good first generic approximation for non-zero mass fields, for a scalar field, as one obtains from a Yukawa potential. The spin 1 and 2 electromagnetic and gravitational fields are more complex but the idea holds as to what would happen if they were massive. The strong and weak fields are even more complex, but this is the basic simplest way of looking at why those other fileds, other than electromagnetic and gravitational, are short range.

There are also non-elementary contact forces like friction, clearly derived from molecular forces, and which are also short range, due to a complex balancing of electromagnetic forces in atoms and molecules. The basic physics forces from which all else are derived are the ones discussed in the paragraphs above

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  • $\begingroup$ Bob Bee, I know what is a magnetic force and an electric force, in both cases are involved dipoles and the force between them could be attractive or repulsive. Furthermore I know what is EM radiation and what is pressure from this radiation on matter. But what is an EM force? $\endgroup$ Commented Feb 18, 2017 at 16:21
  • $\begingroup$ Sorry @HolgerFiedler. I got a bit lazy. Just means electromagnetic. (I don't see where I used but it's probably there) $\endgroup$
    – Bob Bee
    Commented Feb 19, 2017 at 23:41
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This source defines short-range forces as:

Non-bonded interactions can be divided into two classes; short and long range interactions. Formally a force is defined to be short ranged if it decreases with distance quicker than $r^{-d}$ where $d$ is the dimensionality of the system (usually 3).

And the opposite would then be long-range forces and the answer to your question. I'm not sure how official this definition is, but it is a good pointer.

Usually we talk about contact forces (see here and here). They would be the bonded forces that the above quote mentions, and I believe that the short-range forces normally are placed within this category as well - they only act at so close range that it could basicly be defined as contact.

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    $\begingroup$ "Official" isn't the word you want. There is no committee that decides on the definition of words in science. Instead there is consensus and usual practice, so perhaps "wide spread" or "common" would be better. $\endgroup$ Commented Feb 17, 2017 at 17:57
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The term long range force means that at infinite distance the force never actually becomes zero.

These forces do not require any physical contact to exist. Forces like normal force, frictional force need contact in order to exist but forces like gravitational force and electrostatic force do not need any physical contact to come into existence.

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  • $\begingroup$ "The term long range force means that at infinite distance the force never actually becomes zero." It does become zero. $\endgroup$
    – Yashas
    Commented Feb 17, 2017 at 17:58
  • $\begingroup$ google.co.in/… $\endgroup$
    – Mitchell
    Commented Feb 17, 2017 at 18:00
  • $\begingroup$ This is a question of semantics. As $r$ approaches infinity, the force goes to zero. At infinity, it is zero but infinity is not a real number. It does not exist. $\endgroup$
    – Yashas
    Commented Feb 17, 2017 at 18:02
  • $\begingroup$ physics.stackexchange.com/questions/282928/… $\endgroup$
    – Mitchell
    Commented Feb 17, 2017 at 18:11
  • $\begingroup$ Yes, as $r$ approaches infinity, the force goes to zero. But at infinity, it must be zero? $\lim_{r->0} F_g = 0$. $\endgroup$
    – Yashas
    Commented Feb 17, 2017 at 18:13

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