# What is the maximum distance for the action of the strong nuclear force?

If the strong nuclear force is carried by virtual fermions, they ought to be limited in range by the uncertainty principle. Can anybody calculate the maximum distance and tell me whether the decrease in strength with distance is abrupt at the end, or follows an inverse-square law, or whether some other kind of trailing off happens?

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• Anonymous

The strong nuclear force is not mediated by fermions - it is mediated by gluons, which are gauge vector bosons.

Gluons mediate the strong attraction between quarks. Quarks make up the nucleons that make up nuclei.

Quarks have colour charge, as do gluons themselves. There are three types of colour charge. Hence as quarks exchange gluons they change colour charge accoriding to strict rules. This means there are actually 8 types of gluon.

The force between nucleons (between protons and neutrons) is a residual strong force, caused because the echange of gluons between quarks is not symmetric (its like the polarity of the water molecule caused by the distribution of charge, even though the molecule is neutral).

The strong nuclear force is not inverse square. As you separate quarks the energy between them increases, not decreases. Pulled far enough apart it is energetically better for the field to split into a new quark-antiquark pair.

This means that the residual strong force between nucleons drops off very rapidly with distance - and the force has a total range of about 10^-15 m.