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Suppose they are held 10 nm apart.

What is the dominant interaction between them?

The magnetic dipole interaction or something else?

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up vote 5 down vote accepted

Yes, gravity is negligible, the neutrons are color-neutral so there are only leftover strong-interaction forces which are tiny due to confinement (QCD-related forces between color-neutral objects decrease more quickly than a power law), and the range of the weak force is even shorter (1,000 times shorter than the size of the proton).

At 10 nanometers, a molecular distance scale, only the electromagnetic interaction matters. The fields may be expanded in a multipole expansion. The electrostatic forces are absent because neutrons are neutral, the electric dipoles of the neutrons are tiny due to the CP symmetry (experimentally, they're still compatible with zero), so the magnetic dipoles are the first terms in the expansion that matter.

The electric quadrupole interaction is weaker because it's a higher order i.e. a faster decrease with the distance. But even if we considered the quadrupole, the quadrupole moment of the neutron is zero because a spinor can't be converted to a traceless tensor in this way.

It's a sort of an angular momentum selection rule. The matrix element $$ \langle s_1| T | s_2\rangle$$ between two $j=1/2$ spinors of a $j=2$ operator has to vanish because two $j=1/2$ cannot be combined to yield a (much larger) $j=2$ result.

Quadrupole moments only arise for ellipsoid-shape nuclei that are stretched in one (or two) directions and shrunk in the remaining two (or one) direction. But a neutron cannot have such privileged axes because its spin is just 1/2 and both spin-up and spin-down state correspond to the same "preferred axis". The value $j=1/2$ is just too small and qualitatively similar to $j=0$: when it comes to the quadrupole moments (electric or magnetic), the neutron is exactly spherical.

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