# Spin

Spin (to a physicist, not a PR-merchant) is `the same stuff as' angular momentum, which is (pretty much) the same stuff as action.

## Gravitational dipoles

In the theory of inverse-square central forces (of which the standard examples are gravitation, whose `charges' are all of one sign, and electostatics) a natural idiom is the `dipole'. If one has two equal but opposite charges (somehow being kept) a short distance apart, their net field has zero as its spherically-symmetric part (about the mid-point between the charges) but the slight separation of the charges leaves some residual difference between the electric fields of the two charges: they don't quite cancel one anohter out, and the difference is called a `dipole'. Even when (as with gravitation) the two charges' signs are the same, if the charges aren't equal the net field is well described by combining a central field, that of their total charge centred part-way between them, with a `dipole' describing the asymmetry.

Now, keeping apart a pair of charges that attract one another together (which applies both to a pair of opposite electric charges and to a pair of unequal masses - that is, gravitational charges) requires either an opposing force or a mutual orbit (potentially some combination, of course). At least where mutual orbit is involved, this means that dipoles are wont to spin.

To put that another way, spin is a natural property of dipoles, whether electric or gravitational. As for gravitational ones, I have yet to explore. Now, electic dipoles that spin also radiate electromagnetically - and I am under the impression that gravitational dipoles radiate gravitationally. When such radiation is produced, it must act to reduce the spin-energy of the dipole - whence else may the energy of the radiation come ?

Given that spin is a natural property of dipoles, an obvious question presents itself: is all spin carried by dipoles ?

[Presuming dipoles to be those of differences in an inverse-square law may have deficiencies, but I'll stick with it for now. Intra-nuclear strong or weak dipoles, if any, I'll leave for other folk or other times.]

### The standard dipole

Now, the standard form of dipole field is obtained by considering two equal and opposite charges a short way apart: the field they produce tends to a limit if the charges get bigger while moving closer together at appropriately synchronised rates; they do so everywhere for one particular synchronisation, and the common form resulting is the standard dipole field.

Officially, we consider two equal and opposite charges ±M in a central field with force constant G at positions ±h; with M.h being some fixed (vector) value, though we shall vary M and h (in inverse proportion to one another); as the charges come close together, they get more charged. Next we consider the force per unit charge, F/m, acting on some charge at position r;

• F/m = G.M.(1/((r-h)·(r-h)) - 1/((r+h)·(r+h))

in which · denotes the `vector inner product', aka contraction or `dot'. In regions of space `far from' the dipole, i.e. with r much bigger than h, h·h will be ignorable when added to r·r. For convenience, I'll write (h/r) for (h·r)/(r·r) - when r is `equatorial' to the dipole's axis, this is zero even when h is big, but it's always `small' as long as r is much bigger than h. This reduces (r-h)·(r-h) to r·r - 2.r·h which I now write as r·r(1-2.h/r). This simplifies the above to (using ~ for `is approximately equal to')

• F/m ~ G.M.(1/(1-2.h/r) - 1/(1+2.h/r))/(r·r).

Now, when h/r is small, 1/(1-2.h/r) is approximately 1+2.h/r, give or take some terms of similar size to the h·h/r·r we're already ignoring. This turns our answer into

• F/m ~ G.M.(1 +2.h/r -1 +2.h/r)/(r·r), or
• F/m ~ 4.G.M.(h/r)/(r·r)

Writing μ for 4.M.h (the product of difference in charge, 2.M, and total displacement, 2.h, between the charges) and μ/r for mu·r/r·r, consideration of the errors we ignored reveals that, as h tends to zero, so long as μ remains unchanged, the errors also tend to zero, leaving only the above G.(μ/r)/(r·r): this is the standard dipole field. A pair of equal and opposite charges at a definite (non-zero) separation is then described as having a `dipole moment' (or just plain dipole) given by the difference in charge (twice the positive one) times the total displacement (traversed from negative to positive). Its net field is then given by the dipole field plus a correction, arising from the separation being non-zero, which has `negligible' impact in regions of space where the two charges appear close together.

### The asymmetric case

Now, particularly in the gravitational case, we may want to consider cases where we don't have the pre-conditions of the standard dipole. A slighly more general case is the `two-body' case: a pair of charges, M0 and M1, are separated by a displacement k (which is 2.h in the preceding); suppose M0 to be the `smaller' charge and k to be pointing from it to the other. Take the mid-point of k as origin and use k=2.h as before. The field at position r will now be

• F/m ~ G.(M1.(1+2.h/r) + M0.(1-2.h/r))/r·r, or
• F/m ~ G.(M0+M1 + 2.(M1-M0).(h/r))/r·r

(note: M1 is the bigger charge, so M1-M0 has the same sign as M1+M0). As arranged, this is the sum of two terms:

G.(M1+M0)/r·r

which is the field which would be produced by a single charge, M1+M0, at the mid-point of k

G.(M1-M0).(k/r)/r·r

which is a dipole, again at k's mid-point, with moment (M1-M0).k

in effect, the total field (at least wherever h/r is small) is that of equal and opposite charges ±(M1-M0)/2 at the ends of k, in the presence of the total charge, M1+M0, at its mid-point.

Now, that is good wherever h/r is small - that is, the displacement between the two charges appears `small', formally it subtends a small angle. At close range, we'll get errors we'll notice; however, a little judicious tampering with the origin (as long as we only move it by `small' amounts, like fractions of k) might plausibly make the approximation better into a closer neighbourhood of the two charges. If nothing else, at least for gravity's charges - masses, I really expect the M1+M0 term to behave as if it's centre were, rather than the mid-point of k, a position M1/(M1+M0) of the way along k (starting from M0's end) - the `centre of charge'.

Now, the position in question can't be M1/(M1+M0), at least in the electrostatic case, where M1+M0 could be 0 ... or merely tiny by comparison to either M1 or M0. So allow that our origin is a fraction e of the way along k, starting from M0, so that we have charges M0 at -e.k and M1 at (1-e).k, giving field (at position r relative to our chosen origin)

• F/m = G.(M1/((r-e.k)·(r-e.k)) + M0/((r+(1-e).k)·(r+(1-e).k))), so

If we multiply this by r·r/G, subtract M1+M2 from it and ignore k·k/r·r terms, we're left with

• M1.(-1+1/(1-2.e.k/r)) +M0(-1+1/(1+2.(1-e).k/r)), which is roughly
• 2.M1.e.k/r -2.M0.(1-e).k/r, ignoring (k/r).(k/r) terms, giving
• 2.(M1.e -M0.(1-e)).k/r, or
• 2.(e.(M1+M2) -M0).k/r
Written by Eddy.
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