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...of matter $σ_i$, which is $-\frac{2}{3}(i-1)$ times that potential, and by adding to it the external disturbing potential.
We have now learned how to include the surface action within the potential calculation. If $W_i$ represents the potential of the external disturbing influence, then the effective potential per unit volume at a point within the sphere—now considered free of surface action and of mutual gravitation—is:
$W_i - \frac{2gw(i-1)}{2i+1} \left(\frac{r}{a}\right)^i σ_i = r^i T_i$ (let us assume).
The complete solution to our problem is then found by substituting $r^i T_i$ in place of $W_i$ in Thomson's Lord Kelvin's solution (8).*
However, in order to apply this solution to the case of the Earth, it will be convenient to use polar coordinates. For this purpose, let $w r^i S_i$ represent $W_i$. Let $r$ be the radius vector; $θ$ be the colatitude the angle down from the North Pole, rather than up from the equator; and $φ$ be the longitude. Let $ρ, \varpi, \text{ and } ν$ be the velocities in the radial direction, along the meridian, and perpendicular to the meridian, respectively. The expressions for $ρ, \varpi, \text{ and } ν$ will be exactly the same as those for $α, β, \text{ and } γ$ in equation (8), except that for $\frac{d}{dx}$ we must substitute $\frac{d}{dr}$; for $\frac{d}{dy}$, we must substitute $\frac{d}{r \sin θ dφ}$; and for $\frac{d}{dz}$, we must substitute $\frac{d}{r dθ}$.
Then, after some mathematical simplifications, we have:
$$
\left.
\begin{aligned}
ρ &= \frac{i^2(i+2)a^2 - i(i^2-1)r^2}{2(i-1)[2(i+1)^2+1]v} r^{i-1} T_i \
\varpi &= \frac{i(i+2)a^2 - (i-1)(i+3)r^2}{2(i-1)[2(i+1)^2+1]v} r^{i-1} \frac{dT_i}{dθ} \
ν &= \frac{i(i+2)a^2 - (i-1)(i+3)r^2}{2(i-1)[2(i+1)^2+1]v} \frac{r^{i-1}}{\sin θ} \frac{dT_i}{dφ}
\end{aligned}
\right} \dots \dots (10)\dagger
$$
where $T_i = w \left( S_i - 2g \frac{i-1}{2i+1} \frac{σ_i}{a^i} \right)$.
These equations for $ρ, \varpi, \text{ and } ν$ provide us with the state of internal flow corresponding to the external disturbing potential $r^i S_i$, including the effects of the mutual gravitation of the matter that makes up the spheroid.
* The introduction of the effects of gravitation may also be carried out synthetically by building the solution from individual components, as is done by Sir W. Thomson (section 840, Natural Philosophy original: 'Nat. Phil.'); but the effects of the lagging of the tide-wave make this method feel somewhat artificial, and I prefer to present the proof in the manner given here. Conversely, the elasticity problem may be solved as it is in this text.
$\dagger$ There seems to be a misprint regarding the signs of the $\varpi$ terms in the second and third parts of equations (13) of section 834 of Natural Philosophy (1867). When this is corrected, $μ$ and $ν$ can be reduced to fairly simple forms. It also appears to me that the differentiation of $ρ$ in equation (15) is incorrect, which invalidates the argument in the three following lines. This correction is not, however, important in any significant way.