To Mary Somerville   [9 March 1830]

Slough Mar[ch] 9 1830

Dear Madam,

I lose not a moment in forwarding you a work I have just received from the author. It is a translation of the Mecanique celeste with a running comment, by (apparently) an able hand. How far its appearance at this junction may influence your views, I am of course incapable of judging but it is a matter of fact which you cannot be too early in possession of. I received your packet by Mr Richards. You are indefatigable, & the retouching you have given it has been most effective. Still I am not quite satisfied about the way in which the fundamental equation

$$\delta x\{Pdt-\frac{d^{2}x}{dt}\}+\delta y\{Qdt-\frac{d^{2}y}{dt}\}+\delta z\{Rdt-\frac{d^{2}z}{dt}\}=0$$

is derived. Laplace is ingeniously obscure here, he regards the Force acting on a body in motion, not, as expended in producing its effect, ie motion — but as transferred into the moving body & accumulated in it in the shape of momentum which, in asmuch as it is an affect in the body to proceed forward & if opposed, to displace an obstacle, may be regarded as a moving force capable of being balanced or equilibrated by others.

Would it not be simpler, to proceed there (or at least more satisfactory)

• Let $h$ be perfectly free to obey any forces $P$ $Q$ $R$

• then, there will produce in it velocities $Pdt$ $Qdt$ $Rdt$ proportional to their intensities & in their direction in any instant dt. (by the laws of motion — regarding velocity as an effect of force & as its measure)

• therefore when $h$ is free $d\cdot \frac{dx}{dt}=Pdt$; &c.

& therefore multiplying these equations by $\delta x$ &c.

$$\left(\left(A\right)\right)0=\delta x(Pdt-\frac{d^{2}x}{dt})+\delta y(Qdt-\frac{d^{2}y}{dt})+\delta z(Rdt-\frac{{\partial }^{2}z}{\partial t})$$

and since the quantities $Pdt-\frac{{\partial }^{2}x}{\partial t}$, &c are separately zero, $\delta x$, $\delta y$, $\delta z$ are absolutely arbitrary & independent — and vice versa, if they are so this one eq[uation] will be equivalent to the three separate ones.

Case II. But if $h$ be not free it must be either constrained to move on some curve, or on some surface or be subjected to some resistance, or otherwise subjected to some condition. But matter is not moved otherwise then by force — therefore whatever constrains it or subjects it to conditions is force. If a curve or a surface, or a string tying it, this force is called reaction — if a viscous medium, resistance. If a condition however abstract, (for example that it move in a tautochrone &c ) still this condition, by obliging it to move out of its free course or with an irrational velocity, must resolve itself for [illegible] analysis into Force — only that in this case it is an implicit not explicit Function of the coordinates. It may therefore be considered 1^st as involved in $P$, $Q$, $R$, or 2^nd as added in the resolved forms $P\mathrm{\&\#x2032;}$ $Q\mathrm{\&\#x2032;}$ $R\mathrm{\&\#x2032;}$ to them if we prefer it.

In the first case, if we regard it as involved in $P$, $Q$, $R$, there really constrain an indeterminate function. But the equations $Pdt-\frac{d^{2}t}{dt}=0$; $Q\partial t-\frac{{\partial }^{2}y}{\partial t}=0$ &c still subsist and therefore also the equation ((A)). But these are now not enough to determine $x$, $y$, $z$ in functions of $t$ because of the unknown forms of $P$, $Q$, $R$. But if we superadd to these equations the equation or equations ($u=0$) with all their consequences ($\delta u=0$, $du=0$ &c ) which express the conditions of constraint. These will then be sufficient to determine the problem. Thus our equations are $u=0$; $(P-\frac{d^{2}x}{d{t}^2})=0$; $(Q-\frac{d^{2}y}{d{t}^2})=0$; $(R-\frac{d^{2}z}{d{t}^2})=0$.

$u$ is a function on $x$, $y$, $z$; $P$, $Q$, $R$ and $t$ and $P$, $Q$, $R$ being implicit functions of $x$, $y$, $z$, $t$, $u$ is a function of $x$, $y$, $z$, $t$. Therefore the equation $u=0$ establishes the existence of a relation $p\delta x+q\delta y+r\delta z=0$ between the variations $\delta x$, $\delta y$, $\delta z$ which can no loner be regarded as arbitrary. But the equation ((A)) subsists whether they be so or not & may therefore be used simultaneously with $\delta u=0$. To eliminate one — after which the other two, being really arbitrary their coefficients must be separately zero.

In the second case if we will not regard the forces arising from the conditions of constraint as involved in $P$, $Q$, $R$, Let $\delta u=0$ be that condition and Let $P\mathrm{\&\#x2032;}$, $Q\mathrm{\&\#x2032;}$, $R\mathrm{\&\#x2032;}$ be the unknown forces brought into action by that condition, by which the free action of $P$, $Q$, $R$ is modified. then with the whole forces acting on $h$ be $P+P\mathrm{\&\#x2032;}$ $Q+Q\mathrm{\&\#x2032;}$ $R+R\mathrm{\&\#x2032;}$ and under the influence of these the body will move as a free body and therefore $\delta x$, $\delta y$, $\delta z$, being any variations we have $$0=(P+P\mathrm{\&\#x2032;}-\frac{d^{2}x}{d{t}^2})\delta x+(Q+Q\mathrm{\&\#x2032;}-\frac{{\partial }^{2}y}{\partial t^2})\delta y+(R+R\mathrm{\&\#x2032;}-\frac{{\partial }^{2}z}{\partial t^2})\delta z$$ & this equation is independent of any particular relations between $\delta x$, $\delta y$, $\delta z$ & holds good whether they subsist or not.

But the condition $\delta u=0$ establishes a relation, of the form $p\delta x+q\delta y+r\delta z=0$ where $p=\left(\frac{du}{dx}\right)$; $q=\left(\frac{du}{dy}\right)$ and since this is true it is so when multiplied by any arbitrary quantity $\lambda$ when it becomes $$p\cdot \lambda \delta x+q\cdot \lambda \delta y+r\cdot \lambda \delta z=0.$$

Let this be added to ((B)) and it becomes $$(P+P\mathrm{\&\#x2032;}-p\lambda -\frac{d^{2}x}{d{t}^2})\delta x+(Q+Q\mathrm{\&\#x2032;}-q\lambda -\frac{{\partial }^{2}y}{\partial t^2})\delta y+\dots \left(z\right)=0$$ which is true whatever $\delta x$, $\delta y$, $\delta z$ and $\lambda$ are.

Now since $P\mathrm{\&\#x2032;}$ $Q\mathrm{\&\#x2032;}$ $R\mathrm{\&\#x2032;}$ are forces acting in the directions $x$ $y$ $z$ (tho’ unknown) they may be compounded into one resultant S which must have one direction whose element may be represented by $\delta s$ and since the single force $S$ is resolvable into $P\mathrm{\&\#x2032;}$ $Q\mathrm{\&\#x2032;}$ $R\mathrm{\&\#x2032;}$, we must have

$$P\mathrm{\&\#x2032;}\delta x+Q\mathrm{\&\#x2032;}\delta y+R\mathrm{\&\#x2032;}\delta z=S\delta s$$ so that the above equation becomes $$(P-\frac{d^{2}x}{d{t}^2})\delta x+(Q-\frac{{\partial }^{2}y}{\partial t^2})\delta y+(R-\frac{{\partial }^{2}z}{\partial t^2})\delta z+S\delta s-\lambda \delta u=0$$ and this is true whatever be $\lambda$

But $\lambda$ being thus left arbitrary, we are at liberty to determine it by any convenient condition. Let this condition be then $S\delta s-\lambda \delta u=0$ or $\lambda =S\cdot \frac{\delta s}{\delta u}$ & the equation reduces to ((A)) so that still this equation (when $P$, $Q$, $R$, are only the acting forces explicitly given) suffices to resolve the problem provided we take it in conjunction with the equation $p\delta x+q\delta u+r\delta z=0$ which establishes a relation between $\delta x$ $\delta y$ $\delta z$.

Let us now consider the condition $\lambda =S\cdot \frac{\delta s}{\delta u}$ by which we determined $\lambda$. Since $S$ is the resultant of the forces $P\mathrm{\&\#x2032;}$ $Q\mathrm{\&\#x2032;}$ $R\mathrm{\&\#x2032;}$ its magnitude must be represented by $\sqrt{P{\mathrm{\&\#x2032;}}^2+Q{\mathrm{\&\#x2032;}}^2+R{\mathrm{\&\#x2032;}}^2}$ and since $S\delta s=\lambda \delta u$ or $P\mathrm{\&\#x2032;}\delta x+Q\mathrm{\&\#x2032;}\delta y+R\mathrm{\&\#x2032;}\delta z=\lambda \frac{du}{dx}\delta x+\lambda \frac{du}{dy}\delta y+\lambda \frac{du}{dz}\delta z$

Therefore in order that $\delta x$, $\delta y$, $\delta z$ may remain arbitrary we must have $P\mathrm{\&\#x2032;}=\lambda \cdot \frac{du}{dx}$ $Q\mathrm{\&\#x2032;}=\lambda \cdot \frac{du}{dy}$ $R\mathrm{\&\#x2032;}=\lambda \cdot \frac{du}{dz}$ and consequently $S=\sqrt{P{\mathrm{\&\#x2032;}}^2+Q{\mathrm{\&\#x2032;}}^2+R{\mathrm{\&\#x2032;}}^2}=\lambda \cdot \sqrt{{\left(\frac{du}{dx}\right)}^2+{\left(\frac{du}{dy}\right)}^2+{\left(\frac{du}{dz}\right)}^2}$ and $\lambda =\frac{S}{\sqrt{{\left(\frac{du}{dx}\right)}^2+{\left(\frac{du}{dy}\right)}^2+{\left(\frac{du}{dz}\right)}^2}}$; $P\mathrm{\&\#x2032;}=\frac{\left(\frac{du}{dx}\right)}{\sqrt{{\left(\frac{du}{dx}\right)}^2+{\left(\frac{du}{dy}\right)}^2+{\left(\frac{du}{dz}\right)}^2}}$; &c. ((B))

This is being given in terms of $x$, $y$, $z$; $\lambda$, $P\mathrm{\&\#x2032;}$, $Q\mathrm{\&\#x2032;}$, $R\mathrm{\&\#x2032;}$ are all determined.

If the condition of constraint be pressure against a Surface $S$ is the reaction &c.

I am dear Madam yours very truly J F W Herschel.