Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/197

Rh indeed reaches high-water mark in the paragraphs in which he mentions them. Concerning another notation, $$\Phi^{\times}_{\times} \Phi$$ (defined in Nature, vol. xliii, p. 513) [this vol., p. 160], he exclaims, "Thus burden after burden, in the form of new notation, is added apparently for the sole purpose of exercising the faculty of memory." He would vastly prefer, it would appear, to write with Hamilton $$m\phi '^{-1},$$ "where $$m$$ represents what the unit volume becomes under the influence of the linear operator." But this notation is only apparently compact, since the $$m$$ requires explanation. Moreover, if a strain were given in what Hamilton calls the standard trinomial form, to write out the formula for the operator on surfaces in that standard form by the use of the expression $$m\phi '^{-1}$$ would require, it seems to me, ten (if not fifty) times the effort of memory and of ingenuity, which would be required for the same purpose with the use of $$\tfrac{1}{2}\Phi^{\times}_{\times} \Phi.$$

I may here remark that Prof. Tait's letter of endorsement of Prof. Knott's paper affords a striking illustration of the convenience and flexibility of a notation entirely analogous to $$\Phi^{\times}_{\times} \Phi,$$ viz., $$\Phi : \Phi.$$ He gives the form $$\text{S}\nabla\nabla_{1} \, \text{S}\sigma \sigma_{1}$$ to illustrate the advantage of quaternionic notations in point of brevity. If I understand his notation, this is what I should write $$\nabla \sigma : \nabla \sigma.$$ (I take for granted that the suffixes indicate that $$\nabla$$ applies as differential operator to $$\sigma,$$ and $$\nabla_{1}$$ to $$\sigma_{1},$$ $$\sigma,$$ and $$\sigma_{1}$$ being really identical in meaning, as also $$\nabla$$ and $$\nabla_{1}.$$) It will be observed that in my notation one dot unites in multiplication the two $$\nabla $$'s, and the other the two $$\sigma$$'s, and that I am able to leave each $$\nabla$$ where it naturally belongs as differential operator. The quaternionist cannot do this, because the $$\nabla$$ and $$\sigma$$ cannot be left together without uniting to form a quaternion, which is not at all wanted. Moreover, I can write $$\Phi$$ for $$\nabla \sigma,$$ and $$\Phi : \Phi$$ for $$\nabla \sigma : \nabla \sigma.$$ The quaternionist also uses a $$\phi,$$ which is practically identical with my $$\Phi$$ (viz., the operator which expresses the relation between $$d\sigma$$ and $$d\rho$$), but I do not see how Prof. Knott, who I suppose dislikes $$\Phi : \Phi$$ as much as $$\Phi^{\times}_{\times}$$ would express $$\text{S}\nabla \nabla_{1} \, \text{S}\sigma \sigma_{1}$$ in terms of this $$\phi.$$

It is characteristic of Prof. Knott's view of the subject, that in translating into quatemionic from a dyadic, or operator, as he calls it, he adds in each case an operand. In many cases it would be difficult to make the translation without thia But it is often a distinct advantage to be able to give the operator without the operand. For example, in translating into quaternionic my dyadic or operator $$\Phi \times \rho,$$ he adds an operand, and exclaims, "The old thing!" Certainly, when this expression is applied to an operand, there is no advantage (and no disadvantage) in my notation as compared with the quaternionic. But if the quaternionist wished to express what I would write in the form $$(\Phi \times \rho)^{-1},$$ or $$\left\vert \Phi \times \rho \right\vert,$$ or $$(\Phi \times \rho)_{\text{S}},$$ or $$(\Phi \times \rho)_{\times}$$ he would, I think, find the operand very much in the way.