# Confusion in Maxwell's derivation of Ampere's Force Law

I am reading Maxwell's "a treatise on electricity and magnetism, Volume 2, page 155" about "Ampere's Force Law". I have some confusion in the following pages:  On the top of the page, Maxwell says: (The coordinates of points on either current are functions of s or of s')

Does he mean we parametrize the path of one circuit with variable s and another circuit with variable s'?

Then he says: (If F is any function of the position of a point, then we shall use the subscript$_{(s,0)}$ to denote the excess of its value at P over that at A, thus $$F_{(s,0)}=F_{P}-F_{A}$$ Such functions necessarily disappear when the circuit is closed.)

Does he mean$$\oint F.ds = 0$$ or something else?

If he means something else by "Such functions necessarily disappear when the circuit is closed" what is it? • You must go back to page 511 and therein Figure 29. Image of this page has been added to your question. The variables $\:s\:$ and $\:s^{\prime}\:$ parametrize the two curves and more exactly are arc lengths from reference points $\:A\:$ and $\:A^{\prime}\:$ respectively. – Frobenius May 18 '16 at 9:11
• On the other hand by "If F is any function ..." this function is not necessarily a vector function to use in a curvilinear integral $\:\int_{C} \mathbf{F}\cdot d \mathbf{s} \:$. The "disappearence" of $\:F_{(s,0)}=F_{P}-F_{A}\:$ in case of a closed curve is due simply to the coincidence of the final point $\: P\:$ with the initial point $\: A\;$. – Frobenius May 18 '16 at 10:29
• OK thanks, I understand s and s' parametrize the two curves. – N.G.Tyson May 18 '16 at 10:30
• I understand this. But in page 156, article 517, equation 19 there is a term 2Pr added inside the integral. I was considering it as a conservative vector field when dotted with ds' and integrated around closed path gives zero. Am I right or wrong? Please explain why that term 2Pr is added inside the integral in equation 19. – N.G.Tyson May 18 '16 at 11:00
• On one hand, what is the relation of equation (19) with my comment about the function $\:F\:$ ??? On the other hand, must I read all the Chapter to explain what 2 is, what $\:P\:$ is, why $\:2Pr\:$ is inserted in an integral ?? You are reading the book by yourself. Continue trying to understand the ingenious Maxwell's thoughts and writings. – Frobenius May 18 '16 at 11:12

## 1 Answer  \begin{align} \int_{0}^{s^{\prime}}\dfrac{dP}{ds^{\prime}} \xi^{2}ds^{\prime} & = \biggl[P\xi^{2}\biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}P d\xi^{2}\\ & = \biggl[P\xi^{2}\biggr]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}}2P \xi d\xi\\ & = \biggl[P\xi^{2}\biggr]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}}2P \xi \underbrace{\dfrac{d\xi}{ds^{\prime}}}_{=l ^{\prime}}ds^{\prime}\\ & =\biggl[P\xi^{2}\biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}2P \xi l ^{\prime}ds^{\prime} \end{align}

• While the OP may have found this helpful, to make a lasting, useful resource for future visitors this post does need additional clarification of what you're plotting and calculating, and why. – Emilio Pisanty Jun 1 '16 at 13:47
• What software are you plotting with? – WetSavannaAnimal Jun 1 '16 at 14:02
• @Emilio Pisanty : This answer was a comment at the begining. I tried to help without answering. But later on I answered to an additional question on the same problem here : physics.stackexchange.com/questions/257127/…, a question already on hold. I don't think that these questions and answers are useful to the broader community, and to future users. The present question must be put on hold too. – Frobenius Jun 1 '16 at 17:55
• @ WetSavannaAnimal aka Rod Vance : GeoGebra free amazing software. It accepts text and equations in LateX as you see in the Figures. I saw this software last year as suggested tool in the right column of Physics SE. You will be surprised for its capabilities in a day by day use. – Frobenius Jun 1 '16 at 18:03
• @Frobenius Fair enough. That's plotted with GeoGebra? I might need to give it a closer look, then. – Emilio Pisanty Jun 1 '16 at 19:11