I am reading Maxwell's "a treatise on electricity and magnetism, Volume 2, page 156" about "Ampere's Force Law". I have some confusion in the following pages:

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My question is of two parts:

1. Equation 20, i.e. $P=\dfrac{B+C}{2r}$ is the outcome of special case (i.e. l=1, m=0, n=0)

But in Page 156, Article 517, Maxwell says: "We can now eliminate P, and find the general value of $\dfrac{dX}{ds}$" and uses this formula (i.e. $P=\dfrac{B+C}{2r}$) in the general case.

However in the general case, where 0 < l, m, n < 1, and hence $$\dfrac{d^{2}X}{dsds'}=l\left( \frac{dP}{ds'}\xi^{2}-\dfrac{dQ}{ds'}+(B+C)\dfrac{l'\xi}{r}\right) +m(...)+n(...)\neq0$$ (since direction of X is not in the direction of ds)

therefore, $$\dfrac{dX}{ds}=l\left[ (P\xi^{2}-Q)_{(s',0)}-\int\limits_0^s' (2Pr-B-C)\dfrac{l'\xi}{r}ds'\right] +m\int\limits_0^s'(...)ds'+n\int\limits_0^s'(...)ds'$$ Now in this general case, how can we get $P=\dfrac{B+C}{2r}$.

If $P\neq\dfrac{B+C}{2r}$ in general case, what does Maxwell mean by "We can now eliminate P, and find the general value of $\dfrac{dX}{ds}$"

2. How can one get equation 21 from equation 15. Please give a lengthy derivation.

  • $\begingroup$ 1. What are $A$, $B$, and $C$? If they do not depend on $l$, $m$, $n$, then any property you find with specific values will be true for all values. Just as the equation $P=(B+C)/2r$ has been derived from the closed circuit case, and extended to all cases probably because the terms cannot depend on the circuit shape. 2. Is there anything that prevents you from doing this derivation yourself? $\endgroup$ – L. Levrel May 23 '16 at 19:54
  • $\begingroup$ A, B, and C are functions of r. $\endgroup$ – N.G.Tyson May 24 '16 at 6:27
  • $\begingroup$ If A, B, and C do not depend on l, m and n; how can any property of specific values true for all values? Please explain with an example $\endgroup$ – N.G.Tyson May 24 '16 at 6:32
  • $\begingroup$ I have no idea of how $\dfrac{B-C}{2}$ came into the expression. I also know that when $B+C=0$, and after integration with respect to ds', we can get equation 21 from equation 15. But how come $B+C=0$.If $B+C=0$ is not valid then I have no idea how to get equation 21 from equation 15. $\endgroup$ – N.G.Tyson May 24 '16 at 8:29
  • $\begingroup$ Sorry I didn't understand how the example of the trajectory of a satellite relates to the question. Please explain a bit more clearly in the context of the above question. $\endgroup$ – N.G.Tyson May 24 '16 at 15:23

ANSWER TO QUESTION 1. The distance $r$ between two points P and P' in space does not depend on the frame $Oxyz$. It's invariant.

Suppose that we have two points P and P' in space at a distance $r$ apart and we want to find a scalar function $P(r)$ that satisfies a number of conditions. Let there exists a system of coordinates $Oxyz$ that is more convenient than others and makes my life easy to calculate and determine this scalar function $P(r)$, for example say that I find $P(r)=3r^{3/2}-2$. This result is independent of the choice of system $Oxyz$ since $r$ is the same in all systems. That what Maxwell did : He finds that a coordinate system $Oxyz$ with its axis $Ox$ aligned with the tangent of the unprimed curve $\:\left(l=dx/ds=1, m=dy/ds=0, n=dz/ds=0 \right)\:$ at point P is more convenient and reach to equation (20-in textbook), see below, which is independent of the system.

ANSWER TO QUESTION 2. The key for the solution is to understand the geometry of the problem and to know which variables depend on the lenght parameters $s,s'$ of the curves. \begin{equation} \begin{matrix} \dfrac{dx}{ds}=l, & \dfrac{dy}{ds}=m, &\dfrac{dz}{ds}=n,\\ \dfrac{dx'}{ds'}=l', & \dfrac{dy'}{ds'}=m', &\dfrac{dz'}{ds'}=n', \end{matrix} \tag{2-in textbook} \end{equation}

\begin{equation} \xi = x'-x, \qquad \eta = y'-y, \qquad \zeta = z'-z. \tag{under 12-in textbook} \end{equation}

\begin{align} \dfrac{d^{2}X}{dsds'} & =l\: \:\biggl\{ -\left( A+B \right)\dfrac{1}{r^{2}}\dfrac{dr}{ds'} \xi^{2}+C\dfrac{dr}{ds'}+\left( B+C \right)\dfrac{l'\xi}{r} \biggr\},\\ & +m \biggl\{ -\left( A+B \right)\dfrac{1}{r^{2}}\dfrac{dr}{ds'} \xi\eta+C\dfrac{l'\eta}{r}+B\dfrac{m'\xi}{r} \biggr\},\\ & +n \: \biggl\{ -\left( A+B \right)\dfrac{1}{r^{2}}\dfrac{dr}{ds'} \xi\zeta+C\dfrac{l'\zeta}{r}+B\dfrac{\eta'\xi}{r} \biggr\}. \tag{15-in textbook} \end{align}

\begin{equation} P=\int_{r }^{\infty}\left( A+B \right)\dfrac{1}{r^{2}}dr, \qquad \text{and} \qquad Q=\int_{r }^{\infty}Cdr, \tag{16-in textbook} \end{equation}

\begin{equation} \text{Hence} \qquad \left( A+B \right)\dfrac{1}{r^{2}}= - \dfrac{dP}{dr}, \qquad \text{and} \qquad C= - \dfrac{dQ}{dr}. \tag{17-in textbook} \end{equation}

\begin{equation} P=\dfrac{1 }{ 2r }\left( B+C\right) \tag{20-in textbook} \end{equation}

\begin{align} & \dfrac{dX}{ds} = \biggl\{\dfrac{ B+C}{2} \dfrac{\xi}{r}\left( l\xi+m\eta+n \zeta\right)-Q\biggr\}_{0}^{s'}\\ & +m \int_{0}^{s'} \dfrac{ B-C}{2} \dfrac{ m'\xi -l'\eta}{r}ds'-n\int_{0}^{s'} \dfrac{ B-C}{2} \dfrac{ l'\zeta-n'\xi}{r}ds'. \tag{21-in textbook} \end{align} In the textbook the minus sign of $\:-Q\:$ in the first term of the rhs is probably misprinted as plus $\:+Q\:$.

The unit vector $\left(l,m,n\right)$ tangent to the uprimed curve at point P is dependent on the length parameter $\;s\;$ but is independent of the length parameter $\;s'\;$ of the primed curve.. That's why these variables are outside of the integrals with respect to $\;s'\;$ in the following integrations \begin{align} \dfrac{dX}{ds} & =l\: \: \int_{0}^{s'} \biggl\{ -\left( A+B \right)\dfrac{1}{r^{2}} \xi^{2}dr+Cdr+\left( B+C \right)\dfrac{l'\xi}{r}ds' \biggr\},\\ & +m \int_{0}^{s'} \biggl\{ -\left( A+B \right)\dfrac{1}{r^{2}} \xi\eta dr+C\dfrac{l'\eta}{r}ds'+B\dfrac{m'\xi}{r}ds' \biggr\},\\ & +n \: \int_{0}^{s'} \biggl\{ -\left( A+B \right)\dfrac{1}{r^{2}}\xi\zeta dr +C\dfrac{l'\zeta}{r}ds'+B\dfrac{\eta'\xi}{r} ds'\biggr\}. \tag{A-01} \end{align}

\begin{align} \dfrac{dX}{ds} & =l\: \: \int_{0}^{s'} \biggl\{ \xi^{2}dP-dQ+\left( B+C \right)\dfrac{l'\xi}{r}ds' \biggr\},\\ & +m \int_{0}^{s'} \biggl\{ \xi\eta dP+C\dfrac{l'\eta}{r}ds'+B\dfrac{m'\xi}{r}ds' \biggr\},\\ & +n \: \int_{0}^{s'} \biggl\{\xi\zeta dP +C\dfrac{l'\zeta}{r}ds'+B\dfrac{\eta'\xi}{r} ds'\biggr\}. \tag{A-02} \end{align}

In the following equations we make use of the integration by parts, we have in mind the definitions of the variables $\:l,m,n,l',m',n',\xi,\eta,\zeta \:$ and their dependence on (or independence of) the length parameters $\;s,s'\;$ and the textbook equations (17) and (20), relations for functions $P(r)$ and $C(r)$ necessary to explain the results.
\begin{align} \int_{0}^{s^{\prime}}\xi^{2}dP & = \biggl[\xi^{2}P\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}\\ &= \biggl[\dfrac{B+C}{ 2 } \dfrac{ \xi^{2}}{ r } \biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}\left( B+C \right)\dfrac{ l ^{\prime} \xi }{r}ds^{\prime} \tag{A-03.$\xi\xi$} \end{align}

\begin{align} \int_{0}^{s^{\prime}}\xi\eta dP & = \biggl[\xi\eta P\biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}P d\left(\xi\eta \right)\\ & = \biggl[P\xi\eta\biggr]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}}P \xi d\eta-\int_{0}^{s^{\prime}}P\eta d\xi\\ & = \biggl[P\xi\eta \biggr]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}}P \xi \underbrace{\dfrac{d\eta}{ds^{\prime}}}_{=m ^{\prime}}ds^{\prime} - \int_{0}^{s^{\prime}}P\eta \underbrace{\dfrac{d\xi}{ds^{\prime}}}_{=l ^{\prime}}ds^{\prime}\\ & = \biggl[ P\xi\eta \biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}P\left(m'\xi + l'\eta \right)ds^{\prime}\\ & = \biggl[\dfrac{B+C}{ 2 } \dfrac{ \xi\eta}{ r } \biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}\dfrac{B+C}{ 2 }\dfrac{m'\xi + l'\eta }{r}ds^{\prime} \tag{A-03.$\xi\eta$} \end{align}

\begin{align} \int_{0}^{s^{\prime}}\xi\zeta dP & = \biggl[\xi\zeta P\biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}P d\left(\xi\zeta \right)\\ & = \biggl[P\xi\zeta\biggr]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}}P \xi d\zeta -\int_{0}^{s^{\prime}}P \zeta d\xi \\ & = \biggl[P\xi\zeta \biggr]_{0}^{s^{\prime}}-\int_{0}^{s^{\prime}}P \xi \underbrace{\dfrac{d\zeta}{ds^{\prime}}}_{=n ^{\prime}}ds^{\prime} -\int_{0}^{s^{\prime}}P \zeta \underbrace{\dfrac{d\xi}{ds^{\prime}}}_{=l ^{\prime}}ds^{\prime}\\ & = \biggl[ P\xi\zeta \biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}P\left(n' \xi +l'\zeta \right)ds^{\prime}\\ & = \biggl[\dfrac{B+C}{ 2 } \dfrac{ \xi\zeta}{ r } \biggr]_{0}^{s^{\prime}} -\int_{0}^{s^{\prime}}\dfrac{B+C}{ 2 }\dfrac{n'\xi + l'\zeta }{r}ds^{\prime} \tag{A-03.$\xi\zeta$} \end{align}


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  • $\begingroup$ The term with coefficient $l$ works fine. But simplifying the term with coefficient $m$, $$\int\limits_0^{s'}\left[-(B+C)\dfrac{m'\xi+l'\eta}{r} + B\dfrac{m'\xi}{r} + C\dfrac{l'\eta}{r}\right]ds'$$ $$=\int\limits_0^{s'}\left[-B\dfrac{l'\eta}{r} - C\dfrac{m'\xi}{r}\right]ds'$$ How shall I get $\int\limits_0^{s'}\dfrac{B-C}{2} \dfrac{m'\xi-l'\eta}{r}ds$ from here? $\endgroup$ – N.G.Tyson May 28 '16 at 14:44
  • $\begingroup$ Please give a last explanation here. I have just passed high school and am new to this. You indeed helped a lot. $\endgroup$ – N.G.Tyson May 28 '16 at 16:25
  • $\begingroup$ @faheemahmed400 : I apologize. Take a look at the new corrected equations $(A-03.\xi\eta)$ and $(A-03.\xi\zeta)$. I corrected an error with a coefficient 2 instead of 1. But now I have a problem with the sign of Q in the first braced term of (21). I find $(-Q)_{0}^{s'}$ but in textbook is $(+Q)_{0}^{s'}$. The errata page doesn't refer to error in page 156 for this. May be I am wrong. I'll try to solve this contradiction. $\endgroup$ – Frobenius May 29 '16 at 6:28
  • $\begingroup$ @faheemahmed400 : So, in your above comment we have after the correction $$ \int\limits_0^{s'}\left[-\dfrac{B+C}{2} \dfrac{m'\xi+l'\eta}{r} + B\dfrac{m'\xi}{r} + C\dfrac{l'\eta}{r}\right]ds'=\int\limits_0^{s'}\dfrac{B-C}{2} \dfrac{m'\xi-l'\eta}{r}ds' $$ The same is valid for the term with coefficient $\;n$. $\endgroup$ – Frobenius May 29 '16 at 6:58
  • $\begingroup$ Thanks for the correction. But I think the $+Q$ in equation 21 is a misprint. Even in equation 19 its$ -Q$ $\endgroup$ – N.G.Tyson May 29 '16 at 8:10

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