### The maths behind the trans-Atlantic telegraph cable

In the mid-19th century, the most sought-after feat of engineering and industry was a trans-Atlantic telegraph cable. But when the first successful cable was laid between Britain and America in 1858, the speed of transmission was only about 0.1 words per minute. This post will be an analysis of Thomson's 1855 paper "On the Theory of the Electric Telegraph", published in the journal of the Royal Society of London.

Thomson starts by defining the variables that he uses throughout the work. I shall define the same variables using more modern lettering conventions. Let $C$ be the capacitance of the wire, $R$ the resistance of the wire, $V$ the potential at a point $P$ on the wire at a time $t$, and $I$ the current at the same point in the same instant.

The charge $Q$ at $P$, called "quantity of electricity" by Thomson, is given by

\[Q = VC = It\]

Thus, in an infinitesimal length $dx$ of wire at $P$, we have a charge of $VCdx$ and, in an infinitesimal period of time, the charge that leaves $P$ is equal to

\[dtdI = dt\frac{dI}{dx}dx\]

Using two equations for the electromotive force, Thomson showed that $RI = -\frac{\partial V}{\partial x}$. Combining these formulae gives the all-important simple form of the telegraphist's equation:

\begin{equation}

CR\frac{\partial V}{\partial t} = \frac{\partial^2V}{\partial x^2}

\end{equation}

Thomson notes instantly that this is the heat equation.

In a real wire, there would be some electrical leakage into the surrounding water. By introducing some coefficient $h$ to measure this loss, Thomson recovers the equation:

\[CR\frac{\partial V}{\partial t} = \frac{\partial^2V}{\partial x^2} -hV\]

but with the change of variable:

\[V = e^{-\frac{h}{RC}t}\phi\]

he obtains a PDE in the original form, i.e:

\[CR\frac{\partial \phi}{\partial t} = \frac{\partial^2\phi}{\partial x^2}\]

###
Conclusion

### Derivation of the PDE

Thomson starts by defining the variables that he uses throughout the work. I shall define the same variables using more modern lettering conventions. Let $C$ be the capacitance of the wire, $R$ the resistance of the wire, $V$ the potential at a point $P$ on the wire at a time $t$, and $I$ the current at the same point in the same instant.

The charge $Q$ at $P$, called "quantity of electricity" by Thomson, is given by

\[Q = VC = It\]

Thus, in an infinitesimal length $dx$ of wire at $P$, we have a charge of $VCdx$ and, in an infinitesimal period of time, the charge that leaves $P$ is equal to

\[dtdI = dt\frac{dI}{dx}dx\]

Using two equations for the electromotive force, Thomson showed that $RI = -\frac{\partial V}{\partial x}$. Combining these formulae gives the all-important simple form of the telegraphist's equation:

\begin{equation}

CR\frac{\partial V}{\partial t} = \frac{\partial^2V}{\partial x^2}

\end{equation}

Thomson notes instantly that this is the heat equation.

In a real wire, there would be some electrical leakage into the surrounding water. By introducing some coefficient $h$ to measure this loss, Thomson recovers the equation:

\[CR\frac{\partial V}{\partial t} = \frac{\partial^2V}{\partial x^2} -hV\]

but with the change of variable:

\[V = e^{-\frac{h}{RC}t}\phi\]

he obtains a PDE in the original form, i.e:

\[CR\frac{\partial \phi}{\partial t} = \frac{\partial^2\phi}{\partial x^2}\]

### Solutions

Thomson uses his observation that his derived equation corresponds to the heat equation by stating immediately a solution, obtained by Fourier methods:

\[V = \frac{\tilde{V}}{\pi}\int_0^\infty e^{-zn^\frac{1}{2}}\frac{\sin(2nt-zn^\frac{1}{2})-\sin[(t-T)2n-zn^\frac{1}{2}]}{n}dn\]

where the wire is understood to be infinitely-long, $z:=x\sqrt{RC}, \tilde{V}$ is the voltage to which the end $O$ is instantaneously raised and $T$ is the time for which this potential is maintained. By cleverly noting that:

\begin{align}

\int_{t-T}^t \cos(2n\theta - zn^\frac{1}{2})d\theta &= \left[\frac{1}{2n}\sin(2n\theta-zn^\frac{1}{2})\right]_{t-T}^t \\

&= \frac{1}{n}\left[\sin(2nt-zn^\frac{1}{2})-\sin(2n(t-T)-zn^\frac{1}{2})\right]

\end{align}

and taking $T$ to be infinitesimal, so that $\sin(T) \approx T$ with $t > 0$, Thomson could conclude that:

\begin{align}

V &= T\frac{2\tilde{V}}{\pi}\int_0^\infty e^{-zn^\frac{1}{2}}\cos(2nt-zn^\frac{1}{2})\\

&= T\frac{\tilde{V}z}{2\pi^\frac{1}{2}t^\frac{3}{2}}e^{-\frac{z^2}{4t}}

\end{align}

Thomson now moves to finding the solution to the problem when a charge $Q$ is communicated instantaneously to the wire at $O$. He states the strength of the current $I$ at any point in the wire to be equal to $-\frac{1}{k}\frac{dV}{dx}$ and deduces that the "maximum electrodynamic effect" of an impulse into the wire will therefore be found by finding the value of t making $\frac{\partial V}{\partial z}$ (this being proportional to $\frac{\partial V}{\partial x}$) maximal: with $V$ given by an equation similar to the above:

\[V = \frac{e^{-\frac{z^2}{4t}}}{\sqrt{t}}\frac{Q}{\sqrt{\pi}}\sqrt{\frac{R}{C}}\]

Upon finding $\frac{\partial^2 V}{\partial z^2}$, we find that $\frac{\partial V}{\partial z}$ is maximal when:

\[\frac{1}{2t^\frac{3}{2}} = \frac{z^2}{4t^\frac{5}{2}}\]

giving the value:

\[t = \frac{z^2}{2} = \frac{RCx^2}{2}\]

Note that this is different to the value of t which Thomson gives, being $\frac{z^2}{6}$. This is however entirely irrelevant to the remainder of the exposition; the point of the calculation was to show that the time that one should leave between message pulses due to the "retardations of signals" was proportional to the square of the length of the wire, now taking x to be the length of the cable. This is a deduction which Thomson’s erroneous calculation still makes.

Thomson then goes on to discuss the "velocity of transmission" of the signal. He states that, if the potential at $O$ is varied in a sinusoidal way (namely with $\sin(2nt)$), then this velocity is equal to $2\sqrt{\frac{n}{RC}}$ . It is interesting that the velocity of transmission and the retardations of the signals bear no mathematical relation to one another: showing again that this problem is, perhaps counter-intuitively, one regarding the length of the wire

where the wire is understood to be infinitely-long, $z:=x\sqrt{RC}, \tilde{V}$ is the voltage to which the end $O$ is instantaneously raised and $T$ is the time for which this potential is maintained. By cleverly noting that:

\begin{align}

\int_{t-T}^t \cos(2n\theta - zn^\frac{1}{2})d\theta &= \left[\frac{1}{2n}\sin(2n\theta-zn^\frac{1}{2})\right]_{t-T}^t \\

&= \frac{1}{n}\left[\sin(2nt-zn^\frac{1}{2})-\sin(2n(t-T)-zn^\frac{1}{2})\right]

\end{align}

and taking $T$ to be infinitesimal, so that $\sin(T) \approx T$ with $t > 0$, Thomson could conclude that:

\begin{align}

V &= T\frac{2\tilde{V}}{\pi}\int_0^\infty e^{-zn^\frac{1}{2}}\cos(2nt-zn^\frac{1}{2})\\

&= T\frac{\tilde{V}z}{2\pi^\frac{1}{2}t^\frac{3}{2}}e^{-\frac{z^2}{4t}}

\end{align}

Thomson now moves to finding the solution to the problem when a charge $Q$ is communicated instantaneously to the wire at $O$. He states the strength of the current $I$ at any point in the wire to be equal to $-\frac{1}{k}\frac{dV}{dx}$ and deduces that the "maximum electrodynamic effect" of an impulse into the wire will therefore be found by finding the value of t making $\frac{\partial V}{\partial z}$ (this being proportional to $\frac{\partial V}{\partial x}$) maximal: with $V$ given by an equation similar to the above:

\[V = \frac{e^{-\frac{z^2}{4t}}}{\sqrt{t}}\frac{Q}{\sqrt{\pi}}\sqrt{\frac{R}{C}}\]

Upon finding $\frac{\partial^2 V}{\partial z^2}$, we find that $\frac{\partial V}{\partial z}$ is maximal when:

\[\frac{1}{2t^\frac{3}{2}} = \frac{z^2}{4t^\frac{5}{2}}\]

giving the value:

\[t = \frac{z^2}{2} = \frac{RCx^2}{2}\]

Note that this is different to the value of t which Thomson gives, being $\frac{z^2}{6}$. This is however entirely irrelevant to the remainder of the exposition; the point of the calculation was to show that the time that one should leave between message pulses due to the "retardations of signals" was proportional to the square of the length of the wire, now taking x to be the length of the cable. This is a deduction which Thomson’s erroneous calculation still makes.

Thomson then goes on to discuss the "velocity of transmission" of the signal. He states that, if the potential at $O$ is varied in a sinusoidal way (namely with $\sin(2nt)$), then this velocity is equal to $2\sqrt{\frac{n}{RC}}$ . It is interesting that the velocity of transmission and the retardations of the signals bear no mathematical relation to one another: showing again that this problem is, perhaps counter-intuitively, one regarding the length of the wire

### Application to the cable

As stated before, the main obstacle to the efficiency of communication was the length of the wire used. A cursory glance at the equation above will show that using a wire with a lower capacitance and/or resistance would also help to reduce the time taken to send a message. Attempting to use this part of the relation is, however, futile for two reasons.

Firstly, the difference made by reducing the resistance or capacitance would only achieve results proportional to the square root of shortening the distance.

Secondly, the makers of the wire were already using one of the least resistant metals commonly-available in their cable: copper. Copper has the second lowest resistance of any conductive metal; the first, being silver, was far too expensive to use in such quantities. Also, the capacitance of the wire can only be decreased in a meaningful way by decreasing the diameter of the wire, eventually rendering it too fragile to use practically. The makers of the cable came to these conclusions too. The wires of the 1858 and 1865 cables were made with the same materials, with very similar insulation; the only difference was a decrease of 304 nautical miles, or about 350 imperial miles, in the length of the cable. This difference in distance is thus clearly what caused the increased efficiency of the 1865 cables and is a physical proof of the validity of Thomson’s work.

###
Conclusion

The most striking thing about the paper is not the mathematics itself, while ingenious in many parts; it is the surprising and definitive results the mathematics lead to. Based on this, Thomson made the following assertion in his second letter to Stokes:

"We may be

*sure*beforehand that the American telegraph will succeed"
From the point of view of an investor in the early 1860s, this was a very bold claim to make, given the severe limitations of the 1858 model. But, sure enough, by 1866, the mathematics of Thomson had resulted in the laying of two cables, each almost ten times more efficient than the previous.

*This post is adapted from an essay by the author written for a course lectured by Professor Jeremy Gray in November 2017*
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