I. Kinematical Part.Show the German source

II. Electrodynamical Part.Show the German source

§ 10. Dynamics of the (Slowly Accelerated) Electron.Show the German source

Let there move in an electromagnetic field a point-like particle provided with an electric charge $\varepsilon$ (in what follows called “electron”), about whose law of motion we assume only the following:Show the German source

If the electron is at rest at a given epoch, its motion in the next instant of time takes place according to the equationsShow the German source $$\begin{aligned} \mu \frac{d^2 x}{d t^2} &= \varepsilon X \\ \mu \frac{d^2 y}{d t^2} &= \varepsilon Y \\ \mu \frac{d^2 z}{d t^2} &= \varepsilon Z , \end{aligned}$$$\mu$Electron mass$x$Position in the stationary system$t$Time in the stationary system$\epsilon$Electron charge$X$Electric field, stationary system$y$Position in the stationary system$Y$Electric field, stationary system$z$Position in the stationary system$Z$Electric field, stationary systemExplain this equationShow the German source where $x$, $y$, $z$ denote the coordinates of the electron, and $\mu$ the mass of the electron, as long as it moves slowly.Show the German source

Now, secondly, let the electron have the velocity $v$ at a certain epoch of time.Show the German source We seek the law according to which the electron moves in the instant of time immediately following.Show the German source

Without affecting the generality of the consideration, we may and will assume that the electron, at the moment we fix our attention on it, is at the origin of coordinates and is moving along the $X$-axis of the system $K$ with the velocity $v$.Show the German source It is then evident that at the moment in question ($t = 0$) the electron is at rest relative to a coordinate system $k$ moving parallel to itself along the $X$-axis with the constant velocity $v$.Show the German source

From the assumption made above, in combination with the principle of relativity, it is clear that in the time immediately following (for small values of $t$) the electron, regarded from the system $k$, moves according to the equations:Show the German source $$\begin{aligned} \mu \frac{d^2 \xi}{d \tau^2} &= \varepsilon X' , \\ \mu \frac{d^2 \eta}{d \tau^2} &= \varepsilon Y' , \\ \mu \frac{d^2 \zeta}{d \tau^2} &= \varepsilon Z' , \end{aligned}$$$\mu$Electron mass$\xi$Position in the moving system$\tau$Time in the moving system$\epsilon$Electron charge$X^{\prime }$Electric field, moving system$\eta$Position in the moving system$Y^{\prime }$Electric field, moving system$\zeta$Position in the moving system$Z^{\prime }$Electric field, moving systemExplain this equationShow the German source where the symbols $\xi$, $\eta$, $\zeta$, $\tau$, $X'$, $Y'$, $Z'$ refer to the system $k$.Show the German source If we further stipulate that for $t = x = y = z = 0$ we are to have $\tau = \xi = \eta = \zeta = 0$, then the transformation equations of §§ 3 and 6 hold, so that:Show the German source $$\begin{aligned} \tau &= \beta \left( t - \frac{v}{V^2} x \right) , \\ \xi &= \beta (x - v t) , &\qquad X' &= X , \\ \eta &= y , &\qquad Y' &= \beta \left( Y - \frac{v}{V} N \right) , \\ \zeta &= z , &\qquad Z' &= \beta \left( Z + \frac{v}{V} M \right) . \end{aligned}$$$\tau$Time in the moving system$\beta$Lorentz factor$t$Time in the stationary system$v$Speed of the moving system$V$Speed of light$x$Position in the stationary system$\xi$Position in the moving system$X^{\prime }$Electric field, moving system$X$Electric field, stationary system$\eta$Position in the moving system$y$Position in the stationary system$Y^{\prime }$Electric field, moving system$Y$Electric field, stationary system$N$Magnetic field, stationary system$\zeta$Position in the moving system$z$Position in the stationary system$Z^{\prime }$Electric field, moving system$Z$Electric field, stationary system$M$Magnetic field, stationary systemExplain this equationShow the German source

With the help of these equations we transform the above equations of motion from the system $k$ to the system $K$, and obtain:Show the German source $$\left\{ \begin{aligned} \frac{d^2 x}{d t^2} &= \frac{\varepsilon}{\mu} \frac{1}{\beta^3} X , \\ \frac{d^2 y}{d t^2} &= \frac{\varepsilon}{\mu} \frac{1}{\beta} \left( Y - \frac{v}{V} N \right) , \\ \frac{d^2 z}{d t^2} &= \frac{\varepsilon}{\mu} \frac{1}{\beta} \left( Z + \frac{v}{V} M \right) . \end{aligned} \right.$$$x$Position in the stationary system$t$Time in the stationary system$\epsilon$Electron charge$\mu$Electron mass$\beta$Lorentz factor$X$Electric field, stationary system$y$Position in the stationary system$Y$Electric field, stationary system$v$Speed of the moving system$V$Speed of light$N$Magnetic field, stationary system$z$Position in the stationary system$Z$Electric field, stationary system$M$Magnetic field, stationary systemExplain this equationShow the German source

Following the usual line of thought, we now ask for the “longitudinal” and “transverse” mass of the moving electron.Show the German source We write the equations (A) in the formShow the German source $$\begin{aligned} \mu \beta^3 \frac{d^2 x}{d t^2} &= \varepsilon X = \varepsilon X' , \\ \mu \beta^2 \frac{d^2 y}{d t^2} &= \varepsilon \beta \left( Y - \frac{v}{V} N \right) = \varepsilon Y' , \\ \mu \beta^3 \frac{d^2 z}{d t^2} &= \varepsilon \beta \left( Z + \frac{v}{V} M \right) = \varepsilon Z' \end{aligned}$$$\mu$Electron mass$\beta$Lorentz factor$x$Position in the stationary system$t$Time in the stationary system$\epsilon$Electron charge$X$Electric field, stationary system$X^{\prime }$Electric field, moving system$y$Position in the stationary system$Y$Electric field, stationary system$v$Speed of the moving system$V$Speed of light$N$Magnetic field, stationary system$Y^{\prime }$Electric field, moving system$z$Position in the stationary system$Z$Electric field, stationary system$M$Magnetic field, stationary system$Z^{\prime }$Electric field, moving systemExplain this equationShow the German source and note first that $\varepsilon X'$, $\varepsilon Y'$, $\varepsilon Z'$ are the components of the ponderomotive force acting on the electron, and indeed regarded in a system moving at this moment with the electron at the same velocity as it.Show the German source (This force could be measured, for example, with a spring balance at rest in the latter system.)Show the German source If we now call this force simply “the force acting on the electron”, and maintain the equation [mass number × acceleration number = force number]Show the German source $$\text{Massenzahl} \times \text{Beschleunigungszahl} = \text{Kraftzahl}$$Explain this equationShow the German source and if we further stipulate that the accelerations are to be measured in the stationary system $K$, we obtain from the above equations:Show the German source $$\text{Longitudinale Masse} = \frac{\mu}{\left( \sqrt{1 - \left( \frac{v}{V} \right)^2} \right)^3} ,$$$\mu$Electron mass$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source $$\text{Transversale Masse} = \frac{\mu}{1 - \left( \frac{v}{V} \right)^2} .$$$\mu$Electron mass$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source

Naturally, with a different definition of force and acceleration one would obtain other numbers for the masses; one sees from this that one must proceed very cautiously in comparing different theories of the motion of the electron.Show the German source

We remark that these results about the mass also hold for ponderable material points; for a ponderable material point can be made into an electron (in our sense) by adding an arbitrarily small electric charge.Show the German source

We determine the kinetic energy of the electron.Show the German source If an electron moves from the origin of coordinates of the system $K$, with initial velocity 0, steadily along the $X$-axis under the action of an electrostatic force $X$, it is clear that the energy withdrawn from the electrostatic field has the value $\int \varepsilon X d x$.Show the German source Since the electron is to be accelerated slowly and consequently is to give off no energy in the form of radiation, the energy withdrawn from the electrostatic field must be set equal to the energy of motion $W$ of the electron.Show the German source Taking into account that during the whole process of motion considered the first of the equations (A) holds, we therefore obtain:Show the German source $$W = \int \varepsilon X d x = \int_0^v \beta^3 v d v = \mu V^2 \left\{ \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} - 1 \right\} .$$$W$Kinetic energy$\epsilon$Electron charge$X$Electric field, stationary system$x$Position in the stationary system$v$Speed of the moving system$\beta$Lorentz factor$\mu$Electron mass$V$Speed of lightExplain this equationShow the German source

$W$ thus becomes infinitely great for $v = V$.Show the German source Velocities greater than that of light have — as in our earlier results — no possibility of existence.Show the German source

By the argument given above, this expression for the kinetic energy must also hold for ponderable masses.Show the German source

We shall now enumerate the properties of the electron’s motion that result from the system of equations (A) and are accessible to experiment.Show the German source

1. From the second equation of the system (A) it follows that an electric force $Y$ and a magnetic force $N$ have an equally strong deflecting action on an electron moving with the velocity $v$ when $Y = N \cdot v/V$.Show the German source One sees, therefore, that according to our theory the velocity of the electron can be determined, for arbitrary velocities, from the ratio of the magnetic deflectability $A_m$ to the electric deflectability $A_e$, by applying the law:Show the German source $$\frac{A_m}{A_e} = \frac{v}{V} .$$$A_m$Magnetic deflectability$A_e$Electric deflectability$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source

This relation is accessible to experimental test, since the velocity of the electron can also be measured directly, for example by means of rapidly oscillating electric and magnetic fields.Show the German source

2. From the derivation of the kinetic energy of the electron it follows that between the potential difference traversed and the velocity $v$ acquired by the electron the relation must hold:Show the German source $$P = \int X d x = \frac{\mu}{\varepsilon} V^2 \left\{ \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} - 1 \right\} .$$$P$Accelerating voltage$X$Electric field, stationary system$x$Position in the stationary system$\mu$Electron mass$\epsilon$Electron charge$V$Speed of light$v$Speed of the moving systemExplain this equationShow the German source

3. We calculate the radius of curvature $R$ of the path when a magnetic force $N$ acting perpendicular to the velocity of the electron is present (as the only deflecting force).Show the German source From the second of the equations (A) we obtain:Show the German source $$- \frac{d^2 y}{d t^2} = \frac{v^2}{R} = \frac{\varepsilon}{\mu} \frac{v}{V} N \cdot \sqrt{1 - \left( \frac{v}{V} \right)^2}$$$y$Position in the stationary system$t$Time in the stationary system$v$Speed of the moving system$R$Radius of the circular path$\epsilon$Electron charge$\mu$Electron mass$V$Speed of light$N$Magnetic field, stationary systemExplain this equationShow the German source orShow the German source $$R = V^2 \frac{\mu}{\varepsilon} \cdot \frac{\frac{v}{V}}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} \cdot \frac{1}{N} .$$$R$Radius of the circular path$V$Speed of light$\mu$Electron mass$\epsilon$Electron charge$v$Speed of the moving system$N$Magnetic field, stationary systemExplain this equationShow the German source

These three relations are a complete expression of the laws according to which, by the present theory, the electron must move.Show the German source

In conclusion I remark that in working on the problem treated here my friend and colleague M. Besso stood loyally by my side, and that I owe him several valuable suggestions.Show the German source

Bern, June 1905.Show the German source

(Received 30 June 1905.)Show the German source