Special relativity: from clock operations to electrodynamics: §10 · electron force, work, and closing scope

Einstein’s transverse mass, $m\gamma^2$, divides the force measured in the electron’s momentary rest frame by the acceleration measured in the laboratory. Measure both in the laboratory and the ratio is $m\gamma$. The two numbers answer two different questions; the electron has one mass $m$.

Longitudinal laboratory force over acceleration is m gamma cubed. Rest-frame transverse force over laboratory acceleration is m gamma squared. Laboratory transverse force over laboratory acceleration is m gamma.

For an electron moving along $x$, a force across the motion is $\gamma$ times larger in the electron’s momentary rest frame than in the laboratory. Einstein divides the rest-frame force by the laboratory acceleration and gets $m\gamma^2$. Dividing the laboratory force by the same acceleration gives $m\gamma$.

At $0.6c$, $\gamma = 1.25$. The longitudinal mass is $m\gamma^3 = 1.953125\,m$, Einstein’s transverse mass is $m\gamma^2 = 1.5625\,m$, and the all-laboratory transverse ratio is $m\gamma = 1.25\,m$. The electron-dynamics laboratory shows each number with the frames it uses, and leaves Einstein’s result as he printed it.

Return to why there are two transverse coefficients. (included in this file)

Letters: in §10 the paper's electron is a point particle of charge ε and mass μ, where μ is its mass while it moves slowly; X, Y, Z are the electric force and M, N magnetic components, in Gaussian units; β is the modern γ and V the speed of light. This explanation writes m for μ, and F with a frame label for each force.

1. The only assumption about the law of motion: when the electron is at rest at some moment, in the next instant it moves by $\mu\frac{d^2x}{dt^2}=ε X$, and likewise for y and z, ordinary mechanics for slow motion.
2. Now let the electron have speed v along x at the moment t = 0, at the origin of K. It is then at rest in the system k moving at v, so in k, for small times, $\mu\frac{d^2ξ}{d\tau^2}=ε X'$ and likewise for η and ζ.
3. Transform back to K with §3 and §6, using $X'=X$, $Y'=\beta(Y-\frac{v}{V}N)$, $Z'=\beta(Z+\frac{v}{V}M)$. The paper's equations (A): $\frac{d^2x}{dt^2}=\frac{ε}{\mu}\frac{1}{\beta^3}X$, $\frac{d^2y}{dt^2}=\frac{ε}{\mu}\frac{1}{\beta}(Y-\frac{v}{V}N)$, and the same for z.
4. Rewrite them as the paper does: $\mu\beta^3\frac{d^2x}{dt^2}=ε X=ε X'$ and $\mu\beta^2\frac{d^2y}{dt^2}=ε\beta(Y-\frac{v}{V}N)=ε Y'$.
5. Read $εX'$, $εY'$, $εZ'$ as the force on the electron measured in a system moving with it at that moment; the paper says it could be measured with a spring balance at rest in that system.
6. Keep mass times acceleration equal to force, with that force and with accelerations measured in K. Then the ratio of force to acceleration is $\mu\beta^3$ along the motion and $\mu\beta^2$ across it: the paper's longitudinal mass $\mu/(\sqrt{1-(v/V)^2})^3$ and transverse mass $\mu/(1-(v/V)^2)$.
7. Measure the transverse force in K as well: it is smaller than the co-moving force by the factor β, so that ratio is $m\gamma$. The three ratios are below.

Longitudinal laboratory force over acceleration is m gamma cubed. Rest-frame transverse force over laboratory acceleration is m gamma squared. Laboratory transverse force over laboratory acceleration is m gamma.

1. Worked at 0.6c, γ = 1.25: longitudinal 1.953125 m, Einstein's transverse 1.5625 m, and the all-laboratory transverse ratio 1.25 m.
2. The paper's own warning, kept with its result: with other definitions of force and of acceleration, other numbers for these masses would be obtained, so theories of the electron's motion must be compared with great care.
3. The transverse mass $m\gamma^2$ is Einstein's, under his convention: force measured in the co-moving system, acceleration in K. The later habit of defining force as the rate of change of momentum gives $m\gamma$ for the same transverse case; it is a different convention, and it does not correct his result.
4. The paper adds that these results hold for any ponderable material point too, since a vanishingly small charge turns it into an electron in its sense.
5. The electron's mass m is one number; each ratio depends on the direction of the force and on the frames used, and m does not.

Return to why there are two transverse coefficients. (included in this file)

Modern qualifications

Einstein himself warns that other definitions of force and acceleration would give other values for these masses. The modern habit of defining force as the rate of change of momentum gives $m\gamma$ for the transverse ratio. It is a later convention, and this edition does not substitute it for his printed result.

Adding up the work of the force along the motion, from rest to speed $v$, gives the electron’s kinetic energy, $mc^2(\gamma - 1)$. Einstein then gives two more relations a laboratory can test: the voltage that accelerates the electron, and how sharply electric and magnetic fields bend its path.

Integrating the longitudinal work from rest gives m c squared times gamma minus one.

The integral follows from one derivative, $d\gamma/du = \gamma^3 u/c^2$: the integrand $m\gamma^3 u$ is $mc^2$ times it, so the integral is $mc^2(\gamma - 1)$. At low speed this approaches the familiar $mv^2/2$. At $0.6c$ the exact value is $0.25\,mc^2$ and the low-speed formula gives $0.18\,mc^2$, a difference large enough to see.

For acceleration from rest, charge magnitude times voltage magnitude supplies kinetic energy; in a transverse magnetic field the circular radius is gamma m v over charge magnitude B.

The third relation compares bending by the two kinds of field. In an electric field across the motion the path curves with radius $\gamma mv^2/(|q|E_y)$ at that instant; the field also speeds the electron up as the path turns, so the radius keeps changing. In a magnetic field the speed stays constant and the path is a circle of radius $\gamma mv/(|q|B)$. The electron-dynamics laboratory keeps the two cases apart.

Return to the energy it takes to accelerate an electron. (included in this file)

Letters: the paper's electron has charge ε and mass μ (m here); it starts at rest at the origin of K and moves along the X axis under an electrostatic force X; W is its energy of motion, P the potential difference it has passed through, its magnetic and electric deflectabilities $A_m$ and $A_e$, R the radius of curvature of its path, N a magnetic force across its motion, and V the speed of light.

1. The energy taken from the electrostatic field is $\intε X\,dx$. The electron is accelerated slowly and so gives off no energy as radiation, so that energy all becomes its energy of motion W.
2. The first of equations (A) holds throughout: $ε X=\mu\beta^3\frac{d^2x}{dt^2}$. Along the path, acceleration times a small distance is $v\,dv$, so $ε X\,dx=\mu\beta^3v\,dv$.
3. Integrate from rest to speed v, using $\frac{d}{dv}\frac{1}{\sqrt{1-v^2/V^2}}=\frac{v}{V^2}\beta^3$: the paper's result $W=\mu V^2\left\{\frac{1}{\sqrt{1-(v/V)^2}}-1\right\}$. In the explanatory letters, the formula below.

Integrating the longitudinal work from rest gives m c squared times gamma minus one.

1. At low speed, $1/\sqrt{1-ε}$ is about $1+ε/2$, so W approaches $\frac{1}{2}mv^2$.
2. Worked at 0.6c: exact $(1.25-1)mc^2=0.25\,mc^2$; low-speed $0.5\times0.36\,mc^2=0.18\,mc^2$. For an electron, with modern SI constants, $mc^2$ is about 511 kiloelectronvolts, so W is about 127.7 kiloelectronvolts.
3. W becomes infinite at v = V. The paper concludes, as from its earlier results, that speeds greater than light have no possibility of existence.
4. The paper lists three properties of the motion that experiment can reach. First: an electric force Y and a magnetic force N deflect an electron moving at v equally when $Y=N\,v/V$, so the ratio of magnetic to electric deflectability gives the speed, $A_m/A_e=v/V$, and the speed can be measured independently.
5. Second: the potential difference passed through and the speed gained are related by $P=\int X\,dx=\frac{\mu}{ε}V^2\left\{\frac{1}{\sqrt{1-(v/V)^2}}-1\right\}$. In SI, charge times voltage equals W. Worked: to reach 0.6c from rest, an electron needs about 128 kilovolts, with no energy lost to radiation.
6. Third: the radius of curvature R when a magnetic force N across the motion is the only deflecting force. From the second of (A), $v^2/R=\frac{ε}{\mu}\frac{v}{V}N\sqrt{1-(v/V)^2}$, so $R=V^2\frac{\mu}{ε}\frac{v/V}{\sqrt{1-(v/V)^2}}\frac{1}{N}$. In SI that is $r_B=\gamma mv/(|q|B)$, the second formula below.

For acceleration from rest, charge magnitude times voltage magnitude supplies kinetic energy; in a transverse magnetic field the circular radius is gamma m v over charge magnitude B.

1. Worked: an electron at 0.6c in a field of 0.1 tesla circles with radius $1.25\times m\times0.6c/(|q|\times0.1)$, about 1.28 centimetres, with modern SI constants.
2. Keep the magnetic case, a circle at constant speed, apart from bending in an electric field, which also changes the speed and so gives a radius that holds only at one instant.
3. The paper calls these three relations a complete expression of the laws by which, according to the theory, the electron must move. They hold within its assumptions: the fields as specified, and no energy lost to radiation.
4. The kinetic energy formula alone does not give the September paper's result that a body's mass changes with its energy content; that needs the argument from the light complex of §8.

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Modern qualifications

The three relations hold within the model’s assumptions: fields as specified, and no energy lost to radiation. After §10 the paper closes with thanks to M. Besso and the date line, Bern, June 1905. Those belong to the German source text, and this explanation does not repeat them.