Read · Special relativity: from clock operations to electrodynamics

§10 · electron force, work, and closing scope

Follow the introduction and all ten sections, including field transformations, finite light complexes, moving mirrors, charge-current transformations, and the electron-force conventions.

Newly authored explanatory preview in modern notation, with editorial and physics review pending. Both the kinematic and electrodynamic halves are treated, but this is not a German transcription, an aligned translation, or a complete critical edition. The source paragraphs, footnotes, acknowledgment, and date-lines are not claimed to be fully represented or reviewed here. Headings and argument units are editorial.

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§10 · electron force, work, and closing scope

derivation · Within the stated model

A force-to-acceleration ratio needs two frame labels

Why are there two transverse coefficients?

Fxax=mγ3,Fyay=mγ2,Fyay=mγ\frac{F_x}{a_x}=m\gamma^3,\qquad\frac{F_y'}{a_y}=m\gamma^2,\qquad\frac{F_y}{a_y}=m\gamma

Longitudinal laboratory force over acceleration gives m gamma cubed; comoving transverse force over laboratory acceleration gives m gamma squared; laboratory transverse force gives m gamma.

For motion along x, the transverse force in the momentarily comoving frame differs from the laboratory transverse force by γ. Combining it with laboratory acceleration gives the source’s coefficient. Replacing that force with laboratory force changes the ratio.

At 0.6c, γ = 1.25. The longitudinal coefficient is 1.953125m, the printed transverse coefficient is 1.5625m, and the laboratory transverse coefficient is 1.25m. SR-13 displays the frame convention with the coefficient, rather than “repairing” the printed result.

Show every step here: A force-to-acceleration ratio needs two frame labels
  1. Choose motion along the x axis and distinguish longitudinal from transverse components.
  2. Label the frame of the force and acceleration separately.
  3. Use the instantaneous comoving-frame force law.
  4. Transform acceleration and force consistently.
  5. Form the source’s mixed-frame transverse ratio and the separate laboratory ratio.
  6. Keep invariant mass m distinct from either direction-dependent coefficient.
Assumptions and limits: A force-to-acceleration ratio needs two frame labels

Assumed here

  • Start with the electron’s instantaneous rest-frame force law.
  • Specify the frame of each force and acceleration component.

What this does not establish

  • The historical coefficients are not a claim that invariant rest mass changes with observer speed.
  • The ideal slowly accelerated point-particle treatment omits radiation reaction and self-field dynamics.

Earlier step: Electric and magnetic components mix together

Source context: German source · English · Interlinear gloss · Facsimile

derivation · Within the stated model

The work integral has an observable endpoint

What energy is required to accelerate an electron?

K=0vmγ(u)3udu=mc2(γ1)K=\int_0^v m\gamma(u)^3u\,du=mc^2(\gamma-1)

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

Use dγ/du = γ³u/c² to evaluate the integral. Near zero speed, K approaches mv²/2. At 0.6c the exact value is 0.25mc² whereas the quadratic approximation is 0.18mc², so the approximation error is visible.

qU=K,rB=γmvqB|q|U=K,\qquad r_B=\frac{\gamma mv}{|q|B}

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 electric and magnetic deflection, with force and acceleration conventions kept explicit. In a purely transverse electric field the instantaneous curvature radius is γmv²/(|q|E). SR-13 separates that local curvature from a magnetic circular orbit at constant speed.

Show every step here: The work integral has an observable endpoint
  1. Write longitudinal force as mγ³ times acceleration.
  2. Multiply by displacement, using a dx = u du.
  3. Integrate from rest and use the derivative of gamma.
  4. Compare the exact value with the low-speed expansion without calling them equal at 0.6c.
  5. For a voltage estimate, use charge magnitude and state the energy-loss assumption.
  6. For deflection, distinguish magnetic constant-speed curvature from an instantaneous transverse electric curvature.
Assumptions and limits: The work integral has an observable endpoint

Assumed here

  • Use invariant mass m, subluminal speed, and the stated force convention.
  • For the voltage relation, the particle begins at rest and field work becomes kinetic energy.

What this does not establish

  • The ideal treatment excludes radiation loss and material interactions.
  • A kinetic-energy expression alone does not derive the September statement about changing rest energy.

Earlier step: A force-to-acceleration ratio needs two frame labels

Source context: German source · English · Interlinear gloss · Facsimile

References and source status

Newly authored explanatory preview in modern notation, with editorial and physics review pending. Both the kinematic and electrodynamic halves are treated, but this is not a German transcription, an aligned translation, or a complete critical edition. The source paragraphs, footnotes, acknowledgment, and date-lines are not claimed to be fully represented or reviewed here. Headings and argument units are editorial.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content?. Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s reviewed translation or pinned facsimile.

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