Read · Special relativity: from clock operations to electrodynamics
Special relativity: from clock operations to electrodynamics
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.
What should the distant clock read when the flash reflects?
Two stations, A and B, stand apart along a straight track. Each has a clock. A flash leaves A when its clock reads 0, reflects at B, and returns to A when its clock reads 10. What should the clock at B have read at the reflection?
An authored clock-setting example. The readings below are not experimental evidence.
Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it is
Reading
Flash leaves A: A’s clock
0
Flash reflects at B: proposed assignment to B’s clock
Not assigned before an agreement
Flash returns to A: A’s clock
10
What would you say?
Open any answer to examine its assumptions. Nothing must be answered to continue.
0
This uses the departure reading. It would assign no time to the outward trip. Nothing in the two readings requires that assumption.
5
This is the halfway reading. It follows once we agree to assign equal times to the two directions. Without that agreement, the two readings alone do not force it.
10
This uses the return reading at A. It would assign no time to the return trip. Receiving news of an event is not the event itself.
Nothing tells us yet
Exactly: the two local readings do not yet specify how the distant clock should be set. We first need to state a procedure. This is not a failure to find the answer.
State the agreement before assigning the reading
If we agree that the outward trip takes as long as the return trip, assign the reflection the reading halfway between departure and return. This is a stated clock-setting procedure, not a measurement of the separate one-way travel times.
Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it is
Reading
Flash leaves A: A’s clock
0
Flash reflects at B: proposed assignment to B’s clock
Not assigned before an agreement
Flash returns to A: A’s clock
10
Leaves at 0, returns at 10. No distant reading has been assigned because no agreement is selected.
A path sketch, not a scale drawing or an animation measuring travel time. The complete readings are in the table.
The laboratory uses the stated agreement and an ideal stationary separation that reproduces this round trip. No measured separation or independently measured one-way speed is inferred here.
The whole worked route, without controls
Try a second flash leaving at 20 and returning at 30. The same agreement gives a new reading for its reflection, without changing the rule. Both tables here use the equal-travel-time agreement.
Leaves at 0, returns at 10
Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it is
Reading
Flash leaves A: A’s clock
0
Flash reflects at B: proposed assignment to B’s clock
5
Flash returns to A: A’s clock
10
Leaves at 20, returns at 30
Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it is
Reading
Flash leaves A: A’s clock
20
Flash reflects at B: proposed assignment to B’s clock
25
Flash returns to A: A’s clock
30
In the first example, the reflection is assigned 5 while its returning signal reaches A at 10. Those are different events. In the second, the reflection is assigned 25 and the signal returns at 30.
What is getting in the way?
Is the event the same as seeing it?
No. The reflection happens at B. A receives the returning signal later. The time of reception at A is not the reading assigned to the earlier reflection at B.
Why agree on a rule?
The departure and return readings use one clock, at A. They do not by themselves tell us how a distant clock at B is set. The equal-travel-time rule supplies that missing agreement.
Does this measure the two one-way travel times?
No. The round-trip readings are the starting observations. Assigning equal time to each direction is a stated procedure, not a separate measurement of each one-way time.
Suppose a moving pair of stations uses the same agreement with its own flashes. Will it agree with us about which distant events happen at the same moment?
Set a distant clock by an agreed procedure instead of assuming it already agrees with yours.
Section 1 defines what “at the same time” means for distant clocks by this agreement. An event ledger keeps the resulting readings separate from the times when signals arrive.
The linked reading is a draft explanation. A reviewed sentence-aligned original and translation are still in preparation.
Introduction · magnet, conductor, and the postulates
assumption · Within the stated model
One apparatus, two descriptions
Why should changing the observer change the explanation?
A moving conductor in a magnetic field and a moving magnet near a conductor invite different classical explanations. Relativity asks for consistent predictions of the same physical arrangement, not identical numerical field components.
In one description, charges move through a magnetic field. In another, the conductor is at rest and an electric field drives them. The apparent explanatory asymmetry motivates the paper; it is not resolved by declaring that electric and magnetic fields have the same values for every observer.
The two postulates state that the laws have the same form in inertial frames and that light propagates in vacuum at a definite speed independent of the source motion. They are the starting point, not consequences of failed ether-drift experiments alone.
SR-02 compares the descriptions under explicit transformations. At finite relative speed, forces and electromotive quantities must be compared through their transformation laws. The familiar equal-value argument is only a low-speed approximation.
Identify the conductor, magnet, charges, and the observer.
Describe the same interaction in the other inertial frame.
Record which quantities change under that change of description.
State the two postulates before using them.
Return to §6 to see electric and magnetic components transform together.
Show every step here: One apparatus, two descriptions
Identify the conductor, magnet, charges, and the observer.
Describe the same interaction in the other inertial frame.
Record which quantities change under that change of description.
State the two postulates before using them.
Return to §6 to see electric and magnetic components transform together.
A moving conductor in a magnetic field and a moving magnet near a conductor invite different classical explanations. Relativity asks for consistent predictions of the same physical arrangement, not identical numerical field components.
What this relies on
Compare inertial descriptions of the same apparatus and events.
The relativity principle and light-speed postulate are stated physical assumptions.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: One apparatus, two descriptions
Assumed here
Compare inertial descriptions of the same apparatus and events.
The relativity principle and light-speed postulate are stated physical assumptions.
What this does not establish
A simulator cannot prove its own postulates.
The restricted dipole-and-conductor model does not cover every induction experiment.
How can clocks at different places share a time coordinate?
Exchange a light signal, assign the distant reflection the midpoint of the departure and return times, and state the equal-travel-time convention. The event time is not the time its image reaches an observer.
tB−tA=tA,return−tB
The outward assigned travel time equals the return assigned travel time.
tB=2tA+tA,return
The assigned reflection time is the midpoint of departure and return times.
If A sends at time 0 and receives the reflected signal at time 10, the convention assigns time 5 at the reflection event at B. A later reception of a photograph of B is another event, not a replacement for that time assignment.
SR-01 retains departure, reflection, reception, and clock-reading records separately. The equal-time rule defines a coordinate procedure in the chosen rest frame.
Record departure and return on the same clock at A.
Identify the intervening reflection event at B.
Assign its clock reading the arithmetic midpoint.
Check the assigned outward and return intervals.
Keep the time of a remote event separate from reception of its signal.
Show every step here: Time at a distant clock is an operation
Record departure and return on the same clock at A.
Identify the intervening reflection event at B.
Assign its clock reading the arithmetic midpoint.
Check the assigned outward and return intervals.
Keep the time of a remote event separate from reception of its signal.
Exchange a light signal, assign the distant reflection the midpoint of the departure and return times, and state the equal-travel-time convention. The event time is not the time its image reaches an observer.
What this relies on
The clocks and endpoints are at rest in one inertial frame.
The light-based convention assigns equal outward and return travel times.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Time at a distant clock is an operation
Assumed here
The clocks and endpoints are at rest in one inertial frame.
The light-based convention assigns equal outward and return travel times.
What this does not establish
One round-trip clock reading alone does not independently measure both one-way travel times.
Synchronizing clocks is distinct from moving them or changing their rates.
Which endpoint measurements define a moving rod’s length?
A moving rod’s length uses endpoint positions measured at the same time in the measuring frame. Transforming a pair simultaneous in another frame usually does not produce that measurement.
A ruler traveling with the rod measures its rest length. An observer past whom it moves must record both endpoint positions simultaneously in that observer’s frame. Using one endpoint now and the other later includes part of the rod’s motion.
For the standard frame change, two event differences obey the same Lorentz map as coordinates. A pair with Δt = 0 and nonzero Δx generally has nonzero Δt′.
Δt=0:Δt′=−γc2vΔx
Events simultaneous in the unprimed frame have a primed time difference of minus gamma v delta x over c squared.
In SR-03, select a simultaneous slice and inspect the endpoint event identities. A larger transformed separation from an unsuitable pair does not mean the moving rod has become longer.
Choose the frame in which the length is to be measured.
Choose the same coordinate time for its two endpoint events.
Subtract their positions in that frame.
When changing frame, transform the event times as well as positions.
Select a new simultaneous pair in the other frame before calling a separation its measured length.
Show every step here: A length is a specified pair of events
Choose the frame in which the length is to be measured.
Choose the same coordinate time for its two endpoint events.
Subtract their positions in that frame.
When changing frame, transform the event times as well as positions.
Select a new simultaneous pair in the other frame before calling a separation its measured length.
A moving rod’s length uses endpoint positions measured at the same time in the measuring frame. Transforming a pair simultaneous in another frame usually does not produce that measurement.
What this relies on
The rod moves inertially along the boost axis with a fixed rest length.
Each frame uses its own synchronized clocks.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: A length is a specified pair of events
Assumed here
The rod moves inertially along the boost axis with a fixed rest length.
Each frame uses its own synchronized clocks.
What this does not establish
Coordinate length is not the appearance in a photograph.
Comparing different event pairs is essential, not an optional correction.
Linearity and the two light directions constrain space and time to mix. Reciprocity and spatial symmetry then fix a remaining scale, with the positive choice connected to the identity.
The moving origin follows x = vt, so write x′ = a(v)(x − vt). Requiring x′ = ct′ on x = ct and x′ = −ct′ on x = −ct fixes the time combination: t′ = a(v)(t − vx/c²). An overall scale a(v) remains.
x′=a(v)(x−vt),t′=a(v)(t−c2vx)
The light constraints give a common coefficient multiplying x minus vt and t minus vx over c squared.
Applying the inverse with speed −v gives a(v)a(−v)(1 − v²/c²) = 1. Isotropy identifies a(v) with a(−v). Continuity from a(0) = 1 selects the positive square root.
γ=1−v2/c21,x′=γ(x−vt),t′=γ(t−c2vx)
The Lorentz factor is one over the square root of one minus v squared over c squared, giving the standard space and time transformation.
Use the moving origin to determine the combination x − vt.
Substitute both right-moving and left-moving light paths into the linear time map.
Solve the resulting coefficient constraints without setting the scale by hand.
Compose with the inverse transformation.
Use isotropy to equate the positive-speed and negative-speed scales.
Choose the identity-connected positive solution, requiring |v| less than c.
Show every step here: Light constraints leave a scale to determine
Use the moving origin to determine the combination x − vt.
Substitute both right-moving and left-moving light paths into the linear time map.
Solve the resulting coefficient constraints without setting the scale by hand.
Compose with the inverse transformation.
Use isotropy to equate the positive-speed and negative-speed scales.
Choose the identity-connected positive solution, requiring |v| less than c.
Linearity and the two light directions constrain space and time to mix. Reciprocity and spatial symmetry then fix a remaining scale, with the positive choice connected to the identity.
What this relies on
Use aligned inertial axes and coincident origins at zero time.
Assume homogeneous space and time so the coordinate map is linear.
Use the postulates together with reciprocity, spatial isotropy, and continuity at zero speed.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Light constraints leave a scale to determine
Assumed here
Use aligned inertial axes and coincident origins at zero time.
Assume homogeneous space and time so the coordinate map is linear.
Use the postulates together with reciprocity, spatial isotropy, and continuity at zero speed.
What this does not establish
Preserving a light ray alone does not uniquely fix all transformation coefficients.
A spacetime interval or matrix is a later verification aid, not an unannounced 1905 premise.
The longitudinal light constraints do not alone determine transverse scale. Enforcing the isotropic light-speed condition on transverse propagation, together with the inverse and positive identity limit, completes the map.
Write y′ = b(v)y for a transverse axis. On a ray with x = 0, y = ct, the already determined longitudinal map gives x′ = −γvt and t′ = γt. Requiring the transformed ray’s speed to be c gives γ²v² + b²c² = γ²c², hence b² = 1.
The positive, identity-connected choice gives y′ = y and likewise z′ = z. The inverse map replaces v with −v and exchanges primed and unprimed quantities.
x=γ(x′+vt′),t=γ(t′+c2vx′)
The inverse transformation uses plus v and exchanges the two frames.
At v = 0.6c, γ = 1.25. In units where c = 1, the event (t, x) = (2, 1) becomes (1.75, −0.25); the inverse returns (2, 1). SR-04 can check the same map from either direction.
Introduce an independent transverse scale rather than assuming it is one.
Transform a ray originally perpendicular to the boost.
Require the full transformed speed, including its longitudinal component, to equal c.
Solve for the transverse scale and select the positive identity-connected branch.
Apply the inverse to a complete event, not just its spatial coordinate.
Show every step here: The sideways step needs its own condition
Introduce an independent transverse scale rather than assuming it is one.
Transform a ray originally perpendicular to the boost.
Require the full transformed speed, including its longitudinal component, to equal c.
Solve for the transverse scale and select the positive identity-connected branch.
Apply the inverse to a complete event, not just its spatial coordinate.
The longitudinal light constraints do not alone determine transverse scale. Enforcing the isotropic light-speed condition on transverse propagation, together with the inverse and positive identity limit, completes the map.
What this relies on
Use the scale fixed in the longitudinal transformation.
Light has the same speed in all spatial directions, not only along x.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: The sideways step needs its own condition
Assumed here
Use the scale fixed in the longitudinal transformation.
Light has the same speed in all spatial directions, not only along x.
What this does not establish
Checking a single axial ray cannot test transverse normalization.
Matrix determinants and interval checks verify this result but do not replace its stated physical premises.
For an inertially moving clock, its own elapsed time is smaller than the coordinate interval between its ticks. A moving rod’s simultaneous coordinate length is smaller than its rest length; neither result is a photographic distortion.
Δt=γΔτ,L=γL0
Coordinate time between a moving clock’s ticks is gamma times its proper elapsed time; simultaneous moving-rod length is rest length over gamma.
At 0.6c, a clock experiences 0.8 seconds during a one-second coordinate interval. A rod of rest length 1 metre has simultaneous coordinate length 0.8 metre. Their shared factor follows from different event constraints.
For the clock, use ticks at the same place in its rest frame. For the rod, set the endpoint measurement times equal in the measuring frame. SR-03 and SR-05 let the reader inspect those constraints separately.
1−γ1≈21c2v2
At small speed, the clock’s fractional loss per coordinate second approaches one half v squared over c squared.
For a clock’s own ticks, use its worldline and the frame where it is locally at rest.
For a moving length, impose simultaneity in the measuring frame.
At 0.6c compute γ = 1.25 and its inverse 0.8.
Keep the exact loss 0.2 distinct from the second-order approximation 0.18 at that speed.
Show every step here: Use the same measurement protocol
Specify which clock supplies each time reading.
For a clock’s own ticks, use its worldline and the frame where it is locally at rest.
For a moving length, impose simultaneity in the measuring frame.
At 0.6c compute γ = 1.25 and its inverse 0.8.
Keep the exact loss 0.2 distinct from the second-order approximation 0.18 at that speed.
For an inertially moving clock, its own elapsed time is smaller than the coordinate interval between its ticks. A moving rod’s simultaneous coordinate length is smaller than its rest length; neither result is a photographic distortion.
What this relies on
Use an ideal clock and inertial rod, with the appropriate event pairs.
γ uses the relative inertial speed in the specified frame.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Use the same measurement protocol
Assumed here
Use an ideal clock and inertial rod, with the appropriate event pairs.
γ uses the relative inertial speed in the specified frame.
What this does not establish
Reciprocal moving-clock rates compare different distant-clock procedures.
A single transformed event separation is not automatically a length or clock reading.
Can reciprocal rate descriptions settle a reunion?
Reunited clocks compare elapsed times along different paths between shared meetings. Flat-spacetime path accounting does not predict a real equator-versus-pole clock comparison by itself.
Δτ=∫t0t11−c2u(t)2dt
An ideal clock’s elapsed proper time is the path integral of the square root of one minus its speed squared over c squared.
For two idealized inertial legs at speed magnitude 0.6c lasting a total of 10 coordinate seconds, the traveling clock accumulates 8 seconds when the turnaround duration is neglected. A clock remaining at rest accumulates 10. Each piece and its common endpoints must be specified.
SR-05 distinguishes a transported-clock comparison from a statement about reciprocal rates between separated inertial clocks. The equator remark has a different boundary: a real rotating Earth includes gravity and is not decided by the flat-spacetime model alone.
Integrate the ideal elapsed-time law along each path.
For constant-speed legs, multiply each coordinate duration by the appropriate inverse γ.
Do not substitute a one-way visual delay for elapsed proper time.
Leave an Earth comparison outside the model when gravitational inputs are absent.
Show every step here: A reunion compares whole paths
Fix the meetings being compared.
Specify each clock’s entire path between them.
Integrate the ideal elapsed-time law along each path.
For constant-speed legs, multiply each coordinate duration by the appropriate inverse γ.
Do not substitute a one-way visual delay for elapsed proper time.
Leave an Earth comparison outside the model when gravitational inputs are absent.
Reunited clocks compare elapsed times along different paths between shared meetings. Flat-spacetime path accounting does not predict a real equator-versus-pole clock comparison by itself.
What this relies on
Use ideal clocks following specified subluminal paths in flat spacetime.
The comparison fixes the departure and reunion events.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: A reunion compares whole paths
Assumed here
Use ideal clocks following specified subluminal paths in flat spacetime.
The comparison fixes the departure and reunion events.
What this does not establish
Real terrestrial comparisons also require gravitational effects.
Acceleration changes the path; no universal extra acceleration penalty is added to the clock formula.
Why do two ordinary-looking speeds not simply add?
Velocity is a ratio of a spatial difference to a time difference. Transform both differences before dividing, and light speed stays c while collinear subluminal speeds compose below c.
ux′=1−vux/c2ux−v,uy′=γ(1−vux/c2)uy
The transformed longitudinal velocity is u x minus v over one minus v u x over c squared; transverse velocity has the same denominator and an additional gamma.
To describe an object moving at w along the moving frame’s positive axis in the original frame, use the inverse relation U = (v + w)/(1 + vw/c²). With v = w = 0.6c, U is 15c/17, about 0.88235c, not 1.2c.
For a right-moving light ray u_x = c, the passive transformation also gives c. SR-06 shows the difference from Galilean addition without clamping a superluminal arithmetic result.
Differentiate x′ = γ(x − vt) along the trajectory.
Differentiate t′ = γ(t − vx/c²) along that same trajectory.
Divide dx′ by dt′ only after both have been transformed.
For transverse motion use dy′ = dy but retain the transformed time denominator.
Use the inverse relation for the plus-sign composition of two forward speeds.
Distinguish aligned composition from the general non-collinear case.
Show every step here: The denominator changes as well
Differentiate x′ = γ(x − vt) along the trajectory.
Differentiate t′ = γ(t − vx/c²) along that same trajectory.
Divide dx′ by dt′ only after both have been transformed.
For transverse motion use dy′ = dy but retain the transformed time denominator.
Use the inverse relation for the plus-sign composition of two forward speeds.
Distinguish aligned composition from the general non-collinear case.
Velocity is a ratio of a spatial difference to a time difference. Transform both differences before dividing, and light speed stays c while collinear subluminal speeds compose below c.
What this relies on
Use the same two neighboring events on a particle’s trajectory.
Frames are inertial and the boost is along x.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: The denominator changes as well
Assumed here
Use the same two neighboring events on a particle’s trajectory.
Frames are inertial and the boost is along x.
What this does not establish
The plus-sign composition formula and a passive frame change have different stated directions.
Non-collinear boosts generally include a rotation, not only another aligned boost.
The chain rule mixes time and longitudinal derivatives. Grouping the resulting terms suggests transformed electric and magnetic components; covariance alone is not a uniqueness proof of their physical identification.
∂t=γ(∂t′−v∂x′),∂x=γ(∂x′−v∂t′/c2)
The chain rule mixes the time and x derivatives, with gamma and the appropriate powers of light speed.
For a plane wave traveling along x with E_y and B_z, the SI vacuum pair is ∂t E_y = −c² ∂x B_z and ∂t B_z = −∂x E_y. Substitute the two derivative operators above into both equations; do not transform only one equation.
Combining those substituted equations groups E_y − vB_z with B_z − vE_y/c². Multiplication by the common γ produces E′_y and B′_z and restores the corresponding primed equation pair. SR-07 opens the grouping step rather than treating the field law as a typographic replacement.
The source permits a common factor for all transformed fields. Inversion requires its value at v times its value at −v to be one. Spatial symmetry and the positive identity-connected normalization select one.
Apply the chain rule to time and x derivatives of the same field.
Substitute into both members of the coupled plane-wave equation pair.
Group the electric/magnetic combinations, retaining every factor of c.
Check the inverse and spatial symmetry before fixing the common normalization.
Keep physical field identification separate from the algebraic fact that the equations keep their form.
Show every step here: Transform derivatives before naming fields
Name the coordinate map and its inverse.
Apply the chain rule to time and x derivatives of the same field.
Substitute into both members of the coupled plane-wave equation pair.
Group the electric/magnetic combinations, retaining every factor of c.
Check the inverse and spatial symmetry before fixing the common normalization.
Keep physical field identification separate from the algebraic fact that the equations keep their form.
The chain rule mixes time and longitudinal derivatives. Grouping the resulting terms suggests transformed electric and magnetic components; covariance alone is not a uniqueness proof of their physical identification.
What this relies on
Use the Lorentz coordinate map and vacuum Maxwell equations.
The example is expressed in SI with explicit factors of c.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Transform derivatives before naming fields
Assumed here
Use the Lorentz coordinate map and vacuum Maxwell equations.
The example is expressed in SI with explicit factors of c.
What this does not establish
SI factors cannot be obtained by merely renaming the source’s Gaussian field symbols.
A common field scale needs a normalization argument in addition to rearranging equations.
For a boost along x, longitudinal field components stay unchanged and transverse electric and magnetic components mix. Transforming the force consistently does not require equal raw force values in both frames.
The x components stay unchanged; the transverse electric and magnetic components mix with the stated signs and SI factors.
For E = 0 and a positive B_z, an observer moving along positive x measures a negative E′_y. At v = 0.6c the factor γ is 1.25. The transformed electric field is not simply the old field copied into a different diagram.
SR-08 and SR-02 retain the charge and apparatus assumptions. In the charge’s instantaneous rest frame the local force is electric; relating it to the original frame also requires the force transformation, not an assertion of numerical equality.
Fix the direction of the boost and the field unit convention.
Leave E_x and B_x unchanged for this aligned boost.
Mix E_y with B_z and E_z with B_y using the cross-product signs.
Transform magnetic components with the corresponding v/c² electric terms.
Transform the particle’s velocity before comparing force descriptions.
Use the inverse field transformation to recover the original components.
Show every step here: Electric and magnetic components mix together
Fix the direction of the boost and the field unit convention.
Leave E_x and B_x unchanged for this aligned boost.
Mix E_y with B_z and E_z with B_y using the cross-product signs.
Transform magnetic components with the corresponding v/c² electric terms.
Transform the particle’s velocity before comparing force descriptions.
Use the inverse field transformation to recover the original components.
For a boost along x, longitudinal field components stay unchanged and transverse electric and magnetic components mix. Transforming the force consistently does not require equal raw force values in both frames.
What this relies on
Use vacuum SI fields and a boost of speed v along positive x.
Compare forces at corresponding events with transformed particle velocity.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Electric and magnetic components mix together
Assumed here
Use vacuum SI fields and a boost of speed v along positive x.
Compare forces at corresponding events with transformed particle velocity.
What this does not establish
A charge’s momentary rest frame is not one global frame for an accelerated path.
Fields, forces, and electromotive quantities need their own transformation laws.
Why must a frequency change accompany a direction change?
Transforming a plane wave’s phase yields both the frequency ratio and aberration. At 0.6c a collinear ray has frequency ratio one half; a ray transverse in the original frame instead has ratio 1.25.
q=νν′=γ(1−cvcosφ)
The frequency ratio is gamma times one minus v over c times the unprimed direction cosine.
cosφ′=1−(v/c)cosφcosφ−v/c
The aberration direction cosine is cosine phi minus v over c, divided by one minus v over c times cosine phi.
Write the same wave phase as ωt − k_x x − k_y y − k_z z. Substitute the inverse event transformation, then read off the new time and space coefficients. That single substitution determines both formulas; an analogy with sound is not the derivation.
At v = 0.6c and φ = 0, q = 0.5. For φ = 90 degrees, q = 1.25 and cos φ′ = −0.6. SR-09 labels the angle’s frame so these different comparisons cannot be conflated.
Substitute the inverse coordinate map into the phase at one event.
Identify the coefficient of primed time as the new angular frequency.
Identify primed wave-vector components and divide by the new wave number for direction.
Check collinear and transverse cases with their angle frames stated.
Show every step here: One phase fixes frequency and direction
Specify the wave direction in the original frame.
Substitute the inverse coordinate map into the phase at one event.
Identify the coefficient of primed time as the new angular frequency.
Identify primed wave-vector components and divide by the new wave number for direction.
Check collinear and transverse cases with their angle frames stated.
Transforming a plane wave’s phase yields both the frequency ratio and aberration. At 0.6c a collinear ray has frequency ratio one half; a ray transverse in the original frame instead has ratio 1.25.
What this relies on
Use a vacuum plane wave and corresponding events under an inertial Lorentz map.
The angle is between the unprimed propagation direction and positive boost axis.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: One phase fixes frequency and direction
Assumed here
Use a vacuum plane wave and corresponding events under an inertial Lorentz map.
The angle is between the unprimed propagation direction and positive boost axis.
What this does not establish
Transverse in one frame need not mean transverse in the other.
Why does a finite light complex need a volume calculation?
Radiation energy density changes by q², but the volume of the same moving light complex on a simultaneous slice changes by 1/q. Its total energy therefore changes by q, just like frequency.
uu′=q2,VpacketVpacket′=q1,EE′=q
Energy density scales by q squared, packet volume by inverse q, and total energy by q.
The bounding region travels with the light, so changing the simultaneous slice does not give the ordinary material-volume contraction. At 0.6c with a forward collinear ray, q = 0.5: density becomes one quarter, volume doubles, and total energy becomes one half.
Incorrectly using material contraction 0.8 in that case would give total-energy factor 0.2. At an originally transverse angle the volume factors happen to coincide, so that case alone cannot detect the mistake. SR-10 includes the discriminating comparison.
The September mass-energy argument imports exactly this light-energy transformation. It does not first assume E = mc² and does not require localized light quanta.
Transform the wave amplitude and obtain the squared energy-density factor.
Specify the boundary enclosing the same light complex.
Intersect its transformed moving boundary with a simultaneous primed-time slice.
Compute the volume factor 1/q for that slice.
Multiply density and volume factors to obtain q, not q².
Use the collinear 0.6c case to distinguish this result from material contraction.
Show every step here: Density is not the energy of the whole packet
Transform the wave amplitude and obtain the squared energy-density factor.
Specify the boundary enclosing the same light complex.
Intersect its transformed moving boundary with a simultaneous primed-time slice.
Compute the volume factor 1/q for that slice.
Multiply density and volume factors to obtain q, not q².
Use the collinear 0.6c case to distinguish this result from material contraction.
Radiation energy density changes by q², but the volume of the same moving light complex on a simultaneous slice changes by 1/q. Its total energy therefore changes by q, just like frequency.
What this relies on
Compare the same finite light complex on each frame’s own simultaneous slice.
Use the plane-wave field transformation with q defined from the unprimed direction.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Density is not the energy of the whole packet
Assumed here
Compare the same finite light complex on each frame’s own simultaneous slice.
Use the plane-wave field transformation with q defined from the unprimed direction.
What this does not establish
A light complex is not a material rod or a photon rest frame.
The energy law does not need the light-quantum hypothesis or the desired mass-energy conclusion.
Transform into the mirror’s rest frame, reflect there, and transform back. First check that the light catches the surface: its normal speed must exceed the mirror’s receding speed.
ννref=1−(v/c)21−2(v/c)cosφ+(v/c)2
The reflected frequency ratio is one minus twice v over c cosine phi plus v over c squared, divided by one minus v over c squared.
At normal incidence and v = 0.6c, the reflected frequency is one quarter of the incident frequency. The intercepted incident power is 0.4 of the fixed-surface value. With incident energy density u, the pressure is 0.5u. These are different quantities.
SR-11 uses the relative surface-crossing speed, not c by itself, to count intercepted energy. For a ray transverse in the original frame and a mirror receding along positive x, interception fails; the result is not a reflection with zero frequency.
Compare the incident normal light speed c cos φ with the mirror speed.
Refuse a reflection event if the ray cannot reach the surface.
Transform the incident wave into the mirror’s rest frame.
Reflect its normal direction there while preserving frequency in that frame.
Transform the reflected wave back to the original frame.
Use the moving-surface crossing rates when comparing incident power, reflected power, and mechanical work.
Show every step here: Reflection starts with interception
Compare the incident normal light speed c cos φ with the mirror speed.
Refuse a reflection event if the ray cannot reach the surface.
Transform the incident wave into the mirror’s rest frame.
Reflect its normal direction there while preserving frequency in that frame.
Transform the reflected wave back to the original frame.
Use the moving-surface crossing rates when comparing incident power, reflected power, and mechanical work.
Transform into the mirror’s rest frame, reflect there, and transform back. First check that the light catches the surface: its normal speed must exceed the mirror’s receding speed.
What this relies on
Use an ideal planar perfect reflector with subluminal normal speed.
The incident ray must satisfy c cos φ greater than v.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Reflection starts with interception
Assumed here
Use an ideal planar perfect reflector with subluminal normal speed.
The incident ray must satisfy c cos φ greater than v.
What this does not establish
At the interception boundary, a tolerance-limited numerical result is not a valid reflection event.
A reflected ray can still have positive laboratory x velocity and separate from a faster receding mirror.
Charge density and longitudinal current mix under a frame change. Zero charge density with nonzero current is allowed; it cannot be handled by dividing current by charge density to invent one charge velocity.
ρ′=γ(ρ−c2vJx),Jx′=γ(Jx−vρ)
Primed charge density is gamma times rho minus v J x over c squared; primed longitudinal current is gamma times J x minus v rho.
The transverse current components remain unchanged for this aligned boost. For ρ = 0 and J_x nonzero, ρ′ generally is not zero. The individual charged species must be transformed consistently rather than treating J/ρ as meaningful at neutrality.
Charge conservation requires density and current to satisfy the continuity equation together. SR-12 compares these quantities and the total charge of a specified body. Integrating the same body over each frame’s simultaneous slice preserves its total charge; arbitrary open spatial boxes are a different comparison.
Record both charge density and all current components.
Transform rho and J_x as a coupled pair.
Keep J_y and J_z unchanged for the aligned boost.
Do not divide by rho when the original system is neutral.
Check continuity and the boundary of the same charge-carrying body before comparing total charge.
Show every step here: Neutrality and current belong to a frame
Record both charge density and all current components.
Transform rho and J_x as a coupled pair.
Keep J_y and J_z unchanged for the aligned boost.
Do not divide by rho when the original system is neutral.
Check continuity and the boundary of the same charge-carrying body before comparing total charge.
Charge density and longitudinal current mix under a frame change. Zero charge density with nonzero current is allowed; it cannot be handled by dividing current by charge density to invent one charge velocity.
What this relies on
Use a physical charge-current distribution and a boost along x.
Track the same isolated body when comparing its total charge.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
Assumptions and limits: Neutrality and current belong to a frame
Assumed here
Use a physical charge-current distribution and a boost along x.
Track the same isolated body when comparing its total charge.
What this does not establish
Charge density is not total charge.
A neutral two-species current need not have a single common velocity.
A force-to-acceleration ratio needs two frame labels
Why are there two transverse coefficients?
The printed transverse coefficient compares comoving force with laboratory acceleration and equals mγ². Using laboratory force with laboratory acceleration instead gives mγ. These are different definitions, not two conflicting values of invariant mass.
axFx=mγ3,ayFy′=mγ2,ayFy=mγ
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.
Choose motion along the x axis and distinguish longitudinal from transverse components.
Label the frame of the force and acceleration separately.
Use the instantaneous comoving-frame force law.
Transform acceleration and force consistently.
Form the source’s mixed-frame transverse ratio and the separate laboratory ratio.
Keep invariant mass m distinct from either direction-dependent coefficient.
Show every step here: A force-to-acceleration ratio needs two frame labels
Choose motion along the x axis and distinguish longitudinal from transverse components.
Label the frame of the force and acceleration separately.
Use the instantaneous comoving-frame force law.
Transform acceleration and force consistently.
Form the source’s mixed-frame transverse ratio and the separate laboratory ratio.
Keep invariant mass m distinct from either direction-dependent coefficient.
The printed transverse coefficient compares comoving force with laboratory acceleration and equals mγ². Using laboratory force with laboratory acceleration instead gives mγ. These are different definitions, not two conflicting values of invariant mass.
What this relies on
Start with the electron’s instantaneous rest-frame force law.
Specify the frame of each force and acceleration component.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
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.
What energy is required to accelerate an electron?
Integrating the longitudinal force gives kinetic energy mc²(γ − 1), not the total rest energy. Accelerating voltage and curvature provide separate experimental relations under specified fields.
K=∫0vmγ(u)3udu=mc2(γ−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.
∣q∣U=K,rB=∣q∣Bγmv
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.
Write longitudinal force as mγ³ times acceleration.
Multiply by displacement, using a dx = u du.
Integrate from rest and use the derivative of gamma.
Compare the exact value with the low-speed expansion without calling them equal at 0.6c.
For a voltage estimate, use charge magnitude and state the energy-loss assumption.
For deflection, distinguish magnetic constant-speed curvature from an instantaneous transverse electric curvature.
Show every step here: The work integral has an observable endpoint
Write longitudinal force as mγ³ times acceleration.
Multiply by displacement, using a dx = u du.
Integrate from rest and use the derivative of gamma.
Compare the exact value with the low-speed expansion without calling them equal at 0.6c.
For a voltage estimate, use charge magnitude and state the energy-loss assumption.
For deflection, distinguish magnetic constant-speed curvature from an instantaneous transverse electric curvature.
Integrating the longitudinal force gives kinetic energy mc²(γ − 1), not the total rest energy. Accelerating voltage and curvature provide separate experimental relations under specified fields.
What this relies on
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.
The reviewed German, aligned English, gloss, facsimile, and split view for this passage are not yet available. The explanation does not stand in for those source layers.
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.
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, 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.
16 foundation readings sit behind this argument.
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