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Chapter 05 · Part VI · Modern Physics
Revised on ____________________
Chapter 05 · Special Relativity

One postulate about light, and every clock changes

2 sheets
1 diagram
28 results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

I

The postulates and γ

NOTHING WITH MASS REACHES c
I.First postulate  the laws of physics are the same in every inertial frame
II.Second postulate  c is the same in every inertial frame whatever the source does
III.Lorentz factor  γ = 1√(1 − v²/c²) γ ≥ 1, always
IV.At 0.6 c  γ = 1.25 at 0.8c, γ = 5/3; at 0.99c, γ ≈ 7.1
V.Low speeds  γ ≈ 1 + v²2c² Newton is the v ≪ c limit
VI.Invariant  (Δs)² = (cΔt)² − (Δx)² every frame agrees on this
II

Time and length

ONE MULTIPLIES, THE OTHER DIVIDES
I.Time dilation  Δt = γ Δt0 Δt₀ read by one clock at both events
II.So  moving clocks run slow
III.Length contraction  L = L0γ L₀ measured at rest
IV.So  moving rods are short along the motion only
V.Transverse lengths  unchanged
VI.Simultaneity  Δt′ = − γ v Δxc² why mutual dilation is no paradox
0.1 c
γ = 1.005
0.6 c
γ = 1.25
0.8 c
γ = 5/3
0.9 c
γ = 2.29
0.99 c
γ = 7.09
Muon at 0.98c
lives 5× longer
Ask which frame holds a single clock present at both events — that frame reads the proper time. Ask which frame sees the object at rest — that frame reads the proper length. Everything else follows.
III

Transformations

FRAME ROLES DECIDE THE SIGNS
I.Lorentz, position  x′ = γ (x − v t)
II.Lorentz, time  t′ = γ ( t − v xc² )
III.Velocity addition  u = u′ + v1 + u′v/c² backwards ⇒ u′ < 0
IV.Light stays light  u′ = c gives u = c
V.Doppler, approaching  f′ = f √( 1 + β1 − β ) β = v/c
VI.Receding  f′ = f √( 1 − β1 + β )
VII.Closing speed  may reach 2c no object ever does
VIII.Aberration  the direction of light changes with the frame
IV

Momentum and energy

THE ONE TRIANGLE WORTH MEMORISING
I.Momentum  p = γ m v
II.Total energy  E = γ m c²
III.Rest energy  E0 = m c²
IV.Kinetic energy  K = (γ − 1) m c² not ½mv², ever, at high speed
V.Energy–momentum  E² = (p c)² + (m c²)²
VI.Photon  m = 0, so E = p c
VII.Massless limit  always travels at c
VIII.Useful  p cE = vc
one hyperbola holds every particlemassive particles live on the curve; a photon rides the asymptote it can never leavepcEphoton:E = pcmassive particleE² − (pc)² = (mc²)²E = mc² at p = 0 (rest energy)KE = γmc²K = E − mc²E² = (pc)² + (mc²)² · same value in every framem = 0 collapses it to E = pc: a photon has momentum without mass, and no rest frame at all.Use it whenever a problem gives you two of E, p, m and asks for the third.
V

Where the marks go

FOUR ERRORS, EVERY YEAR
Where marks are lost in this chapter
1Not naming the proper time. Δt0 is read by the single clock present at both events.
2Getting γ the wrong way up. Time multiplies by γ, length divides — moving clocks slow, moving rods short.
3Reading mutual time dilation as a contradiction. The frames disagree about simultaneity by v x/c².
4Calling a 1.6c closing speed a velocity. Use the addition law for the relative speed; it stays under c.