Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 4C (Dynamics and Relativity)

Table of contents
- a 4C
  - i a
    - Solution i
  - ii a
    - Solution ii
- b 4C
  - Solution b

a

i

Solution

The Galilean transformation is

c t^{'} = c t, x^{'} = x - u t = x - β c t .

Therefore

A_{N} = (1 - β 01) .

ii

Solution

The one-dimensional Lorentz transformation gives

c t^{'} = γ (c t - β x), x^{'} = γ (x - β c t), γ = \frac{1}{1 - β ^{2}} .

Hence

A_{SR} = γ (1 - β - β 1) .

∣ β ∣ \to 0

,

γ = 1 + O (β^{2})

. The spatial transformation is therefore

x^{'} = x - u t + O (β^{2})

, while

t^{'} = t - \frac{ux}{c ^{2}} + O (β^{2} t) .

Under the stated condition

∣ x ∣ < c ∣ t ∣

, the fractional correction

∣ ux ∣/ (c^{2} ∣ t ∣)

is smaller than

∣ β ∣

and tends to zero. Thus the Lorentz transformation has the nonrelativistic limit given by the Galilean transformation.

b

Solution

For the special-relativistic matrix,

A (1 1) = γ (1 - β) (1 1), A (- 1 1) = γ (1 + β) (- 1 1) .

Thus the eigenvalues and corresponding eigenvectors are

λ_{+} = \frac{1 - β}{1 + β}, v_{+} = (1 1), λ_{-} = \frac{1 + β}{1 - β}, v_{-} = (- 1 1) .

The eigenvector equations are

x = c t

and

x = - c t

. They are the two null directions of the light cone, representing right-moving and left-moving light rays. A Lorentz boost preserves each light-ray direction while rescaling its null coordinate.

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