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Worked 1+1 boost matrix: metric, determinant, inverse and interval invariance

Example

Worked computation in an inertial frame whose spatial direction is $e_{1}$ , with coordinates $(x^{0}, x^{1})$ and $η_{0} = diag (- 1, 1)$ . Take $β = 3 / 5$ , so $γ = (1 - 9 / 25)^{- 1 / 2} = 5 / 4$ and the passive boost of The passive Lorentz boost with parameter beta is the linear map with matrix $B = (\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})$ on the $(e_{0}, e_{1})$ plane (the identity on the other spatial directions of the frame). Then:

(i) $B^{T} η_{0} B = η_{0}$ : the $(0, 0)$ entry is $(5 / 4)^{2} - (3 / 4)^{2} = 25 / 16 - 9 / 16 = 1$ times $- 1$ , the spatial diagonal entry is $25 / 16 - 9 / 16 = 1$ , and the off-diagonal entry $(5 / 4) (3 / 4) + (- 3 / 4) (5 / 4) = 0$ , giving $η_{0}$ ;

(ii) $\det B = (5 / 4)^{2} - (3 / 4)^{2} = 1$ , so $B$ is proper;

(iii) $B e_{0}$ has coordinates $(5 / 4, - 3 / 4)$ and $η_{0} (B e_{0}, B e_{0}) = - 25 / 16 + 9 / 16 = - 1$ , so the future timelike vector $e_{0}$ is mapped to a future timelike vector;

(iv) $B (- 3 / 5) B (3 / 5) = (\begin{array}{cc} 5 / 4 & 3 / 4 \\ 3 / 4 & 5 / 4 \end{array}) (\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ , so the inverse is the boost with the opposite parameter;

(v) the events with coordinates $(0, 0)$ and $(13, 11)$ have images $(0, 0)$ and $(8, 4)$ : $x^{' 0} = (5 / 4) (13 - (3 / 5) 11) = (5 / 4) (32 / 5) = 8$ and $x^{' 1} = (5 / 4) (11 - (3 / 5) 13) = (5 / 4) (16 / 5) = 4$ , and the interval computed in the two frames is the same: $- (8)^{2} + 4^{2} = - 48 = - (13)^{2} + 11^{2}$ in the convention $η_{0}$ ;

(vi) the null lines $x^{1} = \pm x^{0}$ are preserved: $B (1, 1) = (1 / 2, 1 / 2)$ and $B (1, - 1) = (2, - 2)$ are null, and $B (1, - 1) = 4 (1 / 2, - 1 / 2)$ , so the images are exactly the same pair of lines.

Facts & Assumptions

Given: An inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ , the parameter $β = 3 / 5$ , and the boost $B : = B (3 / 5 e_{1})$ .

[L1]

For $∣ β ∣ < 1$ the passive boost in the $1 + 1$ case $β = β e_{1}$ acts on the coordinates of the $(e_{0}, e_{1})$ plane by $x^{' 0} = γ (x^{0} - β x^{1})$ , $x^{' 1} = γ (x^{1} - β x^{0})$ , with $γ = (1 - β^{2})^{- 1 / 2} > 0$ and $γ$ the rational power of the positive base $1 - β^{2}$ ; the matrix in the ordered basis $(e_{0}, e_{1})$ is $(\begin{array}{cc} γ & - γ β \\ - γ β & γ \end{array})$ , and the map fixes every vector with $x^{0} = x^{1} = 0$ , namely every vector in $span (e_{2}, e_{3})$ (The passive Lorentz boost with parameter beta).

[L2]

$B (β)$ is a proper orthochronous Lorentz map with $B (β)^{- 1} = B (- β)$ , and $B (β) e_{0} = γ (e_{0} - β e_{1})$ ; the map $β \mapsto B (β)$ is injective (The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L3]

In an inertial frame $η (e_{a}, e_{b}) = diag (- 1, 1, 1, 1)_{a b}$ and $e_{0} \in C^{+}$ ; one has $η (x^{0} e_{0} + x^{1} e_{1}, x^{0} e_{0} + x^{1} e_{1}) = - (x^{0})^{2} + (x^{1})^{2}$ in the plane coordinates (Inertial frames and inertial charts of the affine Minkowski model).

[L4]

$C^{+}$ is one of the two components of the timelike set and is the set of future timelike vectors; a Lorentz map is orthochronous exactly when it maps $C^{+}$ onto $C^{+}$ (Time orientation, future and past timelike/null/causal vectors, and the closed cones, The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)).

[L5]

A nonzero vector $v$ is null when $η (v, v) = 0$ and timelike when $η (v, v) < 0$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space).

Proof

technique · direct

1.1L1algebra

Evaluate the $2 \times 2$ matrix $B$ : $γ = (1 - 9 / 25)^{- 1 / 2} = (16 / 25)^{- 1 / 2} = 5 / 4$ , so $B = (\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})$ , and $B$ acts as the identity on the coordinate directions $e_{2}, e_{3}$ because $β = 3 / 5 e_{1}$ has vanishing $e_{2}, e_{3}$ components in [L1]. For the product $B^{T} η_{0} B$ , the four entries are $(5 / 4) (- 5 / 4) + (- 3 / 4) (- 3 / 4) = - 25 / 16 + 9 / 16 = - 1$ in the $(0, 0)$ slot, $(5 / 4) (3 / 4) + (- 3 / 4) (5 / 4) = 15 / 16 - 15 / 16 = 0$ and $(- 3 / 4) (- 5 / 4) + (5 / 4) (- 3 / 4) = 15 / 16 - 15 / 16 = 0$ in the two off-diagonal slots, and $(- 3 / 4) (3 / 4) + (5 / 4) (5 / 4) = - 9 / 16 + 25 / 16 = 1$ in the $(1, 1)$ slot; hence $B^{T} η_{0} B = diag (- 1, 1) = η_{0}$ .

1.2L2algebra

The determinant is $\det B = (5 / 4) (5 / 4) - (- 3 / 4) (- 3 / 4) = 25 / 16 - 9 / 16 = 1$ , so $B$ is proper in the sense of the properness condition of [L2].

1.3L2L3L4L5algebra

By [L2], $B e_{0} = γ (e_{0} - β e_{1}) = (5 / 4) (e_{0} - (3 / 5) e_{1})$ , with coordinates $(5 / 4, - 3 / 4)$ ; [L3] gives $η (B e_{0}, B e_{0}) = - (5 / 4)^{2} + (3 / 4)^{2} = - 1$ , so $B e_{0}$ is timelike by [L5]. Since $B$ is orthochronous by [L2] and $e_{0} \in C^{+}$ by [L3], [L4] gives $B e_{0} \in B (C^{+}) = C^{+}$ , so $B e_{0}$ is future timelike: the future timelike vector $e_{0}$ is mapped to a future timelike vector.

1.4L1L2algebra

The product of clause (iv) is $(\begin{array}{cc} 5 / 4 & 3 / 4 \\ 3 / 4 & 5 / 4 \end{array}) (\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array}) = (\begin{array}{cc} 25 / 16 - 9 / 16 & - 15 / 16 + 15 / 16 \\ - 15 / 16 + 15 / 16 & - 9 / 16 + 25 / 16 \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ , and the first factor is the matrix of $B (- 3 / 5)$ by [L1]; this is the instance at $β = 3 / 5$ of $B (- β) B (β) = id$ from [L2], so the inverse is the boost with the opposite parameter.

1.5L1L5algebra

For the two null directions, [L1] gives $B (1, 1) = ((5 / 4) (1 - 3 / 5), (5 / 4) (1 - 3 / 5)) = (1 / 2, 1 / 2)$ and $B (1, - 1) = ((5 / 4) (1 + 3 / 5), (5 / 4) (- 1 - 3 / 5)) = (2, - 2)$ ; both images satisfy $η_{0} = 0$ on their squares: $- (1 / 2)^{2} + (1 / 2)^{2} = 0$ and $- 4 + 4 = 0$ , so they are null by [L5]. Since $B (1, 1) = \frac{1}{2} (1, 1)$ and $B (1, - 1) = 2 (1, - 1)$ , the image of the line $x^{1} = x^{0}$ is the same line and the image of the line $x^{1} = - x^{0}$ is the same line; hence the pair of null lines ${x^{1} = x^{0}} \cup {x^{1} = - x^{0}}$ is mapped onto itself, and the images are exactly that pair.

2.1L1step 1.1algebra

For the event with coordinates $(13, 11)$ , [L1] gives $x^{' 0} = (5 / 4) (13 - (3 / 5) 11) = (5 / 4) (13 - 33 / 5) = (5 / 4) (32 / 5) = 8$ and $x^{' 1} = (5 / 4) (11 - (3 / 5) 13) = (5 / 4) (11 - 39 / 5) = (5 / 4) (16 / 5) = 4$ , while $(0, 0)$ is fixed because the boost is linear. The interval in $η_{0}$ is $- (8)^{2} + (4)^{2} = - 64 + 16 = - 48$ and $- (13)^{2} + (11)^{2} = - 169 + 121 = - 48$ ; this numerical equality is the instance of $B^{T} η_{0} B = η_{0}$ from step 1.1 applied to the coordinate difference $(13, 11)$ .

3.1step 1.1step 1.2step 1.3step 1.4step 2.1step 1.5∎

Steps 1.1-1.4 verify clauses (i)-(iv), step 2.1 verifies clause (v), and step 1.5 verifies clause (vi); every clause of the Example is a checked finite computation with the displayed rational entries.

Depends on

Dependency tree · 2 levels

Sources

B. Crowell, Special Relativity, revision 2020-12-10 (author-published undergraduate textbook), Section 1.4 'The Lorentz transformation' pp. 30-32 including Examples 11-14, Sections 2.3-2.4 'Additional postulates'/'Other axiomatizations' pp. 46-49, and Section 8.3 'Boosts and rotations' pp. 174-176; signature (+---) converted to (-+++) with x^0 = ct restored