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Worked collinear composition and rapidity addition at beta = 3/5

Example

Worked computation. Let $e$ be a unit spatial direction of an inertial frame and let $B (\frac{3}{5} e)$ be the passive boost with parameter $\frac{3}{5}$ (The passive Lorentz boost with parameter beta), so $γ = 5 / 4$ and, in the ordered pair of directions $(e_{0}, e)$ , the boost has the $2 \times 2$ matrix $(\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})$ and is the identity on ${y \in S : δ (e, y) = 0}$ , the $η$ -orthogonal complement of $span (e_{0}, e)$ (Rapidity parametrization of collinear boosts). Then:

(i) $B (\frac{3}{5} e)^{2} = {(\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})}^{2} = (\begin{array}{cc} 17 / 8 & - 15 / 8 \\ - 15 / 8 & 17 / 8 \end{array}) = B (\frac{15}{17} e)$ , because $γ (\frac{15}{17}) = (1 - \frac{225}{289})^{- 1 / 2} = (\frac{64}{289})^{- 1 / 2} = \frac{17}{8}$ and $γ (\frac{15}{17}) \cdot \frac{15}{17} = \frac{17}{8} \cdot \frac{15}{17} = \frac{15}{8}$ . This is the velocity-addition law $\frac{3 / 5 + 3 / 5}{1 + (3 / 5) (3 / 5)} = \frac{6 / 5}{34 / 25} = \frac{15}{17}$ , with $∣ 15 / 17 ∣ < 1$ .

(ii) The rapidities are $r (\frac{3}{5}) = \frac{1}{2} \log (\frac{1 + 3 / 5}{1 - 3 / 5}) = \frac{1}{2} \log 4 = \log 2$ and $r (\frac{15}{17}) = \frac{1}{2} \log (\frac{1 + 15 / 17}{1 - 15 / 17}) = \frac{1}{2} \log 16 = \log 4 = 2 \log 2$ , so the rapidities add under composition: $r (\frac{15}{17}) = r (\frac{3}{5}) + r (\frac{3}{5})$ .

(iii) The two boosts commute: the reversed product $B (\frac{3}{5} e) B (\frac{3}{5} e)$ equals the square computed in (i), and this is the instance $β_{1} = β_{2} = 3 / 5$ of the commutativity of collinear boosts.

Facts & Assumptions

Given: A unit spatial direction $e$ of an inertial frame, the passive boost $B (\frac{3}{5} e)$ , and the collinear family $B (β e)$ , $β \in (- 1, 1)$ .

[L1]

For $∣ β ∣ < 1$ the boost $B (β e)$ acts in the frame coordinates by $x^{' 0} = γ (x^{0} - β x^{1})$ , $x^{' 1} = γ (x^{1} - β x^{0})$ on the $(e_{0}, e)$ plane, with $γ = (1 - β^{2})^{- 1 / 2} > 0$ and $β$ the coefficient of the unit direction; the matrix on the ordered pair $(e_{0}, e)$ is $(\begin{array}{cc} γ & - γ β \\ - γ β & γ \end{array})$ and the map is the identity on ${y \in S : δ (e, y) = 0}$ (The passive Lorentz boost with parameter beta).

[L2]

For a unit spatial direction $e$ and $r \in R$ , the boost $B (\tanh (r) e)$ has on $(e_{0}, e)$ the matrix $(\begin{array}{cc} \cosh r & - \sinh r \\ - \sinh r & \cosh r \end{array})$ ; in particular, with $β = \tanh (r)$ , $γ (β) = \cosh (r)$ and $γ (β) β = \sinh (r)$ (Rapidity parametrization of collinear boosts).

[L3]

For unit $e$ and $β_{1}, β_{2} \in (- 1, 1)$ one has $B (β_{1} e) B (β_{2} e) = B (β e)$ with $β = \frac{β_{1} + β_{2}}{1 + β_{1} β_{2}}$ and $∣ β ∣ < 1$ , and collinear boosts commute (Collinear boost composition and the relativistic velocity-addition law).

[L4]

For $β \in (- 1, 1)$ the rapidity is $r (β) = \frac{1}{2} \log \frac{1 + β}{1 - β}$ , a strictly increasing bijection of $(- 1, 1)$ onto $R$ with $β = \tanh (r (β))$ ; for a fixed unit direction, rapidities add: $B (\tanh (r) e) B (\tanh (s) e) = B (\tanh (r + s) e)$ (Rapidity parametrization of collinear boosts, The natural logarithm as the inverse of the exponential function).

[L5]

For $x, y > 0$ one has $\log (x y) = \log x + \log y$ and $\log (x / y) = \log x - \log y$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm); the exponential satisfies $e^{r} = \exp (r)$ and the hyperbolic functions are $\sinh (r) = (e^{r} - e^{- r}) / 2$ , $\cosh (r) = (e^{r} + e^{- r}) / 2$ (The real exponential function and the number $e$ by a power series).

[L6]

The Euclidean norm of a vector of $R^{3}$ satisfies $∣ β e ∣ = ∣ β ∣$ for a unit direction (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof

technique · direct

1.1L1L2L6algebra

With $β = 3 / 5$ and $∣ β e ∣ = 3 / 5$ by [L6], [L1] gives $γ = (1 - 9 / 25)^{- 1 / 2} = (16 / 25)^{- 1 / 2} = 5 / 4$ , so the matrix of $B (\frac{3}{5} e)$ on the ordered pair $(e_{0}, e)$ is $(\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})$ , and the map is the identity on the $η$ -orthogonal complement of $span (e_{0}, e)$ . This is the matrix form recorded in [L2], with the rapidity fixed by $\tanh r = 3 / 5$ ; the numerical value of that rapidity is computed below.

1.2L3

By the commutativity clause of [L3], $B (\frac{3}{5} e) B (\frac{3}{5} e) = B (\frac{3}{5} e) B (\frac{3}{5} e)$ , i.e. the reversed order gives back the same product; clause (iii) is this instance, and no separate computation is needed.

2.1L3step 1.1algebra

Squaring the matrix of step 1.1 in the $2 \times 2$ block, $(5 / 4)^{2} + (3 / 4)^{2} = 25 / 16 + 9 / 16 = 34 / 16 = 17 / 8$ on the diagonal and $- (5 / 4) (3 / 4) - (3 / 4) (5 / 4) = - 15 / 8$ off the diagonal, while the identity block on the complement squares to itself; hence $B (\frac{3}{5} e)^{2}$ has matrix $(\begin{array}{cc} 17 / 8 & - 15 / 8 \\ - 15 / 8 & 17 / 8 \end{array})$ on $(e_{0}, e)$ and is the identity on the complement. By [L3], $B (\frac{3}{5} e)^{2} = B (β e)$ with $β = \frac{3 / 5 + 3 / 5}{1 + (3 / 5) (3 / 5)} = \frac{6 / 5}{34 / 25} = \frac{15}{17}$ , and $∣ 15 / 17 ∣ < 1$ . For this $β$ , $γ (β) = (1 - 225 / 289)^{- 1 / 2} = (64 / 289)^{- 1 / 2} = 17 / 8$ and $γ (β) β = \frac{17}{8} \cdot \frac{15}{17} = \frac{15}{8}$ ; by [L1] the matrix of $B (\frac{15}{17} e)$ is therefore $(\begin{array}{cc} 17 / 8 & - 15 / 8 \\ - 15 / 8 & 17 / 8 \end{array})$ on $(e_{0}, e)$ , matching the square. This proves clause (i) and the velocity-addition law.

3.1L4L5step 2.1algebra

By [L4] and [L5], $r (\frac{3}{5}) = \frac{1}{2} \log \frac{8 / 5}{2 / 5} = \frac{1}{2} \log 4$ , and $\log 4 = \log 2 + \log 2$ by the product law, so $r (\frac{3}{5}) = \frac{1}{2} (2 \log 2) = \log 2$ ; similarly $r (\frac{15}{17}) = \frac{1}{2} \log \frac{32 / 17}{2 / 17} = \frac{1}{2} \log 16$ and $\log 16 = \log 4 + \log 4 = 4 \log 2$ , so $r (\frac{15}{17}) = \frac{1}{2} (4 \log 2) = 2 \log 2 = \log 4$ . With $\log 2 = \log 2$ this gives $r (\frac{15}{17}) = r (\frac{3}{5}) + r (\frac{3}{5})$ : the rapidity of the composed boost in step 2.1 is the sum of the two equal rapidities, in agreement with the addition statement of [L4].

4.1step 1.1step 1.2step 2.1step 3.1∎

Steps 1.1-2.1 prove clauses (i) and (iii) and step 3.1 proves clause (ii); every entry is an exact rational or logarithmic identity justified by the displayed suppliers.

Worked collinear composition and rapidity addition at beta = 3/5

Example

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources