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Worked covector and vector component transformation under a beta = 3/5 boost

Example

Worked computation. In an inertial frame with coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ consider the passive boost $Λ : = B (\frac{3}{5} e_{1})$ (The passive Lorentz boost with parameter beta), so that on the $(x^{0}, x^{1})$ plane $Λ = (\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})$ and $Λ^{- 1} = B (- \frac{3}{5} e_{1}) = (\begin{array}{cc} 5 / 4 & 3 / 4 \\ 3 / 4 & 5 / 4 \end{array})$ , and let $x^{'}$ be an inertial chart with $x' = Λ x + b$ (Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition). Take the covector $k$ with components $(k_{0}, k_{1}) = (- 1, 1)$ and the vector $w$ with components $(w^{0}, w^{1}) = (1, 1)$ , with transverse components zero. A covector $k \neq 0$ is called null here when $η^{- 1} (k, k) = 0$ , where the inverse form is defined by $η^{- 1} (k, l) : = η (k^{♯}, l^{♯})$ and $η (k^{♯}, v) = k (v)$ for every $v \in V$ . In this orthonormal basis the unique vector $k^{♯}$ has coordinates $(- k_{0}, k_{1}, k_{2}, k_{3})$ , verifying both existence and uniqueness directly; hence $η^{- 1} (k, k) = - k_{0}^{2} + k_{1}^{2} + k_{2}^{2} + k_{3}^{2}$ . This uses the nondegenerate Lorentz form of Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space. Then:

(i) $k_{a}^{'} = (Λ^{- 1})^{b}_{a} k_{b}$ gives $k_{0}^{'} = (5 / 4) (- 1) + (3 / 4) (1) = - 1 / 2$ and $k_{1}^{'} = (3 / 4) (- 1) + (5 / 4) (1) = 1 / 2$ , so $k^{'} = (- 1 / 2, 1 / 2)$ .

(ii) $w^{' a} = Λ^{a}_{b} w^{b}$ gives $w^{' 0} = (5 / 4) (1 - (3 / 5) \cdot 1) = 1 / 2$ and $w^{' 1} = (5 / 4) (1 - (3 / 5) \cdot 1) = 1 / 2$ , so $w^{'} = (1 / 2, 1 / 2)$ .

(iii) $k$ is null for the inverse form and $w$ is null for $η_{0}$ : $- 1 + 1 = 0$ , and so are $k^{'}$ and $w^{'}$ : $- 1 / 4 + 1 / 4 = 0$ .

(iv) The contraction is invariant: $k_{a}^{'} w^{' a} = (- 1 / 2) (1 / 2) + (1 / 2) (1 / 2) = 0 = (- 1) (1) + (1) (1) = k_{a} w^{a}$ .

The example shows all three features of the component law of Transformation of tensor, covector, derivative and current components under affine Lorentz maps: the covariant slot transforms with $Λ^{- 1}$ , the contravariant slot with $Λ$ , and the contraction is unchanged. The common factor $1 / 2$ on the null covector is the numerical form of the transformation law; no frequency or Doppler interpretation is attached to it here (that belongs to the light-and-doppler page).

Facts & Assumptions

Given: An inertial frame with coordinates $(x^{0}, x^{1}, x^{2}, x^{3})$ and $η_{0} = diag (- 1, 1, 1, 1)$ , the passive boost $Λ = B (\frac{3}{5} e_{1})$ on the $(x^{0}, x^{1})$ plane, an inertial chart $x'$ with $x' = Λ x + b$ , the covector $k$ with components $(- 1, 1, 0, 0)$ and the vector $w$ with components $(1, 1, 0, 0)$ .

[L1]

For $∣ β ∣ < 1$ the boost $B (β e_{1})$ acts on the $(x^{0}, x^{1})$ coordinates 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 its matrix on $(e_{0}, e_{1})$ is $(\begin{array}{cc} γ & - γ β \\ - γ β & γ \end{array})$ ; $B (β e_{1})^{- 1} = B (- β e_{1})$ (The passive Lorentz boost with parameter beta).

[L2]

For every proper orthochronous linear Lorentz map $L$ and every event $c$ there is an inertial frame whose chart $x_{L}$ satisfies $x_{L} (p) = M_{L}^{- 1} (x (p) - x (c))$ , where $M_{L} = E^{- 1} \circ L \circ E$ ; the transition $x^{'} \circ x^{- 1}$ of inertial charts $x, x^{'}$ is affine with linear part a proper orthochronous Lorentz matrix (Inertial charts are related by proper orthochronous affine Lorentz maps, and every such map is a chart transition).

[L3]

Let $x, x^{'}$ be inertial charts with transition $x^{'} = Λ x + b$ , $Ω$ open, and let $J$ , $k$ , $T$ be $C^{ℓ}$ vector, covector and type- $(r, s)$ tensor fields with components $J^{a}$ , $k_{a}$ , $T^{a_{1} \dots a_{r}}_{b_{1} \dots b_{s}}$ in $x$ and primed components in $x^{'}$ . Then (i) $J^{' a} = Λ^{a}_{b} J^{b}$ ; (ii) $k_{a}^{'} = (Λ^{- 1})^{b}_{a} k_{b}$ ; (iii) every contraction of one contravariant with one covariant index is invariant (Transformation of tensor, covector, derivative and current components under affine Lorentz maps, A type $(r, s)$ tensor on a finite-dimensional vector space).

[L4]

A nonzero vector is null when $η (v, v) = 0$ , and in the frame $η (v, v) = - (v^{0})^{2} + (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}$ (Timelike, null, spacelike and causal vectors of a Lorentz vector space, Inertial frames and inertial charts of the affine Minkowski model).

Proof

technique · direct

1.1L1L2algebra

By [L1] with $β = 3 / 5$ , $γ = (1 - 9 / 25)^{- 1 / 2} = 5 / 4$ , so the matrix of $Λ = B (\frac{3}{5} e_{1})$ on the $(x^{0}, x^{1})$ plane is $(\begin{array}{cc} 5 / 4 & - 3 / 4 \\ - 3 / 4 & 5 / 4 \end{array})$ and its inverse is the matrix of $B (- \frac{3}{5} e_{1})$ , namely $(\begin{array}{cc} 5 / 4 & 3 / 4 \\ 3 / 4 & 5 / 4 \end{array})$ ; the existence of the chart $x^{'}$ with this linear part is [L2].

2.1L3step 1.1algebra

Clause (i): by L3, $k_{a}^{'} = (Λ^{- 1})^{b}_{a} k_{b}$ , where $(Λ^{- 1})^{b}_{a}$ is the matrix entry in row $b$ and column $a$ of $Λ^{- 1}$ from step 1.1. Hence $k_{0}^{'} = (5 / 4) (- 1) + (3 / 4) (1) = - 5 / 4 + 3 / 4 = - 1 / 2$ and $k_{1}^{'} = (3 / 4) (- 1) + (5 / 4) (1) = - 3 / 4 + 5 / 4 = 1 / 2$ .

2.2L3step 1.1algebra

Clause (ii): by L3, $w^{' a} = Λ^{a}_{b} w^{b}$ , so $w^{' 0} = (5 / 4) (1) + (- 3 / 4) (1) = 1 / 2$ and $w^{' 1} = (- 3 / 4) (1) + (5 / 4) (1) = 1 / 2$ , which agrees with the displayed forms $(5 / 4) (1 - 3 / 5) = 1 / 2$ ; the components in the transverse directions are unaffected.

3.1L3L4step 2.1step 2.2algebra

Clause (iii): the inverse form defined in the Example gives $η^{- 1} (k, k) = - (- 1)^{2} + 1^{2} = 0$ , while [L4] gives $η (w, w) = - (1)^{2} + 1^{2} = 0$ . Both objects are nonzero, so the covector and vector are null in their respective spaces. In the primed orthonormal frame the inverse metric matrix is again $diag (- 1, 1, 1, 1)$ , the inverse of its metric matrix; the computed components give $η^{- 1} (k, k) = - (- 1 / 2)^{2} + (1 / 2)^{2} = 0$ and $η (w, w) = - (1 / 2)^{2} + (1 / 2)^{2} = 0$ . Thus the primed components describe the same null covector and null vector; no covector is inserted into the form on $V \times V$ .

3.2L3step 2.1step 2.2algebra

Clause (iv): by L3 the contraction $k_{a} w^{a}$ is invariant under the chart change, so $k_{a}^{'} w^{' a} = k_{a} w^{a}$ ; directly, $k_{a}^{'} w^{' a} = (- 1 / 2) (1 / 2) + (1 / 2) (1 / 2) = - 1 / 4 + 1 / 4 = 0$ and $k_{a} w^{a} = (- 1) (1) + (1) (1) = 0$ , consistent with the invariant statement.

4.1step 1.1step 2.1step 2.2step 3.1step 3.2∎

Steps 1.1-2.2 prove clauses (i)-(ii), step 3.1 proves clause (iii) and step 3.2 proves clause (iv); every entry is an exact rational matrix computation from the displayed law, and no frequency, wave or Doppler interpretation is attached to the null covector.

Worked covector and vector component transformation under a beta = 3/5 boost

Example

Facts & Assumptions

Proof

Depends on

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