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Worked affine transition between two inertial charts

Example

Worked computation of the transition between two inertial charts. Let $(A, V, η, C^{+})$ be an oriented time-oriented affine Minkowski space with inertial frames $(o; e_{0}, e_{1}, e_{2}, e_{3})$ and $(o^{'}; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ and charts $x (p) = E^{- 1} (p - o)$ , $x^{'} (p) = E^{' - 1} (p - o^{'})$ , and write $e_{a}^{'} = \sum_{b} M_{a}^{b} e_{b}$ with $M$ the change-of-basis matrix. Then for every event $p$ ,

$x^{'} (p) = M^{- 1} (x (p) - x (o^{'})) = M^{- 1} x (p) + b, b : = - M^{- 1} x (o^{'}),$

so the transition is an affine map of $R^{4}$ with invertible linear part $M^{- 1}$ .

Restriction to the plane of $e_{0}, e_{1}$ . Fix an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ with $o$ the origin of $A$ used in the charts, and a real parameter $r$ , and put

$e_{0}^{'} : = \cosh r e_{0} + \sinh r e_{1}, e_{1}^{'} : = \sinh r e_{0} + \cosh r e_{1}, e_{2}^{'} : = e_{2}, e_{3}^{'} : = e_{3},$

with $o^{'} : = o$ . Then $(o^{'}; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is again an inertial frame: the matrix of the change of basis is the block matrix $M = diag ((\begin{array}{cc} \cosh r & \sinh r \\ \sinh r & \cosh r \end{array}), I_{2})$ with $\det M = \cosh^{2} r - \sinh^{2} r = 1$ , and

$M^{- 1} = diag ((\begin{array}{cc} \cosh r & - \sinh r \\ - \sinh r & \cosh r \end{array}), I_{2}) .$

Since $M$ has determinant $1$ , the two bases are positively oriented together; the new timelike vector satisfies $η (e_{0}^{'}, e_{0}^{'}) = - \cosh^{2} r + \sinh^{2} r = - 1 < 0$ with $\cosh r > ∣ \sinh r ∣$ , so it is future timelike; hence the primed tuple is an inertial frame. The transition values are $x^{'} (o) = 0$ and $x^{'} (x^{- 1} (1, 0, 0, 0)) = (\cosh r, - \sinh r, 0, 0)$ . Inertial frames exist at every event by Existence of future-oriented inertial frames at every event, so the computation applies with any chosen origin.

Facts & Assumptions

Given: The inertial frames and charts of the Example, with $e_{a}^{'} = \sum_{b} M_{a}^{b} e_{b}$ , and the restricted basis with the hyperbolic parameter $r$ .

[L1]

An inertial frame's ordered basis is $η$ -orthonormal, future and positively oriented; its chart is $x (p) = E^{- 1} (p - o)$ , so $x (o) = 0$ ; and for another inertial frame with chart $x^{'}$ the transition law $x^{'} = M^{- 1} (x - x (o^{'}))$ holds (Inertial frames and inertial charts of the affine Minkowski model).

[L2]

$\cosh u = (e^{u} + e^{- u}) / 2 > 0$ and $\sinh u = (e^{u} - e^{- u}) / 2$ (The six hyperbolic functions and their natural domains), and $\cosh^{2} u - \sinh^{2} u = 1$ for every real $u$ (Addition formulas, identities, parity, and derivatives of the hyperbolic functions).

[L3]

In an $η$ -orthonormal basis the matrix of $η$ is $diag (- 1, 1, 1, 1)$ , so $η (v, w) = - v^{0} w^{0} + v^{1} w^{1} + v^{2} w^{2} + v^{3} w^{3}$ (Inertial frames and inertial charts of the affine Minkowski model).

[L4]

A timelike vector $v$ with $η (v, v) < 0$ and $v^{0} > ∣ v_{v e c} ∣$ is future timelike (Timelike, null, spacelike and causal vectors of a Lorentz vector space, Time orientation, future and past timelike/null/causal vectors, and the closed cones, The timelike set has exactly two convex components, described in orthonormal coordinates).

[L5]

The change-of-basis map between two ordered bases has $\det > 0$ exactly when the bases lie in the same positive class; a positive class is an orientation and a basis in it is positively oriented (Orientation of a finite-dimensional real vector space).

[L6]

The change-of-basis matrix between two ordered bases is invertible (Invertible linear maps, linear isomorphisms, and inverse linear maps), and an expression $u \mapsto M^{- 1} u + b$ is an affine map of $R^{4}$ (Affine spaces over a real vector space and displacement vectors).

Verification

technique · direct computation with the transition law and the hyperbolic identity

1.1L1L6algebra

For a general pair of inertial frames, substituting $E^{'} (u) = E (M u)$ , equivalently $E^{'} = E \circ M$ , into $x^{'} = E^{' - 1} (p - o^{'})$ and using $p - o^{'} = E (x (p) - x (o^{'}))$ gives $x^{'} (p) = M^{- 1} (x (p) - x (o^{'})) = M^{- 1} x (p) - M^{- 1} x (o^{'})$ , an affine map with invertible linear part $M^{- 1}$ by [L1] and [L6].

1.2L1L2L3algebra

For the restricted basis, $η (e_{0}^{'}, e_{0}^{'}) = - \cosh^{2} r + \sinh^{2} r = - 1$ and $η (e_{1}^{'}, e_{1}^{'}) = - \sinh^{2} r + \cosh^{2} r = 1$ by [L3] and [L2], while $η (e_{0}^{'}, e_{1}^{'}) = - \cosh r \sinh r + \sinh r \cosh r = 0$ ; also $η (e_{i}^{'}, e_{j}^{'}) = δ_{i j}$ for $i, j \in {2, 3}$ and the mixed entries vanish. Hence $(e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is $η$ -orthonormal by [L1].

1.3L2algebra

The matrix $M$ of the restricted change of basis is the displayed block matrix, and $\det M = \cosh^{2} r - \sinh^{2} r = 1$ by [L2]; its inverse is the displayed block matrix, since multiplying the two blocks gives the identity using $\cosh^{2} r - \sinh^{2} r = 1$ .

2.1step 1.2step 1.3L1L2L4L5algebra

By $\cosh^{2} r - \sinh^{2} r = 1$ and $\cosh r > 0$ one has $\cosh r = \sqrt{1 + \sinh^{2} r} > ∣ \sinh r ∣$ for every real $r$ ; with $η (e_{0}^{'}, e_{0}^{'}) = - 1 < 0$ from step 1.2, the vector $e_{0}^{'}$ is future timelike by [L4]. Since $\det M = 1 > 0$ by step 1.3, the primed basis lies in the same positive class as the positively oriented original basis by [L5]; hence $(o^{'}; e_{0}^{'}, e_{1}^{'}, e_{2}^{'}, e_{3}^{'})$ is an inertial frame by [L1].

3.1step 1.1step 1.3L1L2algebra∎

With $o^{'} = o$ and $x (o) = 0$ by [L1], the transition law of step 1.1 gives $x^{'} (o) = M^{- 1} (x (o) - x (o^{'})) = 0$ , and for the event $p = x^{- 1} (1, 0, 0, 0)$ , whose coordinates in the first chart are $x (p) = (1, 0, 0, 0)$ , it gives $x^{'} (p) = M^{- 1} (1, 0, 0, 0) = (\cosh r, - \sinh r, 0, 0)$ by the inverse computed in step 1.3. This completes the worked computation of the transition and its two displayed values.

Worked affine transition between two inertial charts

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources