ExampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6.1-sol)

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A passive boost that brings a moving particle to rest

Example

In the setting of Four-momentum and kinetic energy at speed three-fifths c, let $B (β)$ be the passive boost of The passive Lorentz boost with parameter beta with parameter $β = 3 / 5$ along $e_{1}$ , and let $P^{'}$ be the components of $P$ after the coordinate change $x^{'} = B (β) x$ with $v = 3 c / 5$ . Then $P^{'} = (m c, 0, 0, 0)$ : the energy becomes $E^{'} = m c^{2}$ and the spatial momentum vanishes, and the mass shell gives $E^{' 2} - c^{2} ∣ p^{'} ∣^{2} = m^{2} c^{4}$ . The observer energy of the same particle thus changes from $5 m c^{2} / 4$ to $m c^{2}$ while the invariant mass stays $m$ ; in particular $T_{e_{0}^{'}} (P) = 0$ in the new frame, the frame of the particle's rest.

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; $m > 0$ ; the inertial frame and particle of the $β = 3 / 5$ example, so that $w = (3 c / 5, 0, 0)$ ; the passive boost $B (β)$ with $β = 3 / 5$ along $e_{1}$ and $v = 3 c / 5$ .

[F1]

(Four-momentum and kinetic energy at speed three-fifths c, Example): in the original frame $P = m U$ with $U = γ_{w} (c, w)$ , $γ_{w} = 5 / 4$ , and $P$ has components $(5 m c / 4, 3 m c / 4, 0, 0)$ , observer energy $5 m c^{2} / 4$ and spatial momentum $(3 m c / 4, 0, 0)$ .

[F2]

(The passive Lorentz boost with parameter beta and The passive boost is a proper orthochronous Lorentz map with inverse B(-beta)): the boost with parameter $β = (v / c) s$ , $∣ v ∣ < c$ , has $γ_{v} = (1 - v^{2} / c^{2})^{- 1 / 2}$ and matrix $B^{0}_{0} = γ_{v}$ , $B^{0}_{i} = B^{i}_{0} = - γ_{v} (v / c) s_{i}$ ; it is a proper orthochronous Lorentz map and is invertible.

[F3]

(Transformation of tensor, covector, derivative and current components under affine Lorentz maps, clause (i)): under a coordinate change $x^{'} = Λ x + b$ the components of a vector transform as $P^{' a} = Λ^{a}_{b} P^{b}$ ; applied to $Λ = B (β)$ this gives $E^{'} = γ_{v} (E - v p_{∥})$ , $p_{∥}^{'} = γ_{v} (p_{∥} - v E / c^{2})$ , $p_{⊥}^{'} = p_{⊥}$ with $p_{∥}$ the component of $p$ along $s$ .

[F4]

(Four-momentum components, energy-velocity relation and passive boost transformation in an inertial frame, clauses (a) and (c)): in an inertial frame with stationary observer $e_{0}$ the components are $(E_{e_{0}} (P) / c, q^{1}, q^{2}, q^{3})$ , and the combination $E^{2} - c^{2} ∣ p ∣^{2}$ is invariant with value $m (P)^{2} c^{4}$ for future timelike $P$ .

[F5]

(Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum, Statement (b)): for $P = m U$ with $m > 0$ , $U$ future timelike with $η (U, U) = - c^{2}$ , and $n$ future unit, $T_{n} (P) = m c^{2} (γ - 1)$ with $γ = - η (U / c, n) \geq 1$ , and $T_{n} (P) = 0$ if and only if $U = c n$ .

Verification

1.1F1F2

By [F1], $E = E_{e_{0}} (P) = 5 m c^{2} / 4$ and $p = (3 m c / 4, 0, 0)$ , so with the boost direction $s = e_{1}$ the parallel component is $p_{∥} = 3 m c / 4$ and $p_{⊥} = 0$ ; by [F2], $β = 3 / 5$ gives $v = 3 c / 5$ and $γ_{v} = (1 - 9 / 25)^{- 1 / 2} = 5 / 4$ .

2.1F3step 1.1

Apply [F3]: $E^{'} = γ_{v} (E - v p_{∥}) = \frac{5}{4} (\frac{5 m c^{2}}{4} - \frac{3 c}{5} \cdot \frac{3 m c}{4}) = \frac{5}{4} (\frac{25 m c^{2} - 9 m c^{2}}{20}) = \frac{5}{4} \cdot \frac{16 m c^{2}}{20} = \frac{5}{4} \cdot \frac{4 m c^{2}}{5} = m c^{2}$ , and $p_{∥}^{'} = γ_{v} (p_{∥} - v E / c^{2}) = \frac{5}{4} (\frac{3 m c}{4} - \frac{3 c}{5} \cdot \frac{5 m c^{2}}{4 c^{2}}) = \frac{5}{4} (\frac{3 m c}{4} - \frac{3 m c}{4}) = 0$ , while $p_{⊥}^{'} = p_{⊥} = 0$ ; hence $P^{'} = (m c, 0, 0, 0)$ .

3.1F4step 2.1

The mass shell in the new frame: by [F4] applied in the primed inertial frame, $E^{' 2} - c^{2} ∣ p^{'} ∣^{2} = m^{2} c^{4}$ ; substituting $E^{'} = m c^{2}$ , $p^{'} = 0$ by [step 2.1] gives $m^{2} c^{4} - 0 = m^{2} c^{4}$ , and $m (P) = m$ since the invariant combination is unchanged.

4.1F4F5step 2.1step 3.1

The observer energy of the particle in the new frame is $E_{e_{0}^{'}} (P) = c P^{' 0} = m c^{2}$ by [F4] and [step 2.1], so the kinetic energy $T_{e_{0}^{'}} (P) = E_{e_{0}^{'}} (P) - m c^{2} = 0$ ; by [F5], $T_{e_{0}^{'}} (P) = 0$ if and only if $U = c e_{0}^{'}$ , so the particle is at rest in the new frame, whose time vector is the particle's four-velocity up to $c$ ; in particular the observer energy changed from $5 m c^{2} / 4$ to $m c^{2}$ while the invariant mass stayed $m$ .

5.1step 2.1step 3.1step 4.1∎

All claims of the example are verified: the boost with parameter $3 / 5$ along $e_{1}$ sends $P = (5 m c / 4, 3 m c / 4, 0, 0)$ to $P^{'} = (m c, 0, 0, 0)$ with $E^{'} = m c^{2}$ and vanishing spatial momentum, the mass shell reads $m^{2} c^{4}$ in both frames, and the new frame is the particle's rest frame.

A passive boost that brings a moving particle to rest

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources