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Four-momentum and kinetic energy at speed three-fifths c

Example

Let $(A, V, η, C^{+})$ be the oriented time-oriented affine Minkowski model (Oriented time-oriented affine Minkowski space (the spacetime primitives)), $c > 0$ , $m > 0$ , and let $(o; e_{0}, e_{1}, e_{2}, e_{3})$ be an inertial frame (Inertial frames and inertial charts of the affine Minkowski model) in which a massive particle has four-velocity $U = γ_{w} (c, w)$ (Four-velocity and four-acceleration of a future timelike worldline, Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(e)) with $w = (3 c / 5, 0, 0)$ , so that $γ_{w} = 5 / 4$ . Then its four-momentum is $P = m U$ with components $(5 m c / 4, 3 m c / 4, 0, 0)$ . Relative to the stationary observer $n = e_{0}$ : $E_{n} (P) = 5 m c^{2} / 4$ , $q_{n} (P)$ has coordinates $(3 m c / 4, 0, 0)$ , the kinetic energy is $T_{n} (P) = m c^{2} / 4$ , and the mass shell $E_{n} (P)^{2} - c^{2} ∣ q_{n} (P) ∣^{2} = m^{2} c^{4}$ reads $25 m^{2} c^{4} / 16 - 9 m^{2} c^{4} / 16 = m^{2} c^{4}$ . The exact kinetic energy $m c^{2} / 4 = 0.25 m c^{2}$ exceeds the Newtonian expression $\frac{1}{2} m ∣ w ∣^{2} = 9 m c^{2} / 50 = 0.18 m c^{2}$ ; the comparison is a numerical specialisation, not a limiting claim and not a measurement.

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; $m > 0$ ; an inertial frame in which $w = (3 c / 5, 0, 0)$ and $γ_{w} = (1 - ∣ w ∣^{2} / c^{2})^{- 1 / 2}$ .

[F1]

(Norms of four-velocity and four-acceleration, and the observer rest-space decomposition, clause (e)): in the chart the four-velocity is $U = γ_{w} (c, w)$ with $γ_{w} = (1 - ∣ w ∣^{2} / c^{2})^{- 1 / 2}$ , and the frame components are read in $(e_{0}, e_{1}, e_{2}, e_{3})$ (Inertial frames and inertial charts of the affine Minkowski model).

[F2]

(The Euclidean spatial structure and cross product of an inertial frame, Definition): the spatial part of a frame vector lies in $S = e_{0}^{⊥}$ with $∣ w ∣^{2} = \sum_{i} (w^{i})^{2}$ for $w = \sum_{i} w^{i} e_{i}$ , and $w = (3 c / 5, 0, 0)$ means $w^{1} = 3 c / 5$ , $w^{2} = w^{3} = 0$ .

[F3]

(Laws of rational exponents and Rational powers $a^{r}$ of a positive base): for positive reals the rational-power laws hold, so $(1 - r)^{- 1 / 2}$ for $0 \leq r < 1$ is the positive number whose square is $(1 - r)^{- 1}$ ; in particular $(16 / 25)^{- 1 / 2} = 5 / 4$ since $(4 / 5)^{2} = 16 / 25$ and $(4 / 5)^{- 1} = 5 / 4$ .

[F4]

(Four-momentum components, energy-velocity relation and passive boost transformation in an inertial frame, clause (b)): $E_{n} (P) = m γ_{w} c^{2}$ and $q_{n} (P) = \sum_{i} m γ_{w} w^{i} e_{i}$ for $P = m U$ .

[F5]

(Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum, Definition): $T_{n} (P) = E_{n} (P) - m c^{2}$ for $P = m U$ with $m > 0$ and $η (U, U) = - c^{2}$ .

[F6]

(Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum, Statement (a)(ii)): for future timelike $P$ the mass-shell identity $E_{n} (P)^{2} = m (P)^{2} c^{4} + c^{2} ∣ q_{n} (P) ∣^{2}$ holds, and $m (P) = m$ for $P = m U$ .

Verification

1.1F1F2F3

The Lorentz factor: by [F2], $∣ w ∣^{2} = (3 c / 5)^{2} = 9 c^{2} / 25$ , so $∣ w ∣^{2} / c^{2} = 9 / 25$ and $1 - ∣ w ∣^{2} / c^{2} = 16 / 25$ ; hence $γ_{w} = (16 / 25)^{- 1 / 2} = 5 / 4$ by [F1] and [F3].

1.2F1F2

The four-momentum: [F1] gives $U = γ_{w} (c, w) = (5 c / 4, 3 c / 4, 0, 0)$ in the frame components, so $P = m U$ has components $(5 m c / 4, 3 m c / 4, 0, 0)$ .

2.1F4F5step 1.1

The observer quantities: by [F4], $E_{n} (P) = m γ_{w} c^{2} = 5 m c^{2} / 4$ and $q_{n} (P) = (3 m c / 4) e_{1}$ , with spatial coordinates $(3 m c / 4, 0, 0)$ ; by [F5], $T_{n} (P) = E_{n} (P) - m c^{2} = 5 m c^{2} / 4 - m c^{2} = m c^{2} / 4$ .

3.1F6step 1.2step 2.1

The mass shell: by [step 2.1], $E_{n} (P)^{2} - c^{2} ∣ q_{n} (P) ∣^{2} = 25 m^{2} c^{4} / 16 - c^{2} (9 m^{2} c^{2} / 16) = 25 m^{2} c^{4} / 16 - 9 m^{2} c^{4} / 16 = 16 m^{2} c^{4} / 16 = m^{2} c^{4}$ , in agreement with [F6] and $m (P) = m$ .

3.2F2F3step 1.1step 2.1

The Newtonian comparison: $\frac{1}{2} m ∣ w ∣^{2} = \frac{1}{2} m (9 c^{2} / 25) = 9 m c^{2} / 50$ , while [step 2.1] gives $T_{n} (P) = m c^{2} / 4$ ; since $1 / 4 = 25 / 100$ exceeds $9 / 50 = 18 / 100$ , the exact kinetic energy exceeds the Newtonian expression at this speed, the comparison being a numerical specialisation of the two formulas and not a limit claim.

4.1step 1.2step 2.1step 3.1step 3.2∎

The components of $P$ are $(5 m c / 4, 3 m c / 4, 0, 0)$ , $E_{n} (P) = 5 m c^{2} / 4$ , $q_{n} (P) = (3 m c / 4) e_{1}$ , and $T_{n} (P) = m c^{2} / 4$ . The mass shell is $E_{n} (P)^{2} - c^{2} ∣ q_{n} (P) ∣^{2} = m^{2} c^{4}$ , and $T_{n} (P) > 9 m c^{2} / 50$ ; no measurement or limiting statement is asserted.

Four-momentum and kinetic energy at speed three-fifths c

Example

Facts & Assumptions

Verification

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Dependency tree · 2 levels

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