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Energy Momentum — Examples

1 · Prerequisites

2 · Summary

These worked computations accompany sr-energy-momentum. Four-momentum and kinetic energy at speed three-fifths c evaluates the four-momentum at coordinate speed $3 c / 5$ , obtaining $γ_{w} = 5 / 4$ , components $(5 m c / 4, 3 m c / 4, 0, 0)$ , kinetic energy $m c^{2} / 4$ and the mass shell $E^{2} - c^{2} ∣ p ∣^{2} = m^{2} c^{4}$ , and compares the exact kinetic energy with the Newtonian expression $9 m c^{2} / 50$ . A passive boost that brings a moving particle to rest applies the passive boost with parameter $3 / 5$ along $e_{1}$ to the same particle, finding $P^{'} = (m c, 0, 0, 0)$ and exhibiting the frame in which the particle is at rest while the observer energy changes and the invariant mass does not.

Two equal masses in head-on motion: system mass exceeds the sum of the masses adds the four-momenta of two equal masses in head-on motion, finding total $(5 m c / 2, 0, 0, 0)$ , invariant mass $5 m / 2 > 2 m$ and a mass excess equal to the total kinetic energy divided by $c^{2}$ ; the strictness is checked against the equality case of the collision necessary conditions. A conserving two-into-one collision datum and the non-uniqueness of its final state exhibits a conserving two-into-one datum with the same total and shows that the same incoming side also admits an outgoing side of two particles at rest with masses $m$ and $3 m / 2$ , so conservation fixes only the total. Collective mass of a pair of oppositely directed massless momenta computes the collective mass $2 \sqrt{E_{1} E_{2}} / c^{2}$ of two oppositely directed massless momenta, the equal special case $2 E / c^{2}$ , and the parallel case in which the total is future null with $M = 0$ and no center-of-momentum frame. Threshold energy for a projectile on a stationary target substitutes into the threshold inequality for a projectile on a co-moving target, obtaining $E \geq m c^{2} + 2 μ c^{2} + μ^{2} c^{2} / (2 m)$ and, at $μ = m$ , $E \geq 7 m c^{2} / 2$ .

The two counterexamples delimit the definitions: A massless four-momentum is not obtained from the massive formula at m = 0 shows that setting $m = 0$ in $P = m U$ yields the zero vector, not a nonzero future null four-momentum, so the this substitution cannot produce a nonzero future null momentum; and Kinetic energy depends on the observer even though the invariant mass does not exhibits one massive particle with kinetic energy $m c^{2} / 4$ for one future unit observer and $0$ for another, while the invariant mass stays $m$ . All items are exact computations or refutations in the model; no measurement is asserted.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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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.

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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.

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Two equal masses in head-on motion: system mass exceeds the sum of the masses

Example

Let $m > 0$ and let two massive particles at a common event have four-momenta $P_{1} = m γ (c, w)$ and $P_{2} = m γ (c, - w)$ with $w = (3 c / 5, 0, 0)$ and $γ = 5 / 4$ . Their total is $Q = P_{1} + P_{2} = (5 m c / 2, 0, 0, 0)$ , which is future timelike with invariant mass $M = 5 m / 2$ . Since the first basis vector $e_{0}$ already equals $Q / (M c)$ , the displayed frame is a center-of-momentum frame; moreover $M = 5 m / 2 > 2 m = m_{1} + m_{2}$ , the mass excess is $M - 2 m = m / 2$ , and the corresponding excess rest energy $(M - 2 m) c^{2} = m c^{2} / 2$ equals the total kinetic energy of the two particles relative to $e_{0}$ ( $m c^{2} / 4$ each). Thus the rest mass of the system is not the sum of the rest masses of its parts. The computation is exact at $γ = 5 / 4$ and asserts no collision.

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; $m > 0$ ; an inertial frame in which $P_{1} = m γ (c, w)$ , $P_{2} = m γ (c, - w)$ with $w = (3 c / 5, 0, 0)$ and $γ = 5 / 4$ .

[F1]

(Four-momentum and kinetic energy at speed three-fifths c, Example): at $w = (3 c / 5, 0, 0)$ one has $γ_{w} = 5 / 4$ , and $P = m U$ with $U = γ_{w} (c, w)$ has components $(5 m c / 4, 3 m c / 4, 0, 0)$ , observer energy $E_{e_{0}} (P) = 5 m c^{2} / 4$ , spatial momentum $q_{e_{0}} (P)$ with coordinates $(3 m c / 4, 0, 0)$ and kinetic energy $T_{e_{0}} (P) = m c^{2} / 4$ .

[F2]

(Inertial frames and inertial charts of the affine Minkowski model and Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space): the frame basis $(e_{0}, e_{1}, e_{2}, e_{3})$ is $η$ -orthonormal with $η (e_{0}, e_{0}) = - 1$ and $η (e_{i}, e_{i}) = 1$ , $η$ is bilinear, components of a sum of vectors add, and for a vector with components $(v^{0}, v^{1}, v^{2}, v^{3})$ one has $η (v, v) = - (v^{0})^{2} + (v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}$ .

[F3]

(Total four-momentum, invariant mass and center-of-momentum frame of a finite system, Definition): the total $Q$ of a finite list is the finite sum, and if $Q$ is future timelike then its invariant mass is $M = \sqrt{- η (Q, Q)} / c$ .

[F4]

(Center-of-momentum frames of a future timelike four-momentum and their unique time vector, Statement (a)): for future timelike $Q$ with invariant mass $M$ an inertial frame with $e_{0} = Q / (M c)$ exists, and in it $q_{e_{0}} (Q) = 0$ and $E_{e_{0}} (Q) = M c^{2}$ .

[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, $E_{n} (P) = m c^{2} γ$ and $T_{n} (P) = m c^{2} (γ - 1)$ with $γ = - η (U / c, n) \geq 1$ , and $T_{n} (P) = 0$ if and only if $U = c n$ .

[F6]

(Invariant mass, center-of-momentum frame and necessary conditions for conserving collision data, Statement (a)): for a mass-specified conserving causal on-shell collision datum (Collision data and four-momentum conservation) with future timelike total $Q$ of invariant mass $M$ , each side satisfies $M \geq \sum m_{ℓ}$ over its massive legs, with equality if and only if the side is massive-only and every massive leg is at rest in the center-of-momentum frame.

Verification

1.1F1F2

With $w = (3 c / 5, 0, 0)$ and $γ = 5 / 4$ , [F1] applied to $P_{1} = m γ (c, w)$ gives the components $P_{1} = (5 m c / 4, 3 m c / 4, 0, 0)$ ; replacing $w$ by $- w$ leaves $γ$ unchanged and changes the sign of the spatial part, so $P_{2} = m γ (c, - w)$ has components $(5 m c / 4, - 3 m c / 4, 0, 0)$ . Both $P_{i}$ are $m$ times the four-velocity $γ (c, \pm w)$ of a massive particle, hence future timelike with $η (P_{i}, P_{i}) = - m^{2} c^{2}$ .

1.2F2F3

By [F2] components add, so $Q = P_{1} + P_{2} = (5 m c / 2, 0, 0, 0)$ ; then $η (Q, Q) = - (5 m c / 2)^{2} < 0$ , so $Q$ is future timelike ( $Q^{0} > 0$ ), and [F3] gives $M = \sqrt{- η (Q, Q)} / c = (5 m c / 2) / c = 5 m / 2$ .

2.1F4F5step 1.2

The displayed frame is a center-of-momentum frame: $Q / (M c) = (5 m c / 2, 0, 0, 0) / ((5 m / 2) c) = (1, 0, 0, 0) = e_{0}$ , so by [F4] $q_{e_{0}} (Q) = 0$ and $E_{e_{0}} (Q) = M c^{2} = 5 m c^{2} / 2$ ; by [F5] each particle has $T_{e_{0}} (P_{i}) = m c^{2} (γ - 1) = m c^{2} / 4$ , so the total kinetic energy relative to $e_{0}$ is $m c^{2} / 2$ .

3.1F5F6step 1.1step 2.1

Strictness of the mass inequality: the triple $(p; (P_{1}, P_{2}); (P_{1}, P_{2}))$ with any event $p$ is a mass-specified conserving causal on-shell collision datum with nonzero legs, so [F6] gives $M \geq m_{1} + m_{2} = 2 m$ for the common total; equality would require every massive leg to satisfy $T_{e_{0}} (P_{i}) = 0$ , hence $U_{i} = c e_{0}$ by [F5], contradicting $T_{e_{0}} (P_{i}) = m c^{2} / 4 > 0$ from [step 2.1]. Therefore $M > 2 m$ , in agreement with $M = 5 m / 2$ from [step 1.2].

4.1step 1.2step 2.1step 3.1∎

Mass excess: writing $T_{t o t a l} : = T_{e_{0}} (P_{1}) + T_{e_{0}} (P_{2}) = m c^{2} / 2$ from [step 2.1], one has $(M - 2 m) c^{2} = (5 m / 2 - 2 m) c^{2} = m c^{2} / 2 = T_{t o t a l}$ . Since $c > 0$ , division by $c^{2}$ gives $M - 2 m = m / 2 = T_{t o t a l} / c^{2} > 0$ ; hence the invariant mass of the system exceeds the sum of the rest masses of its parts by the total kinetic energy in the center-of-momentum frame divided by $c^{2}$ .

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A conserving two-into-one collision datum and the non-uniqueness of its final state

Example

In the setting of Two equal masses in head-on motion: system mass exceeds the sum of the masses, the datum with incoming side $(P_{1}, P_{2})$ and the single outgoing vector $Q = (5 m c / 2, 0, 0, 0) = (5 m / 2) c e_{0}$ is an on-shell conserving collision datum: $Q$ is massive with mass $5 m / 2$ and four-velocity $c e_{0}$ , and $P_{1} + P_{2} = Q$ . The outgoing rest mass $5 m / 2$ exceeds the incoming sum $2 m$ by $m / 2$ , and the incoming total kinetic energy $m c^{2} / 2 = (M - 2 m) c^{2}$ has become rest energy. The final state is not unique: the same total $Q$ is also the sum of two outgoing massive legs at rest with masses $m$ and $3 m / 2$ (four-velocities $c e_{0}$ ), so four-momentum conservation fixes only the total, not the number or masses of the outgoing particles. The datum is a mathematical configuration and no interaction is asserted.

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; $m > 0$ ; the inertial frame and the incoming four-momenta $P_{1}, P_{2}$ of the equal-mass example.

[F1]

(Two equal masses in head-on motion: system mass exceeds the sum of the masses, Example): in the displayed frame $P_{1}, P_{2}$ are the four-momenta of two massive particles of mass $m$ with components $(5 m c / 4, \pm 3 m c / 4, 0, 0)$ , their total is $Q = P_{1} + P_{2} = (5 m c / 2, 0, 0, 0)$ , $Q$ is future timelike with invariant mass $M = 5 m / 2$ , the displayed frame is a center-of-momentum frame, each particle has kinetic energy $m c^{2} / 4$ there, and $(M - 2 m) c^{2} = m c^{2} / 2$ is the total kinetic energy.

[F2]

(Collision data and four-momentum conservation, Definition): a collision datum records an event and the incoming and outgoing legs with side totals formed as finite sums, and conserves when the totals are equal; a leg $P$ is massive with mass $m^{'} > 0$ if $P = m^{'} U^{'}$ for some future timelike $U^{'}$ with $η (U^{'}, U^{'}) = - c^{2}$ ; a mass-specified datum is on-shell with recorded masses, and a causal datum has all legs in $c l (C^{+})$ .

[F3]

(Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space and Inertial frames and inertial charts of the affine Minkowski model): the frame basis is $η$ -orthonormal with $η (e_{0}, e_{0}) = - 1$ , components of sums and scalar multiples are computed componentwise, and for $λ > 0$ the vector $λ (c e_{0})$ has components $(λ c, 0, 0, 0)$ .

[F4]

(Invariant mass, center-of-momentum frame and necessary conditions for conserving collision data, Statement (a)): for a mass-specified conserving causal on-shell datum with future timelike total $Q$ of invariant mass $M$ , each side satisfies $M \geq \sum m_{ℓ}$ over its massive legs, with equality if and only if the side is massive-only and every massive leg is at rest in the center-of-momentum frame.

[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 $U^{'} = c n$ ; in particular a massive leg with four-velocity $c e_{0}$ is at rest in the frame of $e_{0}$ .

Verification

1.1F1F2F3

The outgoing leg: by [F3], $Q = (5 m c / 2, 0, 0, 0) = (5 m / 2) (c e_{0})$ with $η (c e_{0}, c e_{0}) = c^{2} η (e_{0}, e_{0}) = - c^{2}$ , and $5 m / 2 > 0$ ; hence by the massive-leg clause of [F2] the vector $Q$ is a massive leg with mass $5 m / 2$ and four-velocity $c e_{0}$ , future timelike, and its recorded mass equals the invariant mass $5 m / 2$ computed in [F1].

2.1F1F2step 1.1

The datum: $(p; (P_{1}, P_{2}); (Q))$ with any event $p$ is mass-specified (recorded masses $m, m, 5 m / 2$ ), on-shell (all three legs massive by [F1] and [step 1.1]), causal (all legs future timelike), and conserving because $P_{1} + P_{2} = Q$ by [F1] and [F2]; its common total is $Q$ with invariant mass $M = 5 m / 2$ .

3.1F1F4F5step 2.1

The mass inequality and the rest-energy interpretation: by [F4] applied to the incoming side, $M = 5 m / 2 \geq m_{1} + m_{2} = 2 m$ , with equality only if both incoming massive legs are at rest in the center-of-momentum frame; by [F1] each has kinetic energy $m c^{2} / 4 \neq 0$ there, so [F5] gives $U_{i} \neq c e_{0}$ and equality fails, $5 m / 2 > 2 m$ . The excess $(M - 2 m) c^{2} = m c^{2} / 2$ equals the incoming total kinetic energy by [F1], and in the datum it is the rest energy of the single outgoing leg beyond the incoming rest energies.

4.1F1F2F3step 3.1

Non-uniqueness of the final state: define the alternative outgoing side $(Q_{1}^{'}, Q_{2}^{'})$ with $Q_{1}^{'} : = m (c e_{0}) = (m c, 0, 0, 0)$ and $Q_{2}^{'} : = (3 m / 2) (c e_{0}) = (3 m c / 2, 0, 0, 0)$ . By [F2] and [F3] both are massive legs with four-velocities $c e_{0}$ and masses $m$ and $3 m / 2$ , so they are at rest in the center-of-momentum frame, and $Q_{1}^{'} + Q_{2}^{'} = (5 m c / 2, 0, 0, 0) = Q$ ; hence $(p; (P_{1}, P_{2}); (Q_{1}^{'}, Q_{2}^{'}))$ is a second conserving mass-specified causal on-shell datum with the same incoming side and the same total but a different number and masses of outgoing legs.

5.1step 3.1step 4.1∎

Summary: the two-into-one datum with outgoing leg $Q$ conserves and has outgoing rest mass $5 m / 2 > 2 m$ , the mass excess being $m / 2$ , whose rest energy $(M - 2 m) c^{2} = m c^{2} / 2$ equals the incoming total kinetic energy; and the same total $Q$ is realised by the alternative outgoing pair of masses $m$ and $3 m / 2$ at rest, so four-momentum conservation alone fixes only the total, not the number or masses of the outgoing particles.

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Collective mass of a pair of oppositely directed massless momenta

Example

Let $e_{0}, e_{1}$ be the first two vectors of an inertial frame, $E_{1}, E_{2} > 0$ , and let $P_{1} = (E_{1} / c) (e_{0} + e_{1})$ , $P_{2} = (E_{2} / c) (e_{0} - e_{1})$ . Both are nonzero future null vectors with $E_{e_{0}} (P_{i}) = E_{i}$ , and their total is $Q = ((E_{1} + E_{2}) / c) e_{0} + ((E_{1} - E_{2}) / c) e_{1}$ , a future timelike vector with invariant mass $M = 2 \sqrt{E_{1} E_{2}} / c^{2}$ ; for $E_{1} = E_{2} = E$ this gives $Q = (2 E / c) e_{0}$ and $M = 2 E / c^{2}$ . If instead $P_{2} = (E_{2} / E_{1}) P_{1}$ (parallel momenta), then $Q$ is future null, $M = 0$ by convention and there is no center-of-momentum frame. In the opposite-direction case with $E_{1} \neq E_{2}$ the first basis vector is not the center-of-momentum vector; the center-of-momentum frame exists by Center-of-momentum frames of a future timelike four-momentum and their unique time vector but differs from the displayed frame. The arithmetic is exact and no measurement is asserted.

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; the first two vectors $e_{0}, e_{1}$ of an inertial frame; $E_{1}, E_{2} > 0$ ; the vectors $P_{1} = (E_{1} / c) (e_{0} + e_{1})$ and $P_{2} = (E_{2} / c) (e_{0} - e_{1})$ .

[F1]

(Inertial frames and inertial charts of the affine Minkowski model and Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space): the frame basis is $η$ -orthonormal with $η (e_{0}, e_{0}) = - 1$ , $η (e_{1}, e_{1}) = 1$ , $η (e_{0}, e_{1}) = 0$ ; $η$ is bilinear, so $η (P_{1}, P_{2}) = (E_{1} E_{2} / c^{2}) (η (e_{0}, e_{0}) - η (e_{1}, e_{1}))$ and components add.

[F2]

(The timelike set has exactly two convex components, described in orthonormal coordinates, Statement): in an $η$ -orthonormal basis whose first vector is timelike, $c l (C^{+}) = {v : v^{0} \geq ∣ v_{s} ∣}$ for the naming of $C^{+}$ containing the first vector, and $C^{+}$ is closed under multiplication by positive reals.

[F3]

(Timelike, null, spacelike and causal vectors of a Lorentz vector space and Time orientation, future and past timelike/null/causal vectors, and the closed cones): a vector is null when $η (v, v) = 0$ and $v \neq 0$ , timelike when $η (v, v) < 0$ , and future null when it is null and lies in $c l (C^{+})$ .

[F4]

(Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum, Definition): $E_{n} (P) = - c η (P, n)$ for $P \in c l (C^{+})$ and future unit $n$ .

[F5]

(Total four-momentum, invariant mass and center-of-momentum frame of a finite system, Definition): the total of a finite list is the finite sum; the invariant mass of a future timelike total is $M = \sqrt{- η (Q, Q)} / c$ ; for a future null total $M = 0$ by convention.

[F6]

(Center-of-momentum frames of a future timelike four-momentum and their unique time vector, Statement (a)–(c)): a future timelike $Q$ with invariant mass $M$ has an inertial frame with $e_{0} = Q / (M c)$ and $q_{e_{0}} (Q) = 0$ ; if an inertial frame satisfies $q_{e_{0}} (Q) = 0$ then $e_{0} = Q / (M c)$ ; a nonzero future null vector has no center-of-momentum frame.

[F7]

(Invariant mass, center-of-momentum frame and necessary conditions for conserving collision data, Statement (d)): for future null $P_{i} = (E_{i} / c) (e_{0} + s_{i})$ with unit $s_{i}$ and $σ = s_{1} \cdot s_{2}$ , the total is future timelike for $σ < 1$ with $M = \sqrt{2 E_{1} E_{2} (1 - σ)} / c^{2}$ , and for $σ = 1$ ( $θ = 0$ , parallel spatial directions) it is future null with $M = 0$ and no center-of-momentum frame.

[F8]

(Laws of rational exponents and Rational powers $a^{r}$ of a positive base): for $a > 0$ the nonnegative square roots multiply, so $\sqrt{4 E_{1} E_{2} / c^{2}} = 2 \sqrt{E_{1} E_{2}} / c$ for $E_{1}, E_{2} > 0$ and $c > 0$ .

Verification

1.1F1F2F3

Both vectors are nonzero future null: by [F1] and [F2], $P_{1}$ has components $(E_{1} / c) (1, 1, 0, 0)$ , so $v^{0} = E_{1} / c = ∣ v_{s} ∣$ and $η (P_{1}, P_{1}) = (E_{1} / c)^{2} (- 1 + 1) = 0$ with $P_{1} \neq 0$ ; [F3] then makes $P_{1}$ future null, and the same computation with the sign of the second component reversed applies to $P_{2} = (E_{2} / c) (e_{0} - e_{1})$ .

1.2F1F4

The observer energies are the stated ones: by [F4] and [F1], $E_{e_{0}} (P_{1}) = - c η (P_{1}, e_{0}) = - c (E_{1} / c) (η (e_{0}, e_{0}) + η (e_{1}, e_{0})) = E_{1}$ and likewise $E_{e_{0}} (P_{2}) = - c (E_{2} / c) (- 1) = E_{2}$ .

1.3F1F5F8

The total: components add by [F1], so $Q = P_{1} + P_{2} = ((E_{1} + E_{2}) / c) e_{0} + ((E_{1} - E_{2}) / c) e_{1}$ ; then $η (Q, Q) = - (E_{1} + E_{2})^{2} / c^{2} + (E_{1} - E_{2})^{2} / c^{2} = - 4 E_{1} E_{2} / c^{2} < 0$ , so $Q$ is future timelike with $Q^{0} > 0$ , and [F5] and [F8] give $M = \sqrt{4 E_{1} E_{2} / c^{2}} / c = 2 \sqrt{E_{1} E_{2}} / c^{2}$ .

2.1F6step 1.3

Center-of-momentum frame and the equal-energy case: by [F4]-type computation in the frame, $q_{e_{0}} (Q) = Q + η (Q, e_{0}) e_{0} = ((E_{1} - E_{2}) / c) e_{1}$ , so the displayed frame is a center-of-momentum frame exactly when $E_{1} = E_{2}$ ; for $E_{1} = E_{2} = E$ one has $Q = (2 E / c) e_{0}$ , $M = 2 E / c^{2}$ and $Q / (M c) = e_{0}$ , while for $E_{1} \neq E_{2}$ the center-of-momentum vector $Q / (M c)$ has a nonzero $e_{1}$ component and differs from $e_{0}$ , the frame existing by [F6].

2.2F2F3F5F6F7step 1.1

The parallel case: if $P_{2} = (E_{2} / E_{1}) P_{1}$ then $Q = (1 + E_{2} / E_{1}) P_{1}$ is a positive multiple of the nonzero future null $P_{1}$ , hence future null by [F2] and [F3]; the case $σ = 1$ of [F7] (the two vectors as the two legs of the trivially conserving datum $(p; (P_{1}, P_{2}); (P_{1}, P_{2}))$ ) gives the same conclusion, and [F5] then sets $M = 0$ while [F6] shows that no center-of-momentum frame exists.

3.1step 1.1step 1.2step 1.3step 2.1step 2.2∎

All claims of the example are verified: the future null vectors $P_{1}, P_{2}$ have observer energies $E_{1}, E_{2}$ ; their total is future timelike with $M = 2 \sqrt{E_{1} E_{2}} / c^{2}$ , equal to $2 E / c^{2}$ when $E_{1} = E_{2} = E$ ; the displayed frame is the center-of-momentum frame exactly in that equal-energy case and otherwise differs from it; and parallel momenta give a future null total with $M = 0$ and no center-of-momentum frame.

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Threshold energy for a projectile on a stationary target

Example

Let $m > 0$ , $μ > 0$ and consider a conserving causal on-shell collision datum whose incoming side is two massive legs, a projectile of mass $m$ with energy $E$ measured by the rest observer $n_{2}$ of the target, and a target of mass $m$ at rest in that observer's frame, and whose outgoing side is massive with masses $m$ , $m$ and $μ$ . By Invariant mass, center-of-momentum frame and necessary conditions for conserving collision data,(c) with $m_{1} = m_{2} = m$ and $\sum_{l} μ_{l} = 2 m + μ$ ,

$E \geq \frac{c^{2}}{2 m} [(2 m + μ)^{2} - 2 m^{2}] = m c^{2} + 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m},$

equivalently $T = E - m c^{2} \geq 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m}$ . For $μ = m$ this reads $E \geq 7 m c^{2} / 2$ and $T \geq 5 m c^{2} / 2$ . These are necessary kinematic thresholds under the adopted balance of the datum; the datum need not exist for any concrete interaction and no process is asserted.

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; $m > 0$ ; $μ > 0$ ; a conserving causal on-shell collision datum with the stated incoming and outgoing legs, the incoming legs being $P_{i} = m U_{i}$ , $i \in {1, 2}$ , and $n_{2} : = U_{2} / c$ the rest observer of the target.

[F1]

(Invariant mass, center-of-momentum frame and necessary conditions for conserving collision data, Statement (c)): for a mass-specified conserving causal on-shell datum with nonzero legs whose incoming side is two-leg with $P_{i} = m_{i} U_{i}$ , $m_{i} > 0$ , and with $n_{2} = U_{2} / c$ , $γ = - η (U_{1} / c, n_{2}) \geq 1$ , $E = E_{n_{2}} (P_{1}) = m_{1} c^{2} γ$ and invariant mass $M$ of the future timelike total, one has $M^{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{2} E / c^{2}$ ; if the outgoing side is mass-specified with massive masses $μ_{1}, \dots, μ_{L}$ (possibly with null legs), then the datum can conserve only if $E \geq \frac{c^{2}}{2 m_{2}} [(\sum_{l} μ_{l})^{2} - m_{1}^{2} - m_{2}^{2}]$ and $T : = E - m_{1} c^{2} \geq \frac{c^{2}}{2 m_{2}} [(\sum_{l} μ_{l})^{2} - (m_{1} + m_{2})^{2}]$ .

[F2]

(Collision data and four-momentum conservation, Definition): a collision datum records an event and the incoming and outgoing legs with side totals formed as finite sums; it conserves when the totals are equal; a mass-specified datum is on-shell with recorded masses, and a causal datum has all legs in $c l (C^{+})$ .

[F3]

(Four-velocity and four-acceleration of a future timelike worldline, Definition): $U_{2}$ is the four-velocity of a future timelike worldline with $η (U_{2}, U_{2}) = - c^{2}$ , and $n_{2} = U_{2} / c$ is future unit; $E_{n_{2}} (P_{1})$ is the projectile energy measured by the observer comoving with the target, so the target is at rest in that observer's frame.

[F4]

(Rational powers $a^{r}$ of a positive base and Laws of rational exponents): for positive reals the rational-power laws hold, so the displayed algebraic rearrangements of positive expressions are legitimate; in particular $μ^{2} c^{2} / (2 m) = (c^{2} / 2 m) μ^{2}$ for $2 m > 0$ .

Verification

1.1F1F2F3

The hypotheses of [F1] are met by the stated datum: by [F2] it is a conserving causal on-shell datum with the two massive incoming legs $P_{1} = m U_{1}$ (projectile) and $P_{2} = m U_{2}$ (target), and by [F3] $n_{2} = U_{2} / c$ is the future unit observer comoving with the target, in whose frame the target is at rest and $E = E_{n_{2}} (P_{1})$ is the projectile energy; so the datum can conserve only if the threshold inequality of [F1] holds.

2.1F1step 1.1

Substituting $m_{1} = m_{2} = m$ and $\sum_{l} μ_{l} = 2 m + μ$ into the threshold of [F1] gives $E \geq \frac{c^{2}}{2 m} [(2 m + μ)^{2} - m^{2} - m^{2}]$ and $T = E - m c^{2} \geq \frac{c^{2}}{2 m} [(2 m + μ)^{2} - 4 m^{2}]$ .

3.1F4step 2.1

Expanding $(2 m + μ)^{2} - 2 m^{2} = 2 m^{2} + 4 m μ + μ^{2}$ and dividing by $2 m > 0$ gives $m + 2 μ + μ^{2} / (2 m)$ ; multiplying by $c^{2}$ and using [F4] yields $E \geq m c^{2} + 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m}$ .

4.1step 3.1

The kinetic-energy form: $T = E - m c^{2} \geq (m c^{2} + 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m}) - m c^{2} = 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m}$ ; for $μ = m$ this gives $T \geq 2 m c^{2} + \frac{m c^{2}}{2} = \frac{5 m c^{2}}{2}$ and, adding $m c^{2}$ , $E \geq \frac{7 m c^{2}}{2}$ .

5.1step 3.1step 4.1∎

Summary: for a conserving causal on-shell datum with a projectile and a co-moving target of equal mass $m$ and an outgoing massive side of total mass $2 m + μ$ , the projectile energy in the target frame must satisfy $E \geq m c^{2} + 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m}$ , equivalently $T \geq 2 μ c^{2} + \frac{μ^{2} c^{2}}{2 m}$ , reducing to $E \geq 7 m c^{2} / 2$ and $T \geq 5 m c^{2} / 2$ at $μ = m$ ; these are necessary kinematic thresholds under the adopted balance, not a claim that any process occurs.

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A massless four-momentum is not obtained from the massive formula at m = 0

Statement refuted

The four-momentum of an ideal massless particle is obtained by setting $m = 0$ in $P = m U$ , where $U$ is a future timelike four-velocity with $η (U, U) = - c^{2}$ .

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; an inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ ; a future timelike $U$ with $η (U, U) = - c^{2}$ (Four-velocity and four-acceleration of a future timelike worldline); the observer quantities $E_{n}$ of Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum for a future unit observer $n$ .

Here massless four-momentum means a nonzero future null vector of $V$ ; the refutation is purely algebraic and asserts no physical interpretation.

[F1]

(Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Definition): $η$ is a symmetric bilinear form on $V$ , so $η (v, 0) = 0$ and $η (0, 0) = 0$ for all $v$ , and $η (λ v, μ w) = λ μ η (v, w)$ for real $λ, μ$ ; in an $η$ -orthonormal ordered basis $η (e_{0}, e_{0}) = - 1$ , $η (e_{1}, e_{1}) = 1$ and $η (e_{0}, e_{1}) = 0$ .

[F2]

(Timelike, null, spacelike and causal vectors of a Lorentz vector space, Definition): $v$ is timelike when $η (v, v) < 0$ , null when $v \neq 0$ and $η (v, v) = 0$ , spacelike when $η (v, v) > 0$ , and the zero vector is none of timelike, null or spacelike.

[F3]

(Inertial frames and inertial charts of the affine Minkowski model, Definition): an inertial frame's basis $(e_{0}, e_{1}, e_{2}, e_{3})$ is $η$ -orthonormal with $e_{0} \in C^{+}$ , so $e_{0}$ is future timelike and $η (e_{0}, e_{0}) = - 1$ .

[F4]

(The timelike set has exactly two convex components, described in orthonormal coordinates, Statement): in an $η$ -orthonormal basis whose first vector is timelike, $c l (C^{+}) = {v : v^{0} \geq ∣ v_{s} ∣}$ for the naming of $C^{+}$ containing the first vector, where $v = v^{0} e_{0} + v_{s}$ and $∣ v_{s} ∣ = \sqrt{(v^{1})^{2} + (v^{2})^{2} + (v^{3})^{2}}$ .

[F5]

(Time orientation, future and past timelike/null/causal vectors, and the closed cones, Definition): a vector is future null when it is nonzero, $η (v, v) = 0$ and $v$ lies in $c l (C^{+})$ , and future causal when it is future timelike or future null.

[F6]

(Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum, Definition): $E_{n} (P) = - c η (P, n)$ for $P \in c l (C^{+})$ .

[F7]

(Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum, Statement (a)): for nonzero future causal $P$ , $E_{n} (P) > 0$ and, for future null $P$ , $E_{n} (P) = c ∣ q_{n} (P) ∣$ .

Counterexample

1.1F1F2F6

The case $m = 0$ : for the future timelike $U$ one has $0 \cdot U = 0$ , so the massive formula at $m = 0$ produces the zero vector; by [F1] and [F6], $η (0, 0) = 0$ and $E_{n} (0) = - c η (0, n) = 0$ , and by [F2] the zero vector is not a nonzero future null vector since it is not null (nullity requires $P \neq 0$ ).

1.2F1F3F4F5F6F7

The witness of a genuine massless momentum: in the inertial frame of [F3] choose $E > 0$ and set $P : = (E / c) (e_{0} + e_{1})$ . Then [F1] gives $η (P, P) = (E / c)^{2} (η (e_{0}, e_{0}) + 2 η (e_{0}, e_{1}) + η (e_{1}, e_{1})) = (E / c)^{2} (- 1 + 0 + 1) = 0$ and $P \neq 0$ ; its coordinates are $(v^{0}, v^{1}, v^{2}, v^{3}) = (E / c) (1, 1, 0, 0)$ , so $v^{0} = E / c = ∣ v_{s} ∣$ and [F4] places $P$ in $c l (C^{+})$ ; hence $P$ is future null by [F5]. Moreover $E_{e_{0}} (P) = - c η (P, e_{0}) = - (E) (η (e_{0}, e_{0}) + η (e_{1}, e_{0})) = E > 0$ by [F1] and [F6], in agreement with [F7].

2.1F1F2step 1.1step 1.2

The case $m > 0$ : for every $m > 0$ , [F1] gives $η (m U, m U) = m^{2} η (U, U) = - m^{2} c^{2} < 0$ , so $m U$ is timelike by [F2]; if $m U = P$ then $η (m U, m U) = η (P, P) = 0$ , a contradiction with $m U$ timelike. Therefore no $m > 0$ produces the future null witness $P = m U$ , and by [step 1.1] no $m = 0$ does either: no value $m \geq 0$ in the massive formula $P = m U$ produces the massless momentum $P$ .

3.1step 1.1step 1.2step 2.1∎

Combining the two cases, the statement is refuted: setting $m = 0$ in $P = m U$ gives the zero vector, which is not a nonzero future null four-momentum and has zero energy in every frame, while a nonzero future null four-momentum such as the witness of [step 1.2] is not of the form $m U$ for any $m \geq 0$ and any future timelike $U$ with $η (U, U) = - c^{2}$ . This refutes direct substitution with a defined normalized timelike $U$ ; the argument leaves mass-dependent limits and physical formulation choices open.

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Kinetic energy depends on the observer even though the invariant mass does not

Statement refuted

The kinetic energy $T_{n} (P) = E_{n} (P) - m c^{2}$ of a massive particle is a property of the particle alone, independent of the observer $n$ .

Facts & Assumptions

Given: the affine Minkowski model $(A, V, η, C^{+})$ ; $c > 0$ ; $m > 0$ ; $w = (3 c / 5, 0, 0)$ ; $U = γ (c, w)$ with $γ = 5 / 4$ ; $P = m U$ ; the observers $n_{1} = e_{0}$ of the displayed inertial frame and $n_{2} = U / c$ .

[F1]

(Four-momentum and kinetic energy at speed three-fifths c, Example): with $w = (3 c / 5, 0, 0)$ and $γ = 5 / 4$ the four-momentum $P = m U$ has $E_{e_{0}} (P) = 5 m c^{2} / 4$ , spatial momentum $(3 m c / 4, 0, 0)$ and kinetic energy $T_{e_{0}} (P) = m c^{2} / 4$ .

[F2]

(A passive boost that brings a moving particle to rest, Example): in the frame obtained by the passive boost with parameter $3 / 5$ along $e_{1}$ one has $E^{'} = m c^{2}$ , $p^{'} = 0$ , the mass shell $E^{' 2} - c^{2} ∣ p^{'} ∣^{2} = m^{2} c^{4}$ , and that frame is the particle's rest frame, its four-velocity being $c e_{0}^{'}$ .

[F3]

[F4]

(Observer energy, spatial momentum, invariant mass, rest energy and kinetic energy of a four-momentum, Definition): $T_{n} (P) = E_{n} (P) - m c^{2}$ and $E_{n} (P) = - c η (P, n)$ ; the invariant mass $m (P)$ depends on the scalar $η (P, P)$ alone and hence on no observer.

[F5]

(Four-velocity and four-acceleration of a future timelike worldline, clause (c)): a point observer of the model at an event is a future unit timelike vector $n$ with $η (n, n) = - 1$ , and the four-velocity of the observer's own worldline is $c n$ ; the four-velocity $U$ of the particle satisfies $η (U, U) = - c^{2}$ (Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(a)).

[F6]

(Lorentz forms of inertia (3,1,0) on a 4-dimensional real vector space, Definition): $η$ is bilinear, so $η (n_{2}, n_{2}) = c^{- 2} η (U, U) = - 1$ for $n_{2} = U / c$ .

Counterexample

1.1F5F6

Two observers: $n_{1} = e_{0}$ is future unit by definition of an inertial frame, and $n_{2} = U / c$ satisfies $η (n_{2}, n_{2}) = c^{- 2} η (U, U) = - 1$ by [F5] and [F6] and is future timelike; both are point observers of the particle in the sense of [F5], and $n_{1} \neq n_{2}$ because $U = γ (c, w) \neq c e_{0}$ has nonzero spatial part.

1.2F1F3

Kinetic energies: in the frame of $n_{1} = e_{0}$ , [F1] gives $T_{n_{1}} (P) = m c^{2} / 4 \neq 0$ ; for $n_{2} = U / c$ one has $U = c n_{2}$ , so [F3] gives $γ = - η (n_{2}, n_{2}) = 1$ and $T_{n_{2}} (P) = m c^{2} (γ - 1) = 0$ .

1.3F2F4

Energies and mass: $E_{n_{1}} (P) = 5 m c^{2} / 4$ by [F1] and $E_{n_{2}} (P) = m c^{2}$ by [F2]; the invariant mass is $m (P) = m$ in both cases, depending on the scalar $η (P, P)$ alone by [F4].

2.1step 1.2step 1.3

The same particle and the same vector $P$ therefore have kinetic energy $m c^{2} / 4$ for the observer $n_{1}$ and kinetic energy $0$ for the observer $n_{2}$ , while $P$ and the invariant mass $m$ are unchanged; hence the displayed statement is false: $T_{n} (P)$ depends on the observer $n$ , and so does $E_{n} (P)$ ( $5 m c^{2} / 4$ versus $m c^{2}$ ), whereas the invariant mass does not.

3.1step 1.1step 1.2step 1.3step 2.1∎

The refutation is complete: two future unit observers of the same massive particle give different values of $T_{n} (P)$ and of $E_{n} (P)$ , so neither quantity is a property of the particle alone, while the invariant mass $m (P) = m$ is the same in both frames.

5 · Examples, experiments and counterexamples

None yet.

Energy Momentum — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

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

Example

Facts & Assumptions

Verification

A passive boost that brings a moving particle to rest

Example

Facts & Assumptions

Verification

Two equal masses in head-on motion: system mass exceeds the sum of the masses

Example

Facts & Assumptions

Verification

A conserving two-into-one collision datum and the non-uniqueness of its final state

Example

Facts & Assumptions

Verification

Collective mass of a pair of oppositely directed massless momenta

Example

Facts & Assumptions

Verification

Threshold energy for a projectile on a stationary target

Example

Facts & Assumptions

Verification

A massless four-momentum is not obtained from the massive formula at m = 0

Statement refuted

Facts & Assumptions

Counterexample

Kinetic energy depends on the observer even though the invariant mass does not

Statement refuted

Facts & Assumptions

Counterexample

5 · Examples, experiments and counterexamples

Sources