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Exact trajectory in a uniform electric field and the failure of the unscoped Newtonian extrapolation

Physical scope

An exact conditional solution example for one prescribed uniform external electric field ( $E \equiv E_{0} e_{1}$ , $B \equiv 0$ , $E_{0} > 0$ in $V m^{- 1}$ ) acting on a particle of constant charge $q$ and external rest mass $m > 0$ under the framework's adopted covariant coupling: the trajectory, its subluminality at all finite times, the work-energy identity, and the comparison with the unscoped Newtonian constant-acceleration extrapolation. Excluded: the particle's own field, radiation reaction and self-force, the production of the field, measurement, and any empirical claim.

Example

Let $q \in R$ be a constant charge parameter, $m > 0$ an externally supplied rest mass and $E_{0} > 0$ a field strength in $V m^{- 1}$ , and consider the prescribed exterior fields $E, B : R^{3} \times R \to R^{3}$ of The electric and magnetic fields as fixed-time spatial vector fields with $E \equiv E_{0} e_{1}, B \equiv 0,$ a source-free constant solution of the smooth vacuum Maxwell branch. Put $h : = q E_{0} / (m c)$ , carrying the unit $s^{- 1}$ , and for $r_{0} \in R^{3}$ define $r : R \to R^{3}$ and its chart reading $t \mapsto (c t, r (t))$ by $r (t) : = r_{0} + \frac{c}{h} (\sqrt{1 + h^{2} t^{2}} - 1) e_{1} (h \neq 0), r (t) : = r_{0} (h = 0),$ $v (t) : = r^{'} (t) = \frac{c h t}{\sqrt{1 + h^{2} t^{2}}} e_{1} (h \neq 0), γ (t) : = \sqrt{1 + h^{2} t^{2}}, p (t) : = m γ (t) v (t) .$ Then: (i) [Exact solution] for $h \neq 0$ the curve is a future timelike $C^{\infty}$ curve with coordinate velocity $v (t)$ , $∣ v (t) ∣ < c$ for every finite $t$ and $∣ v (t) ∣ \to c$ as $t \to \pm \infty$ , and $p (t) = q E_{0} t e_{1}$ , so that $\frac{d p}{d t} = q E_{0} e_{1} = q (E + v \times B), \frac{d (γ m c^{2})}{d t} = q E \cdot v$ hold identically, as required by The four-force equation reproduces the coordinate Lorentz force and power (a); at $h = 0$ (equivalently $q = 0$ ) the curve is the constant $r \equiv r_{0}$ with $v \equiv 0$ , $p \equiv 0$ and the same equations hold trivially. (ii) [Energy] $γ (t) m c^{2} = m c^{2} \sqrt{1 + h^{2} t^{2}}$ and $\frac{d}{d t} (γ m c^{2}) = q E_{0} v_{1} (t)$ : the field delivers power $q E \cdot v$ to the particle. (iii) [Unscoped Newtonian extrapolation] For $h \neq 0$ , the Newtonian constant-acceleration prediction $v_{N} (t) : = (q E_{0} / m) t e_{1}$ satisfies $v (t) = v_{N} (t) / \sqrt{1 + h^{2} t^{2}}$ , so the two agree only to first order in $∣ h t ∣$ and $∣ v_{N} (t) ∣ = \sqrt{1 + h^{2} t^{2}} ∣ v (t) ∣$ exceeds $c$ for $∣ t ∣ > 1 / ∣ h ∣$ , whereas the exact solution of the adopted coupling remains subluminal at every finite time; for this explicit solution the approximation is quantified directly. Put $r_{N} (t) = r_{0} + \frac{1}{2} h c t^{2} e_{1}$ . For $h \neq 0$ , $∣ v - v_{N} ∣ \leq \frac{c}{2} ∣ h t ∣^{3}, ∣ r - r_{N} ∣ \leq \frac{c}{8 ∣ h ∣} ∣ h t ∣^{4} .$ Thus on $∣ h t ∣ \leq δ$ the velocity error is at most $c δ^{3} / 2$ and the displacement error at most $c δ^{4} / (8 ∣ h ∣)$ ; these are special-model comparisons, not a general trajectory-stability consequence of The four-force equation reproduces the coordinate Lorentz force and power (c). When $h = 0$ both trajectories and velocities coincide, and the Newtonian superluminal-time condition containing $1 / ∣ h ∣$ is not used. (iv) [Scope] the field is prescribed and external, the particle's own field and any radiation reaction or self-force are excluded, and no measurement or empirical claim is made; the solution is exact for the stated coupling with $E_{0} > 0$ and arbitrary constant $q$ .

Facts & Assumptions

Given: The constant charge parameter $q \in R$ , the rest mass $m > 0$ , the field strength $E_{0} > 0$ , the prescribed fields $E \equiv E_{0} e_{1}$ , $B \equiv 0$ on $R^{3} \times R$ , the number $h = q E_{0} / (m c)$ and the origin $r_{0} \in R^{3}$ ; the chart reading $t \mapsto (c t, r (t))$ with the functions of the statement; and, where used, the real parameter $h \neq 0$ .

[F1]

The fields are those of The electric and magnetic fields as fixed-time spatial vector fields on the whole-space laboratory region, with $E$ carrying $V m^{- 1}$ ; the chart reading $t \mapsto (c t, r (t))$ of a massive worldline satisfies $∣ r^{'} (t) ∣ < c$ and has four-velocity $U = γ (c, v)$ with $γ = (1 - ∣ v ∣^{2} / c^{2})^{- 1 / 2}$ , and a vector $(c, w)$ with $∣ w ∣ < c$ is future timelike (Worldlines, four-velocity, chart readings and the frame/observer distinction (a), (c), Framework-local SI units, the vacuum constants and the exact/measured status of their values (c)).

[F2]

The smooth branch of The microscopic vacuum Maxwell equations: smooth and weak branches with $ρ = 0$ , $J = 0$ consists of $div E = 0$ , $div B = 0$ , $curl E = - \partial_{t} B$ and $curl B = c^{- 2} \partial_{t} E$ , with the divergence and curl of Divergence and curl of a $C^{1}$ vector field applied at fixed time; a constant field has every first partial derivative equal to $0$ , so all four equations hold for $(E, B)$ .

[F3]

The force and power identities of The four-force equation reproduces the coordinate Lorentz force and power (a): for a solution of the adopted coupling with coordinate velocity $v$ , Lorentz factor $γ$ , coordinate momentum $p = m γ v$ and prescribed field $(E, B)$ , $d p / d t = q (E + v \times B)$ and $d (γ m c^{2}) / d t = q E \cdot v$ ; the coordinate derivative $r^{'}$ and the vector-valued derivative are those of The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral. The raised-index operator $Φ$ of Antisymmetric contractions vanish and $m U^{'} = q Φ (U)$ preserves $η (U, U)$ is defined by $η (Φ (u), w) = F (w, u)$ and has components $Φ (u)^{μ} = η^{μ α} F_{α β} u^{β}$ ; for the prescribed field $F^{01} = E_{0} / c$ , $F^{10} = - E_{0} / c$ and all other components zero (The electromagnetic two-form, its observer decomposition and the four-current (a)), so $Φ (u)^{0} = E_{0} u^{1} / c$ , $Φ (u)^{1} = E_{0} u^{0} / c$ and $Φ (u)^{2} = Φ (u)^{3} = 0$ , and the adopted equation of motion is $m d U / d τ = q Φ (U)$ .

[F4]

Powers and roots: for $a > 0$ the map $x \mapsto x^{α}$ is continuous and differentiable on $(0, \infty)$ with $(x^{α})^{'} = α x^{α - 1}$ , and real powers agree with rational powers at rational exponents, so $x^{1 / 2} = \sqrt{x}$ on $(0, \infty)$ with derivative $\frac{1}{2} x^{- 1 / 2}$ ; the rational powers satisfy $a^{r + s} = a^{r} a^{s}$ and $a^{- r} = (a^{r})^{- 1}$ , and $\sqrt{\cdot}$ is the nonnegative square root of Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ (Continuity and derivatives of positive-base real powers, The exponential definition of real powers agrees with the existing rational powers, Rational powers $a^{r}$ of a positive base, Laws of rational exponents).

[F5]

Chain rule and product rule: composites and finite sums and products of differentiable functions are differentiable with the usual rules (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ). The Euclidean norm and inner product are those of The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , the cross product is that of The cross product in $R^{3}$ , and for $a, b \geq 0$ , $a \leq b$ exactly when $a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives).

Verification

technique · direct differentiation of the displayed functions, the identity $(1+s^2)^{1/2}\ge1$ and rationalisation of the square root; the special-model error bounds are proved by elementary inequalities

1.1givenF1F2algebra

The prescribed field is source-free: $E \equiv E_{0} e_{1}$ and $B \equiv 0$ are constant, so every first partial derivative of every component vanishes; by the component formulas of [F2], $div E = div B = 0$ , $curl E = curl B = 0$ and $\partial_{t} E = \partial_{t} B = 0$ , so the four source-free smooth-branch equations hold on $R^{3} \times R$ .

1.2givenF4F5algebra

The trajectory and its velocity ( $h \neq 0$ ): put $g (t) : = \sqrt{1 + h^{2} t^{2}} = (1 + h^{2} t^{2})^{1 / 2}$ , a composition of the positive polynomial $t \mapsto 1 + h^{2} t^{2}$ with the power $x \mapsto x^{1 / 2}$ ; by [F4] and the chain rule [F5], $g$ is differentiable with $g^{'} (t) = \frac{1}{2} (1 + h^{2} t^{2})^{- 1 / 2} \cdot 2 h^{2} t = h^{2} t / g (t)$ , and since every derivative is a rational expression with nonzero denominator $g$ , iterating the same rules shows $g$ and hence $r (t) = r_{0} + \frac{c}{h} (g (t) - 1) e_{1}$ are $C^{\infty}$ on $R$ . Then $r^{'} (t) = \frac{c}{h} g^{'} (t) e_{1} = \frac{c h t}{g (t)} e_{1} = v (t)$ by [F5], so $∣ v (t) ∣ = c ∣ h t ∣ / g (t) < c$ because $h^{2} t^{2} < 1 + h^{2} t^{2}$ for all real $t$ [F5], and $∣ v (t) ∣ \to c$ as $∣ t ∣ \to \infty$ because $∣ h t ∣ / g (t) \to 1$ .

1.3givenF3algebra

The branch $h = 0$ : here $q = 0$ since $E_{0} > 0$ , $m > 0$ and $h = q E_{0} / (m c)$ ; the definition gives $r \equiv r_{0}$ , $v \equiv 0$ , $γ \equiv 1$ and $p \equiv 0$ , so $d p / d t = 0 = q (E + v \times B)$ and $d (γ m c^{2}) / d t = 0 = q E \cdot v$ hold trivially, and the Newtonian comparison trajectory of clause (iii) is $r_{N} \equiv r_{0}$ as well, with $v_{N} \equiv 0$ ; the formulas containing $1 / h$ are not used in this branch.

2.1givenstep 1.2F3F5algebra

Force and power identities ( $h \neq 0$ ): by step 1.2, $γ (t) = g (t)$ equals $(1 - ∣ v (t) ∣^{2} / c^{2})^{- 1 / 2}$ , since $1 - ∣ v ∣^{2} / c^{2} = 1 - \frac{h^{2} t^{2}}{1 + h^{2} t^{2}} = \frac{1}{g (t)^{2}}$ ; hence $p (t) = m γ (t) v (t) = m g (t) \cdot \frac{c h t}{g (t)} e_{1} = m c h t e_{1} = q E_{0} t e_{1}$ , because $m c h = q E_{0}$ . Therefore $d p / d t = q E_{0} e_{1} = q (E + v \times B)$ , since $E = E_{0} e_{1}$ , $B = 0$ and $v \times 0 = 0$ [F5]. For the power identity, $d (γ m c^{2}) / d t = m c^{2} g^{'} (t) = m c^{2} h^{2} t / g (t) = q E_{0} \cdot v_{1} (t)$ because $q E_{0} = m c h$ and $v_{1} (t) = c h t / g (t)$ . These are the two coordinate identities of [F3]. The covariant equation $m d U / d τ = q Φ (U)$ of [F3] also holds: with $U^{0} = g (t) c$ , $U^{1} = c h t$ and $U^{2} = U^{3} = 0$ one has $d U^{0} / d τ = g d (g c) / d t = c g g^{'} = c h^{2} t$ and $d U^{1} / d τ = g d (c h t) / d t = c g h$ , so $m d U^{0} / d τ = m c h^{2} t = q E_{0} h t = q Φ (U)^{0}$ and $m d U^{1} / d τ = m g c h = q g E_{0} = q Φ (U)^{1}$ , while the components $2, 3$ vanish on both sides; hence the displayed curve is an exact solution of the adopted coupling in this special model.

2.2givenstep 1.2step 1.3F1F5algebra

The curve is a future timelike worldline: by step 1.2 the curve is $C^{\infty}$ with coordinate velocity $v (t)$ and $∣ v (t) ∣ < c$ ; its tangent $(c, v (t))$ has $η$ -square $- c^{2} + ∣ v (t) ∣^{2} < 0$ and positive time component $c > 0$ , so it is future timelike at every $t$ , and the chart reading is that of a massive worldline in the sense of [F1]. At $h = 0$ the constant curve has tangent $(c, 0)$ , also future timelike.

2.3givenstep 1.2F5algebra

The unscoped Newtonian extrapolation ( $h \neq 0$ ): define $v_{N} (t) : = (q E_{0} / m) t e_{1} = h c t e_{1}$ and $r_{N} (t) : = r_{0} + \frac{1}{2} h c t^{2} e_{1}$ . Since $q E_{0} / m = h c$ , step 1.2 gives $v (t) = \frac{c h t}{g (t)} e_{1} = \frac{v_{N} (t)}{g (t)} = \frac{v_{N} (t)}{\sqrt{1 + h^{2} t^{2}}}$ , and therefore $∣ v_{N} (t) ∣ = g (t) ∣ v (t) ∣ = c ∣ h t ∣$ , which exceeds $c$ exactly when $∣ h t ∣ > 1$ , while $∣ v (t) ∣ < c$ for every finite $t$ by step 1.2. The two velocities agree only at $t = 0$ ; $v_{N}$ is the Newtonian constant-acceleration prediction of the unscoped law.

3.1givenstep 1.2step 2.3F5algebra

Error bounds ( $h \neq 0$ ): put $s : = h t$ and $g = \sqrt{1 + s^{2}} \geq 1$ . Rationalisation gives $g - 1 = \frac{s^{2}}{g + 1} \leq \frac{s^{2}}{2}$ and $1 - \frac{1}{g} = \frac{g - 1}{g} \leq \frac{s^{2}}{2}$ ; hence $∣ v - v_{N} ∣ = ∣ v_{N} ∣ ∣ 1 - \frac{1}{g} ∣ = c ∣ s ∣ (1 - \frac{1}{g}) \leq \frac{c}{2} ∣ s ∣^{3}$ , which is the velocity bound. For the displacement, $r (t) - r_{N} (t) = \frac{c}{h} (g - 1 - \frac{s^{2}}{2}) e_{1}$ , and the elementary identity $1 + \frac{s^{2}}{2} - g = \frac{(g - 1)^{2}}{2}$ gives $∣ g - 1 - \frac{s^{2}}{2} ∣ = \frac{(g - 1)^{2}}{2} \leq \frac{s^{4}}{8}$ by the first inequality; multiplying by $c / ∣ h ∣$ gives $∣ r - r_{N} ∣ \leq \frac{c}{8 ∣ h ∣} ∣ s ∣^{4}$ . Consequently on $∣ h t ∣ \leq δ$ the velocity error is at most $c δ^{3} / 2$ and the displacement error at most $c δ^{4} / (8 ∣ h ∣)$ ; these are special-model comparisons for this explicit solution and do not import a general trajectory-stability consequence of The four-force equation reproduces the coordinate Lorentz force and power (c).

4.1givenstep 1.1step 1.2step 1.3step 2.1step 2.2step 2.3step 3.1F1F3∎

Closing and scope: the prescribed field is a source-free constant smooth-branch solution (step 1.1); for $h \neq 0$ the displayed curve is $C^{\infty}$ , future timelike and subluminal at every finite time, and it satisfies the force and power identities with $p = q E_{0} t e_{1}$ (steps 1.2, 2.1, 2.2); the Newtonian comparison and its exact superluminal condition are step 2.3, and the special-model error bounds are step 3.1; the branch $h = 0$ with $q = 0$ is step 1.3. The field is prescribed and external, the particle's own field, radiation reaction and any self-force are excluded, no production of the field is described, and no measurement or empirical claim is made. The argument used the field and worldline objects [F1], the smooth-branch equations [F2], the force and power identities [F3], the power and root calculus [F4] and the chain and product rules [F5]; no choice principle is used.

Source notes

Tong §5.4.2 gives the trajectory for a particle initially at rest in a uniform electric field and its subluminal limiting speed. Fitzpatrick nodes 126–127 give the relativistic momentum, force and power relations; Beisert §9.4 gives the general coordinate force and power equations (9.70)–(9.71), without solving the uniform-field trajectory. The verification above differentiates the displayed closed-form trajectory in the framework's chart ( $x^{0} = c t$ , SI units, $h = q E_{0} / m c$ ), checks the coordinate force and power identities of the framework's adopted covariant coupling, and quantifies the failure of the unscoped Newtonian constant-acceleration prediction by explicit elementary bounds; the conversion from the sources' conventions is carried out here rather than assumed.

Exact trajectory in a uniform electric field and the failure of the unscoped Newtonian extrapolation

Example

Facts & Assumptions

Verification

Source notes

Depends on

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