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Worked low-speed correspondence at $∣ w ∣ = c / 5$

Physical scope

Numerical specialisation of pthm-sr-newtonian-p-e-correspondence for a single massive particle state with |w|=c/5 in a named inertial frame of the flat model, in SI conventions. The displayed decimals are rounded values of exact radical expressions, not measurements; no experimental value, uncertainty or precision claim is made, and no other state or frame is covered.

Example

Specialise Specified low-speed momentum-energy correspondence between the relativistic measured quantities and the Newtonian comparison quantities to one massive particle of mass $m > 0$ whose coordinate velocity $w$ in a named inertial frame satisfies $∣ w ∣ / c = 1 / 5$ , so that $z = 1 / 25$ and $z_{0} = z$ . Then $γ_{w} = (1 - 1 / 25)^{- 1 / 2} = 5 / \sqrt{24} = 5 \sqrt{6} / 12 \approx 1.0206207262$ and $K (z_{0}) = \frac{3}{8} (25 / 24)^{5 / 2} \approx 0.4152916366$ (Rational powers $a^{r}$ of a positive base, Laws of rational exponents); all decimals below are rounded and all inequalities are between the exact expressions.

(a) [Rate of the expansion at $z = 1 / 25$ ] $γ_{w} - 1 - \frac{z}{2} = 5 \sqrt{6} / 12 - 51 / 50 \approx 0.0006207262$ , while the bound of The uniform second-order Taylor remainder of $z \mapsto (1 - z)^{- 1 / 2}$ on $[0, z_{0}]$ is $K (z_{0}) z^{2} = \frac{3}{8} (25 / 24)^{5 / 2} / 625 \approx 0.0006644666$ , so the remainder bound holds. Precisely, the fractional overestimate of the positive remainder is $K (z_{0}) z^{2} / R (z) - 1 \approx 0.0704665951$ , about $7 %$ .

(b) [Energy and kinetic energy] $E_{n} (P) - m c^{2} = m c^{2} (5 \sqrt{6} / 12 - 1) \approx 0.0206207262 m c^{2}$ while $N_{K} (m, w) = \frac{1}{50} m c^{2} = 0.02 m c^{2}$ ; the deviation $\approx 0.0006207262 m c^{2}$ obeys the bound $K (z_{0}) z m ∣ w ∣^{2} = m c^{2} K (z_{0}) z^{2} \approx 0.0006644666 m c^{2}$ (all in units of $m c^{2}$ ).

(c) [Momentum] $∣ q_{n} (P) ∣ = m γ_{w} ∣ w ∣ = \frac{\sqrt{6}}{12} m c \approx 0.2041241452 m c$ while $∣ N_{p} (m, w) ∣ = m c / 5 = 0.2 m c$ ; the deviation is $\approx 0.0041241452 m c$ against the bound $(\frac{1}{2} + K (z_{0}) z) z m c / 5 \approx 0.0041328933 m c$ .

(d) [Momentum-to-velocity inversion] $∣ w - q_{n} (P) / m ∣ / ∣ w ∣ = γ_{w} - 1 = 5 \sqrt{6} / 12 - 1 \approx 0.0206207262$ , within the bound $(\frac{1}{2} + K (z_{0}) z) z \approx 0.0206644666$ of clause (d) of the theorem.

The state is admissible: $z = 1 / 25 < 1$ , and $w$ of norm $c / 5$ is realized in a frame chart by the hypothesis; the case $∣ w ∣ = c$ is not considered, and the numbers carry no empirical content.

Facts & Assumptions

Given: A massive particle state $P = m U$ of Specified low-speed momentum-energy correspondence between the relativistic measured quantities and the Newtonian comparison quantities with constant $m > 0$ in a named inertial frame, with coordinate velocity $w$ satisfying $∣ w ∣ / c = 1 / 5$ , so $z = ∣ w ∣^{2} / c^{2} = 1 / 25$ and the speed bound $z_{0} = z$ .

[L1]

The correspondence theorem at the state, for $z_{0} = z = 1 / 25$ and $K (z_{0}) = \frac{3}{8} (1 - z_{0})^{- 5 / 2}$ : $E_{n} (P) = m c^{2} + N_{K} (m, w) + R_{E}$ with $∣ R_{E} ∣ \leq K (z_{0}) z m ∣ w ∣^{2}$ ; $T_{n} (P) = N_{K} (m, w) + R_{E}$ with $∣ T_{n} (P) - N_{K} (m, w) ∣ \leq 2 K (z_{0}) z N_{K} (m, w)$ ; $q_{n} (P) = N_{p} (m, w) + R_{q}$ with $∣ R_{q} ∣ \leq (\frac{1}{2} + K (z_{0}) z_{0}) z ∣ N_{p} (m, w) ∣$ ; $w = q_{n} (P) / (m γ_{w})$ and $∣ w - q_{n} (P) / m ∣ = (γ_{w} - 1) ∣ w ∣ \leq (\frac{1}{2} + K (z_{0}) z_{0}) z ∣ w ∣$ (Specified low-speed momentum-energy correspondence between the relativistic measured quantities and the Newtonian comparison quantities,(a)-(d)).

[L2]

The uniform Taylor remainder: $γ_{w} = (1 - z)^{- 1 / 2}$ , $γ_{w} - 1 = \frac{z}{2} + R (z)$ and $∣ R (z) ∣ \leq K (z_{0}) z^{2}$ (The uniform second-order Taylor remainder of $z \mapsto (1 - z)^{- 1 / 2}$ on $[0, z_{0}]$ ).

[L3]

The observer identities: $E_{n} (P) = m c^{2} γ_{w}$ , $T_{n} (P) = m c^{2} (γ_{w} - 1)$ and $∣ q_{n} (P) ∣ = m c \sqrt{γ_{w}^{2} - 1} = m γ_{w} ∣ w ∣$ (Positivity, mass shell and kinetic energy of the observer quantities of a four-momentum,(b)).

[L4]

The comparison quantities: $N_{K} (m, w) = \frac{1}{2} m ∣ w ∣^{2}$ and $N_{p} (m, w) = m w$ , so $∣ N_{p} (m, w) ∣ = m ∣ w ∣$ (Newtonian comparison momentum and kinetic energy in a named inertial frame).

[L5]

The Lorentz factor of the state's coordinate velocity: $γ_{w} = (1 - ∣ w ∣^{2} / c^{2})^{- 1 / 2}$ (Norms of four-velocity and four-acceleration, and the observer rest-space decomposition,(e)).

[L6]

Rational powers: $(a / b)^{- 1 / 2} = (b / a)^{1 / 2} = b^{1 / 2} a^{- 1 / 2}$ for $a, b > 0$ , $(25 / 24)^{5 / 2} = (25 / 24)^{2} (25 / 24)^{1 / 2}$ , and $(24 / 25)^{1 / 2} = \sqrt{24} / 5 = 2 \sqrt{6} / 5$ (Rational powers $a^{r}$ of a positive base, Laws of rational exponents).

Verification

technique · direct

1.1givenL5L6algebra

The numeric data. With $z = 1 / 25$ and $z_{0} = z$ , [L5] and [L6] give $γ_{w} = (\frac{24}{25})^{- 1 / 2} = (\frac{25}{24})^{1 / 2} = \frac{5}{\sqrt{24}} = \frac{5}{2 \sqrt{6}} = \frac{5 \sqrt{6}}{12},$ and $K (z_{0}) = \frac{3}{8} (25 / 24)^{5 / 2}$ ; the statement records the ten-place rounded values of these exact expressions.

2.1step 1.1L2L5algebra

Clause (a). By [L2] with $z = 1 / 25$ , the exact remainder and the exact bound are $γ_{w} - 1 - \frac{z}{2} = R (z) = \frac{5 \sqrt{6}}{12} - \frac{51}{50}, K (z_{0}) z^{2} = \frac{3}{8} (\frac{25}{24})^{5 / 2} \frac{1}{625},$ the remainder is positive because $γ_{w}^{2} = 25 / 24 > (51 / 50)^{2}$ (cross multiplication gives $62500 > 62424$ ) and both quantities are positive by [L5], so $(γ_{w} - 51 / 50) (γ_{w} + 51 / 50) > 0$ with positive second factor, hence $γ_{w} > 51 / 50$ ; its absolute value is at most the bound by [L2], and the statement records the ten-place rounded values of the two exact expressions and of the fractional overestimate $K (z_{0}) z^{2} / R (z) - 1$ .

2.2step 1.1L1L3L4algebra

Clause (c). By [L3] and [L4], $∣ q_{n} (P) ∣ = m γ_{w} ∣ w ∣ = m \cdot \frac{5 \sqrt{6}}{12} \cdot \frac{c}{5} = \frac{\sqrt{6}}{12} m c$ and $∣ N_{p} (m, w) ∣ = m ∣ w ∣ = m c / 5$ , so the exact deviation is $m c (\sqrt{6} / 12 - 1 / 5)$ ; the bound of [L1] clause (c) is $(\frac{1}{2} + K (z_{0}) z) z \frac{m c}{5} = (\frac{1}{2} + \frac{K (z_{0})}{25}) \frac{1}{25} \frac{m c}{5},$ and it dominates the deviation; the statement records the ten-place rounded values of both exact expressions.

2.3step 1.1L1algebra

Clause (d). By [L1], $w = q_{n} (P) / (m γ_{w})$ exactly and $∣ w - q_{n} (P) / m ∣ = (γ_{w} - 1) ∣ w ∣$ , so the relative deviation is $γ_{w} - 1 = 5 \sqrt{6} / 12 - 1$ , which by [L1] clause (d) is at most $(\frac{1}{2} + K (z_{0}) z) z = (\frac{1}{2} + K (z_{0}) / 25) / 25$ ; the statement records the ten-place rounded values of these two exact expressions.

3.1step 1.1step 2.1L1L3L4algebra

Clause (b). By [L1] and [L3], $E_{n} (P) - m c^{2} = m c^{2} (γ_{w} - 1) = m c^{2} (5 \sqrt{6} / 12 - 1)$ and $N_{K} (m, w) = \frac{1}{2} m (c / 5)^{2} = m c^{2} / 50$ , so the exact deviation is $m c^{2} (γ_{w} - 1 - \frac{1}{50}) = m c^{2} (5 \sqrt{6} / 12 - 51 / 50)$ , which is the remainder of step 2.1 and obeys the bound $K (z_{0}) z m ∣ w ∣^{2} = m c^{2} K (z_{0}) z^{2}$ of [L1] clause (a); the statement records the ten-place rounded values of the deviation and of this bound.

4.1step 2.1step 3.1step 2.2step 2.3given∎

Conclusion of the worked instance. The state satisfies the hypotheses of Specified low-speed momentum-energy correspondence between the relativistic measured quantities and the Newtonian comparison quantities with $z = 1 / 25 < 1$ and $z_{0} = z$ ; clauses (a)-(d) have been evaluated at the exact radical values $γ_{w} = 5 \sqrt{6} / 12$ and $K (z_{0}) = \frac{3}{8} (25 / 24)^{5 / 2}$ , every displayed inequality of the statement is the corresponding exact inequality of The uniform second-order Taylor remainder of $z \mapsto (1 - z)^{- 1 / 2}$ on $[0, z_{0}]$ or of the correspondence theorem at this data, and the decimals are the ten-place roundings of the displayed exact expressions. The numbers are a numerical illustration of the conditional correspondence for this one state and carry no empirical content or precision claim.

Worked low-speed correspondence at $∣ w ∣ = c / 5$

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Worked low-speed correspondence at ∣w∣=c/5∣w∣=c/5

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Worked low-speed correspondence at $∣ w ∣ = c / 5$