Irodov 1.375: Momentum from Kinetic Energy Alone – Amit & Deepali Sinha Foundation

Problem Irodov 1.375

A particle of rest mass $m_0$ moves with kinetic energy $T$ . Find its momentum $p$ as a function of $T$ and $m_0$ , and then evaluate that momentum for a proton whose kinetic energy is $T = 500$ MeV.

The energy-momentum right triangle: the legs are pc and the rest energy m_0c^2, the hypotenuse the total energy T+m_0c^2. Finding p from T is finding one leg from the hypotenuse and the other. Drawn to scale for this problem's proton. — The energy-momentum right triangle: the legs are $pc$ and the rest energy $m_0c^2$ , the hypotenuse the total energy $T+m_0c^2$ . Finding $p$ from $T$ is finding one leg from the hypotenuse and the other. Drawn to scale for this problem’s proton.

Given

$m_0$
$T$
proton with $T = 500$ MeV
$m_pc^2 = 938$ MeV

Find

The function

p(T)

, and its value for that proton

Solution

Step 1The invariant tying energy to momentum

Momentum and total energy both carry the same Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2}$ , so a combination exists from which $\gamma$ and the speed disappear together. Forming that combination from $p = \gamma m_0v$ and $E = \gamma m_0c^2$ ,

\begin{align*}E^2-p^2c^2 &= \gamma^2m_0^2c^4-\gamma^2m_0^2v^2c^2 \\ &= \gamma^2m_0^2c^4\left(1-\frac{v^2}{c^2}\right) \\ &= m_0^2c^4\end{align*}

because $\gamma^2$ and $1-v^2/c^2$ are reciprocals. Rearranged for the momentum, this is the theorem of Pythagoras on the triangle above.

p^2c^2 = E^2-\left(m_0c^2\right)^2

(1)

Step 2Trade total energy for kinetic energy

The kinetic energy is what the total energy carries beyond the rest energy, so $E = T+m_0c^2$ is the substitution that puts (1) in terms of the given quantity. Expanding the square, the rest energy term cancels against the one already there.

\begin{align*}p^2c^2 &= \left(T+m_0c^2\right)^2-\left(m_0c^2\right)^2 \\ &= T^2+2Tm_0c^2+\left(m_0c^2\right)^2-\left(m_0c^2\right)^2\end{align*}

The two rest energy terms cancel and what is left has $T$ as a common factor, so the invariant now reads entirely in the given quantity.

p^2c^2 = T\left(T+2m_0c^2\right)

(2)

Taking the square root of this and dividing by $c$ is all that stands between it and the momentum itself.

Step 3Evaluate for a proton of 500 MeV

Both factors on the right of (2) are energies and its left side is the square of $pc$ , so the arithmetic can be done entirely in MeV and the answer read off as a momentum in MeV $/c$ . With $T = 500$ MeV and $m_pc^2 = 938$ MeV,

\begin{align*}T+2m_pc^2 &= 500+1876 = 2376\ \text{MeV} \\ T\left(T+2m_pc^2\right) &= 500\times2376 = 1.188\times10^6\ \text{MeV}^2\end{align*}

The square root of that is $1090$ MeV, which is the product $pc$ ; dividing by $c$ leaves the momentum.

p = \frac{\sqrt{T\left(T+2m_0c^2\right)}}{c} = 1.09\ \text{GeV}/c

(3)

The momentum that fixes, against kinetic energy, both in units of the rest energy, with the marked point this problem's proton. The dashed gold curve is the Newtonian sqrt(2m_0T), the royal line the ultrarelativistic T/c; the exact curve lies above both. — The momentum that (2) fixes, against kinetic energy, both in units of the rest energy, with the marked point this problem’s proton. The dashed gold curve is the Newtonian $\sqrt{2m_0T}$ , the royal line the ultrarelativistic $T/c$ ; the exact curve lies above both.

Check

Cross-check the number by the independent route through the speed. The Lorentz factor is $\gamma = 1+T/m_0c^2$ , which is $1.533$ here, and the triangle’s angle gives the speed as $\beta = pc/E$ , that is $1090/1438 = 0.758$ ; then $p = \gamma m_0 v$ rebuilt from these, $1.533\times938\times0.758$ , returns $1090$ MeV $/c$ , the same momentum. The two limits are right as well: for $T \ll m_0c^2$ the bracket on the right of (2) tends to $2m_0c^2$ , so $p^2c^2 \to 2m_0c^2T$ and $p \to \sqrt{2m_0T}$ , the Newtonian result, while for $T \gg m_0c^2$ that bracket tends to $T$ , so $p^2c^2 \to T^2$ and $p \to T/c$ , the photon-like relation.

Remark

The proton at $500$ MeV sits squarely between those two limits, which is why the figure’s triangle has two comparable legs: the Newtonian formula would give $\sqrt{2m_0T} = 969$ MeV $/c$ , low by $11$ percent, and the ultrarelativistic one $500$ MeV $/c$ , low by more than half. It is worth noticing what the units are doing here. Quoting momentum in MeV $/c$ and mass in MeV $/c^2$ turns (2) into pure arithmetic on energies, both sides of it alike, with no powers of $c$ left to carry, which is exactly why accelerator physics adopted them.