Relativistic Momentum Calculator

Momentum is still mass times velocity, but with the Lorentz factor attached:

$p = \gamma m v, \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

• $p$ — momentum, kg·m/s (or MeV/c in particle physics)
• $m$ — invariant rest mass, kg
• $v$ — speed in the observer's frame, m/s; $c = 2.998 \times 10^8$ m/s

Kinetic energy is not $\tfrac12 mv^2$ with a $\gamma$ bolted on. It is the total energy minus the rest energy:

$K = (\gamma - 1)mc^2, \qquad E = \gamma mc^2$

both in joules, or in MeV if you use $mc^2$ in MeV.

When it applies: any inertial frame, any speed. Both formulas reduce to the Newtonian ones as $\beta \to 0$ — expand $\gamma \approx 1 + \tfrac12\beta^2$ and $K \to \tfrac12 mv^2$ falls out.

The assumption people forget: $p = \gamma m v$ diverges as $v \to c$, which is the real content of the speed limit. A photon has $m = 0$ but still carries momentum $p = E/c$, because $\gamma m$ is an indeterminate form, not a zero.

Eliminating $v$ between $E = \gamma mc^2$ and $p = \gamma mv$ gives the single most useful identity in relativistic mechanics:

$E^2 = (pc)^2 + (mc^2)^2$

Every term is an energy, so work in eV throughout: $1\ \text{eV} = 1.602 \times 10^{-19}$ J, $m_ec^2 = 0.511$ MeV, $m_pc^2 = 938.3$ MeV. Momentum then comes out in MeV/c, which is a perfectly good unit — divide by $c$ only if the question wants SI.

The relation is frame-independent: $E$ and $p$ both change between frames, but $E^2 - (pc)^2$ is always $(mc^2)^2$. Two consequences worth memorising:

$\beta = \frac{v}{c} = \frac{pc}{E}, \qquad \gamma = \frac{E}{mc^2}$

Those two let you recover speed and Lorentz factor from an energy measurement without ever touching a square root of $1 - \beta^2$.

The assumption people forget: $E$ here is the total energy, not the kinetic energy. Substituting $K$ for $E$ is the single most common error on this topic.

• Using $K = \tfrac12 mv^2$ above about $0.1c$ — at $0.99c$ the classical value is twelve times too small.
• Putting $K$ into $E^2 = (pc)^2 + (mc^2)^2$ — that slot takes total energy, $E = K + mc^2$.
• Writing $p = mv/\sqrt{1-\beta^2}$ but forgetting to square $\beta$ — at $0.9c$ the denominator is $\sqrt{0.19}$, not $\sqrt{0.1}$.
• Mixing MeV with kg·m/s mid-line — pick natural units or SI and convert only at the end.
• Treating $\gamma m$ as the mass in every formula — it works for $p = (\gamma m)v$ by coincidence and fails for $K$, where $\tfrac12(\gamma m)v^2$ is simply wrong.
• Assuming momentum caps out — energy and momentum both grow without bound; only $v$ is bounded by $c$.