Relativistic momentum and the energy–momentum invariant

Momentum also changes in relativity. For a particle of mass $m$ moving with velocity $\vec{v}$, Newton’s simple $m\vec{v}$ no longer holds; instead it is corrected by the Lorentz factor:

$\ev{\vec{p} = \gamma\,m\,\vec{v}}$

The presence of $\gamma$ has a spectacular consequence: as $v \to c$, the factor $\gamma \to \infty$ and so $|\vec{p}| \to \infty$. The particle’s “effective” inertia diverges — an infinite push would be needed to accelerate it even a little more — and this is another profound reason why no massive object can reach the speed of light.

Combining energy and momentum reveals a relation of striking beauty. If in a given reference frame the particle has energy $E$ and momentum $p$, the combination $E^2 - (pc)^2$ does not change when passing from one frame to another: it is an invariant.

Principle — Energy–momentum invariant

$\ev{E^2 - (p\,c)^2 = (m\,c^2)^2}$ The right-hand side depends only on the rest mass $m$ and is the same in every reference frame.

This is extremely powerful from a practical standpoint: by measuring $E$ and $p$ in any reference frame — however fast the particle is moving — one can deduce its rest mass, an intrinsic property that does not depend on the state of motion. This is how particles are identified from their tracks at accelerators.

Key formula

$E = \gamma\,mc^2$ $p = \gamma\,mv$ $E^2 - p^2c^2 = m^2c^4$ Photon: $m = 0$, hence $E = pc$.

The limiting case of the photon is particularly instructive. Setting $m = 0$ in the invariant, the right-hand side vanishes and we are left with $E = pc$: a massless particle can still carry energy and momentum, provided it travels at exactly $c$. Light is not “light matter” — it is something qualitatively different, and the energy–momentum invariant captures this in a single line.

Links

Topics: Relatività ristretta Concepts: Quantità di moto relativistica

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