13 Jul 2025

Relativistic Mass, Mass–Energy Equivalence, and Massless Particles

Momentum-based derivation of velocity-dependent relativistic mass, total energy, rest energy, and the massless energy-momentum relation.

bsc semester-i special-relativity relativistic-mass mass-energy-equivalence massless-particles

Let $m_0$ denote invariant rest mass. For a massive particle with $\lvert v\rvert<c$, relativistic momentum is

\[\boxed{\mathbf p=\gamma m_0\mathbf v}, \qquad \gamma=\frac{1}{\sqrt{1-v^2/c^2}}.\]

In the older velocity-dependent terminology, the relativistic mass is

\[\boxed{m(v)=\gamma m_0 =\frac{m_0}{\sqrt{1-v^2/c^2}}},\]

so $\mathbf p=m(v)\mathbf v$. It equals $m_0$ at rest and increases without bound as $v\to c$. Modern usage normally reserves “mass” for the invariant $m_0$ and puts the velocity dependence in energy and momentum.

Work and total energy

For one-dimensional motion, $dW=F\,dx$ and $F=dp/dt$, hence

\[dW=\frac{dp}{dt}\,dx=v\,dp.\]

Now $p=\gamma m_0v$. Differentiate $\gamma$:

\[\frac{d\gamma}{dv} =\frac{d}{dv}(1-v^2/c^2)^{-1/2} =\frac{v}{c^2}(1-v^2/c^2)^{-3/2} =\frac{\gamma^3v}{c^2}.\]

Therefore

\[\begin{aligned} \frac{dp}{dv} &=m_0\left(\gamma+v\frac{d\gamma}{dv}\right)\\ &=m_0\left(\gamma+\frac{\gamma^3v^2}{c^2}\right)\\ &=m_0\gamma^3, \end{aligned}\]

because $\gamma+\gamma^3\beta^2=\gamma^3[\gamma^{-2}+\beta^2]=\gamma^3$. Thus

\[dW=m_0\gamma^3v\,dv.\]

Since $d\gamma=\gamma^3v\,dv/c^2$,

\[K=\int_0^v dW =m_0c^2\int_1^\gamma d\gamma =\boxed{(\gamma-1)m_0c^2}.\]

Adding the rest energy $E_0=m_0c^2$ gives

\[\boxed{E=\gamma m_0c^2=m(v)c^2}, \qquad \boxed{E_0=m_0c^2}.\]

This is mass–energy equivalence: rest mass is a form of energy, and a change $\Delta m_0$ of a system’s rest mass corresponds to

\[\boxed{\Delta E_0=\Delta m_0c^2}.\]

Energy–momentum relation

Using $E=\gamma m_0c^2$ and $p=\gamma m_0v$, where $p=\lvert\mathbf p\rvert$,

\[\begin{aligned} E^2-p^2c^2 &=\gamma^2m_0^2c^4-\gamma^2m_0^2v^2c^2\\ &=\gamma^2m_0^2c^4(1-v^2/c^2)\\ &=m_0^2c^4. \end{aligned}\]

Hence

\[\boxed{E^2=p^2c^2+m_0^2c^4}.\]

Using a massive particle's rest energy as the plotting scale, its energy curve begins at $m_0c^2$, while the massless relation is the exact line $E=pc$.

For a massless particle, $m_0=0$. For positive energy, the energy-momentum relation becomes

\[\boxed{E=pc}.\]

It cannot have a rest frame: putting $p=0$ would also give $E=0$, not a nonzero-energy particle. Its dispersion relation $E(p)=pc$ gives the propagation speed in vacuum,

\[\boxed{v=\frac{dE}{dp}=c}.\]

In the older mass convention, a massless particle can still have the nonzero relativistic mass $m_{\rm rel}=E/c^2=p/c$; it is its rest mass that is zero. Conversely, a massive particle cannot reach $c$ because $\gamma$, $E$, and $p$ diverge as $v\to c$.

The momentum derivative, work-energy relation, energy-momentum invariant, and massless limit are verified in the Maxima worksheet; every printed residual is zero.