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 $v\to c$.

Solved Problems

1. Energy and momentum of a relativistic proton

A proton moves at $v=+0.800c$ in the laboratory. Given $m_pc^2=938.272\ \mathrm{MeV}$, calculate its relativistic mass ratio, total energy, kinetic energy, and momentum, then verify the energy–momentum relation.

At $\beta=0.800$,

\[\gamma=\frac{1}{\sqrt{1-0.800^2}}=\frac53.\]

The older velocity-dependent mass ratio is therefore

\[\frac{m(v)}{m_p}=\gamma=\boxed{1.6667}.\]

The total and kinetic energies are

\[E=\gamma m_pc^2 =\frac53(938.272) =\boxed{1563.79\ \mathrm{MeV}},\] \[K=(\gamma-1)m_pc^2 =\frac23(938.272) =\boxed{625.515\ \mathrm{MeV}}.\]

Because the motion is along $+x$, the momentum is positive:

\[p_xc=\gamma\beta m_pc^2 =\frac53(0.800)(938.272) =1251.03\ \mathrm{MeV},\]

or

\[\boxed{p_x=1251.03\ \mathrm{MeV}/c}.\]

Finally,

\[E^2-(p_xc)^2 =(1563.79)^2-(1251.03)^2 \simeq(938.272)^2\ \mathrm{MeV^2},\]

which recovers the invariant rest energy within rounding. As $v\to0$, $E\to m_pc^2$, $K\to0$, and $p_x\to0$; as $v\to c$, all three velocity-dependent quantities diverge.

2. Energy, momentum, and mass convention for a photon

A photon propagates along $-x$ with wavelength $\lambda=500\ \mathrm{nm}$ in vacuum. Find its energy, signed momentum, and velocity-dependent mass $E/c^2$. Use $h=6.62607015\times10^{-34}\ \mathrm{J\,s}$ and $c=2.99792458\times10^8\ \mathrm{m\,s^{-1}}$.

For a massless particle, $E=pc$ in magnitude and $E=h\nu=hc/\lambda$. Thus

\[E=\frac{hc}{\lambda} =\boxed{3.97289\times10^{-19}\ \mathrm J} =\boxed{2.47968\ \mathrm{eV}}.\]

The propagation direction fixes the momentum sign:

\[p_x=-\frac Ec=-\frac h\lambda =\boxed{-1.32521\times10^{-27}\ \mathrm{kg\,m\,s^{-1}}}.\]

In the older convention,

\[m_{\rm rel}=\frac{E}{c^2} =\boxed{4.42044\times10^{-36}\ \mathrm{kg}},\]

but the invariant rest mass remains exactly zero. Indeed,

\[E^2-p_x^2c^2=0.\]

The negative momentum records direction; energy and $m_{\rm rel}$ are positive. No rest-frame limit exists for the photon, while increasing $\lambda$ makes $E$, $\lvert p_x\rvert$, and $m_{\rm rel}$ tend to zero.

Descriptive Questions

1. Starting from relativistic momentum and the work integral, derive $K=(\gamma-1)m_0c^2$.
2. Explain the distinction between invariant rest mass and the older velocity-dependent relativistic mass.
3. Derive $E^2=p^2c^2+m_0^2c^4$ from the definitions of relativistic energy and momentum.
4. Use the energy–momentum relation to explain why a massless particle has no rest frame and propagates at $c$.

Numerical Problems

1. Find the ratio $m(v)/m_0$ for a particle moving at $0.600c$. Answer: $m(v)/m_0=\gamma=1.25$.
2. A massive particle has momentum $p=(3/4)m_0c$. Use the energy–momentum relation to find its total energy, kinetic energy, and speed. Answer: $E=(5/4)m_0c^2$, $K=(1/4)m_0c^2$, and $v=(3/5)c=0.600c$.
3. A particle has total energy $E=2m_0c^2$. Find its speed. Answer: $v=(\sqrt3/2)c\simeq0.866c$.
4. A particle has total energy $5.00\ \mathrm{GeV}$ and rest energy $3.00\ \mathrm{GeV}$. Find its momentum. Answer: $p=4.00\ \mathrm{GeV}/c$.
5. A system loses $1.00\ \mathrm{mg}$ of rest mass. Find the released energy using $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$. Answer: $\Delta E=9.00\times10^{10}\ \mathrm J$.
6. A massless particle has momentum components $p_x=3.00\ \mathrm{MeV}/c$ and $p_y=4.00\ \mathrm{MeV}/c$. Find its momentum magnitude, energy, and velocity components. Answer: $p=5.00\ \mathrm{MeV}/c$, $E=5.00\ \mathrm{MeV}$, $v_x=0.600c$, and $v_y=0.800c$.

The derivations and all problem values are checked in the Maxima worksheet; every printed residual is zero.

References

1. “Mass–energy equivalence,” Wikipedia.
2. A. P. French, Special Relativity, 1st ed., MIT Introductory Physics Series, W. W. Norton, 1968, Chapter 7, “More about Relativistic Dynamics.”
3. David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017, Chapter 12, §12.2, “Relativistic Mechanics.”