# What is the correct expression for phase velocity of a relativistic free particle?

According to most books,the relativistic free particle with velocity $$v$$ has phase velocity, $$v_p=\frac{c^2}{v}\tag{1}$$

But, a free particle also means potential energy $$V=0$$.Therefore we might as well use, $$v_p=\frac{T}{p}$$ where relativistic kinetic energy=$$T=m_0c^2(\gamma -1)$$ and relativistic momentum = $$p=m_0v\gamma$$, giving $$v_p=\frac{c^2}{v}(1-\frac{1}{\gamma})\tag{2}$$

Equating with the more familiar $$c^2/v$$,we get $$1=0$$ !So clearly these two are not the same.

I have reasons to believe in (2) because---

Non-relativistic formulas must reappear at low speeds.Non-relativistically $$v_p=\frac{v}{2}$$,and this can be easily obtained by binomial expansion of the term within parentheses in (2),but not from (1).Also if (1) is correct,then for low speeds phase velocity is extremely high,and cannot be $$v/2$$.Indeed, if we equate $$v/2=c^2/v$$,particle velocity($$v=\sqrt{2}c$$) exceeds speed of light,which cannot be true.But for (2),the phase velocity is not only approximately $$v/2$$ for low velocity,but equating (2) with $$v/2$$ gives exactly $$v=0$$ also as expected.

Which of (1) and (2) is correct and why?

• I'm suspicious of this "most books" you refer to. Can you give an example of a textbook that says the velocity of a particle is $c^2/v$? – knzhou Apr 25 at 2:18
• @knzhou I’ve seen it before too. It’s not very commonly cited, but it was part of De Broglies original derivation of his well-known relations. – Thatpotatoisaspy Apr 25 at 2:26

When you say "velocity" and "phase velocity", I assume you mean "group velocity" and "phase velocity" of matter waves, where $$v_g = \frac{d\omega}{dk} = \frac{dE}{dp}, \quad v_p = \frac{\omega}{k} = \frac{E}{p}$$ where the second equalities are from the de Broglie relations. For a relativistic particle, $$E^2 = p^2 + m^2$$ which implies that $$E \, dE = p \, dp$$ or equivalently, $$v_p v_g = 1$$ which is the first equation you cite.

This equation is correct, but as you notice, the nonrelativistic limit is more subtle. The reason is that in nonrelativistic physics, we don't work with the full energy $$E$$ above, but instead subtract out the rest mass energy, giving $$T = E - m$$. This subtraction changes the phase velocity, $$v_g' = \frac{dT}{dp} = v_g, \quad v_p' = \frac{T}{p} \neq v_p.$$ This is why most modern textbooks don't talk much about the phase velocity of matter waves; it changes under things like shifts in energy, so it's not a very useful idea.

In any case, we have $$(T+m)^2 = p^2 + m^2$$ which implies that $$(T+m) \, dT = p \, dp$$ from which we conclude $$(v_p' + m/p) \, v_g = (v_p' + 1/\gamma v_g) v_g = 1$$ which simplifies to give $$v_p' v_g = 1 - \frac{1}{\gamma}$$ in agreement with your second equation. So both equations are right. They're just using different definitions of the phase velocity, coming from different definitions of the energy.