Velocity of massless particles

Question

Due to the equation $E = \frac12{mv}^2$, can a massless particle travel at an infinite speed in a vacuum (as its mass would be $0$ so its energy would also be $0$)?

Answer 1

The equation you give:

$$ T = \tfrac{1}{2}mv^2 $$

is the non-relativistic equation for the kinetic energy. The relativistic equation for the total energy is:

$$ E^2 = p^2c^2 + m^2 c^4 $$

where $p$ is the momentum of the particle and $m$ is the rest mass. For a massless particle like a photon, where $m=0$, this equation simpifies to:

$$ E = pc $$

The energy is $E=h\nu$ and the momentum is $p = h/\lambda$, and substituting these in our equation we get:

$$ h\nu = \frac{hc}{\lambda} $$

or:

$$ \nu\lambda = c $$

But $\nu\lambda$ is the velocity, so we find that the velocity of our massless particle is $c$ i.e. the speed of light.

Answer 2

That is the equation for the kinetic energy in classical mechanics, you can't extrapolate it to massless particles, you need relativity for that.

So, in relativity massless particles have $v=c$ where $c$ is the speed of light and the equation for the energy of a particle is $$ E= \sqrt{m^2c^4+p^2c^2} $$ Where $p$ is the momentum. For massless particles ($m=0$) the equation reduces to

$$ E=pc $$

So massless particles have energy and momentum, and both are finite.

Answer 3

Since the question is about kinetic energy in classical mechanics, I'll answer in terms of classical mechanics where kinetic energy is $E = \frac{1}{2}mv^2$. The speed $v$ is normally defined as a non-negative real number, and the same goes for its square $v^2$. Multiplying any real number with $0$ gives $0$, so the energy is always zero when $m=0$.

Infinity is not a number however, so this doesn't say anything about what happens at infinite speed. We first have to define what infinite speed means. An obvious definition is that a particle has infinite speed if it's at two places at once, i.e. $\Delta t=0$ for two events where the particle is located at $x_1$ and $x_2$, where $x_1\ne x_2$. Kinetic energy can then be calculated as a limit:

$$ \lim_{\Delta t\, \to \,0} \frac{1}{2}m\Big(\frac{x_2-x_1}{\Delta t}\Big)^2$$

This limit is $0$ when $m=0$, but it can be any number if you take the limit of $m\to 0$ and $\Delta t \to 0$ at the same time. In that case, the answer depends on how you pose the question.

Note: in the real world the classical formula for kinetic energy breaks down in the relativistic domain where $v\ll c\,$ no longer holds, so the above limits don't actually apply to real massless particles.