# Velocity of massless particles

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 $$0$$)?

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.

• I gave an answer that is essentially a duplicate of yours, some minutes after you posted yours. What's the procedure in this situation? Should i delete mine? Nov 15, 2018 at 16:43
• @Runlikehell leave your question in place. The OP might prefer your answer! :-) Nov 15, 2018 at 16:46

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.

• The first $c$ under the square root should be raised to the 4th power, not squared. Nov 15, 2018 at 17:05
• And the sentence “in relativity particles have $v=c$” is wrong. Only massless particles do. Nov 15, 2018 at 17:08
• @G.Smith You are right, i corrected. I've made other mistakes but i didn't notice those, thanks Nov 15, 2018 at 17:17
• Why do you write that massless particles have $v=c$? Can’t they have $v<c$? If not, why not? Nov 29, 2021 at 1:15
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.