# The relation between mass and speed

Massless particles (or luxons) have no (rest) mass ($m = 0$) and a speed equal to the speed of light ($v = c$). Massive particles (or bradyons) have mass ($m > 0$) and a speed lower than the speed of light ($v < c$). So in other words, the closer a particle's mass is to 0, the closer the it's speed is to the speed of light. So, what is the equation for this relation between mass $m$ and speed $v$?

• "the closer a particle's mass is to 0, the closer the it's speed is to the speed of light." This is an incorrect simplification. Which moves faster: a jumbo jet, or a cyclist? – Asher Jul 4 '15 at 8:58
• This question doesn't make sense. Massless objects move at the speed of light, massive objects don't. There's no further relation, and "the closer a particle's mass is to 0, the closer the it's speed is to the speed of light" is simply false. – ACuriousMind Jul 4 '15 at 9:33
• The question may not make complete sense, but hey, we get to questions that do make sense by asking a few that don't. And I think I know what you're asking, because I think I had this thought a long time ago. See whether my answer does anything to clear your ideas up. – WetSavannaAnimal Jul 4 '15 at 13:33
• What does "the particle's speed" mean? – WillO Jul 5 '15 at 3:38

There's no one-to-one relationship. With zero rest mass, a particle must always be observed to move at $c$. A particle with nonzero rest mass, on the other hand, can move at any speed in $[0,\,c)$ (note the closed-open interval).

At the risk of putting words in your mouth, I think I can recall the exact same question in my mind and it went something like this: I think you're trying to ask something like "how does the limiting case of $m_0=0$ constrain all observed velocities to be $c$ for a massless particle, yet allow all other speeds for massive particles?". Motor cars and tennis balls go at speeds determined by their momentum states. So you need to explore how the speed's dependence on the particle's state variables changes as $m\to0$: what is it that switches this dependence off as $m\to 0$?

Let's begin with the defining relationship for rest mass (the square norm of the momentum four-vector):

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

This is also an equation whence we can get the particle's dispersion relationship, and thus its group velocity. Writing its total energy as $\hbar\,\omega$ and its momentum as $\hbar k$ we get:

$$\hbar^2\,\omega^2 = \hbar^2\,k^2\,c^2 + m^2 c^4$$

and then the particle's group velocity is $v = d\omega/dk$, which is:

$$v = c\frac{p}{\sqrt{p^2+m^2 c^2}} = c^2\frac{p}{E}$$

This equation should give you some more insight into what's happenning. In general, the speed depends on the momentum state of the particle: something that applies to motor vehicles and tennis balls too! In the singular case $m=0$, we get $v=c$ always. If $m$ is very small, the momentum needs to be very small to observe a speed significantly different from $c$. Thus the particle's wavelength in such a state might become impracticably long to prepare the state and its momentum may be too small to trigger detection events. Very small momentums also mean very high uncertainty in particle position (Heisenberg uncertainty relationship) and in such a state the group velocity is not a good measure of the particle's velocity. So it becomes very hard to detect such a particle at any speed other than $c$: conditions for other observations are too seldom and too hard to achieve experimentally.

Indeed, this is exactly the problem we face at the moment with neutrinos. We know neutrinos have a nonzero mass indirectly through the phenomenon of flavor oscillation: but we don't know what it is and indeed so far all neutrino speed measurements (including the infamous 2011 OPERA measurements) have yielded $c$ for their speed, to within experimental error.

In closing, it's instructive to plot our $v$ versus $p$ relationship. If we normalize our momentum units and plot the normalized speed $\frac{v}{c}$ against normalized momentum $\tilde{p} = \frac{p}{m\,c}$ we get the curve below. There is always behavior like this: for very massive particles, the region where $v$ differs significantly from $c$ is wide: for very light particles, the same region is very concentrated around $p=0$, but its the same shapen curve, just stretched or shrunken in $p$-scale:

• Reasonable guess to address a confused question. – Floris Jul 6 '15 at 2:22
• @Floris Eeerily enough I think I know exactly what the OP was asking, because I clearly recall, about 30 years ago, spending a whole afternoon in a state of confusion exactly like the OP's, even worse. As evening came, I saw a pair of car headlights going by out the window and only then realized my confusion: cars are massive and can go any speed they want to!! (duh) it was then I realized my question was really about limiting behaviors and how the horizontal axis on the curve above zooms out infinitely for a zero mass particle. – WetSavannaAnimal Jul 6 '15 at 3:05
• Right. I am glad you answered rather than voting-to-close. This is more helpful. For OP, and for other visitors. Thanks. – Floris Jul 6 '15 at 3:07

Assuming constant total energy, $E^2 = (m_0^2c^2)^2 + (pc)^2 = \gamma^2 m_0^2c^4$ leads to the equation $E^2 = \frac{m^2c^6}{c^2-v^2}$ and then $m = \frac{E}{c^3} \sqrt{c^2 - v^2}$, which also yields the quarter of an ellipse: $v^2 +\frac{c^6}{E^2}m^2 = c^2$, producing the following graph (ignore the number):

where the vertical axis is rest mass which peaks at zero velocity when the rest mass of the system accounts for all the relativistic mass of the total energy; the horizontal is velocity, which peaks at the speed of light with no rest mass.

Only equation I know is $$m = \frac{m_0}{\sqrt{1 - (\frac{v}{c})^2}}$$