2

This seems obvious but it is confusing since I know that in MKS (i.e. SI) system, we use Joule with Meter with KG.

But if we're using electron volt (1.6e-19 J) for energy, should we change the mass and distance units too?

student1
  • 594

1 Answers1

4

It depends on what you mean by "should." You can, if you'd like, quote all of your energies in $\mathrm{eV}$, all of your masses in $\mathrm{kg}$, and all of your distances in $\mathrm{m}$ without any inconsistency.

However, you're probably referring to the fact that joules are naturally suited to kilograms, meters, and seconds in the sense that $$ \mathrm J = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2} $$ without any weird dimensionless prefactors on either side of the equation, and you want to know what mass and length units, lets call them $\mathrm M_\mathrm{eV}$ and $\mathrm L_\mathrm{eV}$ give you a relationship similar to the relationship for $\mathrm{J}$; \begin{align} \mathrm{eV} = \frac{\mathrm M_\mathrm{eV}\cdot \mathrm L_\mathrm{eV}^2}{s^2}\qquad ?\tag{$\star$} \end{align} If so, then note that there isn't a unique pair that does this. As you noted, there is a dimensionless constant $a$ relating Joules and electron volts; $$ a\cdot \mathrm{eV} = \mathrm J $$ This means that we can use the first relation above to write $$ a \cdot\mathrm{eV} = \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{s}^2} $$ Now we notice that both of the following pairs of choices will lead to an expression like $(\star)$: for choice 1 take $$ \mathrm M_\mathrm{ev} = a^{-1}\cdot \mathrm{kg}, \qquad \mathrm L_\mathrm{eV} = \mathrm m $$ and for choice 2 take $$ \mathrm M_\mathrm{ev} = \mathrm{kg}, \qquad \mathrm L_\mathrm{eV} = a^{-1/2}\cdot \mathrm m $$

joshphysics
  • 57,120
  • What about choosing units such that $c=1$, so both mass and energy would be measured in $\mathrm{eV}$? I'm no particle physicist, but that would seem the most sensible way to me. – N. Virgo Mar 15 '13 at 02:34
  • @Nathaniel Yea. I mean, let's face it, choosing which system of units is "sensible" is almost entirely a matter of taste. – joshphysics Mar 15 '13 at 02:40
  • Yes. Taste and practicality, the latter being very dependent on the application domain. – N. Virgo Mar 15 '13 at 03:04
  • Couldn't agree more. – joshphysics Mar 15 '13 at 03:06
  • 1
    @Nathaniel that's exactly what we do in particle physics. Though it comes with a danger of forgetting that mass and energy aren't the same thing. (Just look around and see how many websites quote the mass of the Higgs boson as 126 GeV, with no factor of $1/c^2$!) – David Z Mar 15 '13 at 05:07
  • @joshphysics, So with energy in eV, mass in Kg, then distance is in m/sqrt(1.6e-19). – student1 Mar 15 '13 at 21:49
  • @ali8 If you want to write a relation between your units like in $(\star)$ then, unless I'm making a simple mistake here, it would be $\mathrm m\cdot \sqrt{1.6\times 10^{-19}}$; note I defined $\mathrm J = a\cdot\mathrm{eV}$ not $\mathrm{eV} = a\cdot \mathrm J$. – joshphysics Mar 15 '13 at 23:24
  • Alright, so with eV = 1.6e-19 J, my distances is in sqrt(1.6e19) meters. – student1 Mar 16 '13 at 03:36