Memo #4: energy units

By Chip Brock, on March 25th, 2012

There is a considerable simplification in using quantities of mass, energy and momentum once relativity is taken into account. Really. You’ll see that by being careful with factors of $c$ we can make the equations simpler and the numbers more manageable.

First, let me remind you of the energy unit electron Volts. If I were to take a 9 V battery and use that voltage to accelerate an electron, it would gain “9 electron volts” of energy, or 9 eV. That’s just a standard that’s convenient, albeit with a conversion factor that’s sort of unwieldy:

(1) $\begin{align*} 1eV = 1.6 \times 10^{-19}\mbox{~Joules.} \end{align*}$

That number might be familiar—by defining the electron volt this way, it brings in the fundamental charge unit and that’s the actual number $1.6 \times 10^{-19}$ .

Okay. So, we can convert from kg to a unit involving electron volts by remembering that the rest mass of an object is related to the rest energy of that object by the proportion factor of $c^2$ . So, the rest energy of an electron is calculated this way:

(2) $\begin{align*} E_m &= mc^2 \ &= (9.1 \times 10^{-31}\mbox{~kg})(3 \times 10^{8}\mbox{m/s})^2 \nonumber \ &= 8.2 \times 10^{-14} \frac{\mbox{kg m}^2}{\mbox{s}^2} \nonumber \end{align*}$

The combination $\frac{\mbox{kg m}^2}{\mbox{s}^2}$ is the equivalent to $1$ J.

The conversion to electron volts is straightforward from here. Using Equation 2, we can calculate the rest energy in Joules to the rest energy in electron volts:

(3) $\begin{align*} E_m &= (8.2 \times 10^{-14} \mbox{~J})\left( \frac{1 \mbox{eV}}{1.6 \times 10^{-19}\mbox{J}}\right) \nonumber \ E_m &= 5.1 \times 10^{5}\mbox{~eV} \nonumber \end{align*}$

Remember the names for the powers of 10? For numbers that we’ll care about, they are listed in Table~??. So, we can quickly convert the mass of the electron from a number with lots of zeros, to something simpler:

(4) $\begin{align*} 5.1\times 10^{5}\mbox{ eV}\left(\frac{1 \mbox{ MeV}}{10^{6}\mbox{ eV}}\right) = 0.51 \mbox{ MeV} \nonumber \end{align*}$

The powers of 10, their nicknames, and how they are used with electron volts.
power of 10	nickname	in energy units
$10^3$	kilo	keV
$10^6$	mega	MeV
$10^9$	giga	GeV
$10^{12}$	tera	TeV

Of course, Equation 2 is just the defining relationship for rest energy and makes use of the mass of the electron in kilograms, $9.1 \times 10^{-31}\mbox{ kg}$ . If you go to Wikipedia and search for “electron” you’ll find this mass listed this way, but also another way that seems to be an equation and not just a mass: 0.511 MeV/c $^2$ . What’s that about?

From Equation 2, we can obviously see that mass could just as well have been written as:

(5) $\begin{equation*} m = \frac{E_m}{c^2} \end{equation*}$

and the trick here is to continue to use energy units, keeping the '' as a symbol and not dividing it out all the way. In this fashion, we're done. We would quote the mass of the electron as . Yes, explicitly like that. You'd say that the mass of the electron is :0.511 MeV over c squared.”

This is how I solved some of the Tablet problems. You see, there are three sets of units treated in this way:

Energy, momentum, and mass units in “c units.”
quantity	nickname	comments
energies?	keV, MeV, GeV, TeV	the energy of Fermilab’s accelerator was ~2 GeV while the energy of the LHC is 7 TeV and eventually will be 14 TeV
masses?	keV/ $c^{2}$ , MeV/ $c^{2}$ , GeV/ $c^{2}$	the mass of an electron is about half MeV/c $^2$ and the mass of a proton is about 1 GeV/c $^2$
momenta?	keV/ $c$ , MeV/ $c$ , GeV/ $c$	GeV

So, now let’s use it:

Example: Calculate the momentum of an electron whose velocity is $0.8c$ . What is the total energy? Do the calculation in natural units. What is the kinetic energy of the electron? What is its relativistic mass?

Relativistic momentum is $m_{0}\gamma v$ and for $v=0.8c$ , $\gamma$ is:

(6) $\begin{align*} \gamma &= \frac{1}{\sqrt{1-v^2/c^2}} \nonumber \ &= \frac{1}{\sqrt{1-0.8^2}} \nonumber \ \gamma &= 1.7 \nonumber \end{align*}$

Now, keep the $c$ ‘s and insert what we know:

(7) $\begin{align*} p &= (1.7)m_0v \nonumber \ m &= 0.511 \mbox{~MeV}/c^{2} \nonumber \ p &= (1.7)\left(0.511 \mbox{~MeV}/c^{2}\right)(0.8c) \nonumber \end{align*}$

Notice that in the units part, that the $c^2$ in the denominator and the $c$ in the numerator can cancel, leaving the units of momentum $\mbox{~MeV}/c$ . So, we never have to divide by $c = 3\times 10^8 \mbox{~m/s}$ , we just leave it with the $c$ ‘s.

(8) $\begin{align*} p &= (1.7)\left(0.511 \mbox{~MeV}/c^{2}\right)(0.8c) \nonumber \ p &= 0.7 \mbox{~MeV}/c \nonumber \end{align*}$

and we say that the momentum is “zero point seven MeV over c.”

In fact, there’s a further simplification that’s sort of a shorthand notation…like a colloquial notation. We just refer to everything in MeV’s…and forget about the $c$ ‘s altogether. That’s equivalent to just defining $c=1$ . Then if we keep track of what the quantity is, we just add in the $c$ ‘s at the end to make the units be right. This latter, uber-simplified notation is called Natural Units.'' I've invented my own name for keeping the 's as $c$ units.”

Continuing with the rest of the problem, now using “ $c$ units,” we need to calculate the total energy. We can actually do that two different ways, since we have two equivalent forms for the total energy. Using the momentum quantity:

(9) $\begin{align*} E_T^2 &= E_m^2 + (pc)^2 \ E_T^2 &= \left(m c^2\right)^2 + (pc)^2 \nonumber \ &= \left(0.51 \mbox{~MeV}/c^{2}\right)^2 c^4 + \left(0.7 \mbox{~MeV}/c\right)^2 \nonumber \ &= 0.261 \frac{\mbox{~MeV}^2}{c^{4}}c^4 + 0.49 \mbox{~MeV}^2 \nonumber \ &= 0.75 \mbox{~MeV}^2 \nonumber \ E_T &= 0.87 \mbox{~MeV} \nonumber \end{align*}$

The second way to characterize the total energy is:

(10) $\begin{align*} E_T^2 &= E\left(m \gamma c^2\right)^2 \ &= \left(0.51 \mbox{~MeV}/c^{2}\right)^2(1.7)^2 c^4 \nonumber \ &= (0.51)^2(1.7)^2 \mbox{~MeV}^2 \nonumber \ &= 0.75 \mbox{~MeV}^2 \nonumber \ E_T &= 0.87 \mbox{~MeV} \nonumber \end{align*}$

See how the $c$ ‘s just don’t matter in this way of writing. In “Natural Units,” we would just go straight to the near-bottom line.

(11) $\begin{align*} E_T^2 &= E_m^2 + (pc)^2 \ E_T^2 &= \left(m c^2\right)^2 + (pc)^2 \nonumber \ &= \left(0.51 \mbox{~MeV}\right)^2 + \left(0.7 \mbox{~MeV}\right)^2 \nonumber \ &= 0.261 \mbox{~MeV}^2 + 0.49 \mbox{~MeV}^2 \nonumber \ &= 0.75 \mbox{~MeV}^2 \nonumber \ E_T &= 0.87 \mbox{~MeV} \nonumber \end{align*}$

The kinetic energy is:

(12) $\begin{align*} T &= mc^2\left(\gamma - 1\right) \mbox{~MeV} \nonumber \ &= (0.511)\left(1.7 - 1\right) \mbox{~MeV}\nonumber \ &= 0.36 \mbox{~MeV} \nonumber \end{align*}$

Notice, that I’ve used Natural Units and just ignored $c$ ‘s. Sort of uncomfortable at first, but easy after you get used to it!

Finally, the relativistic mass of this electron is:

(13) $\begin{align*} m_R &= m \gamma \nonumber \ &= \left(0.51 \mbox{~MeV}/c^{2}\right)(1.7) \nonumber \ m_R &= 0.87 \mbox{~MeV}/c^{2} \nonumber \end{align*}$