How to convert this expression from natural units to those used experimentally?

Question

I cannot for the life of me do the correct conversion of the following expression I derived for two flavour neutrino oscillation. I need to convert the expression from natural units ($c = \hbar = 1$ ) to those used in practice.

$$ \frac{\Delta m_{21}^{2}L}{4E} $$

to

$$ 1.27\frac{\Delta m_{21}^{2}[eV^2]L[km]}{E[GeV]} $$

I have found a multitude of resources showing different ways to do this calculation but I cannot understand any of them so don't hesitate to be overly explicit because I think that's the only way it will sink into my thick brain.

Answer 1

In elementary particle physics, once we fix $c=\hbar=1$, all other kinematic units can be expressed in terms of units of energy.

As Modern Particle Physics by Mark Thomson confirms, we can think about:

• $\Delta m_{21}$ (mass) where the natural units are $\text{eV}/c^2$
  • This comes from the equivalence of mass and energy by $E=mc^2$.
• $L$ (length) where the natural units are $(\text{eV}/(\hbar c))^{-1}$
  • We know $\hbar$, the reduced Planck constant, has units of energy multiplied by time. But time is itself distance divided by speed. Distance, therefore, is the speed of light multiplied by the Planck constant and divided by energy; that is, the units are $\hbar c/\text{eV}$.
• $E$ (energy) where the natural unit is, yes, $\text{eV}$

Assuming we aren't measuring tiny mass changes of the order of $\text{eV}/c^2$, we have, in natural units, the dimensions $$\frac{(\text{GeV}/c^2)^2\cdot(\text{eV}/(\hbar c))^{-1}}{\text{eV}}$$

But we want $$\frac{(\text{eV}/c^2)^2\cdot\text{km}}{\text{GeV}}.$$

Important: we don't want $\dfrac{\text{eV}^2\cdot\text{km}}{\text{GeV}},$ because for the purposes of this problem mass is not exactly energy. Again, $E=mc^2$.

Accordingly, we've got to do the following to the first expression, $\dfrac{\Delta m_{21}^2L}{4E}$:

• Multiply by $10^{18}$ to get rid of two giga's in $\text{GeV}^2$, converting from $\text{GeV}^2$ to $\text{eV}^2$
• Divide by $\hbar c$ to get rid of $((\hbar c)^{-1})^{-1}$. Modern Particle Physics tells us $\hbar c=0.197\cdot10^{-24}\text{ eV}\cdot\text{m}^2$.
• Divide by $10^3$ to convert from m to km
• Multiply by $10^9$ to convert from eV to GeV (denominator)

When this is done, we get $1.27\dfrac{\Delta m_{21}^2\text{[eV$/c^2$]}L\text{[km]}}{E\text{[GeV]}}$.