I am new to relativity and particle physics and I am struggling to understand the conventions for the dimensions that are used in particle physics. I have read this and as I understand it: $$\sqrt{s}=\text{Total energy of collision bewteen two particles in the Centre of Mass frame}=E_{\mathrm{cm}}$$

Now suppose I need to calculate the energy requried to make a $Z$ particle of mass $91\mathrm{GeV/c{^2}}$ in the COM frame (by colliding positrons and electrons both with mass $0.511\mathrm{MeV/c{^2}}$; but this detail is not needed for this question). I'm pretty sure that I use the relation at the top of this post; $$E_{\mathrm{cm}}=\sqrt{s}=91\mathrm{GeV}\tag{1}$$

But there is one serious problem with equation $(1)$ I just wrote above: The mass of the $Z$ particle has units of $\mathrm{GeV/c{^2}}$. But I know that $E_{\mathrm{cm}}=\sqrt{s}$ must have units of energy ($\mathrm{GeV}$) so from where I'm sitting the only way to make this dimensionally correct is to write $$E_{\mathrm{cm}}=\sqrt{s}=91\frac{\mathrm{GeV}}{\mathrm{c^2}}\times \mathrm{c^2}$$ But I can't just multipy by $c^2$ just because I 'feel like it' (to give it units of energy). Unless there is something else going on that I am missing could someone please explain to a lost and confused person why multiplying by $c^2$ is justified here?

Is there perhaps some formula that eludes me here?

By the way, I read this related post but unfortunately it still doesn't answer my question here.

  • $\begingroup$ "Is there perhaps some formula that eludes me here?" yes, $c=1$. $\endgroup$ Aug 7, 2017 at 9:58
  • 3
    $\begingroup$ @AccidentalFourierTransform Hmmm, last time I checked $c=299792458\mathrm{ms^{-1}}\ne 1$. Don't get me wrong I appreciate your response but if you are going to give that little information without reasoning I will never understand what's going on. Would you please consider making an answer elaborating on what you just wrote in that comment, because right now I am more confused than ever? Thanks :-) $\endgroup$
    – BLAZE
    Aug 7, 2017 at 10:05
  • $\begingroup$ You may want to read How can the speed of light be a dimensionless constant? and the linked posts. $\endgroup$ Aug 7, 2017 at 10:09

1 Answer 1


These special units you are referring to are often called 'natural units'. For this, we use natural constants that matter in this specific field and set them to one. In particle physics (which uses special relativity a lot), the common unit system involves $c = 1$, $\hbar = 1$ and $k_B = 1$. Because of this, we obtain seemingly 'funny' units for e.g. time and space ($ \frac{1}{eV} $) and mass ($eV$). The equivalence of the units of time/space and mass/energy also illustrates the close relation of these quantities (spacetime, mass-energy equivalence).


Not the answer you're looking for? Browse other questions tagged or ask your own question.