# What is the temperature of an LHC Bunch?

From various places, I've found that an LHC bunch seems to be about 100mm long, 0.1mm wide, and contains about $10^{11}$ protons or anti-protons. The de Broglie wavelength is about $10^{-18}$ meters, so this is close to a plane wave for the purposes of using the S-matrix formalism. Each bunch is separated from the next bunch by about 10 meters (the beam is about 27km long, there are are 2808 bunches).

If I was co-moving with a bunch, what is the temperature (at the center of the bunch, say)? This I have not been able to find. A reference if possible.

• this is an awesome question. I once asked a bunch of LHC beam physicists this, and they just said something about emittance, how it was related to temperature, shrugged. – Shep Feb 14 '15 at 19:41

## 1 Answer

I guess you may wish to look for the value of the beam emittance, as it is related to beam temperature (https://cas.web.cern.ch/cas/Holland/PDF-lectures/Moehl/Moehl-final.pdf , slide 12)

EDIT: there is some data on LHC emittance at http://www.lhc-closer.es/1/4/18/0

EDIT: there are some definitions and data at http://www.lhcportal.com/Portal/Info/LHCGlossaryDef.pdf (look for "normalized transverse emittance")

• Your second link gives $\beta^*=0.55\,\rm m$ and $\epsilon_n = 3.75\,\rm \mu m$, but is a little unclear on the meaning of the "normalized" emittance. This is nearly enough information to answer the question. – rob Feb 14 '15 at 20:54
• I have from the Moehl lecture, $k_BT\approx m_p(\beta\gamma c)^2\epsilon/\beta_x\approx m_pc^2\beta\gamma\epsilon_n/\beta_x$, where $m_pc^2\beta\gamma$ is very nearly the beam energy per proton, so $T\approx\mbox{(7e12eV)}*\mbox{(1.6e-19joule/eV)}*\mbox{(2.5e-6m)}/\mbox{(0.44m)}/\mbox{(1.38e-23joule/K)}$ $=\mbox{4.6e11K}$, taking numbers from rdemaria.web.cern.ch/rdemaria/www/hllhc11. I wasn't expecting something quite that high!? – Peter Morgan Feb 16 '15 at 18:48
• @Peter Morgan: On the other hand, this is five orders of magnitude less than the beam energy per proton - this seems reasonable to me. – akhmeteli Feb 16 '15 at 22:37
• @PeterMorgan Your computation is fine in the transverse plane but this approach does not work in the longitudinal direction. You need to compute the velocity spread from the energy spread. Also note the beta function at the Interaction Point is in the cm scale, but it goes to km scale in other parts of the machine... – DarioP Feb 19 '15 at 14:19
• @DarioP thanks. So I find somewhere that the energy spread is about $\mbox{1e-4}*\mbox{7TeV}=\mbox{700MeV}=m_p\gamma (\Delta v)^2$, where $\gamma\approx \frac{\mbox{7TeV}}{\mbox{1GeV}}=7000$, so $m_p(\Delta v)^2\approx 0.1MeV$ and $T\approx \mbox{(1e5eV)}*\mbox{(1.6e-19joule/eV)}/\mbox{(1.38e-23joule/K)}$ $\approx \mbox{1e9K}$. Within a couple of orders of magnitude of the transverse temperature, slightly smaller than the transverse temperature at the IP where there's greater transverse pressure but slightly larger elsewhere. – Peter Morgan Feb 20 '15 at 16:08