# What happens to Hydrogen-Boron plasma at 3 billion Kelvin?

A recent popular report states (beware, I believe it is not peer reviewed and this slightly smells like a PR stunt) that a private company TriAlpha has made some notable progress towards Hydrogen-Boron fusion, a version of aneutronic fusion, i.e. fusion where you do not have to care about those pesky uncharged neutrons flying out from your confining magnetic cage and destroying your expensive apparatus.

My concern is the following: they also state that you need temperatures around $3\times10^9\,{\rm K}$ to achieve this kind of fusion. This seems a lot more than we have direct experience with. In usual tokamaks, we are usually talking up to $10^8\,{\rm K}$ and in the Sun it's usually like $10^6-10^7\,{\rm K}$. Speaking in terms of typical energies, the desired $3 \times 10^9\,{\rm K}$ is $250\,{\rm keV}$ which is about the half of the mass of the electron. This means, for instance, that we will have reasonably large corrections to the behaviour of photons due to (electron-positron mediated) photon-photon scattering!

So my question is, have there been detailed inquiries into what changes at such temperatures for a plasma? Does the MHD approximation still even apply? Is the electron-positron soup ($\to$photon-photon scattering) going to have any non-negligible effects?

• One would need to know the number density of the ionized particles, the average charge state, etc. to determine the limits of MHD. I do not think it's impossible to heat a lab plasma to $10^{9}$ K, though I am not sure... I do know that the electron gas would need to be treated relativistically, thus we automatically need to care about $\partial \mathbf{E}/\partial t$, which kind of negates MHD approximations... – honeste_vivere Aug 26 '15 at 23:41
• I forgot to mention that typical "temperatures" in particle colliders can be much higher than this, so it is not unheard of in actual experiments... – honeste_vivere Oct 27 '15 at 22:01

In my understanding, one of the main challenges are the losses due to bremsstrahlung: the power lost through bremstrahlung $P_{\rm br}$ scales like $$P_{\rm br} \propto Z^2 n_i n_e T_e^{1/2},$$ where $Z$ is the charge number, $n_i$ and $n_e$ are respectively the ion and electron density, and $T_e$ the electron temperature. The squared scaling with the charge number is the main problem here. It leads to much higher losses than the typical D-T fusion reaction (since boron has a higher charge number).