Theoretically, if we just create a high pressure with low (room 20C) temperature, at some point nuclear fusion can be started.

Is there any research on this topic, how high should be this pressure for different type of reaction? Maybe someone has some numbers in mind, how many GPa should we have to achieve it?


I made my own calculation after all comments which I got.

I got a number $10^{21}$ Pa from thermodynamics of ideal gas. Of course it is approximation. Let's say for Deuterium-tritium we have to give energy 0.1 MeV for 2 atoms to start fusion, which means from electrostatic point of view distance between nucleus $1.44\times10^{-14}$ (14 femtometers) as @John Rennie answered. If I calculated how many atoms will fit in 1 $m^{3}$, if distance between atoms will be 14 femtometers, I will get $N=3.4\times10^{41}$ atoms. Then from $PV=\frac{N}{N_{a}}RT$ if I assume $V=1 m^{3}$ I get $P=1.4\times10^{21}$ Pa and density $2.24\times10^{21} \frac{kg}{m^{3}}$. Which is still $10^{5}$ times more than pressure in the core of the Sun. And if we consider real gas, might be number will be bigger.

Maybe in the future, if they will find another reaction with much less energy (less then 0.1MeV) it would be possible. Might be quantum tuneling can help a little bit :)

  • 2
    $\begingroup$ Pressure is not a function of temperature at very high densities and cold temperatures. Essentially you are asking what densities are required. $\endgroup$
    – ProfRob
    Sep 19, 2016 at 12:57
  • $\begingroup$ You might find the concept of the Lawson criterion to be helpful in addressing your question. $\endgroup$ Sep 19, 2016 at 18:54
  • $\begingroup$ If we put a 3-layer Russian-Doll set of pressure vessels inside pressure vessels capable of 1000 atmospheres each at the bottom of the ocean which has 1000 atmospheres. That could create the 1 trillion atmosphere pressure required for the addition of a neutron energy emitter at its center to start and initiate fusion of solid hydrogen, am I missing anything here except engineering details? $\endgroup$ Mar 10, 2021 at 1:36
  • 1
    $\begingroup$ @EricLeschinski, yes, the inner vessel would have an inner pressure of 4000 atmospheres, not one trillion atmospheres. (They can sustain a particular pressure differential, not pressure ratio!) $\endgroup$
    – obadz
    Jul 31, 2021 at 22:50

2 Answers 2


From memory the potential barrier for deuterium tritium fusion peaks at around $3$ femtometres.

Suppose we take the deuterium-tritium distance as $r$ then the electrostatic force between the nuclei is:

$$ F = \frac{ke^2}{r^2} $$

and we get can a pressure by dividing this by the area of a sphere with radius $r$ to get:

$$ P = \frac{ke^2}{4\pi r^4} $$

You should regard this as a very rough estimate, but it should be immediately apparent the the $r^{-4}$ dependence is going to be a killer because it rises very rapidly for small $r$. If we take $r$ to be $10$ femtometres we get a pressure of about $10^{28}$ Pa. This is so ridiculously large that even given the rough nature of our estimate it's obvious that this approach is not going to work. The pressure at the centre of the Sun is only around $3\times 10^{16}$ Pa.

  • $\begingroup$ Very good point. I am looking now how to calculate it from thermodynamics point, like internal energy of gas equals to 0.1 MeV. Just to compare with your calculations :) $\endgroup$
    – Zlelik
    Sep 19, 2016 at 14:10
  • 5
    $\begingroup$ To be clear, $10^{28}$ Pa is near the Chandrasekhar limit (the pressure limit of a star supported not by heat but by electron degeneracy). So the fact that the electrons will be pushing back pretty seriously prior to this. I think if you push much more, whatever you are pushing together is going to form neutronium on you? I guess that is a form of fusion. $\endgroup$
    – Yakk
    Sep 19, 2016 at 19:07
  • 1
    $\begingroup$ Pyconuclear reactions in a crystalline lattice may occur before neutronisation. But the order of magnitude forthe pressure looks good to me. $\endgroup$
    – ProfRob
    Sep 19, 2016 at 20:47
  • $\begingroup$ I'm curious how much energy you need to squeeze two helium atoms this close. Maybe this alone would cause some interesting effects? $\endgroup$ Sep 20, 2016 at 0:03
  • $\begingroup$ If this 10^28 Pa is a pressure in the neutron star, that means actual fusion should start much early. I still want to calculate internal gas energy from some real gas model like Vam der Waals. $\endgroup$
    – Zlelik
    Sep 20, 2016 at 7:04

As @John Rennie said, you will need a very high pressure if you want you atoms to be forced to be close enough to actually touch their respective barriers. What you describe sounds more as what happens in a neutron star in superextreme conditons.

To start fusion in a more conventional way, each nucleus must have an energy of 0.1 MeV. Acoding to Boltzmann, $E_p=k_B*T$ is the energy of each particle at a certain temperature. At room temperature this energy is 25meV, several orders of magnitude lower than that.

However, if you increase the pressure of this gas at room temperature and keep the temperature constant, each particle energy remains the same. The fact that you have more pressure means more particles are contained in the gas, but on average each particle will have the same energy (25meV) as stated by Boltzmann's law, and won't be able to break the barrier of fusion.

  • 1
    $\begingroup$ But this works for ideal gas, where potential energy of atoms is not considered. In ideal gas internal energy equals to the sum of kinetic energy of all atoms and potential energy of atoms = 0. For real gas internal energy will be sum of kinetic and potential energy. For example for Van der Waals gas potential energy already not zero. As I understand Van der Waals model is not good for high pressure gas. What is the good model for high pressure gas? $\endgroup$
    – Zlelik
    Sep 19, 2016 at 14:14
  • 1
    $\begingroup$ @Zlelik That depends on the gas you're modeling, and how much of a tradeoff you want to make between accuracy and detail. A fairly simple model for monoatomic gases which includes both attractive and repulsive effects would be a gas in a (quantum) Lennard-Jones pair potential and then bound in some fixed volume $V$ by a particle-in-a-box type potential. But even something as simple as that can't be fully understood in closed form. $\endgroup$
    – Ian
    Sep 19, 2016 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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