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?
UPDATE:
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 🙂
Best Answer
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.