# How big would a star made of pure deuterium have to be to start fusion?

$$D$$-$$D$$ fusion happens much more easily than proton-proton fusion, so I imagine a star made of pure deuterium could be much smaller and still have fusion. Are precise size limits known? How does having a much more reactive fuel affect the star's lifespan?

Edit: Ok so 100% deuterium might not work, but what about a cloud that has enough heavy elements to radiate heat as it collapses?

• Bigger than a 50:50 mix of D and T... Sep 21, 2022 at 22:28
• It would be very difficult for a cloud of pure deuterium to collapse. See en.wikipedia.org/wiki/Jeans_instability Sep 21, 2022 at 22:48
• It's difficult for a gas cloud with no heavy elements to shed the heat formed from gravitational collapse, so the very first stars were much larger than the Sun, possibly hundreds or even thousands of solar masses. And it's possible that they were quasi-stars, with a black hole at the core, rather than a fusion-powered core. See astronomy.stackexchange.com/q/33397/16685 & astronomy.stackexchange.com/q/37135/16685 Sep 21, 2022 at 23:07
• I asked a related question about the maximum metallicity of star formation: physics.stackexchange.com/q/168437/25301 Sep 28, 2022 at 20:19

I've done a similar calculation for a "star" made of water. Minimum size of a "water star"

The essence of the calculation is that deuterium ignites, only if the central temperature reaches the ignition temperature before the core becomes degenerate. The central temperature can be determined from the virial theorem and the perfect gas law and depends on the mean particle mass $$\mu$$ and $$M/R$$.

The radius at which a mass $$M$$ becomes electron-degenerate depends on $$\mu$$ and the mean mass per electron $$\mu_e$$.

The minimum mass is found by setting $$R$$ to be the radius at which degeneracy sets in and assuming the central temperature at that point equals the deuterium ignition temperature, which we can assume is the central temperature of a "star" of normal composition that ignites deuterium at about 13 Jupiter masses (the result given by detailed models).

A normal composition has something like $$\mu = 16/27$$ atomic mass units and $$\mu_e=8/7$$ atomic mass units (for an ionised hydrogen/helium mixture). Ionised deuterium has $$\mu = 2/3$$ and $$\mu_e =2$$.

The minimum mass turns out to be $$\propto \mu^{-3/2}\mu_e^{-1/2}$$. So scaling 13 Jupiter masses for the differing compositions we get a new minimum mass of 14.4 Jupiter masses.

This is a factor of a few lower than the hydrogen (protium) ignition threshold, principally because the ignition temperature for deuterium is also a few times lower than that for hydrogen.

I have seen this calculation in an older book - from the 1950s. I could swear it was Bishop's tome on Project Sherwood, but I just checked my copy and it's not in there.

The calculation, going just on memory here, was basically a comparison of the blackbody radiation from the surface of a sphere at the peak D-D temperature (150 MK?) compared to the reaction rate. As you increase the radius, the reaction rate goes with r^3 and the losses with r^2, so eventually, you get to equilibrium.

• There is no deuterium "flash" near the minimum mass for deuterium burning, and the deuterium burns for of order $10^8$ years. Sep 28, 2022 at 18:55