$D$-$D$ 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?


3 Answers 3


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.

  • $\begingroup$ Except this value is greater than the minimum for whiter dwarfs, and deuterium-deuterium is easier than hydrogen -deuterium. And brown data’s only have a small percentage of deuterium. $\endgroup$ Commented Jul 16, 2023 at 17:53
  • $\begingroup$ @blademan9999 Why would D-D fusion be easier to initiate than p-D fusion? Greater than the the minimum what for white dwarfs and why is that relevant? $\endgroup$
    – ProfRob
    Commented Jul 16, 2023 at 19:14
  • $\begingroup$ Oops, brown dwarfs I meant. Anyway en.m.wikipedia.org/wiki/…. it notes the relative rate of D-D and D-P fusion when catalysed by muons $\endgroup$ Commented Jul 16, 2023 at 19:28
  • $\begingroup$ The physics of the Gamow peak does not involve muon catalysis. Do you have another source? @blademan9999 $\endgroup$
    – ProfRob
    Commented Jul 17, 2023 at 7:19

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.

The answer was "about the size of the moon".

I'll try to track it down.


This is a fun question, although completely inapplicable to the real universe, since the concentration of deuterium in our universe started out low and has been decreasing since then. For the same reason, I don't think anybody bothers doing detailed calculations for such a scenario.

For realistic deuterium concentrations, you get a deuterium flash when a protostar is in the early stages of gravitational collapse, while it is still inside a protostellar nebula and therefore difficult to observe. At this stage, calculations seem to show that the star is unstable, so you would expect to see it being highly variable, but because the evolutoinary process is so fast, the difficulty of observing such a thing is compounded.

By extrapolation, I would guess that if you started with a pure-deuterium nebula and let it collapse gravitationally, it would undergo a deuterium flash and immediately become unstable and blow itself apart. I imagine that this would probably happen before the star had reached any equilibrium state comparable to the main sequence, and before it had fused any significant fraction of its deuterium.

  • $\begingroup$ So even if there was a deuterium gas cloud (didn't think there could be one), it might not even collapse to form a star? $\endgroup$
    – zucculent
    Commented Sep 22, 2022 at 1:21
  • 1
    $\begingroup$ There is no deuterium "flash" near the minimum mass for deuterium burning, and the deuterium burns for of order $10^8$ years. $\endgroup$
    – ProfRob
    Commented Sep 28, 2022 at 18:55

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.