I recently learned that hydrogen fusion (i.e., hydrogen to helium) experiments on Earth have been successful at temperatures in excess of 100,000,000 degrees Celsius. However, I also learned that hydrogen fusion in the core of the Sun takes place at 15,000,000 degrees Celsius.

My question: Why the difference? Is it ONLY because the pressure of Earth's atmosphere is so much less than the pressure in the Sun's core, or are there other factors to be considered?


  • $\begingroup$ First, you might consider providing a link to the experiments that you've read about if they are available, as it would be helpful for those of use are aren't familiar with them. Second, these experiments are very likely performed under controlled conditions (they can control the pressure), so the earth's atmospheric pressure is not likely to factor into this. $\endgroup$
    – tmwilson26
    Dec 22, 2015 at 15:22
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    $\begingroup$ The sun is a woefully slow fuser. I gets away with that by being sodding enormous (as in the active core is of a size with the entire planet). You can't afford the wait in something you build. $\endgroup$ Dec 22, 2015 at 15:43
  • $\begingroup$ According to the data on Wikipedia, the sun central temperature is 15.7e6 Kelvin, not 15e3 Kelvin. en.wikipedia.org/wiki/Sun#Core $\endgroup$
    – James
    Dec 22, 2015 at 16:34
  • $\begingroup$ @James-- You are absolutely correct. Thank you for catching that! I have edited the typo in the op. $\endgroup$
    – E.N.B.
    Dec 22, 2015 at 18:23
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    $\begingroup$ @tmwilson26 The excerpt I'm referring to is on p. 43 of "Living in the Environment: Principles, Connections, and Solutions", by Miller & Spoolman. I also found this helpful blurb online: uni.edu/morgans/astro/course/Notes/section2/fusion.html Finally, this article from Live Science seems relevant to the topic at hand: livescience.com/40246-new-boron-method-nuclear-fusion.html $\endgroup$
    – E.N.B.
    Dec 22, 2015 at 18:38

3 Answers 3


The deuterium-tritium fusion reaction cross-section is highly temperature dependent and peaks at temperature of about $8\times 10^{8}$ K, so I suppose these are the temperatures to aim for in a controlled nuclear fusion experiment. In fact according to this, the operating temperatures are at least $10^{8}$ K.

The density of the fusion plasma is a factor - the reaction rate will be proportional to the product of the densities of the two reactants. In fusion reactors the density is of order $10^{20}$ m$^{-3}$. At the centre of the Sun the particle densities are 12 orders of magnitude higher, so partly the increased temperatures in a fusion reactor are to compensate for the lower densities. However, it is also worth remembering that the Sun is not a particularly intense fusion reactor. It only produces about 250 W per cubic metre in its core. A bigger compensatory factor is that the Deuterium-tritium fusion cross-section is about 25 orders orders of magnitude greater than that for proton-proton fusion in the Sun.

In this question I have posted an answer that estimates the energy release per unit volume in typical reactor conditions versus the Sun. I find (order of magnitude) that you get $10^{4}$ times more energy per unit volume out of a reactor than the core of the Sun. So $\sim 10^{6}$ W m$^{-3}$, which I guess is what you will need to make it commercially viable. If you dropped the temperature at all it would rapidly become unviable as a significant power source without absolutely enormous reactors.

  • $\begingroup$ What's the pressure in NF experiments? $\endgroup$
    – Gert
    Dec 22, 2015 at 16:02
  • $\begingroup$ @Gert $P=nkT \simeq 100$ atmospheres ? $\endgroup$
    – ProfRob
    Dec 22, 2015 at 16:05
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    $\begingroup$ It's interesting to note that a living person outputs several times more power per unit volume than the core of the Sun. The comparison I've seen is that the Sun is more like a giant compost heap. $\endgroup$
    – user10851
    Dec 22, 2015 at 18:07
  • $\begingroup$ @RobJeffries Thanks a lot, Rob. That really helped. $\endgroup$
    – E.N.B.
    Dec 22, 2015 at 18:29
  • $\begingroup$ Is that density $10^{20}$ particles per cubic metre? $\endgroup$
    – PM 2Ring
    Aug 12, 2020 at 18:54

This is really just a footnote to Rob's answer.

The Sun is an absolutely terrible fusion reactor. It uses a reaction $p + p \rightarrow d$ that is hopelessly inefficient. The $d + t \rightarrow He + n$ reaction that we use in fusion reactors is (up to) 26 orders of magnitude faster. As Rob says in his answer, the power produced per cubic metre in the Sun is embarrassingly low. However the Sun has the big advantages that its core (where the fusion occurs) is very, very big and very, very dense. The fusion reactors we've managed to make so far are small and the plasma is little different to a vacuum - the particle number density is about one millionth of the density of air.

All this means that our fusion reactors need all the help they can get if they're going to produce a useful amount of power. Adjusting the temperature to maximise the $d + t$ cross section is one of the ways we can boost the power output. The Sun isn't at the optimum temperature for fusion but, well, it's big enough and dense enough that it doesn't care.

  • $\begingroup$ You know, I recall reading something to that effect, somewhere (i.e, the Sun isn't terribly efficient at fusion reactions)! Thanks John, $\endgroup$
    – E.N.B.
    Dec 22, 2015 at 18:31
  • $\begingroup$ You should also add that fusion reactions on earth have an extremely low duty cycle (the reactor is turned on for a tiny fraction of a second, off for a long time, ...), whereas the sun keeps it going continuously. $\endgroup$ Dec 22, 2015 at 18:44

As you mention all research seems to focus on the d+t = He + n reaction. Much of the energy is unrecoverable in the neutron kinetic energy. High energy neutrons cause many problems which will probably make fusion uneconomical. Only aneutronic fusion should be funded, but it requires even higher temperatures. See work on p+B11 = 3He reaction at LPP Fusion. The same can be said of fast nuclear reactors since they gain high temperature efficiency by reducing moderation of neutrons. Breeder reactors are particularly bad because they generate more high level waste as a side stream. Molten salt reactors may be a compromise design.


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