For the $p$-$p$ fusion reaction chain occurring in the sun, why is it that the main limiting factor placed on the time scale of the initial proton proton fusion into helium due to the fact it is a weak force interaction?

I would've thought it is because the thermal energy available in the core of the sun is much smaller than the Coulomb barrier felt by the protons. So the fact the protons need to QT to fuse is the reason the fusion occurs so slowly.

How slow are weak force interactions that they are the dominant reason here and why?

  • $\begingroup$ why is it that the main limiting factor placed on the time scale of the initial proton proton fusion into helium due to the fact it is a weak force interaction? What makes you think this is true? The reaction rate depends on both. The product of two numbers doesn't have one number or the other as the main factor. $\endgroup$
    – user4552
    Jan 22, 2020 at 19:56
  • $\begingroup$ Related: physics.stackexchange.com/q/77053/123208 FWIW, at solar core temperatures, most of the time when 2 protons fuse into diprotium it immediately fissions back into 2 protons rather than transmuting to a deuteron. The odds of the transmutation happening are on the order of 1 in $10^{26}$. $\endgroup$
    – PM 2Ring
    Jan 23, 2020 at 9:55

2 Answers 2


For the initial steps of nuclear fusion you need to get two protons together and then turn one into a neutron via the weak force interaction.

Both things are important to the overall rate. The weak flavour change of an up quark to a down quark in a proton is a hindrance to the process and means that you have to get a large population of unstable diprotons together in order to create a stable deuteron. i.e. Making the di-proton is easier than getting the di-proton to change into a bound deuteron, so in that sense the weak interaction controls the overall timescale at a given temperature.

Why is forming the deuteron a bottleneck? Well, it's a weak interaction responsible for changing a proton to a neutron, much weaker than the strong nuclear force responsible for allowing the two protons to get near each other in the first place.

It is the hindrance of the weak interaction step that requires the interior temperature of the Sun to be so high. If it were possible for two protons to fuse to a stable state, then much lower temperatures would be required. This is demonstrated by the order of magnitude lower temperatures at which deuterium or lithium fuse with protons, neither of which require any weak interactions but have the same or higher Coulomb barrier between the reactants.

It is an interesting hypothetical quation to ask what would happen if the weak interaction was much stronger, such that a deuteron was always formed when two protons were brought close together? In such a universe I think stars would have much lower internal temperatures, be larger for a given mass (since $T \propto M/R$ from the virial theorem) and much more luminous and short-lived.


I am copying from this link.

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Note the "two protons fuse and usually fall apart" due to the electromagnetic repulsion. The only reason to fuse is the strong quantum mechanical force which gives the probability for the two protons to be close enough to call it fusion.

There is enough energy in the plasma though to bridge the energy difference between the proton mass and the neutron mass, so the proton can turn into a neutron through the weak interaction, for the tail of the proton kinetic energy distribution, but the creation of a neutron goes through the weak interaction. It is called weak because its coupling is $10^{-6}$ of the strong force. That is why the diprotons fuse and decay, being unstable. Once a neutron forms, there are deuterons, which then scattering with another proton form the tritium nuclei which eventually lead to helium.

It is not a simple reaction but a series.

The main control is the weak interaction of a proton turning into a neutron , because of the small probability due to the coupling constant.


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