Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The source of Sun's incessant energy is hydrogen; which is continuously converting to helium through nuclear fusion reaction releasing energy. Why does not all hydrogen convert into helium in one big explosion just like a man-made bomb or cracker explodes?

share|improve this question
add comment

2 Answers 2

The reaction rate doesn't increase that quickly with temperature, but pressure does. If you perturb a solar model, making a zone near the core marginally hotter, the increased pressure will rapidly (at roughly the soundspeed divided by a characteristic length) cause it to expand. That lowers the pressure and temperature enough to substantially quench the reaction. So for normal stars during the long early phases of their evolution, the stability of thermonuclear burning is actually very high. There are some cases in stellar eveolution where pressure becomes insensitive to temperature, generally when the density is so high that electrons average kinetic energy is (almost) independent of temperature. In which case, fusion reactions can accelerate dramatically, but in most cases once the temperature is high enough to increase the pressure, the affected parts of the star rapidly expand and quench the reaction. When a sunlike star intitiates Helium fusion such a runaway reaction referred to as a Helium flash ocurrs, but it is easily contained. On the other hand, when a white dwarf accumulates a critical mass of material to intiate carbon burning, the reaction isn't contained and the star is destroyed in a class 1-a supernova. As dmkee points out, this differs from the core collapse supernova of massive stars.

A useful thing to remember is the virial theorem. If the star (or a portion of it) expands because of an increase in temperature, then once the system reaches hydrodynamic equilibrium the temperature must decrease, as the virial theorem determines that kinetic energy is proportional to gravitational binding energy. Thus unless the reaction=>heating=>reaction feedback is faster than the time to reach hydrostatic equilibrium, the star should be stable.

share|improve this answer
    
It might be worth putting the supernova you mention here into it's correct class (Type I?), because it differs rather a lot from the core-collapse mechanism of very massive stars. –  dmckee Jun 28 '11 at 0:38
    
@Omega Centauri Quote:"When a sunlike star intitiates Helium fusion such a runaway reaction referred to as a Helium flash ocurrs, but it is easily contained." <br> This doesn't explain why is it easily contained. <br> Quote: "Thus unless the reaction=>heating=>reaction feedback is faster than the time to reach hydrostatic equilibrium, the star should be stable"<br> Here again it is not clear why the feedback is not faster for the star in the question (i.e. Sun). –  pongapundit Jul 2 '11 at 18:30
    
@pongapundit Essentially what happens is that degenerate matter, which is matter that is so dense at a given temperature that the pressure is only very weakly dependent upon temperature, and also the amount of heat needed to raise the temperature is low, so runaway can ocurr for a while. In the case of the Helium flash, the temperature rises until the core is no longer degenerate, than it expands on a sonic timescale, and the reaction rate is damped. The current sun is very far from being degenerate. –  Omega Centauri Jul 5 '11 at 15:28
add comment

The temperature & pressure required to sustain nuclear fusion reaction are very high and is only present at a small diameter sphere (in comparison to sun's diameter) at its core. Due to massive convection currents in and out of this sphere a fair amount of hydrogen keeps coming to this sphere to sustain the nuclear fusion.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.