So I've been reading this about white dwarfs, and various other sites about white dwarf stars. In all of them, they say that we can find the radius of a white dwarf by minimizing its total energy. I know that for almost all physics problems we do this, but why is it so? Does it have something to do with variational calculus or something else entirely...?

Grateful for any help!

  • $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Aug 2 '17 at 16:21

Things aren't always in states of minimum energy. This is something that applies to equilibrium states. Simple examples of this idea are certainly familiar to you already - a ball comes to rest at the bottom of a valley, not partway down the side.

If we want to find the equilibrium state for a white dwarf, there must be no forces on it. If there are no forces, then small changes to the white dwarf don't change its energy, since the change in energy is force*distance. That means the energy should be a "stationary point", where the energy curve is flat. (In this case, we're looking at a curve that describes energy as a function of radius.) We want a minimum so that the equilibrium will be stable.

You can apply different reasoning to get to the same result. For example, the white dwarf is in a cold universe, so thermodynamics says it will give away energy as much as it can since this increases the total entropy. This only stops when the white dwarf is at a minimum energy, or at least gets down to a few degrees K.

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  • $\begingroup$ A missing bit of reasoning here is that the "minimum energy" state of a white dwarf is almost reached before it has cooled down, because the internal energy of a degenerate gas (that dominates the equation of state) is independent of temperature (the minimum internal temperatures of any WDs is about $10^{6}$ K!). In fact the radii of new-born white dwarfs are larger than the radii of older white dwarfs precisely because they have yet to get into a (nearly) completely degenerate state. @georginaelston $\endgroup$ – Rob Jeffries Oct 15 '15 at 12:58

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