DarMM said:
Also, the fermions do not repel each other, in the sense that they cause changes in each other's momentum. In fact they do not repel at all. Rather for the star to keep collapsing any two given fermions would have to occupy the same position. However this isn't a possible two particle state because of the half-integer spin of the particles. So there is no repulsion, it's just that for the star to keep collapsing it would have to move into a state which doesn't exist, hence it doesn't and so remains as it is. Which means the star is supported.
OK - so they don't repel each other, in the sense that they don't cause changes in each other's momentum. Fine. Let's ask QM itself to confirm this, shall we?
You're talking about fermions as if they are continuously existing particles, so they must have trajectories. If the classical force is the only force acting on them (as you effectively state) then they must be following Newtonian trajectories, right?
Since we don't know precisely where the particles are, we'll have to work with a statistical distribution \rho of particles with unknown positions. They're obeying ordinary Newtonian dynamics, but just for convenience instead of F=ma I will use the entirely equivalent Hamilton-Jacobi equation -\frac{\partial S}{\partial t} = \frac{(\nabla S)^2}{2m} + V - where S is related to the 'action' - to calculate the trajectories.
Furthermore, the probability distribution of the particles must obey the usual continuity equation - \partial \rho / \partial t = -\nabla\cdot(\rho {\bf v}) - in order that it remains normalized as it changes shape over time (here v is the velocity of the particle).
Just to play a trick, let's combine the classical Hamilton-Jacobi equation and the continuity equation (two
real equations, note) into a single
complex equation. To do this, we introduce a general complex function \Psi = r e^{i\theta}=\sqrt{\rho} e ^{\frac{iS}{\hbar}} with \hbar an arbitrary constant giving a dimensionless exponent. After a small amount of algebra, we find that the complex equation that results is:
i\hbar \frac{\partial \Psi}{\partial t} = \left( -\frac{\hbar^2}{2m}\nabla^2 + V - Q\right)\Psi .
Well, well - that's the time-dependent Schrodinger equation - straight out of quantum mechanics - with something ('Q ') that looks like a potential subtracted off the Hamiltonian, and \Psi has the same interpretation as in QM: a probability density of particle positions. So what this is telling us is that if we want the particles to follow Newtonian trajectories, we need to subtract off an extra 'quantum force' -\nabla Q - the negative gradient of the `quantum potential' - from the usual dynamic equation of quantum mechanics. This extra force is what makes quantum systems different from classical ones. Is doing this, the momentum of the particle is clearly changed, in contradiction to what you state.
If you analyze the electron-degeneracy pressure situation (in a star, say) then one finds that the gravitational field is balanced precisely by a term involving the quantum force at the radius of the white dwarf, and this force has none of the characteristics of the four regular forces. So if you actually want to answer the question in the terms that the guy asked it, then that's the answer.
If you want, you can say "The particles are fermions and thus they can't be in the same state" but that's simply restating the question in different words. What the above analysis is doing is showing why they can't be in the same state. Because the force exerted by the wave field on the particles keeps them apart.
Just for the record, the formula for Q turns out to be :
Q=-\sum_{i}\frac{\hbar^{2}}{2m_{i}}\frac{\nabla_{i}^{2}\left\vert \Psi\right\vert }{\left\vert \Psi\right\vert }
and the natural interpretation of this is that the wave field (the objectively existing field represented by the wave function \Psi) is 'pushing' the particles, and this force is larger at points in configuration space where the curvature of the wave field is large.
Look, I'm only trying to be clear about this, and in the end, I'm the one doing the mathematics, and you guys are the ones doing the philosophy. You probably think it's the other way around. Well, so be it. But at the end of the day saying that the answer to the guy's question is 'Fermi-Dirac statistics' is doing no better than the chaps in Brazil who Feynman was complaining about, when in answer to the question 'what causes sugar crystals to emit light when you crush them with a pair of pliers?' they said '
triboluminescence makes them emit light'. Any the wiser? I thought not.
… this is highly non-standard, it's the pilot-wave interpretation of quantum theory.
