Which of the four forces is responsible for degeneracy pressure?

[cortiver]

Okay - i don't dispute that anything you say is not true.

But i still don't get it. I agree that there is no inflow of energy and maybe no work is done, but i still think there must be a force or energy that prevents gravitational collapse. Maybe you can make me understand via this example:

Let's say i am suspended on a bungee jumping rope, hanging from a bridge. I'm not talking about the situation when you are falling, but after you fell. You're not moving; the axial force of the rope is in equilibrium with the opposite gravitational force. No work is being done.

Now, if my rope is cut, the only way to stop me from falling down is to introduce an opposite force. For instance, a rocket on my back... or by falling onto the ground, where the EM force will stop (and probably kill) me from moving. This last scenario does not involve adding energy, but in every situation, there is an opposing force. There has to be; Newton's law.

So what force in a white dwarf opposes the gigantic gravitational pull? I guess you've already answered that there can be pressure without any force, but i don't understand this. By definition, if there is acceleration, then there is a force... how can Newton's law be violated?

The thing is, Newton's law, F = ma, fundamentally applies to single particles. It can also be generalized to apply to groups of particles whose membership is constant, i.e. the same particles are in the group forever. The set of particles making up a person is, to all intents and purposes, constant, so you can apply Newton's law to a bungee-jumper.

I guess the problem you're having then, is you're trying to apply Newton's Law to a little section of a star, as if it were a little block of solid. But a star is not a solid, and the set of particles contained within a given region does not remain the same over time - because of the kinetic energy of the particles, they are moving around and there will always be particles entering and leaving the region. Hence Newton's Law cannot be applied to a region of gas!

Now, you can apply Newton's Law to an individual gas particle. In this case, since the only force acting on it is gravity, it will indeed be accelerating towards the center. But this does not mean that, averaged over long periods of time, it has to get any closer to the center, as you can see by considering the orbit of the Earth around the sun. So you have lots and lots of particles, all accelerating towards the center, but the overall density of the star at every point is unchanged, so it doesn't collapse.

 

[zenith8]

Now, you can apply Newton's Law to an individual gas particle. In this case, since the only force acting on it is gravity, it will indeed be accelerating towards the center. But this does not mean that, averaged over long periods of time, it has to get any closer to the center, as you can see by considering the orbit of the Earth around the sun. So you have lots and lots of particles, all accelerating towards the center, but the overall density of the star at every point is unchanged, so it doesn't collapse.

Hi Cortiver,

OK, but I showed in my post #22 that, under the sole assumption that particles exist and that they have trajectories, then the equations of quantum mechanics themselves predict the existence of an extra force -\nabla Q on each particle. It is quite clear in your above posts that you are making the same implicit assumption that particles exist. For your argument to stand up, then, you need to get rid of this force. How do you propose to do that, other than by ignoring it?

Cheers,
Zenith

 

[LennoxLewis]

The thing is, Newton's law, F = ma, fundamentally applies to single particles. It can also be generalized to apply to groups of particles whose membership is constant, i.e. the same particles are in the group forever. The set of particles making up a person is, to all intents and purposes, constant, so you can apply Newton's law to a bungee-jumper.

I guess the problem you're having then, is you're trying to apply Newton's Law to a little section of a star, as if it were a little block of solid. But a star is not a solid, and the set of particles contained within a given region does not remain the same over time - because of the kinetic energy of the particles, they are moving around and there will always be particles entering and leaving the region. Hence Newton's Law cannot be applied to a region of gas!

Good points and i agree.

Now, you can apply Newton's Law to an individual gas particle. In this case, since the only force acting on it is gravity, it will indeed be accelerating towards the center. But this does not mean that, averaged over long periods of time, it has to get any closer to the center, as you can see by considering the orbit of the Earth around the sun. So you have lots and lots of particles, all accelerating towards the center, but the overall density of the star at every point is unchanged, so it doesn't collapse.

But the Earth around the Sun is a special case of free fall. You can't tell me that every particle in a region as dense as a white dwarf (or even a neutron star) is constantly in free fall? Because two particles can't have the same state when their wave functions overlap, they are repulsed (?) "by", or as a consequence of the Pauli Exclusion principle?


One unrelated sidenote: what about the Pauli Exclusion principle behind the event horizon of a black hole? Does our physics break down at a singularity?

 

[cortiver]

zenith8 and LennoxLewis, you both seem to have misunderstood my post. I was trying to clarify how a star can be supported by pressure without the need for an actual force. To avoid quantum complications, I was considering a star made up of classical particles. Of course, the motion of quantum particles is governed by the Schrödinger equation, and Newton's Law doesn't apply directly, at least not without the introduction of the quantum force.

However, for a real white dwarf star, the de Broglie wavelength of the particles (which, in a degenerate gas, is on the order of the average distance between neighbouring particles) is much smaller than the characteristic length scale of variation of the star, and hence the semiclassical approximation (use Fermi-Dirac distribution, but otherwise treat particles as classical) gives correct results for macroscopic quantities. Hence, under this picture (which admittedly is not really the true one), we can talk about the particles as moving under Newton's Law without a quantum force.

Edit: the following argument is wrong! Please ignore it.

I can also show directly that the quantum force should be negligible compared to the pressure 'force' in such a situation:
The quantum force is given by
<br /> \begin{align*}<br /> F_q &amp;= \sum_{i}\nabla_i \left(\frac{\hbar^2}{2m} \frac{\nabla_i^2 |\Psi|}{|\Psi|}\right) \\<br /> &amp;\sim \sum_{i}\frac{\hbar^2}{2m} \frac{1}{R^3} \\<br /> &amp;\sim N\frac{\hbar^2}{2m}<br /> \end{align}<br />
where R is the radius of the star, N is the average density of particles in the star, and the last line follows because R^3 is on the order of the volume of the star. The quantum force density is therefore
<br /> f_q = NF_q \sim N^2\frac{\hbar^2}{2m}<br />

On the other hand, the degeneracy pressure is on the order of the kinetic energy density. So P \sim NE, where E is the average kinetic energy of particles in the star, and the pressure 'force' density is
f_p = -\nabla P \sim \frac{NE}{R}
The ratio of the two forces is therefore
<br /> \begin{align*}<br /> \frac{f_p}{f_q} &amp;\sim \frac{N\hbar^2 R}{2mE} \\<br /> &amp;\sim N\lambda_B^2 R<br /> \end{align*}<br />
where \lambda_B is the average de Broglie wavelength of the particles. Since in a degenerate gas, \lambda_B \sim N^{-1/3}, where N^{-1/3} is approximately the average separation between neighbouring particles, we have
<br /> \frac{f_p}{f_q} \sim N^{1/3} R<br />
Since there are obviously many particles in a white dwarf star, we have R \gg N^{-1/3}, and hence:
f_p \gg f_q
So the quantum force is negligible compared to the pressure 'force'.

End of wrong argument.

I note also that the definition of the quantum force is the same for bosons and fermions, so how could it explain repulsion between fermions?

 

[LennoxLewis]

Isn't that fq over fp that you calculated? fp over fq should, other terms aside, be proportional to one over N (before simplifying the terms).

 

[cortiver]

Isn't that fq over fp that you calculated? fp over fq should, other terms aside, be proportional to one over N (before simplifying the terms).

Oops, you're right. Clearly I composed that argument in too much of a hurry - looking at it now, I see it doesn't make much sense. Consider it retracted. I still maintain that the semiclassical approximation should be applicable for white dwarf stars, so the quantum force should be negligible. I'll keep thinking whether it is possible to show this directly.

 

[zenith8]

zenith8 and LennoxLewis, you both seem to have misunderstood my post.

I can also show directly that the quantum force should be negligible compared to the pressure 'force' in such a situation

Your maths is just wrong, as Lennox has already pointed out.

I note also that the definition of the quantum force is the same for bosons and fermions, so how could it explain repulsion between fermions?

Remember that the quantum force is supposed to be a force exerted by the objectively existing wave field (mathematically represented by the wave function) on the particles. It is not directly an interparticle force.

Remember also that the wave fields for fermions and bosons have very different characteristics, the most important of which is the existence of a nodal surface in the fermionic case (i.e. a set of points in configuration space where the wave field has the value zero). Particles trajectories are actively repelled from these surfaces, and effectively, cannot pass through them. This is exactly what is required to produce 'fermionic repulsion' in the sense you state.

 

[cortiver]

Remember that the quantum force is supposed to be a force exerted by the objectively existing wave field (mathematically represented by the wave function) on the particles. It is not directly an interparticle force.

Remember also that the wave fields for fermions and bosons have very different characteristics, the most important of which is the existence of a nodal surface in the fermionic case (i.e. a set of points in configuration space where the wave field has the value zero). Particles trajectories are actively repelled from these surfaces, and effectively, cannot pass through them. This is exactly what is required to produce 'fermionic repulsion' in the sense you state.

Fair enough. I had got the impression somehow that you were claiming that the quantum force could explain the exclusion principle without any additional boundary conditions, but looking back at the thread it looks like I was mistaken and you never made any such claim.

Anyway, all treatments I've seen of the equation of state in a white dwarf star only mention the standard degeneracy pressure
P = \frac{(3\pi^2)^{2/3}}{5} \frac{\hbar^2}{m} N^{5/3}
(or its relativistic generalization), which comes simply from the kinetic energy via the relation P = (2/3)u, where u is calculated from the Fermi-Dirac distribution at T = 0. This must surely mean that the additional effect of the quantum force is considered negligible. I don't know enough to prove this myself, but I do not believe that all those authors are wrong.