Which of the four forces is responsible for degeneracy pressure?

[kakaz]

In other words, you don't know what force causes the Exclusion principle? It just is? Isn't that a bit strange for a physicist to say? 80 years of science ever since, there must have been some publications about the principle that drives the Pauli Exclusion?

I will wrote for You once more what I wrote last time: You do not have any forces in order to have pressure in system. You push with some force system with fermion particles. You want to lover of volume in such gas for example. But as the whole energetic states are occupied, then after You "press the wall " particles has to look for new energetic states. But there is no one available except higher energetic ones, because lower energies are taken by other particles. As there is no other states, the higher energetic ones will be occupied, and energy for this is given by You, when You press the wall. So: You feel lake You energy is given somewhere, and You think it was to break some force: in fact You pump additional energy into system. You may call it "breaking Pauli force" although it was only pumping additional energy for higher energetic states occupied.

Thanks for the extended and well written post... not to come across as unthankful, but i know what fermions and bosons are, and their wavefunctions. But what you don't talk about is which force acts to make sure no same states are occupied and cause degeneracy pressure!

Ha! Here You are: there is no force needed: wavefunctions has some symmetries, and as usually there is no way to destroy it (symmetry), until we do not reach new ways of particle reactions for example nuclear one. So: there is no force needed: as long as there is no other way for wavefunction to behave, it has internal symmetry or antisymmetry and there is degeneracy pressure. Possible ways of breaking symmetry usually are located much higher than normal energetic levels available for system. In fact such types of behaviors usually are connected with destroying system at all!

In order to destroy this internal symmetry You have to pump as much energy into system as there will be another possibilities of decay for system accessible.

For example for gas of nucleons You have to break nuclear forces which contract against nuclear fusion of particles in system. In this case You may think that answer for "which force acts" the answer is "nuclear forces". But in other cases there may be other forces: when as fermionic particles You will consider ions the answer would be "molecular forces" etc. So the proper answer is: there is no one force at all, it depends on case.

 

[kakaz]

OK, this is a nice post as far as it goes, but as Lennox has said you're not actually answering his question, you're merely restating it. You're saying that fermions can't be in the same state. He's effectively asking why.

I understand. In 3-dimensional space, there are 2 different families of representation od SO(3) - gerenal group of rotation. One family has spin which is natural number whils second one family has multiplicity of 1/2. This is mathematics, not physics.
Question of Your friend may be, rephrased acordingly to situation, changed as follows:
1. why there are two families of representations SO(3)?
2. Why real objects has to be governed by one of them?

Answer to first one is unimportant: as long as w accept that SO(3) is important for us in describing real world objects. It is feature os group of rotation. No more o no less.

Answer to second question is interesting one: as long as system has some symmetry, we may choose our description in such way, that states of that system are pure eigenstates of some representations of group of symmetry which system has.

For example, as system has transnational symmetry, You may choose to use momentum states as basis in Hilbert space of wavefunctions.

Other way, as we have symmetry of translation in time, our systems states has well defined energies ( and are in states which are eigenstates for some energy operators which in fact gives us translation in time etc.)

This is the real meaning of symmetry in system. In order to look as it works please take any Quantum Mechanics book with chapter about Representation Group Theory and its relation to Noether Theorem.

 

[LennoxLewis]

The flow of momentum into the region due to particles passing through an element d\mathbf{A} of the surface (oriented outwards), turns out to be (I am ignoring relativistic effects here):
-N \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle d\mathbf{A},
where N is the density of particles, \mathbf{n} is the unit vector pointing in the direction of d\mathbf{A}, \mathf{v} is the velocity of the particles, and the angle brackets denote averaging over all nearby particles. Now, assuming isotropy, \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle will be the same regardless of \mathbf{n}, so we can define the pressure by

P = N \langle m(\mathbf{v}\cdot{\mathbf{n})^2\rangle

In particular we have
<br /> \begin{align*}<br /> P &amp;= N \langle m(\mathbf{v}\cdot\mathbf{n})^2 \rangle \\<br /> &amp;= \frac{N}{3}\left(\langle mv_x^2 \rangle + \langle mv_y^2 \rangle + \langle mv_z^2 \rangle\right) \\<br /> &amp;= \frac{2N}{3} \left\langle \frac{1}{2}m|\mathbf{v}|^2\right\rangle \\<br /> &amp;= \frac{2}{3} u<br /> \end{align}<br />
where u is the kinetic energy density.

Isn't that the flow of kinetic energy? I know that doesn't really make sense as kinetic energy isn't a vector, but momentum is m.v, while m.v.v, which you are using, is related to kinetic energy?

Okay, of course the force the particles exert on the wall is electromagnetic. But that's not what I'm talking about. Pressure is a local property of the individual locations in the gas (take an infinitesimal surface element at a particular point - how much momentum flows through it in unit time?). It doesn't depend on what's happening at the exterior of the gas.

I'm not used to that definition, but for the moment i'll accept it as being true.


Maybe my question becomes more clear when i re-phrase it:
Pressure is related to kinetic energy of the particles. In a normal sequence star, nuclear fusion provides this kinetic energy, leading to a pressure that opposes gravitational collapse. When fusion runs out, the gas loses kinetic energy and with it the pressure needed to retain size. The star collapses until degeneracy pressure equals the gravitational pressure. Where does the energy needed for this pressure come from? The Pauli exclusion principle provides an endless source of energy when inter atomic distances become small enough?

Well, the whole point of degeneracy pressure is that the exclusion principle forbids this. At high densities, the only way you can avoid having two particles in the same state is by filling up lots of very high energy states. So even at absolute zero the particles continue to have lots of kinetic energy.

So where does this energy come from? It never depletes, if the Pauli Principle is universal and timeless (not taking big crunches etc into account).

 

[cortiver]

Isn't that the flow of kinetic energy? I know that doesn't really make sense as kinetic energy isn't a vector, but momentum is m.v, while m.v.v, which you are using, is related to kinetic energy?

The flow of kinetic energy would be
N\left\langle \frac{1}{2}mv^2 \mathbf{v}\cdot d\vect{A} \right\rangle
(though for an isotropic velocity distribution this just evaluates to zero). To calculate the flow of a quantity through a surface, you have to take into account that the faster particles are moving perpendicular to the surface, the more particles pass through per unit time, i.e. you multiply the quantity you're calculating the flow for by the velocity of the particles perpendicular to the surface. What I've shown is that the flow of momentum and the density of kinetic energy are closely related. There are similar results in other parts of physics, e.g. in electromagnetism the http://en.wikipedia.org/wiki/Poynting_vector" tells you both the flow of energy and the density of momentum.

Maybe my question becomes more clear when i re-phrase it:
Pressure is related to kinetic energy of the particles. In a normal sequence star, nuclear fusion provides this kinetic energy, leading to a pressure that opposes gravitational collapse. When fusion runs out, the gas loses kinetic energy and with it the pressure needed to retain size. The star collapses until degeneracy pressure equals the gravitational pressure. Where does the energy needed for this pressure come from?

The original main sequence star had kinetic and gravitational potential energy. When the core collapses to form a white dwarf, some of the energy is radiated but some of it remains in the white dwarf. This is where the kinetic energy of the particles in a white dwarf 'comes from'. The energy then remains in the white dwarf forever - there is no need for a source of energy. (Obviously since white dwarfs are 'white', they are still radiating energy. This is surplus energy, on top of the minimum energy mandated by the exclusion principle. Eventually this surplus energy is exhausted, and the white dwarf becomes a http://en.wikipedia.org/wiki/Black_dwarf" )

 

[LennoxLewis]

The flow of kinetic energy would be
N\left\langle \frac{1}{2}mv^2 \mathbf{v}\cdot d\vect{A} \right\rangle
(though for an isotropic velocity distribution this just evaluates to zero). To calculate the flow of a quantity through a surface, you have to take into account that the faster particles are moving perpendicular to the surface, the more particles pass through per unit time, i.e. you multiply the quantity you're calculating the flow for by the velocity of the particles perpendicular to the surface. What I've shown is that the flow of momentum and the density of kinetic energy are closely related. There are similar results in other parts of physics, e.g. in electromagnetism the http://en.wikipedia.org/wiki/Poynting_vector" tells you both the flow of energy and the density of momentum.

Fair enough. When seeing the entire derivation, it would probably make more sense, but when i see a factor proportional to v squared, my instinct tells me it's related to energy, regardless of whether it's per unit time and/or multiplied by a unit vector.

The original main sequence star had kinetic and gravitational potential energy. When the core collapses to form a white dwarf, some of the energy is radiated but some of it remains in the white dwarf. This is where the kinetic energy of the particles in a white dwarf 'comes from'. The energy then remains in the white dwarf forever - there is no need for a source of energy. (Obviously since white dwarfs are 'white', they are still radiating energy. This is surplus energy, on top of the minimum energy mandated by the exclusion principle. Eventually this surplus energy is exhausted, and the white dwarf becomes a http://en.wikipedia.org/wiki/Black_dwarf" )

I don't understand why there is no need for a energy source to provide degeneracy pressure. It can withstand a gigantic amount of gravitational force, for eternity... I understand that the gas obtains some, or even a lot of kinetic energy upon the collapse to a white dwarf, but the merely provides a temporary, "normal" pressure that clearly doesn't stop the collapse. It's the pressure from the Pauli Exclusion principle that holds a stop to the collapse, and this has little to do with the kinetic energy of the gas.

In fact, the formula for degeneracy pressure:

Does not contain a temperature term, or even an energy term! It is dependent only on the density and the ratio of electrons to neutrons.

Quoting Wikipedia, about the formula i wrote above:
"This degeneracy pressure is omnipresent and is in addition to the normal gas pressure P = nkT / V."

Continueing, on the Heisenberg Uncertainty principle:
"A material subjected to ever increasing pressure will become ever more compressed, and for electrons within it, the uncertainty in position measurements, Δx, becomes ever smaller. Thus, as dictated by the uncertainty principle, the uncertainty in the momenta of the electrons, Δp, becomes larger. Thus, no matter how low the temperature drops, the electrons must be traveling at this "Heisenberg speed," contributing to the pressure. "

So, where does this gain in momentum come from? What happened to conservation of momentum? Or energy??

 

[cortiver]

I don't understand why there is no need for a energy source to provide degeneracy pressure.

I don't understand why you think there is a need for an energy source... you realize that the degeneracy pressure isn't doing any work? The energy of the white dwarf star (apart from the radiation of surplus energy which I mentioned before) is constant. There is no source of new energy.

I understand that the gas obtains some, or even a lot of kinetic energy upon the collapse to a white dwarf, but the merely provides a temporary, "normal" pressure that clearly doesn't stop the collapse. It's the pressure from the Pauli Exclusion principle that holds a stop to the collapse, and this has little to do with the kinetic energy of the gas.

No, the degeneracy pressure has everything to do with the kinetic energy of the gas. The point of my derivation above was that both degeneracy pressure and thermal pressure come from the kinetic energy, via the relation P = \frac{2}{3}u. The only difference between the two is that thermal pressure is the pressure you get in a classical gas, whereas degeneracy pressure is the pressure you get in a extremely degenerate (i.e. very high density or very low temperature) gas. Because classical and degenerate gas have different thermodynamic properties, the equation of state for the pressure is different in the two cases. In particular, the pressure in a high-density gas is much higher than the pressure you would get if classical mechanics still worked at such high densities.

Does not contain a temperature term, or even an energy term! It is dependent only on the density and the ratio of electrons to neutrons.

Well, the equation P = \frac{2}{3}u still holds. The kinetic energy density of a degenerate gas is itself dependent only on the density and the ratio of electrons and neutrons (because the kinetic energy of a degenerate gas is determined by the exclusion principle - all the lowest possible energy states are filled with one particle each).

Quoting Wikipedia, about the formula i wrote above:
"This degeneracy pressure is omnipresent and is in addition to the normal gas pressure P = nkT / V."

I think this is not really true. If T \neq 0, then the pressure will not be exactly given by the degeneracy pressure formula you quoted above (because that is derived by setting T = 0 in the Fermi-Dirac distribution), but it won't simply be the sum of the P = NkT pressure and the degeneracy pressure - it's more complicated than that.

Continueing, on the Heisenberg Uncertainty principle:
"A material subjected to ever increasing pressure will become ever more compressed, and for electrons within it, the uncertainty in position measurements, Δx, becomes ever smaller. Thus, as dictated by the uncertainty principle, the uncertainty in the momenta of the electrons, Δp, becomes larger. Thus, no matter how low the temperature drops, the electrons must be traveling at this "Heisenberg speed," contributing to the pressure. "

It says the uncertainty in the momentum becomes larger. The momentum itself is conserved. Anyway this explanation is not very good. The fact that the particles still have kinetic energy even at absolute zero is due to the exclusion principle, and can't be satisfactorily explained by the uncertainty principle alone, which applies equally to fermions and bosons (the latter of which don't obey the exclusion principle).

 

[DrChinese]

1. How do you justify it to students? Here's a daring strategy - try actually telling them about it and let them make their own minds up.

2. Scientific points...
(1) Bell's theorem(!).

1. That is not what I meant. I meant: the specific hypothesis that the positions of remote particles completely determines all we see is an ad hoc hypothesis. It could be other attributes, and there could be different (amounts of) influences for different remote particles. And this is clouded by the additional hypothesis that the effect is not observable (QEH). Again, I am not arguing the merits of BM - of which I believe there are many - just the fact that it is additional to standard accepted science. That is the criteria for this board, and hypotheses related to interpretations should be properly labeled as such. If you think BM is the best, great, say so, but don't advance it to others in such a way that they think it is generally accepted.

2. Bell's Theorem was not developed as a result of Bohmian Mechanics. Bell obviously knew of it, but there is nothing in his proof that relates to any element of BM. Further, Bell's work around that time clearly was focused on a variety of mathematical proofs associated with the realistic side of the EPR paradox (such as K-S). There has been a lot of revisionist history associated with BM in recent years, and I think that really speaks poorly for the position in general.

 

[bcrowell]

Now if you have two fermions two the left and right of a common center and then do a 180 degree rotation on both of them about the common center they will have swapped places. So the state is no longer \left | ab \right \rangle, but \left | ba \right \rangle. However since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation, hence the new state is the old one with a minus sign.

It's very alluring to have such a conceptually straightforward sketch of the spin-statistics theorem. I don't follow you, though, when you say, "since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation." If we think of the case of a vector field, which can be treated purely in terms of three-dimensional geometry, a rotation by 180 degrees causes a sign flip, but a 90-degree rotation can't simply be treated as a change of sign or phase. So I don't see why it's valid to cut a 360-degree rotation in half to get a 180-degree one, and then assume that the 180-degree one can just be treated as a change of phase.

 

[DarMM]

It's very alluring to have such a conceptually straightforward sketch of the spin-statistics theorem. I don't follow you, though, when you say, "since you have performed two 180 degree rotations, one on each particle, this has the same effect as a single 360 degree rotation." If we think of the case of a vector field, which can be treated purely in terms of three-dimensional geometry, a rotation by 180 degrees causes a sign flip, but a 90-degree rotation can't simply be treated as a change of sign or phase. So I don't see why it's valid to cut a 360-degree rotation in half to get a 180-degree one, and then assume that the 180-degree one can just be treated as a change of phase.

It's not valid. The full proof of the spin-statistics theorem requires heavy mathematical machinery. My post above is meant as a rough guide. The full proof is given in the monograph of Streater and Wightman "PCT, Spin and Statistics and all that". There are other proofs involving other approaches to QFT, but all are equally difficult.

Remember the Spin and Statistics theorem is a theorem of QFT, it does not follow from QM. So any proof will involve the local and causal properties of QFT.

 

[LennoxLewis]

I don't understand why you think there is a need for an energy source... you realize that the degeneracy pressure isn't doing any work? The energy of the white dwarf star (apart from the radiation of surplus energy which I mentioned before) is constant. There is no source of new energy.


No, the degeneracy pressure has everything to do with the kinetic energy of the gas. The point of my derivation above was that both degeneracy pressure and thermal pressure come from the kinetic energy, via the relation P = \frac{2}{3}u. The only difference between the two is that thermal pressure is the pressure you get in a classical gas, whereas degeneracy pressure is the pressure you get in a extremely degenerate (i.e. very high density or very low temperature) gas. Because classical and degenerate gas have different thermodynamic properties, the equation of state for the pressure is different in the two cases. In particular, the pressure in a high-density gas is much higher than the pressure you would get if classical mechanics still worked at such high densities.


Well, the equation P = \frac{2}{3}u still holds. The kinetic energy density of a degenerate gas is itself dependent only on the density and the ratio of electrons and neutrons (because the kinetic energy of a degenerate gas is determined by the exclusion principle - all the lowest possible energy states are filled with one particle each).


I think this is not really true. If T \neq 0, then the pressure will not be exactly given by the degeneracy pressure formula you quoted above (because that is derived by setting T = 0 in the Fermi-Dirac distribution), but it won't simply be the sum of the P = NkT pressure and the degeneracy pressure - it's more complicated than that.


It says the uncertainty in the momentum becomes larger. The momentum itself is conserved. Anyway this explanation is not very good. The fact that the particles still have kinetic energy even at absolute zero is due to the exclusion principle, and can't be satisfactorily explained by the uncertainty principle alone, which applies equally to fermions and bosons (the latter of which don't obey the exclusion principle).

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?

 

[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.