I Calculating the number of energy states using momentum space

[BvU]

the derivative $\frac{dN}{dp}$ actually shows that the number of quantum states per $dp$ increases exponentially as $p$ gets larger. How can an exponentially increasing number of quantum states per $dp$ fit into an n-shell with decreasing thickness per $dp$

That is not exponentially but quadratically. And the 'delta-thickness' is constant.

 

[JohnnyGui]

That is not exponentially but quadratically. And the 'delta-thickness' is constant.

Apologies, I indeed meant quadratically the whole time. And I expected the thickness should be constant but there's a problem, please see below.

You know how to differentiate $$R = \frac{2Lp}{h}\Rightarrow dR = \frac{ 2L}{h} \;dp$$

This differentiation is consistent with the formula that I wrote in my previous post:
$$dR = \frac{(p+dp) \cdot 2L}{h} - \frac{p \cdot 2L}{h} = \frac{2L}{h} dp$$
This indeed shows a fixed thickness of the n-shells. But when I simply rewrite this equation in terms of the corresponding n-sphere radii...
$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$
...then it shows that the thickness decreases with higher $p$ values.
What is exactly wrong with this rewrite of the formula? Doesn't the n-sphere's radius get increased with a factor of $\frac{p+dp}{p}$ every time an n-shell gets added to it? You can see that by $\frac{p+dp}{p} \cdot \frac{2Lp}{h} = \frac{(p+dp)2L}{h}$

 

[BvU]

I supose you meant to place brackets around $$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$like this$$dR=\frac{p+dp}{p}\cdot\left ( n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} \right ) $$which means $$
dR=\frac{p+dp}{p}\cdot 0 \quad ?$$In short: you forgot to work out $n(|p+dp|)$ for the $n_i$ in the first term.

 

[JohnnyGui]

I supose you meant to place brackets around $$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$like this$$dR=\frac{p+dp}{p}\cdot\left ( n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} \right ) $$which means $$
dR=\frac{p+dp}{p}\cdot 0 \quad ?$$In short: you forgot to work out $n(|p+dp|)$ for the $n_i$ in the first term.

No, that's not how I meant it because only the first term $(n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$ is a factor $\frac{p+dp}{p}$ larger with respect to the second one, in order to get the difference, i.e. the thickness. So the thickness is still decreasing when $p$ increases. Besides, putting them both in brackets would give a thickness of $dR =0$ which is incorrect, right?

 

[BvU]

$n$ depends on $p$

 

[JohnnyGui]

$n$ depends on $p$

So according to that, although the following is correct:
$$\frac{2Lp}{h} = (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$
This still means that the following is incorrect?
$$\frac{p + dp}{p} \cdot \frac{2Lp}{h} =\frac{p + dp}{p} \cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$

 

[BvU]

No, that's correct .

 

[BvU]

But when I simply rewrite this equation in terms of the corresponding n-sphere radii...
$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$
...then it shows that the thickness decreases with higher $p$ values.

No it does not:$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} = \frac{dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$and in #66 your first equation shows that this is equal to $$dR=\frac{2L}{h} dp$$I repeat: bottom line of #60. There's much more interesting stuff ahead.

 

[JohnnyGui]

No it does not:$$dR=\frac{p+dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}- (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}} = \frac{dp}{p}\cdot (n_x^2 + n_y^2+n_z^2)^{\frac{1}{2}}$$and in #66 your first equation shows that this is equal to $$dR=\frac{2L}{h} dp$$I repeat: bottom line of #60. There's much more interesting stuff ahead.

Sorry for the late reply. I finally got it and can't believe I was actually missing something so obvious. I kept considering the $(n^2_x + n_y^2 + n_z^2)$ parameter to be constant no matter what the value of $p$ is.

I am now combining the states density with the Botlzmann statistics to understand the Maxwell-Boltzmann distribution. Sorry if this is a bit off-topic but one thing that bothers me is the following. For the derivation it is assumed that the collisions between particles are perfectly elastic and that the system is in thermal equilibrium. Furthermore, the particles are of 1 gas and thus have same mass.

But if this is the case, how is it assumed that particles in a container can have different kinetic energies? What other factors than elasticity, mass and temperature can change the kinetic energy of a colliding particle?

 

[BvU]

What makes you think they should all have the same kinetic energy ?

What other factors than elasticity, mass and temperature can change the kinetic energy of a colliding particle?

The collisions themselves !

 

[JohnnyGui]

What makes you think they should all have the same kinetic energy ?
The collisions themselves !

I thought that perfectly elastic collisions among identical particles, which is assumed for the derivation, would keep a particle's kinetic energy more or less constant. Please elaborate if this is incorrect.

 

[BvU]

Please elaborate if this is incorrect

Very incorrect !
Experiment with sliding coins over a smooth table

 

[JohnnyGui]

Very incorrect !
Experiment with sliding coins over a smooth table

I think it depends on the starting scenario of the system with a certain equilibrium temperature. If each particle has the same kinetic energy initially at the very start, then I can't conclude other than the kinetic energy of each particle staying constant because of perfectly elastic collisions. If however, if at the very start, each particle differ in kinetic energy (temperature has still yet to reach equilibrium) then I would understand why particles can have different kinetic energies in the system, even in the presence of perfectly elastic collisions.

 

[BvU]

DId you try the coins ? Did the kinetic energy of each and every coin remain constant ?
Did you ever have to do an exercise with hard ball elastic collisions ? What is conserved ?

 

[PeterDonis]

If each particle has the same kinetic energy initially at the very start

Which they won't. A given equilibrium temperature only means the average kinetic energy of the particles is a certain value. It does not mean that every single particle has that kinetic energy.

I think you need to read the article on the kinetic theory of gases more carefully.

 

[JohnnyGui]

Which they won't. A given equilibrium temperature only means the average kinetic energy of the particles is a certain value. It does not mean that every single particle has that kinetic energy.

Two questions arises from this.

1. So if each particle does have the same kinetic energy initiallly at the very start, is it correct that each particle's kinetic energy stays constant after perfect elastic collisions?

2. The reason that they don't have the same kinetic energy at the very start is because the final equilibrium temperature is yet to be reached?

 

[PeterDonis]

if each particle does have the same kinetic energy initiallly at the very start

This is much, much too improbable to have any chance of being observed. Remember we're talking about something like $10^{23}$ particles in a typical container of gas.

is it correct that each particle's kinetic energy stays constant after perfect elastic collisions?

In the center of mass frame of the collision, yes, this will be true. But kinetic energy is frame-dependent, so it will not, in general, be true in the rest frame of the gas as a whole.

The reason that they don't have the same kinetic energy at the very start is because the final equilibrium temperature is yet to be reached?

No. Go read my post #75 again, carefully.

 

[JohnnyGui]

No. Go read my post #75 again, carefully.

I did, but I don't see how this post answers my question. It states that a characteristic of an equilibrium temperature is having an average kinetic energy and not every particle having that same kinetic energy. This is clear to me.

My question is more directed towards why particles don't have the same kinetic energy at the very start even if perfect elastic collisions are considered. I have a hard time grasping "rest frame of the gas as a whole" because a gas consists of particles going in different directions and thus each particle having its own rest frame.

 

[BvU]

Experiment with sliding coins over a smooth table

 

[JohnnyGui]

Experiment with sliding coins over a smooth table

My posted conclusion and question is deduced from this experiment. I have difficulty choosing the starting scenario; in the case of 2 coins, should I make 2 coins have the same velocity before collision or should one stay still? If it's the latter case, then I would conclude that the reason that particles don't have the same kinetic energy at equilibrium temperature is because particles had different kinetic energies before that equilibrium temperature was reached.

 

[BvU]

Either. Only precisely head-on collisions of equal coins with equal but opposite velocities conserve the kinetic energies of both coins. Chance of one in very, very many.

 

[PeterDonis]

It states that a characteristic of an equilibrium temperature is having an average kinetic energy and not every particle having that same kinetic energy. This is clear to me.

Ok, good.

My question is more directed towards why particles don't have the same kinetic energy at the very start even if perfect elastic collisions are considered.

Because elastic collisions conserve the total kinetic energy of the two colliding particles. They don't conserve the kinetic energies of the two particles individually except in the very rare case where the combined momentum of the two particles is zero.

I have a hard time grasping "rest frame of the gas as a whole" because a gas consists of particles going in different directions and thus each particle having its own rest frame.

You're confused about frames. I can pick any frame I like to analyze the situation; there is no need to use a different frame for every particle just because each particle has a different velocity. The rest frame of the gas as a whole is the frame in which the center of mass of the gas as a whole is at rest. When we talk about the temperature of a gas being the average kinetic energy of its particles, we mean the average kinetic energy in that frame, the frame in which the center of mass of the gas is at rest. And in that frame, virtually all collisions will change the kinetic energies of both particles.

 

[JohnnyGui]

@PeterDonis : Thank you for the clear explanation. I think I understand it now.

Either. Only precisely head-on collisions of equal coins with equal but opposite velocities conserve the kinetic energies of both coins. Chance of one in very, very many.

Ah, this explains it for me. I was not aware of this.

So, the number of states $n_s(p)$ for a particular momentum $p$ is given by:
$$n_s(p) =\frac{2\pi p^2 \cdot L^2}{h^2}$$
I have read about Boltzmann’s and Maxwell’s derivations for the number of particles with a particular momentum if the allowed momentums are discrete. If the allowed momentums are very closely packed together, is also correct to deduce that the number of particles $n_i$ having a particular momentum of $p_i$ to be:
$$n_i = F(p_i) \cdot n_s(p_i) = F(p_i) \cdot \frac{2\pi p_i^2 \cdot L^2}{h^2}$$
Where $F(p_i)$ is the number of particles at a particular momentum $p_i$ but per 1 microstate.
I am aware it is usually written in the form of a State Density, but I was wondering if this approach is also correct.

 

[learn.steadfast]

You'll have a hard time finding solutions for the Schroedinger equation in this funny case !

I don't see why that would be difficult in many cases.
Calculating the number of energy states using momentum space

First, solve the Schrodinger equation for a box of Lx, Ly2 ; and record the constants for wavelength 'k' in x, and y; eg: record k for the lowest state of n in each direction.

So long as the differences in length, Lx - Lx2, Ly-Ly2, are multiples of the recorded wavelength ( for each respective axis ) Then I think the same wavelength must correctly solve the extended box in each axis.

The reason is simple, sine wave solutions for standing waves are zero at the walls; and happen to be zero at points where the walls "might" have existed if the box was reduced to dimensions Lx by Ly2.

Therefore, I'm sure any infinite well/rigid box can be extended by an integer multiple of wavelengths at points where the sine-waves are naturally zero, without making solution to the Schrodinger equation impossible.

The notation of 'n', can be confounded by the different lengths of the box ... but the ability to solve the Schrodinger equation is not made impossible because traditional notation can be confounded.

To the contrary! In this container (i.e., the one with rigid boundary conditions) the position is well defined as a self-adjoint operator, but momentum is not. There are thus also no momentum eigenstates.

Vanshees, I pointed out in another thread that your proof appears to depend on over-specifiying the number of boundary conditions to compute the domain of a function. I suspect your complaint is probably a mathematical fiction caused by over-specifying the boundary conditions ??

When we only require that the value at the wall be the same as the opposing wall, psi(+a,y)=psi(-a,y) we have already given enough boundary conditions to determine that the momentum is a self adjoint operator for the specified axis. The boundary condition can be repeated for each axis, showing each one to be independently self-adjoint. That is to say, when we only *require* that the wave function be periodic, and not that it is also zero; we get a bigger domain than if we try to restrict the wave function to having a specific value at a periodic boundary. When we solve the *general* case for *any* value at the periodic boundary (the wall is one such boundary), the proof will come out with psi being self adjoint. But the proof will fail if we try to specify a particular value at the wall (even if we *know* what it should be.)

Again, By analogy --- > we *know* that in any test of Young's double slit experiment ... that if we try to specify mathematically that the particle must have a "probably" of *zero* to be found in one of the slits, we would destroy the solution to the interference pattern that is the well known result of the experiment. eg: You can put in mathematical boundary conditions that you are *sure* are true (when tested), that will destroy the ability of the Schrodinger equation to produce results consistent with experiment.

My understanding of the idea of self-adjointness, is essentially to prove the imaginary part of psi is canceled out when computing expectation values.

Operators work on psi by multiplication after differentiation; and self adjointness is required for the final product(s) to sum up to a purely real expectation value.

If only a single point's product (somewhere on psi) is computed, the idea of self adjointness is demonstrated when given real constants a,b that the complex product on the left side of this next equation is always real:

eg: ( a - ib )^{1/2}* ( a + ib )^{1/2}= ( a^2 + b^2 )^{1/2}

I've chosen to represent psi as the square root of a complex number, because in some sense psi is the momentum of the particle; and it's square is the kinetic energy in classical physics. p^2 / 2m = T

For self adjointness of functions, I am not required that the result of the multiplication be purely real at every point; but only that the *sum* (or integral) of the results cancel out the imaginary portion. However, the condition of self adjointness is trivially met when b=0, everywhere.

Since I can give a time invariant solution to Schrodinger's that has a psi that is purely *real* (b=0), in the case of an infinite well box; Where exactly does your claim of failure to be self adjoint come from?

If I naively compute the momentum operator on an infinite well and get an integral of a product that has a purely real result when evaluated; Why should I believe that self-adjointness is not true? eg: As opposed to believing you've over-specified a problem, and thereby made it insoluble by a mathematical proof that is perhaps flawed in cases having more boundary condition than there are unknowns that *must* be solved for?

To solve for N unknowns, in linear equations; I only need N independent equations. If I put in N+1 equations, depending on the textbook ... the proofs for an algorithm solving a linear set of equations may or may not be valid. We need to know the chain of reasoning used in the proofs whenever working with more equations than we have unknowns to solve for, in order to know the proof is valid.

 

[JohnnyGui]

I have a question about calculating the number of particles at a particular energy level using Boltzmann Statistics in case of discrete energy levels.

For the number of particles $n_i$ at a particular discrete energy level $E_i$, I understand that according to Boltzmann this is given by:
$$n_i = \frac{N}{\sum_{i=0}^\infty e^{\frac{-E_i}{k_B}}} \cdot e^{\frac{-E_i}{k_B}}$$
My question is, does this formula take into account the number of possible quantum states at that particular energy level $E_i$ or does it only give the number of particles for just 1 quantums state at that energy level?

 

[PeterDonis]

I have a question about calculating the number of particles at a particular energy state using Boltzmann Statistics

Boltzmann statistics are classical, not quantum.

does this formula take into account the number of possible quantum states at that particular energy state $E_i$?

No; it can't, because, as above, Boltzmann statistics are classical, not quantum.

 

[JohnnyGui]

Boltzmann statistics are classical, not quantum.
No; it can't, because, as above, Boltzmann statistics are classical, not quantum.

Does this mean that the mentioned formula for $n_i$ can be multiplied by the number of quantum states at that energy level in order to get the "true" number of particles at that energy level?

 

[PeterDonis]

Does this mean that the mentioned formula for $n_i$ can be multiplied by the number of quantum states at that energy level in order to get the "true" number of particles at that energy level?

No. Apparently you didn't grasp what "Boltzmann statistics are classical, not quantum" means. Not only that, but $n_i$ is, by definition, the number of particles with energy $E_i$, as you yourself said in your previous post, so I have no idea why you would think you can get a "true" number of particles by multiplying it by something else.

 

[JohnnyGui]

No. Apparently you didn't grasp what "Boltzmann statistics are classical, not quantum" means. Not only that, but $n_i$ is, by definition, the number of particles with energy $E_i$, as you yourself said in your previous post, so I have no idea why you would think you can get a "true" number of particles by multiplying it by something else.

Because you said it can't take into account the number of quantum states at a particular energy level, letting me think that the classical approach would give an erroneous number of particles in the case of a quantum approach for which it should be corrected somehow. Furthermore, the Boltzmann factor is combined with the number of quantum states to derive a formula when energylevels are considered continuous, making me think that perhaps $n_i(E_i)$ should be corrected that way.

This video shows that (part) of the Boltzmann formula is multiplied by the number of states at a particular energylevel $\rho(\epsilon)$ (the $\rho(\epsilon)$ is discussed in his previous video).

 

[PeterDonis]

Because you said it can't take into account the number of quantum states at a particur energy level

Can you give a specific quote? It's been a while.

letting me think that the classical approach would give an erroneous number of particles

If by "erroneous" you mean "different than the number that quantum statistics would give", of course it does. That's why we don't use Boltzmann statistics when the difference between them and the correct quantum statistics is important.

for which it should be corrected somehow

You don't "correct" Boltzmann statistics if you want correct answers when quantum effects are important. You just use the correct quantum statistics instead.

the Boltzmann factor is combined with the number of quantum states to derive a formula when energylevels are considered continuous

Can you give a reference? (Preferably a written one, not a video; it takes a lot more time to extract the relevant information from a video than it does from a written article or paper.)