Calculating the number of energy states using momentum space

[PeterDonis]

From what I understand, an integral gives a continuous function as a function of energy if the derivative of a cumulative function is integrated.

You're thinking of it backwards. You can integrate any function you like. Once you've done the integral, you can consider the thing you integrated as a "cumulative function" as it relates to the thing you get as a result of the integral. But the process of evaluating the integral doesn't care about any of that and does not depend on it.

This is indeed done for the number of states; the derivative of the volume of a sphere in energy-space is within the integral; ##g(\epsilon)##.

##g(\epsilon)## isn't the derivative of the volume of a sphere in energy space. It's the number of states per unit volume in energy space.

Also, there's only one integral being done, so if you want to consider the function of ##\epsilon## inside the integral as the derivative of the function you get by evaluating the integral, that's fine, but it's the entire integrand that's the derivative of the result of the integral; you can't split it up into pieces.

 

[JohnnyGui]

And none of this has anything to do with the continuous vs. discrete question.

I haven't said it has something to do with the continuous vs discrete question. It's a side note question about the formula to understand the formulation better.

No, I didn't. I denied a different statement, which is not part of what we are currently talking about.

Ok. It wasn't distinguishable whether you also denied that statement or not since that statement was quoted in post ##95## after already saying that the Boltzmann formula is classical about an earlier post of mine.

The link you give doesn't derive Boltzmann's formula for discrete energy levels (equation 12) at all. It just assumes it.

I know, I was talking about a text interpretation about the formula, just like when I found the interpretation in the piece of text about equation 35. Another video of the lecture showed the derivation for discrete energy levels as well but did not tell that interpretation either.

 

[JohnnyGui]

g(ϵ)g(ϵ)g(\epsilon) isn't the derivative of the volume of a sphere in energy space. It's the number of states per unit volume in energy space.

Apologies, I typed it without paying attention. This is indeed what I meant.

Also, there's only one integral being done, so if you want to consider the function of ϵϵ\epsilon inside the integral as the derivative of the function you get by evaluating the integral, that's fine, but it's the entire integrand that's the derivative of the result of the integral; you can't split it up into pieces.

I think this is indeed what I was misunderstanding.

 

[JohnnyGui]

I have found another source that takes the degeneracy (e.g. the number of quantums states of an energy level) into account for the derivation of the Boltzmann statistics formula. I have found some inconsistency with other sources at the step where Langrag's Constants (##\alpha## and ##\beta \cdot \epsilon_j##) are applied to solve the equation for 0.

The time stamp in this video derived the following equation (when degeneracy is not taken into account):
$$ln(n_j) + \alpha + \beta \epsilon_j = 0 → n_j = e^{-\alpha} \cdot e^{-\beta \epsilon_j}$$
Notice that Langrang's Constants (##\alpha## and ##\beta\epsilon_j##) are being added to ##ln(n_j)## and then solved for 0.
The link that takes the degeneracy into account somehow shows on sheet ##14## and ##15## that Langrang's constant ##\beta\epsilon_j## should be substracted, which gives:
$$ln(g_j) - ln(n_j) + \alpha - \beta \cdot \epsilon_j = 0 → n_j = g_j \cdot e^\alpha \cdot e^{-\beta \epsilon_j}$$
Where ##g_j## is the number of degeneracy that energy level ##\epsilon_j## has.

I'm not sure why Langrang's Constant ##\beta \cdot \epsilon_j## should be substracted when degeneracy is taken into account. I would assume it could also be added since the solution should be zero nonetheless. And yet, even if it is added instead of substracted, the end equation is still different from the equation from the video in which degeneracy is not taken into account.
I'd expect it should be the same because when energy is considered continuous afterwards, the very same formula ##e^{-\alpha-\beta\epsilon_j}## of the video is multiplied by the number of states, which is the analogue for taking degeneracy into account.

Not sure what I'm missing here.

 

[PeterDonis]

I'm not sure why Langrang's Constant ##\beta \cdot \epsilon_j## should be substracted when degeneracy is taken into account.

First, it's not the sign of ##\beta## that's being changed, it's the sign of ##\alpha##. Rewrite the first equation as

$$
- ln (n_j) - \alpha - \beta \cdot \varepsilon_j = 0
$$

Now it's the same as the second except that the ##ln (g_j)## term is absent (because the first source doesn't consider degeneracy, which is where that term comes from--if ##g_j = 1##, no degeneracy, then ##ln(g_j) = 0##) and the sign of ##\alpha## is changed. If you go back into how the first formula is derived, you will see that the sign gets flipped during the derivation (an equation with minus something = 0 is changed to just something = 0). The second source simply doesn't do that sign flip.

Second, the choice of the sign of ##\alpha## has nothing to do with degeneracy. It's just an arbitrary choice of signs. The two sources are just making different arbitrary choices. (The first source doesn't go on to discuss the link between ##\alpha## and the chemical potential/Fermi energy; if it did, the sign flip would just end up getting to the version of the Maxwell-Boltzmann distribution on slide 21 of the second source in one less step--the step on slide 21 where the sign of the argument of the exponential gets flipped would not be needed.)

 

[JohnnyGui]

It's just an arbitrary choice of signs. The two sources are just making different arbitrary choices.

I might have missed your point. But don't these arbitrary choices of the signs eventually lead to a different formulation for ##n_j##?

Here's why:

If you go back into how the first formula is derived, you will see that the sign gets flipped during the derivation (an equation with minus something = 0 is changed to just something = 0)

If you're referring to exactly this timestamp of the video, then I noticed that if the minus sign before that summation is kept, this would lead to the formula ##n_j = e^{\alpha} \cdot e^{\beta \epsilon_j}## whereas, when it's removed (just like the lecturer did), this would lead to ##n_j = e^{-\alpha} \cdot e^{-\beta \epsilon_j}##. Don't these differences lead to a different result?
Unless addition or substraction of Langrang's Constants depends on whether or not you have kept the minus sign before, I'm still confused.

Furthermore, the fact that Sheet 21 of the second source says that the minus sign is added based on "intuition", makes me think that they have not applied the signs correctly before, because the formula of the video doesn't need that "intuition" step if the lecturer went on discussing the chemical potential/Fermi energy.

 

[PeterDonis]

But don't these arbitrary choices of the signs eventually lead to a different formulation for ##n_j##?

In terms of ##\alpha## and ##\beta##, yes. But those choices of sign will just change the relationship between ##\alpha## and ##\beta## and the chemical potential (or Fermi energy) and temperature (by flipping the signs there). The final formulas in terms of chemical potential/Fermi energy and temperature, which are what actually have physical meaning, will be the same either way.

the fact that Sheet 21 of the second source says that the minus sign is added based on "intuition"

No, that's not what it says. No minus sign is "added in". The two formulas at the top of that slide are equivalent to each other; they're just expressed in slightly different algebraic form.

The note about "It's intuitive" just means that the formula, now that it's written in that form, matches what you would intuitively expect to be the case. It doesn't mean intuition had to be used to obtain the formula.

 

[JohnnyGui]

The final formulas in terms of chemical potential/Fermi energy and temperature, which are what actually have physical meaning, will be the same either way.

I can't seem to reproduce the same final formula. Here's why:

But those choices of sign will just change the relationship between αα\alpha and ββ\beta and the chemical potential (or Fermi energy) and temperature (by flipping the signs there)

Do you mean that the different signs of ##\alpha## and ##\beta## are "compensated" by flipping the signs of the Fermi energy and temperature? If so, then I can't see that's being done in the source . Sheet 20 still says that ##\alpha = \frac{E_F}{kT}## and ##\beta = \frac{1}{kT}## and according to their formula, the number of particles per quantum state ##n_j## (formula divided by ##g_j##) is ##n_j = e^{\alpha - \beta \epsilon_j}##, which would give as Sheet ##20## says:
$$n_j= e^{\frac{E_F-E_j}{kT}}= e^{\frac{-(E_j-E_F)}{kT}}$$
Flipping the formula from the video like you said in your post #110 would give ##n_j = e^{-(\alpha + \beta E_j)}##. Writing the terms ##\alpha## and ##\beta## out like above would give:
$$n_j = e^{\frac{-(E_F + E_j)}{kT}}=e^{\frac{-E_F - E_j}{kT}}$$
The sign of ##E_F## is still different and therefore those two formulas don't give the same number of particles per quantum state at a particular ##E_F## and ##E_j##. What am I still missing here?

 

[PeterDonis]

Do you mean that the different signs of ##\alpha## and ##\beta## are "compensated" by flipping the signs of the Fermi energy and temperature?

In the slides, you don't flip any signs.

In the video, you would flip just one sign, not both; as noted in post #110, that would be the sign of ##\alpha##. Or, if you don't want to flip the sign of ##\alpha## in order to keep the video formulas as they are, you would flip the sign of ##E_F## in the formula for that in terms of ##\alpha##.

 

[JohnnyGui]

In the slides, you don't flip any signs.

In the video, you would flip just one sign, not both; as noted in post #110, that would be the sign of ##\alpha##. Or, if you don't want to flip the sign of ##\alpha## in order to keep the video formulas as they are, you would flip the sign of ##E_F## in the formula for that in terms of ##\alpha##.

That part is now clear to me. But flipping the signs of ##\alpha## or ##E_F## of the formula in the video would give different outcomes for ##n_j## compared to when it is not flipped. What is the reason that the signs of either ##\alpha## or ##E_F## in the video should be flipped while this is how the lecturer derived it?

I would expect the derivation of the slides and the video to end up the being the same without the need of any sign flipping for which I can't find a reason.

 

[PeterDonis]

flipping the signs of ##\alpha## or ##E_F## of the formula in the video would give different outcomes for njn_j compared to when it is not flipped.

What do you mean by "different outcomes"? If you flip the one sign in the formulas in the video, you get the same formulas as are in the other source you linked to. What's the problem?

What is the reason that the signs of either ##\alpha## or ##E_F## in the video should be flipped while this is how the lecturer derived it?

Um, because you want to get the right answer? Physically, the final formula as it's given in the slides you linked to is obviously correct (and the slides explain why). So any derivation is going to have to end up with that formula.

I have no idea why the lecturer in the video chose to start with the sign choices he did. You'd have to ask the lecturer. Expecting all presentations to be entirely consistent in every choice of sign (not to mention lots of other arbitrary choices) is expecting far too much. As long as you end up with the correct answer, it doesn't matter how you get there.

 

[JohnnyGui]

What do you mean by "different outcomes"? If you flip the one sign in the formulas in the video, you get the same formulas as are in the other source you linked to. What's the problem?

The sheet's source is not the problem here. The problem is that when the sign of ##\alpha## or ##E_F## in the video formula is flipped (to make it the same as the sheet's formula from the other source), the formula is not the same anymore as the original video formula that the lecturer derived. It would give different outcomes for ##n_j## compared to before it was flipped.

If it DID give the same outcomes, then I'd have no problem and no questions.

 

[PeterDonis]

The problem is that when the sign of ##\alpha## or ##E_F## in the video formula is flipped (to make it the same as the sheet's formula from the other source), the formula is not the same anymore as the original video formula that the lecturer derived.

So what?

 

[JohnnyGui]

So what?

Not sure if you have read the rest of my previous post other than what you quoted. The lecturer derived it as:
$$n_j = e^{\frac{-E_F - E_j}{kT}}$$
And, just for the sake of making the formula the same as the other source, he'd have to change it to:
$$n_j= e^{\frac{-(E_j-E_F)}{kT}}$$
Clearly, one of these equations is incorrect here, because they give different outcomes. If the first equation is incorrect, what did the lecturer do wrong in his derivation?

 

[PeterDonis]

The lecturer

Which lecturer? The slides? Or the video? From what I saw in the video, he never got to any formula involving ##E_F## at all. He only gave formulas with ##\alpha## and ##\beta## in them, and never gave an equation for ##E_F## in terms of ##\alpha##.

 

[JohnnyGui]

Which lecturer? The slides? Or the video? From what I saw in the video, he never got to any formula involving ##E_F## at all. He only gave formulas with ##\alpha## and ##\beta## in them, and never gave an equation for ##E_F## in terms of ##\alpha##.

The lecturer from the video. It is correct that he didn't give an equation involving ##E_F##, but I substitued his derived ##\alpha## with ##\frac{E_F}{kT}## since that's what's sheet ##20## states from the slides source.

Perhaps another source that I found would help me, this source (a Powerpoint) derived the formula just as the video, ##n_j = e^{-\alpha - \beta E_j}## and says on sheet ##21## that ##-\alpha = \frac{E_F}{kT}## instead of ##\alpha = \frac{E_F}{kT}## such that it ends up with the same formula as the previous mentioned slides source.

I noticed that whether you derive the formula as ##(1)## ##n_j = e^{-\alpha - \beta E_j}## or ##(2)## ##n_j = e^{\alpha - \beta E_j}## depends on how you apply Lagrange's Constants during the derivation. And that if you have derived it as ##(1)## then you should somehow assume that ##-\alpha = \frac{E_F}{kT}##, whereas when you derived it as ##(2)##, then you should assume that ##\alpha = \frac{E_F}{kT}##. I just don't get how these assumptions about ##\alpha## are made based on how Lagrange Constants are applied.

 

[PeterDonis]

I substitued his derived ##\alpha## with ##\frac{E_F}{kT}## since that's what's sheet 2020 states from the slides source.

And that makes no sense, because the slides started with the opposite sign for ##\alpha##. So if you flip the sign of ##\alpha## in the formulas in the video, you have to also flip the sign of ##\alpha## in the formula for ##E_F## in terms of ##\alpha##. You can't take a piece of one derivation and a piece of another and put them together if they started with opposite sign choices.

Note that the slides don't derive that formula: they say it's in another section. If you look at the derivation of that formula, you should find that it requires the same choice of sign for ##\alpha## that the slides made, which is the opposite of the choice that the video made.

Perhaps another source that I found would help me

Yes as you can see, it requires the opposite sign for ##\alpha## in the formula for ##E_F## in terms of ##\alpha##.

I just don't get how these assumptions about ##\alpha## are made based on how Lagrange Constants are applied.

Huh? ##\alpha## is a Lagrange multiplier, and signs of Lagrange multipliers can always be chosen arbitrarily. That choice of sign isn't an "assumption". It's an arbitrary choice. You just have to make sure the choice is consistent throughout your entire derivation.

 

[PeterDonis]

if you have derived it as ##(1)## then you should somehow assume that ##-\alpha = \frac{E_F}{kT}##, whereas when you derived it as ##(2)##, then you should assume that ##\alpha = \frac{E_F}{kT}##.

There aren't two "assumptions". There is only one. The formula for ##\alpha## in terms of ##E_F## and ##kT## isn't "assumed", it's derived, and its derivation has to start with the same sign choice for ##\alpha## as a Lagrange multiplier as the other derivations that are being done, like the derivation of the formula for ##n_j## in terms of ##\alpha## and ##\beta##.

 

[JohnnyGui]

Huh? αα\alpha is a Lagrange multiplier.

Yes, I know that ##\alpha## is a Lagrange multiplier.

and signs of Lagrange multipliers can always be chosen arbitrarily

This is what I didn't know, so it cleared this out for me. Thanks.

I've got one last question. When degeneracy is taken into account in the case of discrete energylevels, the formula is written as:
$$n(\epsilon_j) = g_j \cdot \frac{N}{Z} \cdot e^{-\frac{\epsilon_j}{kT}}$$
Earlier, I have derived that the number of quantum states ##Q_j## for a certain energy ##\epsilon_j## is equal to:
$$Q_j = \frac{1}{2} \pi \cdot \frac{\epsilon_j \cdot 8mL^2}{h^2}$$
This is based on calculating the surface of an 8th a sphere surface in n-space.
Since this formula gives the number of quantum states for a specific energy ##\epsilon_j##, can ##g_j## be substituted by this formula such that:
$$n(\epsilon_j) = \frac{1}{2} \pi \cdot \frac{\epsilon_j \cdot 8mL^2}{h^2} \cdot \frac{N}{Z} \cdot e^{-\frac{\epsilon_j}{kT}}$$
?

 

[PeterDonis]

can ##g_j## be substituted by this formula

I haven't checked your formula for ##g_j##, but in general, yes, if you have a formula for ##g_j## you can substitute it into the formula for ##n##.

 

[JohnnyGui]

I haven't checked your formula for ##g_j##, but in general, yes, if you have a formula for ##g_j## you can substitute it into the formula for ##n##.

Thanks for verifying.

I do find it a bit peculiar that, when energy is considered discrete, the formula takes the form of ##n(\epsilon_j) = g_j \cdot e^{-\alpha - \epsilon_j\beta}## with ##g_j## being the formula in my previous post, but when energy is considered continuous, then ##g_j## has to be substituted by the density of states formula times ##d\epsilon_j## (previously shown), while the part ##e^{-\alpha - \epsilon_j\beta}## can still be kept for the continuous approach.

Why shouldn't the part ##e^{-\alpha - \epsilon_j\beta}## be substituted by a formula as a function of energy that gives the "density of particle numbers" times ##d\epsilon_j##, just like how it is done for the degeneracy?

 

[PeterDonis]

Why shouldn't the part ##^{-\alpha - \epsilon_j\beta}## be substituted by a formula as a function of energy

Because it already is a formula as a function of energy.

that gives the "density of particle numbers" times ##d\epsilon_j##,

The exponential factor is a function of energy that gives the density of particle numbers. In the case where there is no degeneracy, that's the only factor there is. In the case where there is degeneracy, there is an additional factor of ##g_j## because of the degeneracy. The ##d \epsilon_j## doesn't come from ##g_j## or the exponential; it comes from the fact that you're doing an integral instead of a discrete sum.

 

[JohnnyGui]

The exponential factor is a function of energy that gives the density of particle numbers.

Hold on, this makes me reason the other way round; if the exponential factor ##e^{-\alpha - \epsilon_j\beta}## does give the density of particle numbers, doesn't the summation of that same exponential factor ##e^{-\alpha - \epsilon_j\beta}## (when energy is considered discrete) mean that you're adding particle number densities instead of the real particle numbers in each energy level?

 

[PeterDonis]

if the exponential factor ##e^{-\alpha - \epsilon_j\beta}## does give the density of particle numbers, doesn't the summation of that same exponential factor ##e^{-\alpha - \epsilon_j\beta}## (when energy is considered discrete) mean that you're adding particle number densities instead of the real particle numbers in each energy level?

Yes, of course. The exponential of any negative number is less than 1, so it has to be a density, not a particle number.

 

[JohnnyGui]

Yes, of course. The exponential of any negative number is less than 1, so it has to be a density, not a particle number

And yet, for discrete energy levels ##e^{-\alpha - \epsilon_j\beta} = n_j## in which ##n_j## is the particle number at a certain energylevel, not a density? Sorry but I'm lost...

 

[PeterDonis]

for discrete energy levels ##e^{-\alpha - \epsilon_j\beta} = n_j## in which ##n_j## is the particle number at a certain energy level, not a density?

No, it's a density. More precisely, it's a fraction of particles in that energy level, or a statistical probability of a particle being in that energy level.

 

[JohnnyGui]

No, it's a density. More precisely, it's a fraction of particles in that energy level, or a statistical probability of a particle being in that energy level.

Huh? But in your post #99 you said that the fraction of particles at a paticular energy is ##n_j## divided by ##N##, not just ##n_j##. Here's a snippet of your quote regarding this:

In which ##e^{-\alpha} = \frac{N}{Z}##

 

[PeterDonis]

in your post #99 you said that the fraction of particles at a paticular energy is ##n_j## divided by ##N##, not just ##n_j##.

You're confusing yourself by using the symbol ##n## to refer to different things.

In what you quoted from me from post #99, ##n_j## obviously refers to the exponential of a negative number times ##N##. So that would be an actual number of particles. Dividing that by ##N## just leaves the exponential of a negative number, which, as I said, has to be less than ##1## so it can't be a particle number, it has to be a fraction or a probability.

But in post #126, you used ##n## to refer to just the exponential of a negative number, all by itself. So obviously that can't be a particle number, as above.

 

[JohnnyGui]

You're confusing yourself by using the symbol ##n## to refer to different things.

In what you quoted from me from post #99, ##n_j## obviously refers to the exponential of a negative number times ##N##. So that would be an actual number of particles. Dividing that by ##N## just leaves the exponential of a negative number, which, as I said, has to be less than ##1## so it can't be a particle number, it has to be a fraction or a probability.

But in post #126, you used ##n## to refer to just the exponential of a negative number, all by itself. So obviously that can't be a particle number, as above.

My intention in post #126 with writing it as ##n(\epsilon_j)## instead of ##n_j## is because energy is being considered continuous so that one can not dedicate a certain particle number ##n_j## to a certain discrete energy level ##\epsilon_j##.

My mentioned formula in post #126 for ##n(\epsilon_j) = g_j \cdot e^{-\alpha - \epsilon_j\beta}## is stating that this should give the particle number, because ##e^{-\alpha} = \frac{N}{Z}##. So ##n(\epsilon_j)## must be the particle number and not the density.
If ##e^{-\alpha}## is not ##\frac{N}{Z}## in the continuous approach, then that's a whole different story which I don't understand how it changed formula-wise.

Is ##e^{-\alpha}## in the continuous approach then just a value that has to be further derived and is assumed to give the density when multiplied with ##e^{-\beta \epsilon}##?

 

[PeterDonis]

My intention in post #126 with writing it as ##n(\epsilon_j)## instead of ##n_j## is because energy is being considered continuous so that one can not dedicate a certain particle number ##n_j## to a certain discrete energy level ϵj\epsilon_j.

You're confusing yourself with this too. The difference between continuous and discrete is irrelevant to the issue that you're struggling with.

##e^{-\alpha} = \frac{N}{Z}##.

Ah, so now you're confusing ##e^{- \alpha}## with ##e^{- \alpha - \beta E_j}##. And I see that you've been mixing those together for the last few posts. And you have managed to confuse me as well in the process.

I'm not going to try an disentangle all of that. Instead, let's start with the final distribution function, as given in the slides (from the summary, the next to last slide):

$$
f(E_j) = \frac{N_j}{g_j} = e^{\frac{- \left( E_j - E_F \right)}{k_B T}}
$$

What is this? Well, since it's a fraction ##N_j / g_j##, it's clearly a number of particles--the number in a single state of energy ##E_j## where there is a number ##g_j## of degenerate states with that energy. (This easily simplifies to the non-degenerate case ##g_j = 1##.)

But what about that exponential of a negative number on the right? Well, the numerator of that number (the argument of the exponential) is ##- \left( E_j - E_F \right)##. In other words, it's minus the difference between the energy of the state, ##E_j##, and the Fermi energy. But if there are no chemical reactions taking place--i.e., there is only a single type of particle present, or the temperature is too low for any reactions between multiple particle types to take place--then ##E_j < E_F##! Which means that the argument of the exponential is minus a negative number, i.e., a positive number.

In other words, this is telling us that the average number of particles in a single state of energy ##E_j## is more than one if ##E_j## is below the Fermi energy. If ##E_j## exactly equals the Fermi energy, then there is exactly one particle in that state (on average); and if ##E_j## is greater than the Fermi energy, there is on average less than one particle in that state.

Now, "average number of particles" is easier to interpret if the states are discrete. But the formula itself works just fine if the states are continuous. If you struggle with how to interpret "average number of particles" for continuous states, then you can, as I said in what you quoted from post #99, just divide ##f(E_j)## by ##N##, the total number of particles, to get a formula that looks like this

$$
\frac{f(E_j)}{N} = \frac{N_j}{N g_j} = \frac{1}{N} e^{\frac{- \left( E_j - E_F \right)}{k_B T}}
$$

which has a simple interpretation as the statistical probability that a randomly selected particle will have energy ##E_j##.

Now, you keep on separating that exponential factor into two pieces. You write it in terms of ##\alpha## and ##\beta##, but we know what those actually turn out to equal so there's no need to obfuscate. We can just write:

$$
f(E_j) = \frac{N_j}{g_j} = e^{\frac{E_F}{k_B T}} e^{\frac{- E_j}{k_B T}}
$$

And if you want, you can go back and dig up that ##e^{\frac{E_F}{k_B T}} = N / Z## and rewrite this as

$$
f(E_j) = \frac{N_j}{g_j} = \frac{N}{Z} e^{\frac{- E_j}{k_B T}}
$$

which, again, clearly has an interpretation as "number of particles" (because ##N## appears explicitly on the right and we take the exponential of a negative number to get a fraction, with the partition function as an additional divisor). And the "divide by ##N##" form of this would be

$$
\frac{f(E_j)}{N} = \frac{N_j}{N g_j} = \frac{1}{Z} e^{\frac{- E_j}{k_B T}}
$$

which, again, clearly has an interpretation as "fraction of particles".

I really don't care which one of these you want to look at; they are all valid formulas and they all have well-defined meanings. But I think you need to take care to be clear about which of them you are talking about. Otherwise you will just confuse yourself, and, as I noted above, you will end up confusing others (e.g., me) as well.

 

[JohnnyGui]

@PeterDonis Thank you for the extensive explanation. Everything that you posted makes perfect sense to me. But only to the extent of when energy is considered discrete.

Please let me elaborate. Are all of those formulas also valid when energy is considered continuous? If according to you, yes, then why can't I find the ##Z##, the ##g_j## being substituted by the states density and ##e^{\frac{- E_j}{k_B T}}## being the only parameter that is kept from your formulas for the continuous approach?

 

[PeterDonis]

Are all of those formulas also valid when energy is considered continuous?

Yes, except that there is no well-defined concept of "degeneracy" for a continuous spectrum of energy states, so there is no ##g_j## (basically you have to set ##g_j = 1## so it drops out of the formulas), and that you have to calculate the partition function ##Z## correctly for the continuous spectrum.

why can't I find the ##Z##, the ##g_j## being substituted by the states density and ##e^{\frac{- E_j}{k_B T}}## being the only parameter that is kept from your formulas for the continuous approach?

I have no idea what you are trying to ask here.

 

[JohnnyGui]

Yes, except that there is no well-defined concept of "degeneracy" for a continuous spectrum of energy states, so there is no ##g_j##

This is a good start to ask part of my question you didn't understand. I have difficulty grasping this quote. If the degeneracy for a particular discrete energy level is equal to ##g_j = \frac{4 \pi mL^2 \cdot E_j}{h^2}##, why can't the number of quantum states in the continuous approach be ##g(E) = \frac{4 \pi mL^2 \cdot E}{h^2} \cdot dE##? Why should it be the density of states integrated instead?

 

[PeterDonis]

why can't the number of quantum states in the continuous approach be

There is no such thing as "number of quantum states" if the spectrum is continuous. See below.

Why should it be the density of states integrated instead?

The thing you wrote down as ##g(E)## is a "density of states", not a number of states. (I'm not saying it's necessarily the right density of states formula; I haven't checked it. But anything of the form "some function of ##E## times ##dE##" is a density of states: it's a thing you integrate over a continuous range of ##E##. Whereas a "number of states" is something you add up over a set of discrete values of ##E##.)

 

[JohnnyGui]

The thing you wrote down as g(E)g(E)g(E) is a "density of states", not a number of states. (I'm not saying it's necessarily the right density of states formula; I haven't checked it. But anything of the form "some function of EEE times dEdEdE" is a density of states: it's a thing you integrate over a continuous range of EEE. Whereas a "number of states" is something you add up over a set of discrete values of EEE.)

Ah, the function ##g_j## is formulated as the surface of 1/8th of a sphere in n-dimensions, and thus gives the number of quantum states for a particular energy level. So I assume integrating that function would not give the correct outcome for the continuous approach.

What I'd need for the continuous approach is a derivative of a function for the cumulative number of quantum states (which ##g_j## is not). This would be the derivative of the volume of 1/8th of an n-sphere, which is the density as you said.
$$D(E) = \frac{V \cdot 2^{2.5}\pi \cdot m^{1.5}}{h^3} \cdot \sqrt{E}$$
This is the formula that I should multiply with ##dE## in order to get the number of quantum states between ##E \geq E + dE##.
Now that we have the number of quantum states for the continuous approach: ##D(E) \cdot dE##, I'd need to multiply this with a function ##F(E)## that gives the number of paticles between ##E \geq E + dE## per 1 quantum state, so that I can calculate the number of particles between ##E \geq E + dE##

It seems so that this function is not necessarily deduced by reasoning that the function for the discrete appraoch can also be used for the continuous approach, as you said in your post #99. It is rather reasoned that this function ##F(E)## must be proportional to ##e^{-\beta E}## because the probability density function is also proportional to ##e^{-\beta E}##. This reasoning is given on this wiki and also here. It is apparently a property of statistical mechanics. The constant can then be derived through normalization (integrating the probability density to infinity and normalizing it to 1)

The property of the function being proportional to ##e^{-\beta E}## is my exact problem that I can't seem to understand. I don't know if this is deduced from the discrete approach and how this is done. I have tried reading the explanations in the links but I still have trouble grasping it. Perhaps you could give a better example or explanation for this?