Calculating the number of energy states using momentum space

[PeterDonis]

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.

The stack exchange link answer is basically the same reasoning that was in one of the slides you linked to (the only difference is that that answer considers the possibility of there being other constants of the motion besides energy). Which illustrates that that basic line of reasoning works whether the spectrum of energy states is discrete or continuous.

 

[JohnnyGui]

The stack exchange link answer is basically the same reasoning that was in one of the slides you linked to (the only difference is that that answer considers the possibility of there being other constants of the motion besides energy). Which illustrates that that basic line of reasoning works whether the spectrum of energy states is discrete or continuous.

I'm not sure which slide source you mean, I couldn't find an explanation in the slides about the pobability density being proportional to $e^{-\beta E}$. Furthermore, the fact that $n(E) = \frac{N}{Z} \cdot e^{-\beta E}$ can also be used for the continuous approach while it is initially derived based on the discrete approach still surprises me.

How can a specific value of energy be filled in that formula if energy is continuous in the first place?

 

[PeterDonis]

I'm not sure which slide source you mean

The one you linked to in post #121.

I couldn't find an explanation in the slides about the pobability density being proportional to $e^{-\beta E}$.

Deriving that formula doesn't count as an explanation? The number of particles with energy $E$, divided by the total number of particles, is the probability.

How can a specific value of energy be filled in that formula if energy is continuous in the first place?

Huh? A continuous range of values of $E$ is still a set of possible values of $E$. Just pick one.

 

[JohnnyGui]

Huh? A continuous range of values of EEE is still a set of possible values of EEE. Just pick one.

I thought in case of continuous energy one could only speak of $E \geq E + dE$. Hence the formula for the continuous approach consists at least of the States Density that gets integrated.
It doesn't contain $\frac{N}{Z}$ for the number of particles per 1 quantum state.

 

[PeterDonis]

I thought in case of continuous energy one could only speak of $E \geq E + dE$.

First, this doesn't even make sense if $dE$ is positive, which is the usual assumption.

Second, I don't know why you would think this. You can pick a single value of $E$ out of a continuous set just as you can pick one out of a discrete set. In both cases you're picking a single value out of a set of values.

the formula for the continuous approach consists at least of the States Density that gets integrated.

Yes, because you have to integrate over the entire range of possible values of $E$ in order to normalize it (i.e., in order to find the partition function $Z$ that goes in the denominator).

It doesn't contain $\frac{N}{Z}$ for the number of particles per 1 quantum state.

You're getting mixed up. Go back and look at post #99, where I summarized all of the different formulas we've thrown around in this discussion. The first formula I gave in that post is one for the number of particles with a particular energy. What factor does it have in it?

 

[JohnnyGui]

First, this doesn't even make sense if dEdEdE is positive, which is the usual assumption.

Second, I don't know why you would think this. You can pick a single value of EEE out of a continuous set just as you can pick one out of a discrete set. In both cases you're picking a single value out of a set of values

I'm thinking this because of an analogue with probability density. The probability to have, for example, a particle at exactly location $x$ is practically zero because $x$ is a continuous variable. One could therefore only speak of a probability between $x$ and $x+dx$ which is calculated by multiplying the cumulative probability density with $dx$. Another example is found here which is where I got this way of thinking from.

Just like in this case one could only speak of a number of particles between energy $E$ and $E+dE$ because energy is considered continuous, and therefore one could only calculate the number of particles in this case by multiplying a cumulative particle density function with $dE$ and not use the discrete function.

Please explain why this way of thinking is incorrect.

 

[PeterDonis]

The probability to have, for example, a particle at exactly location $x$ is practically zero because $x$ is a continuous variable.

This is not correct.

Suppose I have a Gaussian probability distribution for position: a particle's probability to be at position $x$ is $(1 / \sqrt{\pi} ) e^{-x^2}$. This is a perfectly normalized probability distribution and I can plug any value of $x$ I want into it and get a valid answer back that is not zero. I don't need to multiply it by $dx$ or anything like that.

The fact that $x$ is a continuous variable does not mean the probability of finding a particle at exactly $x$ is zero. It just means that the integral over all possible values of $x$ of whatver probability distribution function we have must be $1$. The function I gave above meets that requirement (do the integral and see).

I think you have been confused by reading some popular presentations on probability that use sloppy language.

 

[JohnnyGui]

This is not correct.

Suppose I have a Gaussian probability distribution for position: a particle's probability to be at position $x$ is $(1 / \sqrt{\pi} ) e^{-x^2}$. This is a perfectly normalized probability distribution and I can plug any value of $x$ I want into it and get a valid answer back that is not zero. I don't need to multiply it by $dx$ or anything like that.

The fact that $x$ is a continuous variable does not mean the probability of finding a particle at exactly $x$ is zero. It just means that the integral over all possible values of $x$ of whatver probability distribution function we have must be $1$. The function I gave above meets that requirement (do the integral and see).

I think you have been confused by reading some popular presentations on probability that use sloppy language.

Hmm, this makes me wonder though. Why is the formula for the Maxwell-Boltzmann Distribution then always written in terms of a probability density times $dE$ which gives the probability for a particle between energy $E \geq E+dE$, while like you said, one should be able use the previously discussed formula to calculate the probability at a specific $E$, even in the continuous approach?

 

[PeterDonis]

Why is the formula for the Maxwell-Boltzmann Distribution then always written in terms of a probability density times $dE$

It is if it is inside an integral because it's inside an integral. I have never seen it written with a $dE$ if it's not in an integral.

 

[JohnnyGui]

It is if it is inside an integral because it's inside an integral. I have never seen it written with a $dE$ if it's not in an integral.

Someone else answered this by saying that the $dE$ is always needed purely for the Density of States function $D(E)$. In a continuous approach, you'd need a function that gives the number of states over a range, e.g. the volume of quantum states within $dE$. There is no exact quantum states function in the continuous approach without the need to integrate it.
According to him, the formula for the number of particles $n(E)$ in the continuous approach is:
$$n(E) = D(E) \cdot dE \cdot \frac{N}{Z} \cdot e^{-\beta E}$$
I have never seen a number of particles function in the continuous approach without the need to integrate one of its functions.

 

[PeterDonis]

Someone else answered this

StackExchange is not a valid source. If you can find a textbook or peer-reviewed paper that makes this claim, we can discuss it.

I have never seen a number of particles function in the continuous approach without the need to integrate one of its functions.

Yes, whenever you're dealing with a continuous spectrum the only quantities that will be meaningful when comparing with experiment will be integrals.

 

[JohnnyGui]

Does the formula in my previous post agree with what you're saying?

Also, how does this...

I have never seen it written with a dEdEdE if it's not in an integral.

...coincide with your previous statement:

Yes, whenever you're dealing with a continuous spectrum the only quantities that will be meaningful when comparing with experiment will be integrals.

Isn't it then obvious that you won't ever find a function without an integral for the continuous approach?

Where is the particle number function $n(E)$ for the continuous approach that doesn't have any integral, in which the quantum states function $g(E)$ is not written in terms of Density of States times $dE$?

 

[PeterDonis]

Isn't it then obvious that you won't ever find a function without an integral for the continuous approach?

No, because some of the very sources you linked to in this thread show the function without an integral, in their derivations of the function, and make no claim that their derivations only apply to the discrete approach. But they also don't try to link that function to any particular experimental results.

Where is the particle number function $n(E)$ for the continuous approach that doesn't have any integral

We've already been over this. You're making this way harder than it needs to be. You have a function with one argument, $E$. You can plug any number you want into that function and get another number. You can also integrate that function over a range of arguments and get a number; and when you do the integral you need to include the $dE$ because that's how integration works.

in which the quantum states function $g(E)$ is not written in terms of Density of States times $dE$?

You're confusing yourself with sloppy language. The function $g(E)$ is never "written in terms of Density of States times $dE$". It is not a function of $dE$. It's a function of $E$. When you integrate it over a range of values of $E$, you have to include $dE$ in the integrand because that's how integration works. That's really all there is to it.

 

[JohnnyGui]

The function g(E)g(E)g(E) is never "written in terms of Density of States times dEdEdE". It is not a function of dEdEdE. It's a function of EEE. When you integrate it over a range of values of EEE, you have to include dEdEdE in the integrand because that's how integration works. That's really all there is to it.

Apparently I misunderstood the part how $g(E)$ is transformed for the integrand to work. The way $g(E)$ is transformed to include it in the integrand supports the explanation of the replier at StackExchange.

$g(E)$ gives the number of quantum states for a particular energy $E$ and is 1/8th of a sphere surface in n-dimensions:
$$g(E) = \frac{4 \pi mL^2 \cdot E}{h^2} = 4\pi r^2 \cdot \frac{1}{8}$$
From this:
$$r=\sqrt{\frac{8mL^2\cdot E}{h^2}}$$
For the integrand for the continuous approach, one would want to calculate the number of quantum states over a range $dE$, which is the surface of 1/8th of an n-sphere times a thickness $dr$
$$g(E\geq E+dE) = \frac{4 \pi mL^2 \cdot E}{h^2} \cdot d(\sqrt{\frac{8mL^2\cdot E}{h^2}})$$
The $dr$ can be wirtten as:
$$dr = d(\sqrt{\frac{8mL^2\cdot E}{h^2}}) = \frac{\sqrt{2}\cdot mL}{h\cdot \sqrt{mE}}\cdot dE$$
Substuting $dr$ with this in $g(E \geq E + dE)$ and simplifying gives:
$$g(E\geq E+dE) = \frac{2^{2.5}\pi \cdot m^{1.5} \cdot V \cdot \sqrt{E}}{h^3}\cdot dE$$
Since the number of quantum states in the energylevels within $dE$ doesn't change much, one can multiply this $g(E \geq E+dE)$ by the number of particles per 1 quantum state at a particular energy $E$, which is $\frac{N}{Z} \cdot e^{-\beta E}$, giving the formula in my previous post #150 that the replier at StackExchange explained.