Calculating the number of energy states using momentum space

[PeterDonis]

Someone else answered this

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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.