Volume of momentum space of an ideal gas

[LightPhoton]

In *An Introduction to Thermal Physics* by Schroeder, while deriving the multiplicity of an ideal gas makes the following statements (image below):

Even in quantum mechanics, the number of allowed wavefunctions is infinite.
But the number of independent wavefunctions (in a technical sense that’s defined
in Appendix A) is finite, if the total available position space and momentum space
are limited. I like to picture it as in Figure 2.9. In this one-dimensional example,
the number of distinct position states is L/Δ x, while the number of distinct momentum states is L_p /Δp_x The total number of distinct states is the product,


$$\frac{L\,L_p}{\Delta x\,\Delta p_x}=\frac{L\,L_p}{h}$$
according to the uncertainty principle. In three dimensions, the lengths become
volumes and there are three factors of h:
$$\Omega=\frac{V\,V_p}{h^3}$$


Here we have an ideal monoatomic gas's single atom in a cuboidal box, whose momentum is constrained in the following manner

$$p^2_x+p^2_y+p^2_z=2mU$$

Where m is its mass and U is its kinetic energy.

Since this is the equation of a sphere, we can multiply by a small thickness Δp_r to make it a volume.




Now the calculation for the momentum volume in Schroeder seems off to me, in that the volume here is a sphere rather than a cubical box, thus instead of using $\Delta p_x,\Delta p_y,\Delta p_y$, shouldn't $\Delta p_r,\Delta p_\theta,\Delta p_\phi$ (uncertainties in spherical coordinates) be used? this would alter the uncertainty products as well, making the overall constants a bit different.

 

[anuttarasammyak]

Phase space cell of volme h is like a piece of Lego block any shapes of which are allowed keeping volume h. I do not share your concern because I think any figure in phase space can be constructed by the cells , like sphere is made of square Lego blocks.

 

[LightPhoton]

Phase space cell of volme h is like a piece of Lego block any shapes of which are allowed keeping volume h. I do not share your concern because I think any figure in phase space can be constructed by the cells , like sphere is made of square Lego blocks.

Ah so you are saying that the V_p can be sliced into multiple L_px, L_py and L_pz, we divide each by Delta x, y,z and then add up all of them to make a sphere?

 

[anuttarasammyak]

Ah so you are saying that the V_p can be sliced into multiple L_px, L_py and L_pz, we divide each by Delta x, y,z and then add up all of them to make a sphere?

Roughly yes, but be careful that I said about phase space. Coordinate space cell has voume of dimension L^3N. Momentum cell space has volunme of dimension (MLT^-1)^3N. Phase space cell has volume of h^3N whose dimension is (ML^2T^-1)^3N. Here N is number of particle in the system.

 

[LightPhoton]

Roughly yes, but be careful that I said about phase space. Coordinate space cell has voume of dimension L^3N. Momentum cell space has volunme of dimension (MLT^-1)^3N. Phase space cell has volume of h^3N whose dimension is (ML^2T^-1)^3N. Here N is number of particle in the system.

Yeah i understand that, but not in relation to what you said. Are you saying that dimensions of physical space and momentum space aren't same? if yes, then yeah i get that. but are you saying anything else in addition to it?

 

[anuttarasammyak]

I just note that momentum cell has coordinate cell as its counterpart so that their volume product is h^3 for one particle case.　In this context phase space which combine coordinate space and momentum space is convenient to describe the states.

 

[div_grad]

@LightPhoton The explanation given in the book sweeps some important methodology under the rug, this methodology being a standard way to calculate the number of available quantum states of a system - not only for an ideal gas but also for black-body radiation (this gives you the famous Planck formula for its energy density) and virtually any system composed of weakly interacting parts. This standard calculation involves, as mentioned in the book, closing a system in a rectangular box, however at a later stage of the calculation you indeed introduce auxiliary variables which you deal with by "closing" them inside... a spherical box . I suppose that your confusion caused by the book using seemingly "rectangular" as well as "spherical" boxes in the "momentum" as well as "coordinate" spaces is precisely due to it not showing the technique I'm speaking about explicitly, instead arriving at the (correct!) results as quickly as possible to move on to different topics and applications.

The explicit methodology is the following. Consider a large rectangular box with sides of some length $L$ and volume $L^3$. The particles inside the box (your ideal gas) can occupy any positions $\mathbf{r}=(x,y,z)$, while the momentum eigenstates of these particles are described by the usual exponential wave functions
$$
e^{\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r}} = e^{\frac{i}{\hbar}p_x x}e^{\frac{i}{\hbar}p_y y}e^{\frac{i}{\hbar}p_z z} \rm{.}
$$
Now, since you enclosed your system in a rectangular box, the system is no longer translationally-invariant (because of the presence of the walls). The principles of mechanics then tell you that actually the system's momentum is not conserved, so it is rather awkward to work with momentum vectors $\mathbf{p}=(p_x,p_y,p_z)$ as they will not actually be well-defined quantities. To remedy this - and this is one of the important steps - you enforce the so-called periodic boundary conditions, namely you assume that when the particle's position coincides with one of the walls of the box, then this particle "leaves" the box and another identical particle "enters" the box through a wall directly opposite to that through which the "original" particle left. Basically, you artificially impose the translation-invariance back on your system, so that you may still speak about momenta $\mathbf{p}$ while dealing with a finite (but large) box. Since the lengths of all sides of the box are equal, the periodic boundary conditions tell you that the state of the particle located at $(x,y,z)$ is the same as if it was located at $(x+L, y+L, z+L)$. Thus
$$
e^{\frac{i}{\hbar}p_x (x+L)}e^{\frac{i}{\hbar}p_y (y+L)}e^{\frac{i}{\hbar}p_z (z+L)} = e^{\frac{i}{\hbar}p_x x}e^{\frac{i}{\hbar}p_y y}e^{\frac{i}{\hbar}p_z z}
$$
which shows that you must have
$$
p_x = \frac{2\pi\hbar}{L}n_x, \quad p_y = \frac{2\pi\hbar}{L}n_y, \quad p_z = \frac{2\pi\hbar}{L}n_z
$$
where $n_x$, $n_y$ and $n_z$ are integers. Now, since the momentum vectors $\mathbf{p}$ can have arbitrary direction, the three integers $n_i$ can be either positive or negative numbers, and based on the above the momentum has the form
$$
\mathbf{p} = (p_x, p_y, p_z) = \frac{2\pi\hbar}{L}(n_x,n_y,n_z) = \frac{2\pi\hbar}{L}(\pm|n_x|,\pm|n_y|,\pm|n_z|) \rm{.}
$$
So you obtain that to every triple $(|n_x|,|n_y|,|n_z|)$ of positive integers there corrsponds $2^3=8$ momentum vectors $\mathbf{p}$.

Now comes the part I mentioned eariler, with the auxiliary variables which will be "enclosed" in a spherical box. Namely, you see that the problem of counting the number of availalbe states for given maximum value of momentum $\mathbf{p}_\text{max}$ can be reduced, in this case, to counting the number of the integer triples $(|n_x|,|n_y|,|n_z|)$ for given maximum value of these triples. You do this counting by drawing a three-dimensional grid of points in the coordinate system whose three axes are labeled with $|n_x|$, $|n_y|$ and $|n_z|$ (instead of, say, $x$, $y$ and $z$). Now, if there are sufficiently many such points then you can draw a sphere centered in the origin of your system of axes and this sphere will contain many points $(|n_x|,|n_y|,|n_z|)$. If you choose some maximum values of $|n_x|$, $|n_y|$ and $|n_z|$ that you wish to consider, then the number of distinct triples of integers that you need to count will be given by the number of points enclosed by the sphere of radius $\sqrt{n_x^2 + n_y^2 + n_z^2}$. If the sphere is large enough, then this number is given by one eighth (1/8) of the volume of your sphere. The 1/8 factor comes from the fact that you consider only positive integers $|n_x|$, $|n_y|$ and $|n_z|$, so that only one octant of the sphere corresponding to positive semi-axes of your coordinate system contribute to the counting.

Thus the number of distinct triples of positive integers $(|n_x|, |n_y|, |n_z|)$ enclosed by the sphere is
$$
\frac{1}{8} \cdot \frac{4}{3}\pi\left(\sqrt{n_x^2 + n_y^2 + n_z^2}\right)^3
$$
Now, remember that earlier we showed that to each triple of positive integers $(|n_x|, |n_y|, |n_z|)$ there corresponds $2^3=8$ momentum vectors $\mathbf{p}$. Since the above expression is the number of such integer triples, to obtain the corresponding number of momentum vectors $\mathbf{p}$ you need to multiply the above equation by 8. You also know that
$$
p^2 = \mathbf{p}\cdot\mathbf{p} = \left(\frac{2\pi\hbar}{L}\right)^2\left(n_x^2 + n_y^2 + n_z^2\right)
$$
and using this you can finally obtain that the desired number of momentum vectors whose magnitude is at most $p$ is given by
$$
N_p = \frac{4\pi}{3} \frac{L^3}{(2\pi\hbar)^3} p^3 \rm{.}
$$
Now, this is the total number of momenta corresponding to the number of integer triples inside the sphere in the "$n_i$-space". To obtain the number of momentum vectors $\text{d}N_p$ whose magnitude lies between $p$ and $p + \text{d}p$ (as is usually done) you differentiate the above expression with respect to $p$, obtaining:
$$
\text{d}N_p = \frac{\partial N_p}{\partial p}\text{d}p = \frac{L^3}{(2\pi\hbar)^3} 4\pi p^2 \text{d}p \rm{.}
$$
You may recognize the $4\pi$ factor above as just the total solid angle, and in fact
$$
4\pi p^2 \text{d}p = \int_0^\pi \int_0^{2\pi} p^2 \sin\Theta_p \text{d}p \,\text{d}\Theta_p \,\text{d}\phi_p \rm{,}
$$
where $p$, $\Theta_p$ and $\phi_p$ are the spherical components of the vector $\mathbf{p}$. Thus the number $\text{d}N_p$ obtained above is just the expression
$$
\text{d}\eta_\mathbf{p} = \frac{L^3}{(2\pi\hbar)^3} p^2 \sin\Theta_p \text{d}p \,\text{d}\Theta_p \,\text{d}\phi_p = \frac{L^3}{(2\pi\hbar)^3} \text{d}^3p
$$
integrated over all angles. This is the final answer. The number of the available momentum states for which the components of $\mathbf{p}$ lie in the intervals $(p_x, p_x+\text{d}p_x)$, $(p_y, p_y+\text{d}p_y)$ and $(p_z, p_z+\text{d}p_z)$ is $\frac{L^3}{(2\pi\hbar)^3} \text{d}^3p$. From this you get that the density of states, that is the number of states divided by the volume of the box $L^3$, is $\frac{\text{d}^3p}{(2\pi\hbar)^3}$.

Of course since the particles in the box can take any position $\mathbf{r}$, it is obvious that the number of states for which the coordinates of the particles lie between $(x, x+\text{d}x)$, $(y, y+\text{d}y)$ and $(z, z+\text{d}z)$ is just $\text{d}^3r = \text{d}x \,\text{d}y \,\text{d}z$. Combining this with the result we derived above, you simply get that the density of states of your gas (the number of available states per unit volume of the box - see? at the end the presence of the box doesn't matter ) is
$$
\frac{\text{d}^3r \,\,\text{d}^3p}{(2\pi\hbar)^3} = \frac{\text{d}^3r \,\,\text{d}^3p}{h^3}
$$
where I just used the fact that $2\pi\hbar = h$, to be consistent with the notation used by the book.

As a bonus I will add a side comment explaining why the above result is remarkable (at least to me). Observe namely that the formula derived above is dimensionless, because of the presence of the Planck constant $h$ in the denominator. As @anuttarasammyak already pointed out, the "bare" product $\text{d}x \,\text{d}p_x$ of the position-momentum variables has a dimension, namely those of action $J\cdot s$ (Joule times second). Now, in statistical mechanics one of the most important concepts is that of the entropy, which is given as the logarithm of the number of states available to the system. In classical mechanics, where there is no Planck constant, the usual phase-space formulation gives you the density of states precisely as this "bare" product $\text{d}x \,\text{d}p_x$. Therefore the entropy $S = \ln{\left(\Delta x \,\Delta p_x\right)}$ is defined only up to an additive constant, which depends on the choice of units one adopts for energy and time (or position and momentum): $S = \ln{\left(\alpha\left[\Delta x \,\Delta p_x\right]\right)} = \ln{\left(\Delta x \,\Delta p_x\right)} + \ln{\alpha}$. Because of this, in classical mechanics only entropy differences can have any physical meaning, as these additional constant $\ln{\alpha}$ terms cancel each other upon taking the difference. Observe, however, that quantum mechanics makes possible - by introducing the Planck constant - to define the entropy in an absolute sense, $S = \ln{\left(\frac{\Delta x \,\Delta p_x}{h}\right)}$ as in this case the argument of the logarithm does not depend on the choice of units. Just a little neat observation.

 

[LightPhoton]

@LightPhoton The explanation given in the book sweeps some important methodology under the rug, this methodology being a standard way to calculate the number of available quantum states of a system - not only for an ideal gas but also for black-body radiation (this gives you the famous Planck formula for its energy density) and virtually any system composed of weakly interacting parts. This standard calculation involves, as mentioned in the book, closing a system in a rectangular box, however at a later stage of the calculation you indeed introduce auxiliary variables which you deal with by "closing" them inside... a spherical box . I suppose that your confusion caused by the book using seemingly "rectangular" as well as "spherical" boxes in the "momentum" as well as "coordinate" spaces is precisely due to it not showing the technique I'm speaking about explicitly, instead arriving at the (correct!) results as quickly as possible to move on to different topics and applications.

The explicit methodology is the following. Consider a large rectangular box with sides of some length $L$ and volume $L^3$. The particles inside the box (your ideal gas) can occupy any positions $\mathbf{r}=(x,y,z)$, while the momentum eigenstates of these particles are described by the usual exponential wave functions
$$
e^{\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{r}} = e^{\frac{i}{\hbar}p_x x}e^{\frac{i}{\hbar}p_y y}e^{\frac{i}{\hbar}p_z z} \rm{.}
$$
Now, since you enclosed your system in a rectangular box, the system is no longer translationally-invariant (because of the presence of the walls). The principles of mechanics then tell you that actually the system's momentum is not conserved, so it is rather awkward to work with momentum vectors $\mathbf{p}=(p_x,p_y,p_z)$ as they will not actually be well-defined quantities. To remedy this - and this is one of the important steps - you enforce the so-called periodic boundary conditions, namely you assume that when the particle's position coincides with one of the walls of the box, then this particle "leaves" the box and another identical particle "enters" the box through a wall directly opposite to that through which the "original" particle left. Basically, you artificially impose the translation-invariance back on your system, so that you may still speak about momenta $\mathbf{p}$ while dealing with a finite (but large) box. Since the lengths of all sides of the box are equal, the periodic boundary conditions tell you that the state of the particle located at $(x,y,z)$ is the same as if it was located at $(x+L, y+L, z+L)$. Thus
$$
e^{\frac{i}{\hbar}p_x (x+L)}e^{\frac{i}{\hbar}p_y (y+L)}e^{\frac{i}{\hbar}p_z (z+L)} = e^{\frac{i}{\hbar}p_x x}e^{\frac{i}{\hbar}p_y y}e^{\frac{i}{\hbar}p_z z}
$$
which shows that you must have
$$
p_x = \frac{2\pi\hbar}{L}n_x, \quad p_y = \frac{2\pi\hbar}{L}n_y, \quad p_z = \frac{2\pi\hbar}{L}n_z
$$
where $n_x$, $n_y$ and $n_z$ are integers. Now, since the momentum vectors $\mathbf{p}$ can have arbitrary direction, the three integers $n_i$ can be either positive or negative numbers, and based on the above the momentum has the form
$$
\mathbf{p} = (p_x, p_y, p_z) = \frac{2\pi\hbar}{L}(n_x,n_y,n_z) = \frac{2\pi\hbar}{L}(\pm|n_x|,\pm|n_y|,\pm|n_z|) \rm{.}
$$
So you obtain that to every triple $(|n_x|,|n_y|,|n_z|)$ of positive integers there corrsponds $2^3=8$ momentum vectors $\mathbf{p}$.

Now comes the part I mentioned eariler, with the auxiliary variables which will be "enclosed" in a spherical box. Namely, you see that the problem of counting the number of availalbe states for given maximum value of momentum $\mathbf{p}_\text{max}$ can be reduced, in this case, to counting the number of the integer triples $(|n_x|,|n_y|,|n_z|)$ for given maximum value of these triples. You do this counting by drawing a three-dimensional grid of points in the coordinate system whose three axes are labeled with $|n_x|$, $|n_y|$ and $|n_z|$ (instead of, say, $x$, $y$ and $z$). Now, if there are sufficiently many such points then you can draw a sphere centered in the origin of your system of axes and this sphere will contain many points $(|n_x|,|n_y|,|n_z|)$. If you choose some maximum values of $|n_x|$, $|n_y|$ and $|n_z|$ that you wish to consider, then the number of distinct triples of integers that you need to count will be given by the number of points enclosed by the sphere of radius $\sqrt{n_x^2 + n_y^2 + n_z^2}$. If the sphere is large enough, then this number is given by one eighth (1/8) of the volume of your sphere. The 1/8 factor comes from the fact that you consider only positive integers $|n_x|$, $|n_y|$ and $|n_z|$, so that only one octant of the sphere corresponding to positive semi-axes of your coordinate system contribute to the counting.

Thus the number of distinct triples of positive integers $(|n_x|, |n_y|, |n_z|)$ enclosed by the sphere is
$$
\frac{1}{8} \cdot \frac{4}{3}\pi\left(\sqrt{n_x^2 + n_y^2 + n_z^2}\right)^3
$$
Now, remember that earlier we showed that to each triple of positive integers $(|n_x|, |n_y|, |n_z|)$ there corresponds $2^3=8$ momentum vectors $\mathbf{p}$. Since the above expression is the number of such integer triples, to obtain the corresponding number of momentum vectors $\mathbf{p}$ you need to multiply the above equation by 8. You also know that
$$
p^2 = \mathbf{p}\cdot\mathbf{p} = \left(\frac{2\pi\hbar}{L}\right)^2\left(n_x^2 + n_y^2 + n_z^2\right)
$$
and using this you can finally obtain that the desired number of momentum vectors whose magnitude is at most $p$ is given by
$$
N_p = \frac{4\pi}{3} \frac{L^3}{(2\pi\hbar)^3} p^3 \rm{.}
$$
Now, this is the total number of momenta corresponding to the number of integer triples inside the sphere in the "$n_i$-space". To obtain the number of momentum vectors $\text{d}N_p$ whose magnitude lies between $p$ and $p + \text{d}p$ (as is usually done) you differentiate the above expression with respect to $p$, obtaining:
$$
\text{d}N_p = \frac{\partial N_p}{\partial p}\text{d}p = \frac{L^3}{(2\pi\hbar)^3} 4\pi p^2 \text{d}p \rm{.}
$$
You may recognize the $4\pi$ factor above as just the total solid angle, and in fact
$$
4\pi p^2 \text{d}p = \int_0^\pi \int_0^{2\pi} p^2 \sin\Theta_p \text{d}p \,\text{d}\Theta_p \,\text{d}\phi_p \rm{,}
$$
where $p$, $\Theta_p$ and $\phi_p$ are the spherical components of the vector $\mathbf{p}$. Thus the number $\text{d}N_p$ obtained above is just the expression
$$
\text{d}\eta_\mathbf{p} = \frac{L^3}{(2\pi\hbar)^3} p^2 \sin\Theta_p \text{d}p \,\text{d}\Theta_p \,\text{d}\phi_p = \frac{L^3}{(2\pi\hbar)^3} \text{d}^3p
$$
integrated over all angles. This is the final answer. The number of the available momentum states for which the components of $\mathbf{p}$ lie in the intervals $(p_x, p_x+\text{d}p_x)$, $(p_y, p_y+\text{d}p_y)$ and $(p_z, p_z+\text{d}p_z)$ is $\frac{L^3}{(2\pi\hbar)^3} \text{d}^3p$. From this you get that the density of states, that is the number of states divided by the volume of the box $L^3$, is $\frac{\text{d}^3p}{(2\pi\hbar)^3}$.

Of course since the particles in the box can take any position $\mathbf{r}$, it is obvious that the number of states for which the coordinates of the particles lie between $(x, x+\text{d}x)$, $(y, y+\text{d}y)$ and $(z, z+\text{d}z)$ is just $\text{d}^3r = \text{d}x \,\text{d}y \,\text{d}z$. Combining this with the result we derived above, you simply get that the density of states of your gas (the number of available states per unit volume of the box - see? at the end the presence of the box doesn't matter ) is
$$
\frac{\text{d}^3r \,\,\text{d}^3p}{(2\pi\hbar)^3} = \frac{\text{d}^3r \,\,\text{d}^3p}{h^3}
$$
where I just used the fact that $2\pi\hbar = h$, to be consistent with the notation used by the book.

As a bonus I will add a side comment explaining why the above result is remarkable (at least to me). Observe namely that the formula derived above is dimensionless, because of the presence of the Planck constant $h$ in the denominator. As @anuttarasammyak already pointed out, the "bare" product $\text{d}x \,\text{d}p_x$ of the position-momentum variables has a dimension, namely those of action $J\cdot s$ (Joule times second). Now, in statistical mechanics one of the most important concepts is that of the entropy, which is given as the logarithm of the number of states available to the system. In classical mechanics, where there is no Planck constant, the usual phase-space formulation gives you the density of states precisely as this "bare" product $\text{d}x \,\text{d}p_x$. Therefore the entropy $S = \ln{\left(\Delta x \,\Delta p_x\right)}$ is defined only up to an additive constant, which depends on the choice of units one adopts for energy and time (or position and momentum): $S = \ln{\left(\alpha\left[\Delta x \,\Delta p_x\right]\right)} = \ln{\left(\Delta x \,\Delta p_x\right)} + \ln{\alpha}$. Because of this, in classical mechanics only entropy differences can have any physical meaning, as these additional constant $\ln{\alpha}$ terms cancel each other upon taking the difference. Observe, however, that quantum mechanics makes possible - by introducing the Planck constant - to define the entropy in an absolute sense, $S = \ln{\left(\frac{\Delta x \,\Delta p_x}{h}\right)}$ as in this case the argument of the logarithm does not depend on the choice of units. Just a little neat observation.

Thanks for the in-depth answer, really appreciate it! Overall it makes perfect sense and is very satisfying to see the inner workings that were "swept under the rug".

Just one thing though, can you please elaborate on If the sphere is large enough, then this number is given by one eighth (1/8) of the volume of your sphere?

I am not quite sure how we can equate an integer to volume.

 

[div_grad]

I am not quite sure how we can equate an integer to volume.

You can do it because these integer numbers have no (physical) dimension attached to them, no meters or parsecs, etc. You can think of this counting procedure as an analog of finding a volume (in units of, say, $m^3$) of some region of space by chopping-up this region into a large number of small cubes whose volumes (also in $m^3$) are the same. Then the volume of the whole region is obtained by counting the number of such small cubes, and just attaching the common physical unit $m^3$ to the result. So the analogy here is that you can think that your points represent such "elementary cubes with the same elementary volumes" (and not, say, "elementary lengths"), and by counting these points you then obtain the "total volume".