Equations of State in Modern Classical Physics (Thorne/Blandford)

[RobertDSmeets]

TL;DR Summary: Questions regarding the book "Modern Classical Physics" by Thorne/Blandford

Hello,
I'm going through this book and on pg. 127, regarding equations of state, there is a parameter, t (explicitly stated: "not to be confused with time"), that uses hyperbolic functions to relate the Fermi energy and the Fermi momentum (when multiplied with mass, in this case, of an electron). The parameter is just given as a parameter (used as t/4) with no information about what it represents. I am trying to understand what exactly this parameter represents. It is then used to describe particle density, mass-energy density, and pressure for hydrogen gas (relativistically degenerate).

I am not sure what this parameter is, both mathematically and physically. Any help is greatly appreciated.

I figured I'd just start a thread where people can ask questions related to the book.

Thanks!

 

[malawi_glenn]

You will have a much higher chance to get replies if you ask in the physics section, where you type the relevant equation(s) in LaTeX.

Anyway, the parameter $t$ is introduced in equation (3.52a)
$\mathcal{E}_F \equiv m_e \cosh(t/4)$ and $p_F \equiv m_e \sinh(t/4)$

If you want to know what $t$ is in terms of $\mathcal{E}_F$ and $p_F$ just solve for $t$ in the definition above.

 

[Dragon27]

I am trying to understand what exactly this parameter represents.

It's just there for convenience's sake. Since the Fermi energy and the Fermi momentum are related to each other (they're not independent), one may want to use just one parameter instead. And since they are related in the same way hyperbolic functions are (3.50), it seems very attractive to express them like that.

 

[malawi_glenn]

Yes $\mathcal{E}_F{}^2 - p_F{}^2 = m_e{}^2$ and then you use the hyperbolic identity $\cosh^2(x) - \sinh^2(x) = 1$

 

[berkeman]

You will have a much higher chance to get replies if you ask in the physics section, where you type the relevant equation(s) in LaTeX.

Thread moved back to the Classical Physics forum (where it started out).

 

[RobertDSmeets]

Thanks for all the replies!

Sorry, but I am unfamiliar with LaTeX formatting.

regarding this:

It's just there for convenience's sake. Since the Fermi energy and the Fermi momentum are related to each other (they're not independent), one may want to use just one parameter instead. And since they are related in the same way hyperbolic functions are (3.50), it seems very attractive to express them like that.

I'm not exactly sure what the Fermi momentum is, exactly. Eqn. 3.52a explains the mathematical relationship of it to the Fermi energy, but I'm not exactly sure what it is conceptually. Is it the momentum of the particle when the particle is at the highest energy state at 0K?

Though, I do think I have a better grasp on what t represents. Thanks!

 

[berkeman]

Sorry, but I am unfamiliar with LaTeX formatting.

Please see the "LaTeX Guide" link below the Edit window. It's a requirement for posting math here at PF. Thanks.

 

[vanhees71]

It's the description of an ideal relativistic Fermi gas at $T=0$.

Generally one should note that the classical phase-space distribution function $f(t,\vec{x},\vec{p}$ for "on-shell" particles with $(p^0)^2=E^2=m^2+\vec{p}^2$ (using natural units with $\hbar=c=1$) is a scalar field. The particle-number four-current is
$$J^{\mu}=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3} \frac{p^{\mu}}{E} f(t,\vec{x},\vec{p}),$$
which is a four-vector field as it must be. Note that everywhere $p^0=E=+\sqrt{\vec{p}^2+m^2}$.

Similarly the energy-momentum tensor is given by
$$T^{\mu \nu}= \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3} \frac{p^{\mu} p^{\nu}}{E} f(t,\vec{x},\vec{p}).$$
For the ideal Fermi gas at $T=0$ in global thermal equilibrium in the rest frame of the gas
$$f(t,\vec{x},\vec{p})=g \Theta(p_F-|\vec{p}),$$
where $p_F$ is the Fermi momentum, and $g$ is the degeneracy factor (e.g., due to spin, for which $g=2s+1$). It's determined by the given density of the gas, i.e., using spherical coordinates for the integral over $\vec{p}$,
$$n=J^0=\frac{4 \pi g}{(2 \pi)^3} \int_0^{p_F} \mathrm{d} p p^2 =\frac{4 \pi g}{3 (2 \pi)^3} p_F^3.$$
The total energy density is
$$\varepsilon=T^{00}=\frac{4 \pi g}{(2 \pi)^3} \int_{0}^{p_F} \mathrm{d} p p^2 \sqrt{p^2+m^2}.$$
Here it becomes clear that it is convenient to parametrize $p$ with hyperbolic functions. For some reason, I'm not clear about, Thorne chooses the convention
$$p=m \sinh(t/4) \; \Rightarrow \; E=\sqrt{m^2+p^2}=m \cosh(t/4).$$
Setting thus $p_F=m \sinh(t_F/4)$ you get
$$n=\frac{4 \pi g}{3} \frac{m^3}{2 \pi^3} \sinh(t_F/4).$$
Note that this is consistent with the textbook since for electrons $g=2s+1=2$ and if using explicitly $\hbar$, you have to write $m^3/(2 \pi \hbar)^3=m^3/h^3=1/\lambda_c^3$, where $\lambda_c$ is the Compton wavelength. Also since I use $t$ for the parameter for the $p$ integration, I used $t_F$ for the corresponding value, i.e., $p_F=m \sinh(t_F/4)$. So where Thorne writes $t$ in my convention I've to write $t_F$.

For the energy density then you get
$$\varepsilon=\frac{4 \pi g}{(2 \pi)^3} \frac{m^4}{4} \int_0^{t_F} \cosh^2(t/4) \sinh^2(t/4).$$
Now
$$\cosh(t/4) \sinh(t/4)=\frac{1}{4} [\exp(t/2)-\exp(-t/2)]$$
and thus
$$[\cosh(t/4) \sinh(t/4)]^2=\frac{1}{16} [\exp(t)+\exp(-t)-2],$$
and the integral gets
$$\varepsilon=\frac{4 \pi g}{(2 \pi)^3} \frac{m^4}{32} [\sinh(t_F)-t_F]=\frac{\pi g m}{8 \lambda_c^3}[\sinh(t_F)-t_F].$$
Finally the pressure is
$$P=T^{33}=\frac{2 \pi g}{(2 \pi)^3} \int_0^{p_F} \mathrm{d}p \int_0^{\pi} \mathrm{d} \vartheta p^2 \sin \vartheta \frac{p^2 \cos^2 \vartheta}{E}.$$
The $\vartheta$ integral is
$$\int_0^{\pi} \mathrm{d} \vartheta \sin \vartheta \cos^2 \vartheta=-\frac{1}{3} \cos^3 \vartheta|_0^{\pi} = \frac{2}{3}.$$
So we have
$$P=\frac{4 \pi g}{3 (2 \pi)^3} \int_0^{p_F} \mathrm{d} p \frac{p^4}{\sqrt{p^2+m^2}} = \frac{4 \pi g}{3 (2 \pi)^3} \int_0^{t_F} \frac{m^4}{4} \mathrm{d} t \sinh^4(t/4) =\frac{\pi g}{24 \lambda^3} m [3 t_F-8 \sinh(t_F/2) + \sinh(t_F)],$$
which also agrees with the textbook. For the integration I used Mathematica, but of course in principle you can evaluate all these integrals by writing the hyperbolic functions in terms of exponential functions ;-)).