Thermodynamic energy and Hamiltonian

[vsv86]

Hello Everyone

This question is motivated by a small calculation I am doing on polarization of bodies in external electric field.

What I wanted to do is this:

1) Mesh the region
2) Prescribe uniform (and non-changing) positive charge distribution
3) Prescribe (initially) uniform negative charge density (net charge is zero)

4) Vary the negative charge density, i.e. shuffle the charges between mesh elements
5) Calculate the potential ($\phi$) due to distributed charge density ($\rho$)
6) Compute energy $U=\int{d^3 r (\phi+\phi_{ext})\rho}$, where $\phi_{ext}$ is due to applied field
7) If the energy has decreased, keep the new negative charge distribution, otherwise go back to the previous distribution
8) Repeat from step 4 onwards., keep going until it converges

What I realized now is that minimization of energy is justified from thermodynamic considerations, i.e. this is a restatement of the second law. This means however, that I now need to understand what is the relationship between the energy of charges in the electromagnetic field, which I would usually extract from the Hamiltonian, and the energy of the system in the thermodynamic sense. I expect the two must be the same, but I am finding it hard to justify it with non-handwave'y arguments.

Can anyone suggest a some literature (or a simple argument) on the matter?

Kind regards

PS: All of this is for electrostatics

 

[Charles Link]

In the formation of electric dipoles, for a linear material, the dipole is a stretching, like a mass on a spring, to a higher potential energy state. The equation $P=\chi E$ is similar to $F=-k x$. ($P=q d$, where $d$ is the distance the dipole is stretched. The result is $d=\alpha E$ where $\alpha=\frac{\chi}{q}$). $\\$ A compete thermodynamic electrostatic description would need to have this attractive linear response incorporated into it somehow. Then there is of course energy per unit volume $U=-\vec{P} \cdot \vec{E}$. (The vector $\vec{E}$ is the applied electric field). $\\$ Doing the thermodynamics to a mass on a spring in a uniform gravitational field, the equilibrium point is reached when the total potential energy is minimized. $U=-\vec{P} \cdot \vec{E}$ could be considered to be the negative charge moving in the externally applied electric field, and it would be the analog of the mass in the gravitational field, with $U_g=mgh$. ($mg \rightarrow q \vec{E}$, and $h \rightarrow d$). The electrostatic attraction of the negative and positive charges would be analogous to the spring constant force. $U_{spring}=\frac{1}{2}kh^2$ . $U_{total}=U_g+U_{spring }$ is minimized when $\frac{dU_{total}}{dh}=mg+kh=0$, so that $h=-\frac{mg}{k}$.

 

[vsv86]

Dear Charles Link

Thanks for the reply.

A compete thermodynamic electrostatic description would need to have this attractive linear response incorporated into it somehow.

Why would electrostatic description be linear? I know about the dielectric constant and susceptibility approaches to describing response of materials to electromagnetic field, but, as far as I know, this is only an approximation, with no claim to being fundamental. Afterall, where linear response stops, the nonlinear optics begins, and nonlinear optics is very much alive and well.

Doing the thermodynamics to a mass on a spring in a uniform gravitational field, the equilibrium point is reached when the total potential energy is minimized.

This is where I have a problem. What is the definition of energy? I am happy with the definition that says energy is the (scalar) quantity of the system that is conserved when system (i.e. the Lagrangian $L$) has no explicit time-dependence ($\left(\partial{L}/\partial{t}\right)_{\vec{r},\,\dot{\vec{r}},\dots}$) . One can then take the correct Lagrangian for the electromagnetic field + charges (and we know it is the correct one because it generates Maxwell's Equations), and extract the expression for the energy, such as the one you gave. More explicitly:

The charge density for a point-like electric dipole ($\vec{p}$) at the origin is $\rho=-\vec{\nabla}.\vec{p}\delta^{(3)}\left(\vec{r}\right)$, so

$\int d^3 r\, \rho \phi=-\int d^3 r\,\vec{\nabla}.\vec{p}\delta^{(3)}\left(\vec{r}\right) \phi =\vec{p}. \int d^3 r\,\delta^{(3)}\left(\vec{r}\right) \vec{\nabla}\phi = \vec{p}.\vec{\nabla}\phi\left(\vec{0}\right)=-\vec{p}.\vec{E}$

But how do we know that this is the thermodynamic energy? The one that is used to define the temperature for example $dU=TdS+dW$, where $U$ is energy, $T$ is temperature, $S$ is entropy and $W$ is work?

 

[Charles Link]

In a solid, equilibrium is reached because the potential is a minimum. If you work through the above equations, you can write $U_{electrostatic}=\frac{1}{2} \beta P^2$ for some $\beta$. If my algebra is correct, $\beta=\frac{1}{q \chi}$. With no external electric field, the minimum energy occurs when $P=0$. $\\$ From what I learned of solid state physics, there are indeed anharmonic terms in the Hamiltonian, or properties such as thermal expansion would not occur. See also: https://www.docsity.com/en/anharmonic-effects-solid-state-physics-lecture-slides/441110/ $\\$ In solid state physics, they often consider the positive nucleus as the item of interest, and consider the forces on the nucleus as it is displaced from its equilibrium position.