I am reading a german physics book called "Festkörperphysik" written by Gross/Marx.

And I came to following situation: $$\frac{d\langle n\rangle}{dt}=\frac{\partial\langle n \rangle}{\partial t}\bigg |_{\text{Diffusion}}+\frac{\partial\langle n \rangle}{\partial t}\bigg |_{\text{Zerfall}}$$ $$\frac{\partial\langle n \rangle}{\partial t}\bigg |_{\text{Zerfall}}=-\frac{\langle n \rangle - \langle n \rangle ^0}{\tau}$$

The diffusion term is related to the temperature gradient. in a time interval $ \Delta t $ all phonons, which were originally located at the location $ x-v_x \Delta t $, arrive at the location $ x $. We can therefore write: $$\frac{\partial\langle n \rangle}{\partial t}\bigg |_{\text{Diffusion}}=\lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t}[\langle n(x-v_x \Delta t)\rangle - \langle n(x)\rangle ] = -v_x \frac{\partial \langle n \rangle}{\partial x}=-v_x \frac{\partial \langle n \rangle ^ 0}{\partial T}\frac{\partial T}{\partial x}$$

I don't understand the derivation of the diffusion-part, which is coming from diffusing phonon's into a certain domain.

My problem is that I do not understand how the velocity v_x and temperature gradient are getting in there.

I hope someone can give me some tips for my issue please!

More about the heat flux density below

In the formulas below I use the momentum $\vec q$ of a phonon and the count of solutions $r$ from the dispersion relation $\omega(\vec q, r)$. $r$ is then also the number of atoms from the basis of the lattice.

E.g. we want to determine the heat flux density $J_{h,x}$ of a cuboid with length $v_x \tau$ in x-direction. Also the group velocity of certain phonons, which are transporting energy, in x direction $v_x(\vec q, r) = \frac{\partial }{\partial q_x}\omega_{\vec q, r}$.

enter image description here

The heat quantity is $Q=(U/V)\cdot Av_x\tau$ with the internal energy of the phonon system $$\langle U \rangle = U^{eq}+\sum_{\vec q, r}\frac{1}{2}\hbar \omega_{\vec q,r} + \sum_{\vec q, r} \hbar \omega_{\vec q,r}\langle n_{\vec q, r}\rangle$$

The first term is the energy of the lattice-atoms itself, which can be set to zero. The second term is the energy of the oscillation (I think you can say 'zero-point energy') and the last term is the energy coming from thermal excitation.

$\langle n_{\vec q, r}\rangle$ is the occupation number of the photons, which comes from bose-einstein statistics $$\langle n_{\vec q, r}\rangle = \frac{1}{\exp(\frac{\hbar \omega(\vec q,r)}{k_B T})-1}$$

Okay, so the definition of the heat flux density is the ratio between the heat quantity $Q$ and the cross sectional area A times the energy-transport time $\tau$ (see figure above). Which brings us to $$J_{h,x}=(U/V)v_x = \frac{1}{V}\sum_{\vec q, r} \hbar \omega_{\vec q,r} \left( \frac{1}{2}+ \langle n_{\vec q, r}\rangle \right) \frac{\partial }{\partial q_x}\omega_{\vec q, r}$$

For better overview I will not write the indices $q$ and $r$ anymore.

When we have thermal equilibrium $J_h$ will be zero, because $\langle n \rangle$ for positive and negative $q$ will be the same. Also according to symmetry $v_x(q) = -v_x(-q)$.

A finite heat flow exists iff $\langle n \rangle$ deviates from $\langle n \rangle^0$. Now we want to express $J_{h,x}$ with the deviation $\langle n \rangle - \langle n \rangle^0$:

$$J_{h,x} = \frac{1}{V}\sum_{\vec q, r} \hbar \omega\left(\langle n \rangle - \langle n \rangle^0\right)v_x$$

Now $\langle n \rangle$ can deviate in two different ways: Diffusion + Zerfall (see formulas at the very beginning of this post)

And let's assume that we only approach steady states, i.e. for them is $\frac{d\langle n \rangle}{dt}=0$. Therefore the deviation over time from "Zerfall" is zero and we can express the deviation from "Diffusion" with the thermal gradient, which is also mentioned in the formula at the beginning of this post or in the first answer below it.

We can write the heat density flux as followed when we put the deviation's in it and also according to the book:

$$J_{h,x}=-\frac{1}{V}\sum_{\vec q, r} \hbar \omega \tau v_x^2\frac{\partial \langle n \rangle ^ 0}{\partial T}\frac{\partial T}{\partial x}$$

Now a question: I guess $\tau$ comes from the "Zerfall"-deviation. But how does it get there?

This formula above contains the specific heat capicity $$c_V=\frac{1}{V}\sum_{\vec q, r} \hbar \omega \frac{\partial \langle n \rangle ^ 0}{\partial T}$$

According to our assumptions above we have a equilibrium, i.e. $\frac{\partial \langle n \rangle ^ 0}{\partial T}$ should be zero. Mentioning this in the formula above in $J_{h,x}$ is fine, because you theoretically cancel $\partial T$, but what about the specific heat capicity $c_V$ itself? It must be zero, when you only look at $c_V$ and not at $J_{h,x}$ as a whole.

I even checked this derivation in another german book and it says the same. So why are they exaclty using $\langle n \rangle ^ 0$, which apparently is the so-called equilibrium value of phonons?

  • $\begingroup$ Hi, welcome to Physics SE! Please don't post formulae as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. I've edited it here as an example. Look at this Math SE meta post for a quick tutorial. Also, next time ensure the content's in English, though I had fun thinking of the most efficient way to translate a picture from German to English ;) $\endgroup$
    – user191954
    Jul 17, 2018 at 16:10
  • $\begingroup$ Ohh thank you and sorry, chair ^^! Won't happen again in future! $\endgroup$
    – Gamdschiee
    Jul 17, 2018 at 16:29

1 Answer 1


The limit term is just from the definition of the partial time derivative for a difusive process, where the particle moves a distance $v_x \Delta t=\Delta x$ in time $\Delta t$.

You can then rewrite the limit as

$$\lim \limits_{\Delta t \to 0} \frac{v_x}{v_x \Delta t}[\langle n(x-v_x\Delta t)\rangle - \langle n(x) \rangle]$$

$$=\lim \limits_{\Delta x \to 0} \frac{v_x}{\Delta x}[\langle n(x-\Delta x)\rangle - \langle n(x) \rangle]$$

Then we can do $\Delta x \rightarrow -\Delta x$

$$=\lim \limits_{\Delta x \to 0} -\frac{v_x}{\Delta x}[\langle n(x+\Delta x)\rangle - \langle n(x) \rangle]$$

$$=-v_x\frac{\partial \langle n \rangle}{\partial x}$$

Then the final step is just a chain rule:

$$-v_x\frac{\partial \langle n(T) \rangle}{\partial x}=-v_x\frac{\partial \langle n \rangle}{\partial T} \frac{\partial T}{\partial x}$$

Based on your comment, we can now replace $\langle n \rangle$ in the partial derivative with $\langle n \rangle ^0$, thus giving us the part on the right.

  • $\begingroup$ Right, should've mentioned that. I will translate 1:1 from the book: <n> with superscript zero is the value of phonons in equilibrium. It can be replaced with <n> after we placed the temperature gradient in the formular, because we assumed a steady state and also a local equilibrium. $\endgroup$
    – Gamdschiee
    Jul 17, 2018 at 16:23
  • 1
    $\begingroup$ @Gamdschiee I have edited the answer. Is there any part of it that does not make sense? $\endgroup$ Jul 17, 2018 at 16:48
  • $\begingroup$ @Aron Stevens I've also edited my question above. I went into more detail. I read the sites in my book again and wrote all important information down. It leads to a question, which you asked me before: "$\frac{\partial \langle n \rangle ^ 0}{\partial T}=0$" Would be cool, when we can discuss this topic further as I do in my original post, so I can fully understand it. I hope I translated/understood it good. $\endgroup$
    – Gamdschiee
    Jul 17, 2018 at 19:00
  • 1
    $\begingroup$ @Gamdschiee This is not my area of expertise, so I will need to take some time to see if I can go deeper into what you are saying. Although, I might have misspoken when I said that partial derivative is equal to $0$. I was thinking about the derivative with respect to time, not temperature. $\endgroup$ Jul 18, 2018 at 2:58

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