Understanding L&L's argument, constant current in a crystal

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fluidistic
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I'm trying to go through chapter III of the vol.8 of Landau and Lifshitz series. (Fortunately the book is uploaded to the archive.org, I guess it is in the public domain.)
At page 87 ( << Archive.org link deleted by the Mentors because of copyright violation >> ), they speak about the Joule effect and that this effect makes the entropy increase. Namely ##dS/dt = \int (\vec J \cdot \vec E )/T dV## and that this quantity must be positive. Where ##\vec J## is the current density vector, and it's worth ##j_i = \sigma_{ik}E_k##. So far so good.

Then they make a remark that I do not quite understand/buy. They say that the symmetry of the crystal would allow ##j_i = \sigma_{ik}E_k + j_i^{(0)}## where ##\vec J^{(0)}## is a constant vector. But that in reality this term cannot exist, because entropy must increase, and since the term ##\vec J^{(0)}\cdot \vec E## in the integrand could take either sign, ##dS/dt## would not be invariably positive.

I do not quite understand the above logic. I do understand that we must have ##dS/dt>0##. Therefore we must have ##\underbrace{\int (\sigma \vec E) \cdot \vec E / T dV}_{>0. \text{not sure I can assume that now...}} + \int \vec J^{(0)} \cdot \vec E / T dV >0##. Thus the second integral could even be negative, yet the sum of the two integrals be positive and we would have no problem whatsoever. If that second term is positive or zero, no problem either. The problem arises when the second integral is lesser than minus the first integral. I do not see how L&L can fix it to zero with the logical argument they give.

Does someone understand L&L's argument here? If so, how would you explain it better to me?

Thank you!
 
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The first term is quadratic in E, while the second term is linear. So for sufficiently small E, the sign would be dominate by that of the second term.
 
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DrDu said:
The first term is quadratic in E, while the second term is linear. So for sufficiently small E, the sign would be dominate by that of the second term.
Very good point. But then what would be the problem of having a positive second term?
 
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