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But then, no proof nor anything involving $k_BT$ is shown anymore. What's more, it is then showed that the number of electrons involved in electrical conduction is proportional to the applied electric field's magnitude, which makes entirely sense to me. More precisely, that number is proportional to $2eEL_m$ where $L_m$ is the mean free path, which is worth about $10$ nm.

But then, no proof nor anything involving $k_BT$ is shown anymore. What's more, it is then showed that the number of electrons involved in electrical conduction is proportional to the applied electric field's magnitude, which makes entirely sense to me. More precisely, he shows that the difference in energy between the most energetic electrons that create a current and the least energetic ones that also carry a current is worth $2eEL_m$ where $L_m$ is the mean free path, which is worth about $10$ nm. In other words, the width of energy around $E_F$ that electrons that create the current have, has nothing to do with $k_BT$.

Edit regarding Jon Custer's comment:

Ashcroft and Mermin discuss this in their Chapter 13, the Semiclassical Theory of Conduction in Metals. Following a volume of electrons as they move through phase space, their ends up being a factor of the derivative of the Fermi function with energy which is non-zero only within a few kT of the Fermi energy.

I had checked that chapter, and seen that the conductivity can be written as an integral with a term that contains $\partial f/\partial \varepsilon$ which involves $k_BT$ (as I wrote in the previous paragraph), but I fail to see how this implies that the electrons responsible for the current at the ones that are within $k_BT$ of $E_F$. But that is indeed probably the way to go. But still, I would have to see where I go wrong in my reasoning that I exposed above.

It is commonly written in textbooks that in metals the electrons responsible for an electric current are the ones that have an energy about $E_F$ and a few $k_BT$ around that energy. See for example Datta's textbook "Electronic transport in mesoscopic systems" page 37 (book available as PDF from a Google search):

It is easy to see why the current flows entirely within a few $k_BT$ of the quasi-Fermi energy.

But then, no proof nor anything involving $k_BT$ is shown anymore. What's more, it is then showed that the number of electrons involved in electrical conduction is proportional to the applied electric field's magnitude, which makes entirely sense to me. More precisely, that number is worth $2eEL_m$ where $L_m$ is the mean free path, which is worth about $10$ nm.

I can understand that regarding the calculations for the specific heat, it is indeed true that only electrons having an energy about $k_BT$ around the Fermi energy (of the order of $1$ eV for metals) can absorb thermal energy, which is itself of the order of $k_BT$ (so about $10^{-5}eV$ to $10^{-3}eV$). It is easy to realize when one uses the fact that electrons are fermions and that at room temperature a metal is similar to a cold Fermi gas. Thus the electrons are roughly forming a sphere (let's take alkali metals to make things simple) in k-space and all states below the surface are occupied. The surface of the sphere is blurred due to finite temperature, in an energy amount around $k_BT$. So that the electrons that are below the surface by more than $k_BT$ cannot absorb thermal energy because the states above them are all occupied. It is only in that $k_BT$ window-range that electrons can absorb thermal energy.

But when I apply the same logic to an electric current, i.e. we apply an electric field to the metal instead of a temperature, I do not get anything related to $k_BT$ anymore. By considering that we apply $1$ V on a $1$ cm sample, the electric's field magnitude is about $100$ V/m which translate as an energy of about $10^{-6}$ eV. In other words, the electric field is a very tiny perturbation to the system, it is about 40 times smaller than rising the temperature of a metal by 1 K. I would expect then that only electrons having an energy around the Fermi energy $E_F$ with a margin equal to that extremely small $10^{-6}$ eV amount would be able to react to the field and produce a current. This has absolutely nothing to do with $k_BT$ and is in fact proportional to $|\vec E|$, as it intuitively (to me at least) should. I.e. I get something linear in the strength of the perturbation, just like with the case of the thermal energy with its thermal perturbation.

So, I do not see, for the life of me, how to reach to the conclusion that only electrons that have an energy within $k_BT$ of $E_F$ are able to produce a current.

I am well aware of the Fermi-Dirac distribution and how its derivative with respect to energy is non zero only around $E_F$, also of the density of states and how temperature affects it, etc. But I fail to see how it is relevant to answer my question.

It is commonly written in textbooks that in metals the electrons responsible for an electric current are the ones that have an energy about $E_F$ and a few $k_BT$ around that energy. See for example Datta's textbook "Electronic transport in mesoscopic systems" page 37 (book available as PDF from a Google search):

It is easy to see why the current flows entirely within a few $k_BT$ of the quasi-Fermi energy.

But then, no proof nor anything involving $k_BT$ is shown anymore. What's more, it is then showed that the number of electrons involved in electrical conduction is proportional to the applied electric field's magnitude, which makes entirely sense to me. More precisely, that number is proportional to $2eEL_m$ where $L_m$ is the mean free path, which is worth about $10$ nm.

I can understand that regarding the calculations for the specific heat, it is indeed true that only electrons having an energy about $k_BT$ around the Fermi energy (of the order of $1$ eV for metals) can absorb thermal energy, which is itself of the order of $k_BT$ (so about $10^{-5}eV$ to $10^{-3}eV$). It is easy to realize when one uses the fact that electrons are fermions and that at room temperature a metal is similar to a cold Fermi gas. Thus the electrons are roughly forming a sphere (let's take alkali metals to make things simple) in k-space and all states below the surface are occupied. The surface of the sphere is blurred due to finite temperature, in an energy amount around $k_BT$. So that the electrons that are below the surface by more than $k_BT$ cannot absorb thermal energy because the states above them are all occupied. It is only in that $k_BT$ window-range that electrons can absorb thermal energy.

But when I apply the same logic to an electric current, i.e. we apply an electric field to the metal instead of a temperature, I do not get anything related to $k_BT$ anymore. By considering that we apply $1$ V on a $1$ cm sample, the electric's field magnitude is about $100$ V/m which translate as an energy of about $10^{-6}$ eV. In other words, the electric field is a very tiny perturbation to the system, it is about 40 times smaller than rising the temperature of a metal by 1 K. I would expect then that only electrons having an energy around the Fermi energy $E_F$ with a margin equal to that extremely small $10^{-6}$ eV amount would be able to react to the field and produce a current. This has absolutely nothing to do with $k_BT$ and is in fact proportional to $|\vec E|$, as it intuitively (to me at least) should. I.e. I get something linear in the strength of the perturbation, just like with the case of the thermal energy with its thermal perturbation.

So, I do not see, for the life of me, how to reach to the conclusion that only electrons that have an energy within $k_BT$ of $E_F$ are able to produce a current.

I am well aware of the Fermi-Dirac distribution and how its derivative with respect to energy is non zero only around $E_F$, also of the density of states and how temperature affects it, etc. But I fail to see how it is relevant to answer my question.