electron-gas theory of a metal (drude model) - intuition I'm back to studying "Microelectronics" by Millman & Grabel (2nd ed.). The book makes some references to the electron-gas theory, and I found out to have some problems with my intuition.
At each collision with the ions, does an electron randomly change only its direction or also the magnitude of its velocity? That is, is it the following animation correct?
http://youtu.be/07qqC85Qcpg
In the animation there is a weak electric field in the negative-x direction. Notice that the velocity of the red dot (the electron) happens to increase as well as decrease after a collision. Up to now, I've only imagined electrons changing direction at each collision without changing their speed.
Why an electron is said to lose its energy at each collision (page 24 Millman Grabel), if it can also increase its speed after a collision?
Thanks,
Luca
 A: Electrons can give/receive momentum/energy from the moving ions. Indeed, the electrons thus change also their magnitude of velocity.
The reason why there is on average kinetic energy transfer from the electrons is that these are driven by the electric field. Before the collision occurs, the electron's velocity is the sum of some random thermodynamic velocity $$\vec{v_T} = \sqrt{3kT/m_e}$$ and some directed velocity $$\vec{v_E} = q\vec{E}\Delta t/m_e$$ The average kinetic energy of such a configuration is $$m_e \frac{v_T^2 + v_E^2}{2} = \frac{3}{2}kT + \frac{(qE\Delta t)^2}{2m_e}$$ After the collision, the average energy is still $3 kT/2$ (neglecting the change of temperature during that time), which implies there is energy transfer on average (but as you noted correctly, not always).

Edit: Suppose we want to add two vectors $\vec A$ (corresponds to $\vec{v_E}$), which without loss of generality points to the positive x-direction and $\vec B$ (the random thermodynamic part), whose all components are random. We want to calculate the square-mean modulus of the resulting vector (as $v^2$ is proportional to the kinetic energy).
Here I use $\langle x \rangle$ to denote the average (not square-mean average) of $x$. Note that $\langle B_x \rangle = 0$.
Then the result is $$\langle |\vec A + \vec B|^2 \rangle = \langle (A_x + B_x)^2 + B_y^2 + B_z^2\rangle = $$
$$ = \langle A_x^2 + B_x^2 + B_y^2 + B_z^2 \rangle + \langle 2 A_x B_x \rangle = $$
$$ = \langle A_x^2 + B_x^2 + B_y^2 + B_z^2 \rangle + 2 \langle A_x \rangle \langle B_x \rangle = $$
$$ = \langle A_x^2 + B_x^2 + B_y^2 + B_z^2 \rangle = $$
$$ = \langle A_x^2 \rangle + \langle B_x^2 + B_y^2 + B_z^2 \rangle = $$
$$ = \langle |A|^2 \rangle + \langle |B|^2 \rangle $$

Edit2: About why $\langle X \rangle \langle Y \rangle = \langle XY \rangle$.
Let $f(x)$ be the probability distribution for $X$ and $g(y)$ for $Y$.
By definition:
$$\langle XY \rangle = \int\int f(x) g(y) dy dx = $$
$$ = \int f(x) (\int g(y) dy) dx = $$
$$ = \int f(x) \langle X \rangle dx = $$
$$ = \langle X \rangle \int f(x) dx = $$
$$ = \langle X \rangle \langle Y \rangle $$
For more details, see this (note that here the two vectors have independent values).
