Doubling I, what happens to v_d and n?

[fluidistic]

I am wondering (again) what's going on when we apply an electric field to a conductor. In particular if we apply an electric field that creates a current density $\vec J = \sigma \vec E$ (I'm neglecting thermoelectric effects), there is a certain number of electrons (of the order of $v_d / v_F$ per 1 free electron, i.e. about one in ten billion for a reasonable experimental setup) that are affected by the $\vec E$ field, and they move with speeds near $v_F$. Yes, this is in stark contrast to the obsolete Drude's model where every single free electron is affected by the $\vec E$ field and move in average with a velocity of $v_d$. But I do this on purpose, hence the graduate level prefix of this thread.
More precisely, $J = e \left ( \frac{nv_d}{v_F} \right )v_F$, where the quantity in parenthesis denotes the number of electrons that is affected by the $\vec E$ field and which moves with speeds at about $v_F$.

So, this suggest that if we double the current density, since $v_F$ won't be affected, either $n$ (the free electron density number), $v_d$ or both will be affected. At first I thought that the drift velocity would double, meaning that the number of electrons that participate in electrical conduction stays constant when we double the current. But then I thought that the apparent shift of the Fermi sphere is greater when we double the current, so the number of electrons taking part in electrical conduction is impacted by the doubling of current.

So what's going on when we double the current in an idealized (perfectly spherical Fermi surface) metallic conductor? Is the number of electrons that respond to the field doubled, or their velocity doubled, or a mix of both?

 

[Lord Jestocost]

The volume density of the free electrons, n, is given and cannot be changed.

 

[fluidistic]

The volume density of the free electrons, n, is given and cannot be changed.

Ah right, that makes sense. This means $v_d$ is doubled. I missed the fact that a displacement of the Fermi sphere can be done by modifying $v_d$ and does not necessarily involve a different number of electrons that reacted to the applied field.Edit: Nevermind. I finally get it. The number of electrons that react to the applied E field DOES double. This means $v_d$ doubles too. The total number of free electrons remains unchanged though. It's just the number of electrons affected by the E field that increases, and it makes sense because the E field acts like a small perturbation whose effect on the system is linear.