# In electrodynamics, why do we say $\mathbf J = \sigma \mathbf E$ and not $\mathbf J = \sigma (\mathbf E + \mathbf v \times \mathbf B)$?

Griffiths notes it's because charges have an extremely low $$\mathbf v$$, so it's essentially an approximation, but aren't charges meant to be electrons? How can they be moving slowly? I usually think of electrons as moving almost light speed typically.

• It can't be an approximation since its a definition of $\sigma$. I don't have any particular insight into why we define it like that beyond it seems to work out. – jacob1729 May 9 at 14:40
• Why do you usually think of electrons as moving at light speed? – zeldredge May 9 at 16:10
• At least close to it. Their mass is so small any force would produce enormous speeds. – sangstar May 9 at 16:17
• Just have a look at this page en.wikipedia.org/wiki/Drift_velocity#Numerical_example all will be clear – ohneVal May 9 at 16:26

Well, it's just that, an approximation, which experience has proved to suffice in many situations. As Griffiths says, $$\mathbf{J}$$ is proportional to the force per unit charge in many substances. If the only forces relevant to the problem are electromagnetic, we get the formula $$\mathbf{J}=\sigma(\mathbf{E + \mathbf{v}\times\mathbf{B}})$$, as you mentioned, which is more accurate than Ohm's law. So the approximation is warranted if the speed is sufficient small for the magnitude $$\sigma|v|\rm{B}\sin\theta$$ relative to the second term to be ignored. The fact that it often is must be verified by experience, and the formula can be used for materials in which this verification has been made.
Still, there's no reason charges can't move slowly, so I see no problem with the possibility of such an approximation. In fact, if electrons always moved at speeds close to $$c$$, non-relativistic quantum mechanics (which we know to be a very good theory) would be mostly useless. Electrons are "slow" to the point that one can get very good results in quantum theory (like the spectrum of the hydrogen atom) without taking relativistic effects into consideration (of course, there are also MANY situations in which they must be considered, but that has nothing to do with the question. My point is that there's just no reason why $$\mathbf{v}$$ can't be sufficiently small for this approximation to work.). You could even take an inertial frame in which they are at rest, if their velocity is constant.
• That's true, it always depends on what one want to measure, some effects just don't appear in certain approximations. Still, if the product of the relevant components of $B$ and $v_x$ is negligible, the hall potential will be small. If the charges move slowly, one may still observe a considerable potential if $B_z$ is big enough. – Othin May 9 at 18:26