# What effective mass to use for conductivity along non-principal directions?

Say I have an ellipsoidal dispersion:

$$E(k) = \frac{\hbar^2}{2}\left[\frac{k_x^2}{m_x}+\frac{k_y^2}{m_y}+\frac{k_z^2}{m_z}\right]$$

If I want to talk about the conductivity in the x-direction, I need to use $$m_x$$.

But what if I want to talk about the conductivity along the unit vector $$(1,1,1)$$? More generally, what is the effective mass we use when talking about the conductivity along some non-principal directions?

I'm studying is the application of Drude theory in semiconductors, where the conductivity along a principal direction is given by $$ne^2\tau/m_i$$, where $$m_i$$ is the effective mass along the direction.

Say I want to know the conductivity in the direction of the electric field $$\vec{E}$$, given that it does not point along any of the principal axes.

Here is how I analyze the problem:

Due to the electric field, the electron would be moving in all three of the $$x,y,z$$ direction, driven by $$E_x,E_y,E_z$$, respectively.

By this argument, the obvious way is to just write the conductivity $$\sigma$$ as a tensor, as shown in textbooks.

However, I was wondering if there is a way to express $$\sigma$$ as a scalar so that the analysis can be reduced to a one-dimensional problem by simply writing the equation as scalars: $$j=\sigma E$$.

I naively tried to add all the contributions together along with the corresponding cosines and sines of the angles the field makes with the principal axes. What I mean is that, if $$\theta$$ is the angle made by $$\vec{E}$$ with the $$z$$-axis, and $$\phi$$ is the angle made by the $$xy$$-projection with the $$x$$-axis, then

$$\sigma_{\parallel \vec{E}}=ne^2\tau\left[\frac{\sin{\theta}\cos{\phi}}{m_x}+\frac{\sin{\theta}\sin{\phi}}{m_y}+\frac{\cos{\theta}}{m_z}\right]$$

but this is clearly nonsensical.

• The reciprocal mass turns into a tensor, and you evaluate the tensor on the unit vector and obtain a velocity" that is not in the same direction as the momentum, which is fine. Just do it naïvely. Apr 27, 2023 at 9:22
• Could you add more details? Do you consider a metal, a semiconductor? Which conductivity model do you use, and from which equations do you find it? Apr 27, 2023 at 10:55

For the case of anisotropic mass tensor, the conductivity also becomes a tensor, as noted by naturallyInconsistent: $$$$\sigma = ne^2\tau\, \mathrm{diag}(m_x^{-1}, m_y^{-1}, m_z^{-1}).$$$$ In general, when $$m_x \ne m_y \ne m_z$$, the current $$\vec{j} = \sigma \vec{E}$$ is not parallel to the electric field: it will also contain a component perpendicular to the electric field. Therefore, there is no way to express $$\sigma$$ as a scalar and to reduce the problem to the one-dimensional case.
However, one can ask about the projection of the current $$\vec{j}$$ on the unit vector parallel to $$\vec{E}$$: $$$$j_\parallel = \frac{(\vec{j},\vec{E})}{|\vec{E}|} = \frac{(\sigma\vec{E},\vec{E})}{|\vec{E}|} = ne^2\tau|\vec{E}| \left[\frac{\sin^2{\theta}\cos^2{\phi}}{m_x}+ \frac{\sin^2{\theta}\sin^2{\phi}}{m_y} + \frac{\cos^2{\theta}}{m_z}\right] = \sigma_{\parallel\vec{E}}|\vec{E}|.$$$$ The proportionality coefficient $$\sigma_{\parallel\vec{E}}$$ can be called "the conductivity along the direction of $$\vec{E}$$":