# Why does the concept of effective mass fail for linear dispersion relations?

I had a discussion about graphene and stumbled about a conceptual problem of myself about the effective mass of electrons close to the Dirac point in graphene or for any linear dispersion relation in general. My problem is that by calculating the effective mass by:

\begin{align} \frac{1}{m_e^*} = \frac{1}{\hbar^2} \frac{\partial^2\epsilon(k)}{\partial k^2} \end{align} And using $$E(k) = \alpha k$$, linear disperion, the result should be $$m_e^* = \infty$$.

This is contradicted by my knowledge, that the effective mass of the electron should be very small around the Dirac-Point. How does that fit together?

Furthermore, in some paper it was stated that one should only use the given formula for parabolic dispersion relations. Why is that so? I couldn't find any part in the derivation that would limit this formula to parabolic dispersion relations only.

I would appreciate some explanations and clarifications!

The actual problem is that dispersion relation is still not linear: it's

$$\varepsilon(\vec k)=\alpha|\vec k|,$$

which in 1-dimensional case is

$$\varepsilon(k)=\alpha|k|=\alpha k(\theta(k)-\theta(-k)),$$

where $$\theta$$ is the Heaviside step function.

Then the derivative of energy with respect to the wavenumber will be

$$\varepsilon''(k)=2\alpha\delta(k),$$

where $$\delta$$ is the Dirac delta. Roughly speaking, it's infinite at $$k=0$$, so its reciprocal would then be zero.

This becomes more evident if you break some symmetry of the crystal so that a small band gap appears in the band structure, in which case the effective mass becomes nonzero but still quite small. Or just look at the effective mass you get for a sequence of $$\varepsilon_n(k)=\alpha \sqrt{k^2+1/n^2}$$ as $$n\to\infty$$.

Update: The solution lies in the definition of the effective mass. It is derived semiclassically with Newton's 2nd law $$F = ma$$. But in this derivation, the mass is implicitly assumed to be constant. But for non-parabolic Bands, the effective mass changes along the band. So, one cannot use the common definition for the effective mass.