A How does graphene Fermi velocity v_F link to the envelope propagation?

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my questions stemmed from reading the article in Physica E. Vol. 86, 10-16.
(https://www.sciencedirect.com/science/article/pii/S1386947716311365)

Why does the graphene Fermi velocity ##v_F## appear in Eq.(11) in this article,?
Eq.(11) is as follows:
$$
\frac{\partial \Omega_p(z,t)}{\partial z}+\frac{1}{v_F}\frac{\partial \Omega_p(z,t)}{\partial t}=i\alpha\gamma_3\rho_{21}(z,t)
$$
where ##\alpha=\frac{N\omega_1|\mu_{21}\cdot e_p|^2}{2\epsilon_r \hbar v_F \gamma_3}##,
and ##\Omega_p(z,t)=\Omega^0_p\eta (z,t)##; ##\eta(0,\tau)=\Omega^0_p e^{-[(\tau-\sigma)/\tau_0]^2}##.

As is well known, the graphene Fermi velocity ##v_F## comes from the nearest
neighboring carbon atom hopping #t# and their distance #a#, and even if slowly varying envelope
approximation(SVEA) has been considered, the group velocity of the pulse cannot be the Fermi velocity.

Could any professionals provide help, either guide me the derivation of the equation or provide
some effective references which can be used to derive the equation.
 
Last edited:
From the BCS theory of superconductivity is well known that the superfluid density smoothly decreases with increasing temperature. Annihilated superfluid carriers become normal and lose their momenta on lattice atoms. So if we induce a persistent supercurrent in a ring below Tc and after that slowly increase the temperature, we must observe a decrease in the actual supercurrent, because the density of electron pairs and total supercurrent momentum decrease. However, this supercurrent...

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