For an electron, the energy and 3-momentum for a 4-vector:
$$ p_{\mu} = (E/c, \vec p)$$
that satisfies (in all reference frames):
$$ p^{\mu}p_{\mu} = E^2/c^2-p^2c^2 = m^2c^2$$
and in the rest frame ($\vec p = 0$) reduces to:
$$ E = mc^2 $$
If we look at the De Broglie relationship, we can describe a free electron with a 4-wave-vector via:
$$ p_{\mu} = \hbar k_{\mu} =(\hbar \omega/c, \hbar \vec k)$$
So that:
$$ k^{\mu}k_{\mu} = \omega^2/c^2 -||k^2|| = \frac{m^2c^2}{\hbar^2}$$
which gives a dispersion relation:
$$ \omega = \sqrt{(ck)^2 + \frac{m^2c^4}{\hbar^2}}$$
A photon satisfies the same conditions, with $m=0$:
$$ E = pc$$
$$ \omega = ck $$
the latter being the well known relation describing dispersionless waves propagating at the speed of light, $c$.
Now if we consider light propagating in a waveguide, or say an O-wave in a plasma with plasma frequency $\omega_p$, the dispersion relation becomes:
$$ \omega = \sqrt{c^2k^2 + \omega_p^2}$$
Which mean EM waves with $\omega < \omega_p$ don't exist, because their is finite energy $(\hbar \omega_p)$ at zero wavenumber, and this is because the wave couples to the electrons in the plasma.
Note that this form is the same as the dispersion relation for an electron in free space. Because of coupling to the Higgs boson, there is non-zero frequency even at zero wavenumber:
$$ \omega_0 = \frac{mc^2}{\hbar}$$
which corresponds to finite energy at zero momentum, aka mass:
$$ E = \hbar\omega_0 = mc^2$$
So the Higgs acts more like a universal plasma or waveguide, than "molasses".
