We start from the dispersion equation for the longitudinal oscillations
$$\epsilon(\omega,k)=0 (2.1)$$
with the following expression for the permittivity
$$\epsilon(\omega,k)=1-\omega_p^2\langle \gamma^{-3} (\omega-kv)^{-2}\rangle (2.2)$$
It is assumed that the perturbations depend on the coordinates and on the time like ##exp(-iwt + ikz)##.
The direction z is arbitrary in the absence of a magnetic field, and coincides with the direction of the intensity vector
of this field if such a field is present in the plasma.
The symbol <> denotes averaging over the particle momenta, ##<(. . . )> = \int f_0(p )(. . . )dp/n, n = \int f_0,(p)dp## is the plasma density, ##f_0(p)## is the equilibrium particle momentum distribution function (one/dimensional), ##\gamma = (1 +p^2/m^2c^2)^{1/2} ## is the relativistic factor, ##v = c(1 +m^2c^2/p^2)^{-1/2} ## is the particle velocity, m is its mass, ## \omega_p^2: = 4\pi e^2n/m ## is the square of the plasma frequency, and e is the particle
charge.
We confine ourselves to a one-dimensional particle distribution in momentum (the case of greatest interest
for applications to the pulsar problem). We find with the aid of (2.1) and (2.2) that the longitudinal oscillations have a phase velocity equal to that of light, ## \omega/k= c## at ##k = k_0##, where ##k_0##, is defined by the relation ##k_0^2= 2\omega_p^2 <\gamma>/c^2##. In this case ##\omega = \omega_0=ck_0##,. Using this result, we can obtain by successive approximations(cf. Refs. 6 and 7) a solution of (2.1) with wave numbers close to ##k_0##. Assuming ##k+ k_0+\Delta k, \omega =\omega_0+\Delta \omega,## where ##\Delta k## and ##\Delta \omega## are small quantities, and taking into account
the result of the zeroth approximation ##\epsilon(\omega,k)=0## we obtain in the first approximation
$$\Delta \omega = -\bigg ( \frac{\frac{\partial \epsilon}{\partial k}}{\frac{\partial \epsilon}{\partial \omega}}\bigg)_0 \Delta k (2.3)$$
The zero subscript in the right/hand side of this equation means that the corresponding derivatives are taken at ##k = k_0, \omega = \omega_0##. In this case
$$\bigg(\frac{\partial \epsilon}{\partial \omega} \bigg)_0=\frac{2\omega_p^2}{k_0^3 c^3}\bigg\langle \gamma^{3} \bigg(1+\frac{v}{c}\bigg)\bigg \rangle^3, \bigg(\frac{\partial \epsilon}{\partial k} \bigg)_0=-c(1-\alpha)\bigg(\frac{\partial \epsilon}{\partial \omega} \bigg)_0, \alpha=\bigg\langle \gamma \bigg(1+\frac{v}{c}\bigg)^2\bigg \rangle \bigg/ \bigg\langle \gamma^{3} \bigg(1+\frac{v}{c}\bigg)\bigg \rangle^3$$Phew that took 2 hours to type!
I took the ##\partial \epsilon / \partial \omega## to start with, I think I'm missing something with the <>. The ##\bigg(\frac{\partial \epsilon}{\partial \omega} \bigg)_0=...\bigg\langle \gamma^{3} \bigg(1+\frac{v}{c}\bigg)\bigg \rangle^3 ## is not inverted in my case. Please any help would be much appreciated...
