Plasma physics - relativistic derivation

[SMannion]

Hello,

I am trying to work through attached paper, deriving from equation 2.2 to 2.4. I am not familiar with the notation. If I try and get the integral of inside the <> brackets, I end up with a different eqn 2.4.
I need some maths help here :)
Any help would be greatly appreciated.

Sinéadhttp://www.jetp.ac.ru/cgi-bin/dn/e_049_03_0483.pdf

 

[PeterDonis]

If I try and get the integral of inside the <> brackets, I end up with a different eqn 2.4.

What do you get?

 

[jtbell]

And how did you get it? If you show your working, someone may be able to point out any mistakes.

Note that you can write equations using LaTeX, according to https://www.physicsforums.com/help/latexhelp/ .

 

[SMannion]

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...