0
$\begingroup$

I look to solve analytically the Vlasov-Maxwell equations for a magnetized hot plasma $$\frac{\partial f_{s1}}{\partial t}+\vec{v}\cdot\frac{\partial f_{s1}}{\partial \vec{r}}+\frac{q_s}{m_s}\Big(\vec{v}\wedge\vec{B}_0 \Big)\cdot\frac{\partial f_{s1}}{\partial \vec{v}} =-\frac{q_s}{m_s}\Big(\vec{E}_1+\vec{v}\wedge\vec{B}_1 \Big)\cdot \frac{\partial f_{s0}}{\partial \vec{v}}$$ $$\vec{\nabla}\cdot\vec{E}_1=\dfrac{1}{\varepsilon_0}\displaystyle\sum_s q_s \int f_{s1}\,\mathrm{d^3}v$$ $$\vec{\nabla}\cdot\vec{B}_1=0$$ $$\vec{\nabla}\wedge\vec{E}_1=-\frac{\partial \vec{B}_1}{\partial t}$$ $$ \vec{\nabla}\wedge\vec{B}_1=\mu_0 \displaystyle\sum_s q_s \int \vec{v}\; f_{s1}\,\mathrm{d^3}v+\frac{1}{c^2}\frac{\partial \vec{E}_1}{\partial t}$$

With $f_{s0}(\vec{v})\equiv f_{s0}(v_\perp,v_z)=$ the zero order distribution function in cylindrical geometry.

$(f_{s1}, \vec{E}_1, \vec{B}_1)=$ the perturbed parameters of order 1 , $B_{0}=$ the constant magnetic field.

In order to calculate the perturbed distribution function $f_{s1}$, resolution of Vlasov equation is done by the method of characteristics with a cylindrical geometry which involves very difficult calculations. (see Swanson p.91-95)

Instead of this method, other works use the development of $f_{s1}$ on the spherical harmonics $$f_{s1}= \displaystyle\sum_l\sum_m f_l^m(v)\;Y_l^m (\theta,\varphi)$$ then by projection of the $Y_l^m$ on Vlasov equation and the relation of orthogonality of $Y_l^m$ they form an infinit system of coupled equations in $f_l^m$.

So I wanted to apply this method of spherical harmonics in this problem but I couldn't solve the system of coupled equations in $f_l^m$. I therefore concluded that spherical harmonics cannot be applied.

Please, do you know of any other alternative methods to calculate the perturbed distribution function $f_{s1}$ in the same cylindrical geometry?

$\endgroup$

1 Answer 1

1
$\begingroup$

I therefore concluded that spherical harmonics cannot be applied.

I think this is a premature assessment. Dum et al. [1980] provides a nice derivation of the dispersion relation using spherical harmonics. Let's first start with some background: $$ \begin{align} Y_{l}^{m}\left( \theta, \phi \right) & = N_{l}^{m} \ P_{l}^{m}\left( \cos{\theta} \right) \ e^{i \ m \ \phi} \tag{0a} \\ N_{l}^{m} & = \left[ \frac{ \left( 2 l + 1 \right) \left( l - m \right)! }{ 4 \ pi \left( l + m \right)! } \right]^{1/2} \tag{0b} \\ Y_{l}^{-m} & = \left( -1 \right)^{m} \ \left( Y_{l}^{m} \right)^{*} \tag{0c} \\ P_{l}^{m}\left( x \right) & = \left( -1 \right)^{m} \ \left( 1 - x^{2} \right)^{m/2} \ \frac{ 1 }{ 2^{l} \ l! } \frac{ d^{l + m} }{ dx^{l + m} } \left( x^{2} - 1 \right)^{l} \tag{0d} \\ P_{l}\left( \hat{\mathbf{k}} \cdot \hat{\mathbf{v}} \right) & = \frac{ 4 \pi }{ 2 l + 1 } \sum_{m = -l}^{l} \ Y_{l}^{m}\left( \hat{\mathbf{k}} \right) \ \left( Y_{l}^{m} \right)^{*}\left( \hat{\mathbf{v}} \right) \tag{0e} \\ Q_{l}\left( z \right) & = \frac{ 1 }{ 2 } \int_{-1}^{+1} dt \ P_{l}\left( t \right) \frac{ 1 }{ z - t } = \left( -1 \right)^{l + 1} Q_{l}\left( -z \right) \tag{0f} \\ P_{l}\left( z \right) & = \frac{ 1 }{ 2^{l} \ l! } \ \frac{ d^{l} }{ dz^{l} } \left( z^{2} - 1 \right)^{l} \tag{0g} \\ W_{l - 1}\left( z \right) & = \sum_{m = 1}^{l} \ \frac{ 1 }{ m } \ P_{m - 1}\left( z \right) \ P_{l - m}\left( z \right) \tag{0h} \\ Q_{0}\left( z \right) & = \frac{ 1 }{ 2 } \ln{ \frac{ z + 1 }{ z - 1 } } \tag{0i} \\ Q_{l}\left( z \right) & = P_{l}\left( z \right) \ Q_{0}\left( z \right) - W_{l - 1}\left( z \right) \tag{0j} \end{align} $$

where the $P_{l}\left( z \right)$ are Legendre polynomials and $\mu = \hat{\mathbf{k}} \cdot \hat{\mathbf{v}}$ is the cosine of the wave normal angle relative to particle velocity.

We can now write the dispersion relation as: $$ \begin{align} \epsilon_{j}\left( \mathbf{k}, \omega \right) & = \sum_{l,m} Y_{l}^{m}\left( \hat{\mathbf{k}} \right) \left( \frac{ \omega_{j} }{ k } \right)^{2} \ \int_{0}^{\infty} \ dv \ 4 \pi \ v^{2} \Biggl\{ \frac{ 1 }{ 2 } \int_{-1}^{+1} d\mu \\ & \frac{ 1 }{ \hat{\omega} - \mu } \ \frac{ k }{ k \ v } \left[ \mu \frac{ \partial }{ \partial v } + \left( 1 - \mu^{2} \right) \frac{ 1 }{ v } \frac{ \partial }{ \partial \mu } \right] \ P_{l}\left( \mu \right) \ f_{l}^{m}\left( v \right) \Biggl\} \tag{1} \end{align} $$

where $\hat{\omega} = \tfrac{ \omega }{ k \ v }$, and the $f_{l}^{m}\left( v \right)$ is expressed as:

$$ \sum_{l,m} Y_{l}^{m}\left( \hat{\mathbf{k}} \right) \ f_{l}^{m}\left( \frac{ \omega }{ k } \right) = f\left( \frac{ \omega }{ k } \ \hat{\mathbf{k}} \right) \tag{2} $$

This is specific to an unmagnetized plasma, yes, but it allows for electromagnetic waves (so long as they propagate along the magnetic field). This can be generalized to a magnetized plasma with more work.

You can also look at Vinas and Gurgiolo [2009] for a discussion of generating velocity moments from spherical harmonic expansions of the velocity distribution function.

$\endgroup$
4
  • $\begingroup$ @honeste-vivere tank you for the explanation; I remind you that my goal is to calculate first the perturbed distribution function $f_{s1}$ in terms of spherical harmonics and then to find the dispersion relation through the Maxwell-Ampere equation. On the other hand the calculation of Dum is specific to an unmagnetized plasma and for electromagnetic modes propagating along the magnetic field. So is it valid for magnetized plasmas where we are dealing with the dielectric tensor and for arbitrary $f_{s0}(v_\perp,v_\parallel)$? Can you give me your email for a more detailed discussion please? $\endgroup$
    – Gallagher
    Commented Feb 11, 2022 at 15:57
  • 1
    $\begingroup$ @Gallagher - Yes, this is for an unmagnetized plasma but the distribution function is generalized already, i.e., it has 3D velocities. The limitation here in the dielectric tensor is only in $\mathbf{k}$. $\endgroup$ Commented Feb 11, 2022 at 16:04
  • $\begingroup$ @honeste-vivere - Please, any idea or Ref. for propagation in a magnetized plasma ? Can you give me your email ? $\endgroup$
    – Gallagher
    Commented Feb 11, 2022 at 16:12
  • $\begingroup$ @honeste-vivere please take my email address: [email protected] $\endgroup$
    – Gallagher
    Commented Feb 11, 2022 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.