Difference between revisions of "Solution details"

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<math>  (\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})</math>''n'' +  
<math>  D (\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})</math>''n'' +  
<math> D_L \frac {\partial^2{}}{\partial{z^2}}</math> -
<math> D_L \frac {\partial^2{}}{\partial{z^2}}</math> -
<math> W \frac {\partial{}}{\partial{z}}</math> ''n'' = 0
<math> W \frac {\partial{}}{\partial{z}}</math> ''n'' = 0
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<math> n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'}  </math>
<math> n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'}  </math>
<math> n_{dipole}  = -(\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
<math> n_{dipole}  = (\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = </math>
<math> -\frac{\partial}{\partial z}n_{monopole}  = (\frac{\pi}{2\lambda_L })^{1/2} -\frac{\partial}{\partial z}\frac{ e^{-\lambda_L (r'-z)}}{r'}  </math>
<math> -\frac{\partial}{\partial r'}n_{monopole}  = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial r'}\frac{ e^{-\lambda_L (r'-z)}}{r'}  </math>
for <math> r'= \sqrt {\rho'^2 + z^2}</math>
for <math> r'= \sqrt {\rho'^2 + z^2}</math>

Latest revision as of 14:31, 23 September 2014

Asymptotic solution details for Boltzmann equation for a hole has a uniform electric field

[math] D (\frac {\partial^2{}}{\partial{x^2}} +\frac {\partial^2{}}{\partial{x^2}})[/math]n + [math] D_L \frac {\partial^2{}}{\partial{z^2}}[/math] - [math] W \frac {\partial{}}{\partial{z}}[/math] n = 0

Steps to solve Boltzmann equation <ref name="Huxley"> Huxley, L. G. H. Leonard George Holden, The diffusion and drift of electrons in gases, John Wiley and sons, 1974 , call number QC793.5.E628 H89 </ref>

for the previous equation let consider the asymptotic solution has the form:

[math] n(x', y', z') = e^{\lambda_L z'} V(x,y,z) [/math]


[math] \nabla'^2 V = \lambda_L^2 V [/math]


[math] \nabla'^2 V = \frac {\partial^2{}}{\partial{x'^2}} + \frac {\partial^2{}}{\partial{y'^2}} + \frac {\partial^2{}}{\partial{z^2}}[/math]


[math] x' = \frac {D_L}{D} x [/math] [math] y' = \frac {D_L}{D} y [/math]

In spherical coordinates:

[math] \frac {1}{r'^2} \frac{\partial{}}{\partial{r'}}r'^2 \frac{\partial{V}}{\partial{r'}} + \frac {1}{r'^2 sin\theta'} \frac{\partial{}}{\partial{\theta}} sin\theta \frac{\partial{V}}{\partial{\theta}} = \lambda_L^2 V [/math] which is symmetric in [math]\phi[/math] direction.

Assuming [math]V(r',\theta) = R_k(r')P_k(\mu) [/math]the solution of the zenith angle direction is the Legendre polynomial, and can be written as:

[math]\frac {1}{r'sin\theta} \frac{\partial{}}{\partial{\theta}} sin\theta\frac{\partial{V}}{\partial{\theta}} = R_k(r') \frac{d}{d \mu} \left [ (1- \mu^2) \frac{d{P_k(\mu)}}{d{\mu}} \right] [/math]


[math] \frac{d}{d \mu} \left [ (1- \mu^2) \frac{d{P_k(\mu)}}{d{\mu}} \right]= -k(k+1) P_k(\mu) [/math]


[math] \frac {1}{r'^2} \frac{d}{dr'}\left (r'^2 \frac{dR_k}{dr'}\right) - \left [ \frac{k(k+1)}{r'^2} +\lambda_L^2 \right]R_k = \frac{d^2 R_k}{dr'^2} +\frac{2}{r'} \frac{dR_k}{dr'}-\left [ \frac{k(k+1)}{r'^2} +\lambda_L^2 \right]R_k = 0 [/math]

The modified Bessel functions, first and second kind, are the solutions for the previous equation but the boundary conditions determines which one to use, in this case [math] r'\rightarrow 0[/math], [math] n \rightarrow \infty [/math], and [math] n \rightarrow 0 [/math] as [math] r'\rightarrow \infty [/math]. so only the modified Bessel of second kind K_k are the non-zaro terms. so the the general solution for the equation can be written as :

[math] V= R_k (r') Pk(\mu) = \exp{(\lambda_L z)}\sum_{k=0}^{\infty} A_k r'^{-1/2} K_{k+1/2} (\lambda_L r') P_k(\mu) [/math]

Solution Analysis

The general form of the previous equation and its solution are defined as the following:

[math] \frac{d^2 y}{dx} +\frac{1-2\alpha}{x} \frac{dy}{dx}-\left [ \frac{\nu^2 \gamma^2 - \alpha^2}{x^2} + (\beta \gamma x^{\gamma - 1})^2 \right]y = 0 [/math]


[math] y = x^{\alpha} I_{\nu} (\beta x^{\gamma}) [/math] and

[math] y = x^{\alpha} K_{\nu} (\beta x^{\gamma}) [/math]

where [math] I_{\nu}[/math] and [math] K_{\nu}[/math] are the modified Bessel of the first and the second kind.

In case of of solving for the density number outside the hole, then the solution contains only the modified Bessel of the second kind [math] K_{\nu}[/math], Also applying the boundary conditions below [math] n= 0[/math], when [math] z = 0 [/math], and as [math] r \gt a [/math] where a is the hole radius.

implies that [math] P_{k} = 0[/math] if k is odd, so the solution can be written as :

[math] n= R_k (r') P_k(\mu) = \exp{(\lambda_L z)}\sum_{k=0}^{\infty} A_k r'^{-1/2} K_{k+1/2} (\lambda_L r') P_k(\mu) [/math]

In case of GEM preamplifer r >>a, since each hole has a radius of 50 um, and the electron streams appear in a mm scale distance; so [math] A_k [/math] s' values decrease since it is a function of [math] \lambda_L a [/math] ([math] \lambda_L a \lt \lt 1 [/math]), so the higher order terms become negligible compared to the first (monopole) and the second (dipole) term. Then the solution can be written for the dipole term (Anisotropic diffusion) :

[math] n = A_1 \exp{(\lambda_L z)} r'^{-1/2} K_{3/2} (\lambda_L r') P_1(\mu) [/math]

and for the monopole term.(Isotropic diffusion)

[math] n = A_0 \exp{(\lambda_L z)} r'^{-1/2} K_{1/2} (\lambda_L r') P_0(\mu) [/math]


[math] K_{1/2} = (\frac{\pi}{2\lambda_L r'})^{1/2} e^{-\lambda_L r'} [/math]

[math] K_{3/2} = (\frac{\pi}{2\lambda_L r'})^{1/2} (1+\frac{1}{\lambda_L r'})e^{-\lambda_L r'} [/math]

So :

[math] n_{monopole} = (\frac{\pi}{2\lambda_L })^{1/2} \frac{ e^{-\lambda_L (r'-z)}}{r'} [/math]

[math] n_{dipole} = (\frac{\pi}{2\lambda_L})^{1/2} \frac{z}{r'} (1+\frac{1}{\lambda_L r'})\frac{ e^{-\lambda_L (r'-z)}}{r'} = [/math]

[math] -\frac{\partial}{\partial r'}n_{monopole} = -(\frac{\pi}{2\lambda_L })^{1/2} \frac{\partial}{\partial r'}\frac{ e^{-\lambda_L (r'-z)}}{r'} [/math]

for [math] r'= \sqrt {\rho'^2 + z^2}[/math]