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We shall be considering f=fbackground+δf=fbackground+δf^eikxf = f_{background} + \delta f = f_{background} + \hat{\delta f}e^{ikx} to be the perturbation that is introduced

The BGK collision operator is defined as:

C[f]=(ffMB(T(f),ρ(f)))τ\begin{align} C[f] = -\frac{(f - f_{MB}(T(f), \rho(f)))}{\tau} \end{align}

In order to determine fMBf_{MB} as a function of fbackgroundf_{background}, we first need to evaluate the temperature and density distributions by computing the zeroth, and 2nd moments of the distribution function ff

ρ(f)=fdv=(fbackground+δf^eikx)=ρbackground+δf^eikx=ρbackground+δρ\begin{align} \rho(f) = \int f dv = \int (f_{background} + \hat{\delta f}e^{ikx}) = \rho_{background} + \int \hat{\delta f}e^{ikx} = \rho_{background} + \delta \rho \end{align}ρvbulk=vfdv=v(fbackground+δf^eikx)\begin{align} \rho v_{bulk} = \int v f dv = \int v (f_{background} + \hat{\delta f}e^{ikx}) \end{align}(ρbackground+δρ)(vbackground+δv)=ρbackgroundvbackground+vδf^eikx\begin{align} (\rho_{background} + \delta \rho)(v_{background} + \delta v) = \rho_{background} v_{background} + \int v \hat{\delta f}e^{ikx} \end{align}δv=(vδf^eikxvbackgroundδρ)ρbackground\begin{align} \delta v = \frac{(\int v \hat{\delta f}e^{ikx} - v_{background} \delta \rho)}{\rho_{background}} \end{align}ρ(f)T(f)+ρ(f)v(f)2=v2fdv=v2(fbackground+δf^eikx)=ρbackgroundTbackground+ρbackgroundvbackground2+v2δf^eikx\begin{align} \rho(f)T(f) + \rho(f)v(f)^2 = \int v^2 f dv = \int v^2 (f_{background} + \hat{\delta f}e^{ikx}) = \rho_{background}T_{background} + \rho_{background} {v_{background}}^2 + \int v^2 \hat{\delta f}e^{ikx} \end{align}(ρbackground+δρ)(Tbackground+δT)+(ρbackground+δρ)(vbackground+δv)2=v2fdv=v2(fbackground+δf^eikx)\begin{align} (\rho_{background} + \delta \rho)(T_{background} + \delta T) + (\rho_{background} + \delta \rho)(v_{background} + \delta v)^2 = \int v^2 f dv = \int v^2 (f_{background} + \hat{\delta f}e^{ikx}) \end{align}

Since δρδTO(τ2) \delta \rho \delta T \sim O(\tau^2) δT=(v2δf^eikxTbackgroundδρ2ρbackgroundvbackgroundδv)ρbackground\begin{align} \delta T = \frac{(\int v^2 \hat{\delta f}e^{ikx} - T_{background}\delta \rho - 2 \rho_{background} v_{background} \delta v)}{\rho_{background}} \end{align}

Now, evaluating the taylor series expansion of fMBf_{MB}

TBT_B shall be used to reference TbackgroundT_{background} in the code

ρB\rho_B shall be used to reference ρbackground\rho_{background} in the code

vBv_B shall be used to reference vbackgroundv_{background} in the code

fBf_B shall be used to reference fbackgroundf_{background} in the code

v_B, T_B, rho_B, delta_rho, delta_T, delta_v, v, x, m, k, tau = var('v_0, T_0, rho_0, delta_rho, delta_T, delta_v, v, x, m, k, tau')
rho = rho_B + delta_rho T = T_B + delta_T v_b = v_B + delta_v f_MB = rho*sqrt(m/(2*pi*k*T)) * exp(-m*(v - v_b)**2/(2*k*T))
f_B = (rho_B)*sqrt(m/(2*pi*k*T_B)) * exp(-m*(v - v_B)**2/(2*k*T_B))
f_MB
12(δρ+ρ0)mπ(T0+δT)ke((δvv+v0)2m2(T0+δT)k)\displaystyle \sqrt{\frac{1}{2}} {\left(\delta_{\rho} + \rho_{0}\right)} \sqrt{\frac{m}{\pi {\left(T_{0} + \delta_{T}\right)} k}} e^{\left(-\frac{{\left(\delta_{v} - v + v_{0}\right)}^{2} m}{2 \, {\left(T_{0} + \delta_{T}\right)} k}\right)}
expr = taylor(f_MB, (delta_rho, 0), (delta_T, 0), (delta_v, 0), 1) - f_B expr.simplify_full()
(22T0δvm32ρ0v+2δTm32ρ0v2+2δTm32ρ0v02+22T02δρkm(22πT052k32mπT0k(22T022T0δT)km)ρ02(2T0δvm32ρ0+2δTm32ρ0v)v0)e(mv22T0k+mvv0T0kmv022T0k)4πT052k32\displaystyle \frac{{\left(2 \, \sqrt{2} T_{0} \delta_{v} m^{\frac{3}{2}} \rho_{0} v + \sqrt{2} \delta_{T} m^{\frac{3}{2}} \rho_{0} v^{2} + \sqrt{2} \delta_{T} m^{\frac{3}{2}} \rho_{0} v_{0}^{2} + 2 \, \sqrt{2} T_{0}^{2} \delta_{\rho} k \sqrt{m} - {\left(2 \, \sqrt{2} \sqrt{\pi} T_{0}^{\frac{5}{2}} k^{\frac{3}{2}} \sqrt{\frac{m}{\pi T_{0} k}} - {\left(2 \, \sqrt{2} T_{0}^{2} - \sqrt{2} T_{0} \delta_{T}\right)} k \sqrt{m}\right)} \rho_{0} - 2 \, {\left(\sqrt{2} T_{0} \delta_{v} m^{\frac{3}{2}} \rho_{0} + \sqrt{2} \delta_{T} m^{\frac{3}{2}} \rho_{0} v\right)} v_{0}\right)} e^{\left(-\frac{m v^{2}}{2 \, T_{0} k} + \frac{m v v_{0}}{T_{0} k} - \frac{m v_{0}^{2}}{2 \, T_{0} k}\right)}}{4 \, \sqrt{\pi} T_{0}^{\frac{5}{2}} k^{\frac{3}{2}}}
# Bump on tail case: var('n_10, n_20, T_10, T_20, v_d0') n_1 = n_10 n_2 = n_20 T_1 = T_10 T_2 = T_20 v_d = v_d0 f_B = n_1*sqrt(m/(2*pi*k*T_1)) * exp(-v**2/(2*k*T_1)) + n_2*sqrt(m/(2*pi*k*T_2)) * exp(-(v - v_d**2)/(2*k*T_2)) # Need to calculate rho_B, T_B, v_B # Check: rho_B = n_10 + n_20 T_B = T_10 + T_10 v_B = v_d rho = rho_B + delta_rho T = T_B + delta_T v_b = v_B + delta_v f_MB = rho*sqrt(m/(2*pi*k*T)) * exp(-m*(v - v_b)**2/(2*k*T)) expr = taylor(f_MB, (delta_rho, 0), (delta_T, 0), (delta_v, 0), 1) - f_B expr.simplify_full()
(n10\displaystyle n_{10}, n20\displaystyle n_{20}, T10\displaystyle T_{10}, T20\displaystyle T_{20}, vd0\displaystyle v_{d_{0}})
(((πT10δTkm32n10+πT10δTkm32n20)vd02e(v22T10k+v2T20k)2(2πT1032δvkm32n10+2πT1032δvkm32n20+(πT10δTkm32n10+πT10δTkm32n20)v)vd0e(v22T10k+v2T20k)+(8πT1052δρk32m+2(4πT1052πT1032δT)k32mn10+2(4πT1052πT1032δT)k32mn20+(πT10δTkm32n10+πT10δTkm32n20)v2+4(πT1032δvkm32n10+πT1032δvkm32n20)v)e(v22T10k+v2T20k))e(mvvd02T10k)8(2πT103k2n20mπT20ke(mv24T10k+v22T10k+vd022T20k)+2πT103k2n10mπT10ke(mv24T10k+v2T20k))e(mvd024T10k))e(mv24T10kmvd024T10kv22T10kv2T20k)16πT103k2\displaystyle \frac{{\left({\left({\left(\sqrt{\pi} \sqrt{T_{10}} \delta_{T} \sqrt{k} m^{\frac{3}{2}} n_{10} + \sqrt{\pi} \sqrt{T_{10}} \delta_{T} \sqrt{k} m^{\frac{3}{2}} n_{20}\right)} v_{d_{0}}^{2} e^{\left(\frac{v^{2}}{2 \, T_{10} k} + \frac{v}{2 \, T_{20} k}\right)} - 2 \, {\left(2 \, \sqrt{\pi} T_{10}^{\frac{3}{2}} \delta_{v} \sqrt{k} m^{\frac{3}{2}} n_{10} + 2 \, \sqrt{\pi} T_{10}^{\frac{3}{2}} \delta_{v} \sqrt{k} m^{\frac{3}{2}} n_{20} + {\left(\sqrt{\pi} \sqrt{T_{10}} \delta_{T} \sqrt{k} m^{\frac{3}{2}} n_{10} + \sqrt{\pi} \sqrt{T_{10}} \delta_{T} \sqrt{k} m^{\frac{3}{2}} n_{20}\right)} v\right)} v_{d_{0}} e^{\left(\frac{v^{2}}{2 \, T_{10} k} + \frac{v}{2 \, T_{20} k}\right)} + {\left(8 \, \sqrt{\pi} T_{10}^{\frac{5}{2}} \delta_{\rho} k^{\frac{3}{2}} \sqrt{m} + 2 \, {\left(4 \, \sqrt{\pi} T_{10}^{\frac{5}{2}} - \sqrt{\pi} T_{10}^{\frac{3}{2}} \delta_{T}\right)} k^{\frac{3}{2}} \sqrt{m} n_{10} + 2 \, {\left(4 \, \sqrt{\pi} T_{10}^{\frac{5}{2}} - \sqrt{\pi} T_{10}^{\frac{3}{2}} \delta_{T}\right)} k^{\frac{3}{2}} \sqrt{m} n_{20} + {\left(\sqrt{\pi} \sqrt{T_{10}} \delta_{T} \sqrt{k} m^{\frac{3}{2}} n_{10} + \sqrt{\pi} \sqrt{T_{10}} \delta_{T} \sqrt{k} m^{\frac{3}{2}} n_{20}\right)} v^{2} + 4 \, {\left(\sqrt{\pi} T_{10}^{\frac{3}{2}} \delta_{v} \sqrt{k} m^{\frac{3}{2}} n_{10} + \sqrt{\pi} T_{10}^{\frac{3}{2}} \delta_{v} \sqrt{k} m^{\frac{3}{2}} n_{20}\right)} v\right)} e^{\left(\frac{v^{2}}{2 \, T_{10} k} + \frac{v}{2 \, T_{20} k}\right)}\right)} e^{\left(\frac{m v v_{d_{0}}}{2 \, T_{10} k}\right)} - 8 \, {\left(\sqrt{2} \pi T_{10}^{3} k^{2} n_{20} \sqrt{\frac{m}{\pi T_{20} k}} e^{\left(\frac{m v^{2}}{4 \, T_{10} k} + \frac{v^{2}}{2 \, T_{10} k} + \frac{v_{d_{0}}^{2}}{2 \, T_{20} k}\right)} + \sqrt{2} \pi T_{10}^{3} k^{2} n_{10} \sqrt{\frac{m}{\pi T_{10} k}} e^{\left(\frac{m v^{2}}{4 \, T_{10} k} + \frac{v}{2 \, T_{20} k}\right)}\right)} e^{\left(\frac{m v_{d_{0}}^{2}}{4 \, T_{10} k}\right)}\right)} e^{\left(-\frac{m v^{2}}{4 \, T_{10} k} - \frac{m v_{d_{0}}^{2}}{4 \, T_{10} k} - \frac{v^{2}}{2 \, T_{10} k} - \frac{v}{2 \, T_{20} k}\right)}}{16 \, \pi T_{10}^{3} k^{2}}
(n10\displaystyle n_{10}, n20\displaystyle n_{20}, T20\displaystyle T_{20}, vd0\displaystyle v_{d_{0}}, δn1\displaystyle \delta_{n_{1}}, δn2\displaystyle \delta_{n_{2}})
%md Here we determine the conditions necessary to ensure that initially pressure is constant throughout the domain \begin{align} p = (\rho_{background} + \delta \rho)(T + \delta T) \end{align} If I expand this now: \begin{align} p(t, x) & = \rho_B T_B + T_B\delta \rho(t, x) + \rho_B\delta T(t, x) \\ p(t, x) & = \rho_B T_B + T_B\Re(\delta \hat{\rho}(t) e^{ikx}) + \rho_B\Re(\delta \hat{T}(t) e^{ikx}) \\ p(t, x) & = \rho_B T_B + T_B(\delta \hat{\rho_r} cos(kx) - \delta \hat{\rho_i} sin(kx)) + \rho_B(\delta \hat{T_r} cos(kx) - \delta \hat{T_i} sin(kx)) \\ \frac{d p(x, t)}{d x} & = -k(\delta\hat{T_r} + \delta\hat{\rho_r})sin(kx) -k(\delta\hat{T_i} + \delta\hat{\rho_i})cos(kx) \end{align} The background quantities are taken to be at unity

Here we determine the conditions necessary to ensure that initially pressure is constant throughout the domain p=(ρbackground+δρ)(T+δT)\begin{align} p = (\rho_{background} + \delta \rho)(T + \delta T) \end{align}

If I expand this now:

p(t,x)=ρBTB+TBδρ(t,x)+ρBδT(t,x)p(t,x)=ρBTB+TB(δρ^(t)eikx)+ρB(δT^(t)eikx)p(t,x)=ρBTB+TB(δρr^cos(kx)δρi^sin(kx))+ρB(δTr^cos(kx)δTi^sin(kx))dp(x,t)dx=k(δTr^+δρr^)sin(kx)k(δTi^+δρi^)cos(kx)\begin{align} p(t, x) & = \rho_B T_B + T_B\delta \rho(t, x) + \rho_B\delta T(t, x) \\ p(t, x) & = \rho_B T_B + T_B\Re(\delta \hat{\rho}(t) e^{ikx}) + \rho_B\Re(\delta \hat{T}(t) e^{ikx}) \\ p(t, x) & = \rho_B T_B + T_B(\delta \hat{\rho_r} cos(kx) - \delta \hat{\rho_i} sin(kx)) + \rho_B(\delta \hat{T_r} cos(kx) - \delta \hat{T_i} sin(kx)) \\ \frac{d p(x, t)}{d x} & = -k(\delta\hat{T_r} + \delta\hat{\rho_r})sin(kx) -k(\delta\hat{T_i} + \delta\hat{\rho_i})cos(kx) \end{align}

The background quantities are taken to be at unity

%md **NOTE**: The 2D and 3D background distributions are assumed to have zero bulk velocities. Determining the linearized collision operator in the case of 2D perturbed systems: We shall be considering $f = f_{background} + \delta f = f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}$ to be the perturbation that is introduced The BGK collision operator is defined as: \begin{align} C[f] = -\frac{(f - f_{MB}(T(f), \rho(f)))}{\tau} \end{align} In order to determine $f_{MB}$ as a function of $f_{background}$, we first need to evaluate the temperature and density distributions by computing the zeroth, and 2nd moments of the distribution function $f$ \begin{align} \rho(f) = \int f dv = \int (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = \rho_{background} + \int \hat{\delta f}e^{i(k_xx+k_yy)} = \rho_{background} + \delta \rho \end{align} \begin{align} \rho(f)T(f) = \int 0.5(v_x^2 + v_y^2) f dv = \int 0.5(v_x^2 + v_y^2) (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = \rho_{background}T_{background} + \int 0.5(v_x^2 + v_y^2)\hat{\delta f}e^{i(k_xx+k_yy)} \end{align} \begin{align} (\rho_{background} + \delta \rho)(T_{background} + \delta T) = \int 0.5(v_x^2 + v_y^2) f dv = \int 0.5(v_x^2 + v_y^2) (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = \rho_{background}T_{background} + \int 0.5(v_x^2 + v_y^2) \hat{\delta f}e^{i(k_xx+k_yy)} \\ \end{align} Since $ \delta \rho \delta T \sim O(\tau^2)$ \begin{align} \delta T \approx \frac{\int 0.5(v_x^2 + v_y^2) \hat{\delta f}e^{i(k_xx+k_yy)} - T_{background} \int \hat{\delta f}e^{i(k_xx+k_yy)}}{\rho_{background}} \end{align} \begin{align} \rho v_{xbulk} = \int v_x f dv = \int v_x (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = v_{xbackground} + \int v_x \hat{\delta f}e^{i(k_xx+k_yy)} = 0 + \int v_x \hat{\delta f}e^{i(k_xx+k_yy)} \end{align} \begin{align} \rho v_{ybulk} = \int v_y f dv = \int v_y (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = v_{ybackground} + \int v_y \hat{\delta f}e^{i(k_xx+k_yy)} = 0 + \int v_y \hat{\delta f}e^{i(k_xx+k_yy)} \end{align}

NOTE: The 2D and 3D background distributions are assumed to have zero bulk velocities.

Determining the linearized collision operator in the case of 2D perturbed systems:

We shall be considering f=fbackground+δf=fbackground+δf^ei(kxx+kyy)f = f_{background} + \delta f = f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)} to be the perturbation that is introduced

The BGK collision operator is defined as:

C[f]=(ffMB(T(f),ρ(f)))τ\begin{align} C[f] = -\frac{(f - f_{MB}(T(f), \rho(f)))}{\tau} \end{align}

In order to determine fMBf_{MB} as a function of fbackgroundf_{background}, we first need to evaluate the temperature and density distributions by computing the zeroth, and 2nd moments of the distribution function ff

ρ(f)=fdv=(fbackground+δf^ei(kxx+kyy))=ρbackground+δf^ei(kxx+kyy)=ρbackground+δρ\begin{align} \rho(f) = \int f dv = \int (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = \rho_{background} + \int \hat{\delta f}e^{i(k_xx+k_yy)} = \rho_{background} + \delta \rho \end{align}ρ(f)T(f)=0.5(vx2+vy2)fdv=0.5(vx2+vy2)(fbackground+δf^ei(kxx+kyy))=ρbackgroundTbackground+0.5(vx2+vy2)δf^ei(kxx+kyy)\begin{align} \rho(f)T(f) = \int 0.5(v_x^2 + v_y^2) f dv = \int 0.5(v_x^2 + v_y^2) (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = \rho_{background}T_{background} + \int 0.5(v_x^2 + v_y^2)\hat{\delta f}e^{i(k_xx+k_yy)} \end{align}(ρbackground+δρ)(Tbackground+δT)=0.5(vx2+vy2)fdv=0.5(vx2+vy2)(fbackground+δf^ei(kxx+kyy))=ρbackgroundTbackground+0.5(vx2+vy2)δf^ei(kxx+kyy)\begin{align} (\rho_{background} + \delta \rho)(T_{background} + \delta T) = \int 0.5(v_x^2 + v_y^2) f dv = \int 0.5(v_x^2 + v_y^2) (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = \rho_{background}T_{background} + \int 0.5(v_x^2 + v_y^2) \hat{\delta f}e^{i(k_xx+k_yy)} \\ \end{align}

Since δρδTO(τ2) \delta \rho \delta T \sim O(\tau^2) δT0.5(vx2+vy2)δf^ei(kxx+kyy)Tbackgroundδf^ei(kxx+kyy)ρbackground\begin{align} \delta T \approx \frac{\int 0.5(v_x^2 + v_y^2) \hat{\delta f}e^{i(k_xx+k_yy)} - T_{background} \int \hat{\delta f}e^{i(k_xx+k_yy)}}{\rho_{background}} \end{align} ρvxbulk=vxfdv=vx(fbackground+δf^ei(kxx+kyy))=vxbackground+vxδf^ei(kxx+kyy)=0+vxδf^ei(kxx+kyy)\begin{align} \rho v_{xbulk} = \int v_x f dv = \int v_x (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = v_{xbackground} + \int v_x \hat{\delta f}e^{i(k_xx+k_yy)} = 0 + \int v_x \hat{\delta f}e^{i(k_xx+k_yy)} \end{align} ρvybulk=vyfdv=vy(fbackground+δf^ei(kxx+kyy))=vybackground+vyδf^ei(kxx+kyy)=0+vyδf^ei(kxx+kyy)\begin{align} \rho v_{ybulk} = \int v_y f dv = \int v_y (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy)}) = v_{ybackground} + \int v_y \hat{\delta f}e^{i(k_xx+k_yy)} = 0 + \int v_y \hat{\delta f}e^{i(k_xx+k_yy)} \end{align}

T_B, rho_B, delta_rho, delta_T, delta_v_x, delta_v_y, v_x, v_y, x, m, k, tau = var('T_B, rho_B, delta_rho, delta_T, delta_vx, delta_vy, v_x, v_y, x, m, k, tau')
rho = rho_B + delta_rho T = T_B + delta_T v_bx = 0 + delta_v_x v_by = 0 + delta_v_y f_MB = rho*(m/(2*pi*k*T))*exp(-m*((v_x - delta_v_x)**2 + (v_y - delta_v_y)**2)/(2*k*T))
f_B = (rho_B)*(m/(2*pi*k*T_B))*exp(-m*(v_x**2 + v_y**2)/(2*k*T_B))
expr = taylor(f_MB, (delta_rho, 0), (delta_T, 0), (delta_v_x, 0), (delta_v_y, 0), 1) - f_B expr.simplify_full()
(δTm2ρBvx2+δTm2ρBvy2+2TB2δρkm+2(δvxm2ρBvx+δvym2ρBvyδTkmρB)TB)e(mvx22TBkmvy22TBk)4πTB3k2\displaystyle \frac{{\left(\delta_{T} m^{2} \rho_{B} v_{x}^{2} + \delta_{T} m^{2} \rho_{B} v_{y}^{2} + 2 \, T_{B}^{2} \delta_{\rho} k m + 2 \, {\left(\delta_{\mathit{vx}} m^{2} \rho_{B} v_{x} + \delta_{\mathit{vy}} m^{2} \rho_{B} v_{y} - \delta_{T} k m \rho_{B}\right)} T_{B}\right)} e^{\left(-\frac{m v_{x}^{2}}{2 \, T_{B} k} - \frac{m v_{y}^{2}}{2 \, T_{B} k}\right)}}{4 \, \pi T_{B}^{3} k^{2}}

Determining the linearized collision operator in the case of 3D perturbed systems:

We shall be considering f=fbackground+δf=fbackground+δf^ei(kxx+kyy+kzz)f = f_{background} + \delta f = f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} to be the perturbation that is introduced

The BGK collision operator is defined as:

C[f]=(ffMB(T(f),ρ(f)))τ\begin{align} C[f] = -\frac{(f - f_{MB}(T(f), \rho(f)))}{\tau} \end{align}

In order to determine fMBf_{MB} as a function of fbackgroundf_{background}, we first need to evaluate the temperature and density distributions by computing the zeroth, and 2nd moments of the distribution function ff

ρ(f)=fdv=(fbackground+δf^ei(kxx+kyy+kzz))=ρbackground+δf^ei(kxx+kyy+kzz)=ρbackground+δρ\begin{align} \rho(f) = \int f dv = \int (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}) = \rho_{background} + \int \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} = \rho_{background} + \delta \rho \end{align}ρ(f)T(f)=(vx2+vy2+vz2)3fdv=(vx2+vy2+vz2)3(fbackground+δf^ei(kxx+kyy+kzz))=ρbackgroundTbackground+(vx2+vy2+vz2)3δf^ei(kxx+kyy+kzz)\begin{align} \rho(f)T(f) = \int \frac{(v_x^2 + v_y^2 + v_z^2)}{3} f dv = \int \frac{(v_x^2 + v_y^2 + v_z^2)}{3} (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}) = \rho_{background}T_{background} + \int \frac{(v_x^2 + v_y^2 + v_z^2)}{3}\hat{\delta f}e^{i(k_xx+k_yy+k_zz)} \end{align}(ρbackground+δρ)(Tbackground+δT)=(vx2+vy2+vz2)3fdv=(vx2+vy2+vz2)3(fbackground+δf^ei(kxx+kyy+kzz))=ρbackgroundTbackground+(vx2+vy2+vz2)3δf^ei(kxx+kyy+kzz)\begin{align} (\rho_{background} + \delta \rho)(T_{background} + \delta T) = \int \frac{(v_x^2 + v_y^2 + v_z^2)}{3} f dv = \int \frac{(v_x^2 + v_y^2 + v_z^2)}{3} (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}) \\= \rho_{background}T_{background} + \int \frac{(v_x^2 + v_y^2 + v_z^2)}{3} \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} \\ \end{align}

Since δρδTO(τ2) \delta \rho \delta T \sim O(\tau^2) δT(vx2+vy2+vz2)3δf^ei(kxx+kyy+kzz)Tbackgroundδf^ei(kxx+kyy+kzz)ρbackground\begin{align} \delta T \approx \frac{\int \frac{(v_x^2 + v_y^2 + v_z^2)}{3} \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} - T_{background} \int \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}}{\rho_{background}} \end{align} ρvxbulk=vxfdv=vx(fbackground+δf^ei(kxx+kyy+kzz))=vxbackground+vxδf^ei(kxx+kyy+kzz)=0+vxδf^ei(kxx+kyy+kzz)\begin{align} \rho v_{xbulk} = \int v_x f dv = \int v_x (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}) = v_{xbackground} + \int v_x \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} = 0 + \int v_x \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} \end{align} ρvybulk=vyfdv=vy(fbackground+δf^ei(kxx+kyy+kzz))=vybackground+vyδf^ei(kxx+kyy+kzz)=0+vyδf^ei(kxx+kyy+kzz)\begin{align} \rho v_{ybulk} = \int v_y f dv = \int v_y (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}) = v_{ybackground} + \int v_y \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} = 0 + \int v_y \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} \end{align} ρvzbulk=vzfdv=vz(fbackground+δf^ei(kxx+kyy+kzz))=vzbackground+vzδf^ei(kxx+kyy+kzz)=0+vzδf^ei(kxx+kyy+kzz)\begin{align} \rho v_{zbulk} = \int v_z f dv = \int v_z (f_{background} + \hat{\delta f}e^{i(k_xx+k_yy+k_zz)}) = v_{zbackground} + \int v_z \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} = 0 + \int v_z \hat{\delta f}e^{i(k_xx+k_yy+k_zz)} \end{align}

T_B, rho_B, delta_rho, delta_T, delta_v_x, delta_v_y, delta_v_z, v_x, v_y, v_z, m, k, tau = var('T_B, rho_B, delta_rho, delta_T, delta_vx, delta_vy, delta_vz, v_x, v_y, v_z, m, k, tau')
rho = rho_B + delta_rho T = T_B + delta_T v_bx = 0 + delta_v_x v_by = 0 + delta_v_y v_bz = 0 + delta_v_z f_MB = rho*((m/(2*pi*k*T))**(3/2))*exp(-m*((v_x - delta_v_x)**2 + (v_y - delta_v_y)**2 + (v_z - delta_v_z)**2)/(2*k*T))
f_B = (rho_B)*((m/(2*pi*k*T_B))**(3/2))*exp(-m*(v_x**2 + v_y**2 + v_z**2)/(2*k*T_B))
expr = taylor(f_MB, (delta_rho, 0), (delta_T, 0), (delta_v_x, 0), (delta_v_y, 0), (delta_v_z, 0), 1) - f_B expr.simplify_full()
(22TBδvxm52ρBvx+2δTm52ρBvx2+22TBδvym52ρBvy+2δTm52ρBvy2+22TBδvzm52ρBvz+2δTm52ρBvz2+22TB2δρkm32(22π32TB72k52(mπTBk)32(22TB232TBδT)km32)ρB)e(mvx22TBkmvy22TBkmvz22TBk)8π32TB72k52\displaystyle \frac{{\left(2 \, \sqrt{2} T_{B} \delta_{\mathit{vx}} m^{\frac{5}{2}} \rho_{B} v_{x} + \sqrt{2} \delta_{T} m^{\frac{5}{2}} \rho_{B} v_{x}^{2} + 2 \, \sqrt{2} T_{B} \delta_{\mathit{vy}} m^{\frac{5}{2}} \rho_{B} v_{y} + \sqrt{2} \delta_{T} m^{\frac{5}{2}} \rho_{B} v_{y}^{2} + 2 \, \sqrt{2} T_{B} \delta_{\mathit{vz}} m^{\frac{5}{2}} \rho_{B} v_{z} + \sqrt{2} \delta_{T} m^{\frac{5}{2}} \rho_{B} v_{z}^{2} + 2 \, \sqrt{2} T_{B}^{2} \delta_{\rho} k m^{\frac{3}{2}} - {\left(2 \, \sqrt{2} \pi^{\frac{3}{2}} T_{B}^{\frac{7}{2}} k^{\frac{5}{2}} \left(\frac{m}{\pi T_{B} k}\right)^{\frac{3}{2}} - {\left(2 \, \sqrt{2} T_{B}^{2} - 3 \, \sqrt{2} T_{B} \delta_{T}\right)} k m^{\frac{3}{2}}\right)} \rho_{B}\right)} e^{\left(-\frac{m v_{x}^{2}}{2 \, T_{B} k} - \frac{m v_{y}^{2}}{2 \, T_{B} k} - \frac{m v_{z}^{2}}{2 \, T_{B} k}\right)}}{8 \, \pi^{\frac{3}{2}} T_{B}^{\frac{7}{2}} k^{\frac{5}{2}}}
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