\(\bullet\) We live in a small part of the universe where plasmas do not occur naturally; otherwise we would not be alive. The reason for this can be seen from the Saha equation, which tells us the amount of ionization to be expected in a gas in thermal equilibrium. \[\begin{aligned} \label{eq_1} \frac{n_i}{n_n} \simeq 2.4 \times 10^{21} \frac{T^{\frac{3}{2}}}{n_i}e^{\frac{-U_i}{KT}} \end{aligned}\]
where \(n_i\) and \(n_n\) are density(number per \(m^3\)) of ionized atoms and of neutral atoms, respectively. \(T\) is the gas temperature in \(^{\circ}K\), \(K\) is Boltzmann’s constant; \(1.38 \times 10^{-23} [\frac{J}{^{\circ}K}]\), and \(U_i\) is the ionization energy of the gas - that is, the number of joules required to remove the outermost electron from an atom.
In room temperature, \(n_n \simeq 3 \times 10^{25}[m^{-3}]\), \(T \simeq 300 ^{\circ}K\), and \(U_i = 14.5eV\) for nitrogen, where \(1eV = 1.6 \times 10^{-10}[J]\). The fractional ionization \(\frac{n_i}{n_n+n_i}\simeq \frac{n_i}{n_n}\) is rediculously low:
\[\begin{aligned} \frac{n_i}{n_n}\simeq 10^{-122}\end{aligned}\]
Let us define degree of ionization \[\alpha = \frac{n_e}{n_e+n_n}\]. Where \(n_e=n_i\) is the number of electrons or ions per volume [\(cm^3\)], and \(n_n\) is the number of neutrals per volume [\(cm^3\)].
\(\alpha << 1\): weakly ionized situation (low temp)
\(\alpha = 1\): fully ionized (high temp)
Trend of Saha equation tells us how fractional ionization changes depending on the temperature. As the temperature increases, the ionization increases significantly.
\(\frac{n_e}{n_n}\rightarrow 0\) at \(K_BT << U_i\)
\(\frac{n_e}{n_n}>> 0\): as \(K_BT >> U_i\)
where again, \(U_i\) is the ionization energy and \(K_BT: T_e \simeq T_i\)
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\(\bullet\) We are surrounded by plasmas starting from the ionosphere a hundred kilometers above us, which is connected to the sun via the plasma of the solar wind.
\(\bullet\) The very tenuous interstellar space is a plasma and so are the largest objects that emit x-rays in the universe.
\(\bullet\) Plasma is a fundamental state of matter: by heating the mater changes from a solid \(\rightarrow\) liquid \(\rightarrow\) gas \(\rightarrow\) plasma.
\(\bullet\) When gas is heated above a certain temperature, or it is subject to strong electromagnetic fields, it gets ionized, and a transition towards the so-called fourth state of matter(coined by W.crookes in 1879), plasma state, is observed.
\(\bullet\) Compare
Ancient: Universe - Earth, water, air, fire
Modern: Universe - solid, liquid, gas, plasma
\(\bullet\) Plasma: Introduced by Tonks and Irving Langmuir(Nobel Prize winner) in 1928.
\(\bullet\) Plasma: Greek words - moldable substacne; jelly.
\(\bullet\) Plasma ionized gas with \(n_e = n_i\); macroscopically charged neutral. Much more complicated than many single charged particle becuase of collective effects.
\(\bullet\) The ensemble of freely moving charged particles of both signs, i.e., ionized gas, can be considered as plasma if the Debye length is small compared with dimensions of the volume occupied by the gas. (Langmuir)
\(\bullet\) A plasma is a quasi-neutral gas of charged and neutral particles which exhibits collective behavior (F.F. Chen)
\(\bullet\) A plasma is a gas of charged particles, in which the potential energy of a typical particle due to its nearest neighbor is much smaller than its kinetic energy. (D. R. Nicholson)
\(\bullet\) A plasma may be roughly defined as a system containing mobile charges, in which the electric and magnetic interactions between particles play a dominant role in the dynamics of the systems. (J. M. Dawson)
\(\bullet\) A plasma is collection of charged particles, usually of opposite sign, that tends to be electrically neutral. We often describe a plasma as the fourth state of matter. Adding energy to a solid melts it and it becomes a liquid; adding energy to a liquid boils it and ib becomes a gas; adding energy to a gas ionizes it and it becomes a plasma. (J. L. Shohet)
\(\bullet\)The plasma state is a characteriza
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전자공학 학생이 아래 질문을 가지고 질문한 사항에 대해 답변을 하고자 작성하였습니다. 틀렸으면 알려주세요. 제 자신도 공부하기 위함입니다. (I just upload the image from Cheng p122; if there is problem with the copyright please let me know. I am trying to discuss the question that student has)
Problem: Find the energy required to assemble a uniform sphere of charge of radius \(b\) and volume charge density \(\rho_v\)
In terms of total charge
\[\begin{aligned} Q= \frac{4\pi}{3}\rho_v b^3 \end{aligned}\]
Hence we have
\[\begin{aligned} W_e = \frac{3Q^2}{20 \pi \epsilon_0 b}\end{aligned}\]
By looking at the equation below, a lot of people asks why does the half goes away?
\[\begin{aligned} W&= \frac{1}{2}\int_v \rho_v V dv\end{aligned}\]
And the answer is it is not actually going away. Those who might have twice differece might solved problem as below ( used \(V =\frac{q}{4\pi \epsilon_0 r}\))
\[\begin{aligned} W&= \frac{1}{2}\int_v \rho_v V dv \\ &= \frac{1}{2}\int_v \frac{\rho_v^2 \frac{4}{3} \pi r^2}{4 \pi \epsilon_0} r^2 \sin \theta dr d\theta d\phi \\ &=\frac{2}{15}\frac{\rho_v^2 r^5}{\epsilon_0}\\ &=\frac{3}{40}\frac{Q^2}{\epsilon_0 r \pi}\end{aligned}\]
Notice that \(Q = \rho_v \frac{4}{3}\pi r^3\). So there is twice difference. The reason for this is that potential is different inside and outside the effective sphere.
Answer: Think we have sphere with radius R; charge exist in this region. For \(r<R\)
\[\begin{aligned} \oint_s \vec{E} \cdot d\vec{a} &= \frac{\rho}{\epsilon_0}\frac{4}{3}\pi r^3 \\ \vec{E} &= \frac{\rho}{3\epsilon_0} r \\ &=\frac{\rho r}{4\pi \epsilon_0 R^3}\end{aligned}\]
On the last step, we used the fact that charge is inside the \(\frac{4}{3}\pi R^3\). For \(r>R\),
\[\begin{aligned} \vec{E} = \frac{Q}{4\pi \epsilon_0 r^2}\end{aligned}\]
Finally \(E\) field and potential becomes
\[\begin{aligned} \vec{E}(r) = \frac{Q}{4\pi \epsilon_0} \times \begin{cases} \frac{r}{R^3} \quad (r<R) \\ \frac{1}{r^2} \quad (r>R) \end{cases}\end{aligned}\]
\[\begin{aligned} \vec{V}(r) = \frac{Q}{4\pi \epsilon_0} \times \begin{cases} -\frac{1}{2} \frac{r^2}{R^3} + \frac{3}{2R} \\ \frac{1}{r} \end{cases}\end{aligned}\]
Notice that constant term from the potential comes from the Boundary Condition; \(at r=R\), \(V = \frac{Q}{4\pi \epsilon_0 R}\). Finally, let us calculate the energy.
\[\begin{aligned} W &= \frac{1}{2} \int_{0}^{R} \frac{Q}{4\pi \epsilon_0} \rho \left( -\frac{r^2}{2R^3} + \frac{3}{2R} \right) 4\pi r^2 dr \\ &=\frac{1}{2}\int_{0}^{R} \frac{Q \rho}{\epsilon_0}\left( -\frac{r^4}{2R^3} + \frac{3r^2}{2R} \right) dr \\ &= \frac{4}{20}\frac{Q \rho R^2}{\epsilon_0}\\ &= \frac{3}{20}\frac{Q^2}{\pi \epsilon_0 R}\end{aligned}\]
Hence, we get the same answer as David K. Cheng.
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Consider a magnetic field that has curvature.
\(\bullet\) Such \(B\)-field will have gradient which result in the grad-B drift.
\(\bullet\) Once particle has a velocity along the magnetic field, it experiences a centrifugal force due to the field curvature and this force gives a drift.
\[\begin{aligned} \vec{F_{cf}}= m\frac{v_{\parallel}^2}{R_c}\hat{r} = \frac{ m v_{\parallel}^2}{R_c}\frac{\vec{R_c}}{R_c}\end{aligned}\]
where \(\vec{R_c}\) is the radius of curvature. The drift velocity due to \(\vec{F_{cf}}\) (the curvature drift) is then
\[\begin{aligned} \boxed{\vec{v_R} = \frac{mv_{\parallel}^2}{qB^2}\frac{\vec{R_c}\times \vec{B}}{R_c^2}}\end{aligned}\]
Curvature drift is \(\vec{v_R}\). Now, total drift in a curved magnetic field is the combination of the grad-B drift and the drift due to the centrifugal force. From Ampere’s Law, \(\nabla \times \vec{B} = \mu_0 \vec{J}\), we have \(\nabla \times \vec{B}=0\) in a vacuum. In the curvature, B-field is defined as \(\vec{B} = B(r) \hat{\phi}\)
\[\begin{aligned} \nabla \times \vec{B} &= \left[ \frac{1}{r} \frac{\partial B_z}{\partial \phi} - \frac{\partial B_{\phi}}{\partial z } \right]\hat{r} + \left[\frac{\partial B_r}{\partial z} - \frac{\partial B_z}{\partial r} \right] \hat{\phi} + \frac{1}{r}\left[ \frac{\partial}{\partial r}(r B_{\phi}) - \frac{\partial B_r}{\partial \phi} \right]\hat{z} \\ &= \frac{1}{r}\frac{d}{dr}(rB) =0\end{aligned}\]
Hence,
\[\begin{aligned} B(r) \propto \frac{1}{r} = \frac{1}{R_c}\end{aligned}\]
Thus,
\[\begin{aligned} \frac{\nabla B}{B} &= \frac{1}{B}\frac{\partial B}{\partial r} \hat{R_c} = -\frac{\vec{R_c}}{R_c^2}\end{aligned}\]
Then the grad-B drift velocity may be written as
\[\begin{aligned} \vec{\nabla B} = \frac{\frac{1}{2}mv_{\perp}^2}{qB} \frac{\vec{B} \times \nabla B}{B^2} = \frac{\frac{1}{2}mv_{\perp}^2}{qB}\frac{\vec{B}}{B} \times \left( \frac{-\vec{R_c}}{R_c^2} \right) = \frac{\frac{1}{2}mv_{\perp}^2}{qB^2}\frac{\vec{R_c}\times \vec{B}}{R_c^2}\end{aligned}\]
Finally, the total drift in a curved magnetic field is sum of grad-B drift and centrifugal force driven drift.
\[\begin{aligned} \boxed{\vec{v_R} + \vec{v_{\nabla B}} = \frac{m}{q}\frac{\vec{R_c}\times \vec{B}}{R_c^2 B^2} \left( v_{\parallel}^2 + \frac{1}{2}v_{\perp}^2 \right)}\end{aligned}\]
As you see, since the equation depends on the charge, we will have charge separation and leads to the rise of current. Notice that \(\frac{1}{2}v_{\perp}^2\) comes from the grad-B drift and \(v_{\parallel}^2\) comes from the centrifugal driven drift; the curvature drift.
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Let us assume gradient B-field exist in the z-direction with gradient along the y-direction.
\[\begin{aligned} \vec{B}(r) = B(y)\hat{z}\end{aligned}\]
Assume \(B\) is slightly in-homogeneous.
\[r_L\left| \frac{\nabla B}{B} \right|<< 1\]
\(\vec{B}\) can be expressed by the Taylor series.
\[\begin{aligned} \vec{B} = \vec{B_0} + (\vec{r} \cdot \nabla)\vec{B} + ...\end{aligned}\]
so that we have
\[\begin{aligned} \vec{B} = \vec{B_0} + (\vec{r} \cdot \nabla)\vec{B} = \vec{B_0} + y\frac{\partial \vec{B}}{\partial y}\end{aligned}\]
From the Lorentz force, we will compute the average force and determine the guiding center drift.
\[\begin{aligned} \frac{dv_x}{dt} &=\frac{q}{m}v_y B(y) \hat{x}\\ \frac{dv_y}{dt} &=-\frac{q}{m}v_x B(y) \hat{y}\end{aligned}\]
\[\begin{aligned} v_x &= v_{\perp}\cos\omega_c t \\ v_y &= \mp v_{\perp}\sin\omega_c t\end{aligned}\]
\[\begin{aligned} x &= r_L \sin(\omega_c t) \\ y &=\pm r_L \cos \omega_c t\end{aligned}\]
Hence, using above equation, we can approximate the Lorentz force in inhomogeneous B-field.
\[\begin{aligned} F_x &= qv_y B =\mp q v_{\perp}\sin(\omega_c t) \left( B_0 \pm r_L \cos(\omega_c t)\frac{\partial B}{\partial y} \right)\\ F_y &=-qv_x B = -qv_{\perp}\cos\omega_c t \left(B_0 \pm r_L \cos(\omega_c t) \frac{\partial B}{\partial y} \right)\\\end{aligned}\]
Since
\[\begin{aligned} <\sin(\omega t)> &= \frac{1}{T} \int_{0}^{T}\sin \omega t dt = 0\\ <\cos(\omega t)> &= \frac{1}{T} \int_{0}^{T} \cos \omega t dt = 0\\ <\sin^2(\omega t)> &= \frac{1}{T} \int_{0}^{T}\sin^2 \omega t dt = \frac{1}{T} \int_{0}^{T} \frac{1-\cos(2\omega t)}{2}dt = \frac{1}{2}\\ <\cos^2(\omega t)> &= \frac{1}{T} \int_{0}^{T} \cos^2 \omega t dt = \frac{1}{T} \int_{0}^{T} \frac{1+\cos(2\omega t)}{2}dt = \frac{1}{2}\\ <\sin \omega t \cos \omega t> &=\frac{1}{T} \int_{0}^{T} \cos(\omega t) \sin(\omega t)dt = \frac{1}{T} \int_{0}^{T} \frac{1}{2}\sin(2\omega t)dt =0\end{aligned}\]
we get
\[\begin{aligned} <F_x> &=\frac{1}{T}\int_{0}^{T}F_x dt =0\\ <F_y> &=\frac{1}{T}\int_{0}^{T}F_y dt \\ &=\frac{1}{T}\int_{0}^{T}F_y dt \\ &=\mp qv_{\perp}r_L \frac{\partial B}{\partial y}\frac{1}{2}\\ &=-\mu \frac{\partial B}{\partial y}\end{aligned}\]
Notice that \(\mu = \pm \frac{1}{2}q v_{\perp} r_L\). By plugging this into the guiding center drift equation for general case,
\[\begin{aligned} \vec{v}_{\nabla B} &=\frac{\vec{F}\times \vec{B}}{qB^2} \\ &= \frac{F_y \hat{y} \times B \hat{z}}{qB^2}\\ &=\frac{F_y}{qB}\hat{x} \\ &= -\frac{\mu}{qB}\frac{\partial B}{\partial y}\hat{x}\end{aligned}\]
The expression can be generalized as
\[\begin{aligned} \vec{v}_{\nabla B} = \frac{\mu}{q}\frac{\vec{B} \times \nabla B}{B^2} \end{aligned}\]
or
\[\begin{aligned} \boxed{\vec{v}_{\nabla_B} = \frac{\frac{1}{2}mv_{\perp}^2}{qB}\frac{\vec{B} \times \nabla B}{B^2}} \end{aligned}\]
The grad-B drift depends of the charge \(q\), it can cause the plasma currents and charge separation.
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Problem In a strictly steady state situation both the ions and the electrons will follow the Boltzmann relation
\[n_i = n_0 e^{\frac{-q_i \phi }{KT_i}}\]
For the case of an infinite, transparent grid charged to a potential \(\phi\), show that the shielding distance is then given approximately by
\[\lambda_D^{-2} = \frac{ne^2}{\epsilon_0} \left( \frac{1}{KT_e} + \frac{1}{KT_i} \right)\]
Answer
\[\begin{aligned} \nabla \cdot \vec{E} &=\frac{\rho}{\epsilon_0} = \frac{e}{\epsilon_0}(n_i - n_e) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( e^{-\frac{e\phi}{KT_i}} - e^{\frac{e\phi}{KT_e}} \right) \nonumber \\ &= - \frac{e}{\epsilon_0} n_0 \left( 1 -\frac{e\phi}{KT_i} - 1 - \frac{e\phi}{KT_e} \right) \nonumber \\ \frac{d^2 \phi }{dx^2 }&=\frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \phi \nonumber\end{aligned}\]
Hence,
\[\begin{aligned} \frac{1}{\lambda_D^2} = \frac{e^2 n_0}{\epsilon_0}\left( \frac{1}{KT_i} + \frac{1}{KT_e} \right) \nonumber\end{aligned}\]
\[\lambda_D = \sqrt{\left( \frac{\epsilon_0 K T_e T_i}{ne^2 (T_i + T_e)} \right)}\]
\(T_i << T_e \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_i}{ne^2} \right)}\)
\(T_e << T_i \quad \quad \lambda_D \simeq \sqrt{\left( \frac{\epsilon_0 K T_e}{ne^2} \right)}\)
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Problem Consider a particle of charge \(q\) and mass \(m\), initially at rest at (0,0,0), in the presence of a static magntic field \(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\).
(a) Taking \(E, B > 0\), sketch the orbit of the particle when \(q>0\).
(b) Derive an exact expression for the orbit \([x(t),y(t),z(t)]\) or the particle. Express your answer in terms of \(E\), \(B\), and \(\omega_c\)
(c) Find the drift velocity after averaging the motion in time. If there were many particles of various charges and masses present, would there be any net current?
(d) Suppose the electric field were replaced by a force \(F\) in the \(y\)-direction. What would be the drift velocity?
Answer (a)
Answer (b)
\(\vec{B}=B\hat{z}\) and \(\vec{E}=E\hat{y}\). From Lorentz equation,
\[m\frac{dv}{dt}=q(\vec{E} + \vec{v}\times \vec{B})\]
\[\begin{aligned} \dot{v_x}&= \quad \quad \quad \quad \frac{q}{m}v_y B \nonumber \\ \dot{v_y}&= \frac{q}{m}\vec{E_y} - \frac{q}{m}v_xB \nonumber \\ \dot{v_z}&= 0\end{aligned}\]
In z-direction \[z(t) = constant = 0\]
\[\begin{aligned} \frac{dv_x}{dt} &= \frac{qB}{m}v_y = \omega_c v_y \nonumber \\ \frac{dv_y}{dt} &= \frac{qE}{m}-\frac{qB}{m}v_x = \frac{qE_y}{m}-\omega_c v_x \nonumber\end{aligned}\]
\[\begin{aligned} \frac{d^2 v_y}{dt^2} &= -\omega_c^2 v_y \nonumber \\ \frac{d^2 v_x}{dt^2} &= -\omega_c^2 \left(v_x-\frac{E}{B} \right) \nonumber\end{aligned}\]
since \(\frac{E}{B}\) is constant, \[\frac{d^2}{dt^2} \left[ v_x-\frac{E}{B} \right] = \frac{d^2 v_x}{dt^2} = -\omega_c^2 \left(v_x-\frac{E}{B} \right)\]
\[\begin{aligned} v_x &= iv_{\perp}e^{i\omega_c t} + \frac{E}{B}\nonumber \\ v_y &= v_{\perp}e^{i\omega_c t} \nonumber \end{aligned}\]
\[\begin{aligned} x(t) &= x(0) + r_L \cos{\omega_c t} +\frac{E}{B}t \nonumber \\ y(t) &= y(0) + r_L \sin{\omega_c t} \nonumber \end{aligned}\]
Answer (c)
\[<v_d> = \frac{\int_{-\infty}^{\infty}v_d f(v) dv}{\int_{-\infty}^{\infty}f(v) dv} = \frac{E}{B}\]
There will be no current because \(\vec{E} \times \vec{B}\) drifts the ions and electrons in the same direction.
Answer (d)
\[\begin{aligned} \vec{v_f}&= \frac{1}{q} \frac{\vec{F} \times \vec{B}}{B^2} \nonumber \\ &=\frac{FE}{qB}\hat{y}\end{aligned}\]
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Problem 2: Prove \(-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\)
Let \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\]
\[\nabla \frac{1}{|\vec{r}-\vec{r}\prime|} = \frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\frac{1}{|\vec{r}-\vec{r}\prime|}\]
Let us consider only \(\hat{x}\) components. \[\begin{aligned} \frac{\partial}{\partial x}\hat{x}\frac{1}{\zeta^{\frac{1}{2}}} &= \frac{1}{2}\zeta^{-\frac{3}{2}}\zeta\prime \nonumber \\ &=\frac{x_1 - x_2}{\zeta^{\frac{3}{2}}}\hat{x} \nonumber \\ &=\frac{x_1 - x_2}{|\vec{r}-\vec{r}\prime|^3}\hat{x} \nonumber\end{aligned}\]
Same procedure occur for the other two component. When we add all, final result become \[-\nabla \frac{1}{|\vec{r}-\vec{r}\prime|}=\frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}\]
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Problem 1: Prove \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)
Answer:
\[\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{x_1-x_2}{|\vec{\eta}^3|} & \frac{y_1-y_2}{|\vec{\eta}^3|} & \frac{z_1-z_2}{|\vec{\eta}^3|}
\end{vmatrix}\] Where \[\vec{r}=x_1\hat{x} + y_1\hat{y} + z_1\hat{z}\] \[\vec{r}\prime=x_2\hat{x} + y_2\hat{y} + z_2\hat{z}\] \[\vec{\eta} = \vec{r}-\vec{r}\prime\], \[|\vec{\eta}|=\sqrt{(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2}\] \[\zeta =(\vec{x_1}-\vec{x_2})^2+(\vec{y_1}-\vec{y_2})^2+(\vec{z_1}-\vec{z_2})^2\] \[\begin{aligned}
\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3} &= (\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\eta}^3|}))\vec{x} \nonumber \\
& -(\frac{\partial}{\partial x}(\frac{z_1-z_2}{|\vec{\eta}^3|}) - \frac{\partial}{\partial z}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{y} \nonumber \\
& +(\frac{\partial}{\partial x}(\frac{y_1-y_2}{|\vec{\eta}^3|})-\frac{\partial}{\partial y}(\frac{x_1-x_2}{|\vec{\eta}^3|}))\vec{z} \nonumber\end{aligned}\]
Since \[\frac{\partial}{\partial y}(\frac{z_1-z_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(z_1 - z_2)(y_1 - y_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]
\[\frac{\partial}{\partial z}(\frac{y_1-y_2}{|\vec{\zeta}^{\frac{3}{2}}|})=-3\frac{(y_1 - y_2)(z_1 - z_2)}{|\vec{\zeta}^{\frac{5}{2}}|}\]
They cancel out each others. Same procedure occur for the other two components. Hence, \(\nabla \times \frac{\vec{r}-\vec{r}\prime}{|\vec{r}-\vec{r}\prime |^3}=0\)
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