[플라즈마 물리][Plasma Physics]CH1 Introduction - Plasma Frequency

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Plasma Frequency

Consider a hypothetical slab of plasma, where we assume that the ions have infinite mass(immobile) and the electrons can move freely through the ions.

1. Suppose the electron slab is displaced a distance \(x\) to the right of the ion slab and then allowed to move freely.

2. An electric field will be set up, causing the electron slab to be pulled back toward the ions.

3. When the electrons exactly overlap the ions (when \(x=0\)), the net force is zero, but the electron slab overshoots.

4. The net result is harmonic oscillation. The frequency of the oscillation is called the electron plasma frequency.

Derivation for Plasma Frequency

From Gauss’s law,

\[\oint \vec{D} \cdot d\vec{a} = Q\]

we have

\[\epsilon_0 EA = neAx\]

or

\[E=\frac{nex}{\epsilon_0}\]

Since the force is given by

\[F=QE=(-neAx)(\frac{nex}{\epsilon_0})=nAxm_e \frac{d^2x}{dt^2}\]

we obtain the equation of motion

\[\begin{aligned} \frac{d^2x}{dt^2} = - \left( \frac{ne^2}{m_e \epsilon_0} \right) x\end{aligned}\]

or

\[\begin{aligned} \frac{d^2x}{dt^2} +\omega_{pe}^2x=0\end{aligned}\]

where

\[\begin{aligned} \boxed{\omega_{pe}=\sqrt{\frac{ne^2}{m_e \epsilon_0}}}\end{aligned}\]

This is electron plasma frequency.
\(\bullet\) If there are no collisions, the disturbance will oscillate indefinitely.
\(\bullet\) Debye length, thermal velocity, and plasma frequency are inter-related.

\[\frac{v_{th}}{\omega_{pe}}=\frac{\sqrt{\frac{K_BT}{m}}}{\sqrt{\frac{ne^2}{m\epsilon_0}}}=\sqrt{\frac{\epsilon_0 KT}{ne^2}}=\lambda_D\]

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[플라즈마 물리][Plasma Physics]CH1 Introduction - Plasma Parameter

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Plasma Parameter

\(\bullet\) The plasma parameter is defined as

\[\begin{aligned} \boxed{ N_D =n\frac{4}{3}\pi \lambda_D^3 }=1.38 \times 10^6 \frac{T^{\frac{3}{2}}}{n^{\frac{1}{2}}} [T](in^{\circ}K)\end{aligned}\]

which is the number of plasma particles in a Debye sphere.

\(\bullet\) For Debye shielding to occur, and for the description of a plasma to be statistically meaningful, the number of particles in a Debye sphere must be large; that is \(N_D>>1\)

\(\bullet\) Plasma parameter is a measure of the ratio of the mean plasma kinetic energy to potential energy.

\[\frac{K.E.}{P.E.} \simeq \frac{\frac{3}{2}K_BT}{\frac{e^2}{4\pi\epsilon_0 \lambda_D}} \simeq \frac{9}{2}N_D >> 1\]

\(\bullet\) Thus \(N_D>>1\) means that the potential energy of a particle due to its nearest neighbor is much smaller than its kinetic energy. If this were not the case, there would be a strong tendency for electrons and ions to bind together into atoms, thus destroying the plasma.

\(\bullet\) An ideal gas corresponds to zero potential energy between the particles. Since the plasma parameter is large, the plasma may be treated as an ideal gas of charged particles, that is, a gas that can have a charge density and electric field but in which no two discrete particles interact.

\(\bullet\) In deriving the Debye potential, we assumed that the electrostatic energy was small compared to the thermal energy. The largeness of the plasma parameter guarantees the validity of the Debye potential.

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[플라즈마 물리][Plasma Physics]CH1 Introduction - Debye Shielding 디바이 차폐

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Debye Shielding

Consider a test charge \(e\) and electron cloud around it in a singly charged plasma. Assume that

\(\bullet\) The ions are fixed; \(\frac{m_i}{m_e} \rightarrow \infty\)

\[\begin{aligned} \label{eq_6} n_i(r)=n \end{aligned}\]

\(\bullet\) The electrons obey Boltzman relation. In the presence of a potential energy \(q\phi\), the electron distribution function is

\[f(u)=A e^{-\frac{\frac{1}{2}mu^2+q\phi}{KT_e}}\]

Integrate this over \(du\)

\[\begin{aligned} \label{eq_7} n_e(r)=ne^{\frac{e\phi(r)}{K_BT}} \end{aligned}\]

Poisson’s equation in one dimension is

\[\begin{aligned} \epsilon_0 \nabla^2 \phi = \epsilon_0 \frac{d^2 \phi}{dx^2}=-e(n_i - n_e) \end{aligned}\]

substitution in to the Poisson’s equation,

\[\begin{aligned} \nabla^2 \phi(r) = \frac{e}{\epsilon_0}n(e^{\frac{e\phi(r)}{KT}}-1)\end{aligned}\]

assume that \(|\frac{e\phi}{KT}|<<1\), then by using taylor expansion; \(e^{\frac{e\phi(r)}{KT}}=1+\frac{e\phi(r)}{KT} +...\)

\[\begin{aligned} \nabla^2 \phi(r) &= \frac{e^2n}{\epsilon_0 KT}\phi(r)\\ &=\frac{\phi(r)}{\lambda_D^2}\end{aligned}\]

here we define Debye length \(\lambda_D\) as

\[\lambda_D = \sqrt{\frac{\epsilon_0 K T }{ne^2}}\]

In spherical coordinate

\[\begin{aligned} \frac{1}{r^2}\frac{d}{dr}\left( r^2 \frac{d\phi}{dr}\right) -\frac{\phi}{\lambda_D^2}&=0\\ \phi^{\prime \prime} + \frac{2}{r}\phi^{\prime} - \frac{1}{\lambda_D^2}\phi&=0\\ r\phi^{\prime \prime} + 2\phi^{\prime} - \frac{1}{\lambda_D^2}r\phi&=0\end{aligned}\]

Let \(\psi(r)=r\phi\), then \(\psi^{\prime}=\phi+r\phi^{\prime}\) and \(\psi^{\prime \prime}=2\phi^{\prime}+r\phi^{\prime \prime}\) so that

\[\psi^{\prime \prime}-\frac{1}{\lambda_D^2}\psi =0\]

\[\begin{aligned} \psi(r) &= C_1 e^{-\frac{r}{\lambda_D}}+C_2 e^{\frac{r}{\lambda_D}}\\ \phi(r) &= \frac{C_1}{r} e^{-\frac{r}{\lambda_D}}+\frac{C_2}{r} e^{\frac{r}{\lambda_D}}\end{aligned}\]

Applying the boundary conditions,

\[\begin{aligned} \phi &\rightarrow 0 \quad \quad \quad \quad ,\quad \quad r \rightarrow \infty \\ \phi &\rightarrow \frac{e}{4\pi\epsilon_0 r} \quad \quad , \quad \quad r \rightarrow 0\end{aligned}\]

we can find the constants.

\[C_1 = \frac{e}{4\pi \epsilon_0} \quad \quad \quad C_2 = 0\]

Hence, the solution is given by

\[\begin{aligned} \label{eq_debyelength} \boxed{\phi(r) = \frac{e}{4\pi\epsilon_0 r}e^{-\frac{r}{\lambda_D}}}\end{aligned}\]

The quantity \(\lambda_D\), called the Debye length, is a measure of the shielding distance or thickness of the sheath over which the influence of an individual charged particle is dominant.

\(\bullet\) \(\lambda_D\) = how long the shielding is effective.

\(\bullet\) Notice that electron temperature is used to define Debye length because it is more mobile; most of time this is true.

\(\bullet\) Useful forms of Eq.([eq_debyelength]) are

• \(\lambda_D = 69\sqrt{\frac{T_e}{n}}[m]\) \(T_e\) in \(^{\circ}K\)

• \(\lambda_D = 7430\sqrt{\frac{KT_e}{n}}[m]\) \(KT_e\) in \(eV\)

\(\bullet\) Trend

• effective shielding \(\quad n \uparrow\) \(\lambda_D \downarrow\)

• poor shielding \(\quad T \uparrow\) \(\lambda_D \uparrow\)

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[플라즈마 물리][Plasma Physics]CH1 Introduction - Temperature

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Temperature

The one-dimensional Maxwellian distribution is given by

\[\begin{aligned} f(v)=A e^{-\frac{mv^2}{2K_BT}}\end{aligned}\]

Unlike normal distribution gaussian equation can have a form of

\[\begin{aligned} \label{eq_4} f(x) = \frac{n}{\sqrt{2\pi}\sigma}e^{-\frac{(v-v_{\mu})^2}{2\sigma^2}}\end{aligned}\]

Where \(n\) is the number density. \(fdv\) is the number of particles per [\(m^3\)] with velocity between \(v\) and \(v+dv\), \(\frac{1}{2}mv^2\) is the kinetic energy, and \(K_B\) is the Boltzmann’s constant. The density \(n\), or number of particles per [\(m^3\)], is given by

\[\begin{aligned} n=\int_{-\infty}^{\infty}f(v)dv\end{aligned}\]

so that the constant \(A\) is found to be

\[\begin{aligned} A=n\sqrt{\frac{m}{2\pi K_BT}}\end{aligned}\]

Where \[\begin{aligned} \int_{-\infty}^{\infty}e^{-ax^2}dx = \sqrt{\frac{\pi}{a}} \end{aligned}\] is used.

\(\bullet\) meaning of T = Distribution of the particles

The width of the distribution is characterized by the constant \(T\). By computing the average kinetic energy of particles in the distribution, we can see the exact meaning of \(T\).
\[\begin{aligned} \label{eq_5} E_{av}=\frac{\int_{-\infty}^{\infty}\frac{1}{2}mu^2 f(u) du}{\int_{-\infty}^{\infty} f(u) du}\end{aligned}\]

Defining \(v_{th}=\sqrt{\frac{2K_BT}{m}}\) and \(y=\frac{u}{v_{th}}\), 1-D Maxwellian distribution can be written as

\[\begin{aligned} f(u)=Ae^{-\frac{u^2}{v_{th}^2}}\end{aligned}\]

By substitution average kinetic energy becomes

\[\begin{aligned} E_{av}&=\frac{\frac{1}{2}mAv_{th}^3 \int_{-\infty}^{\infty} e^{-y^2}y^2 dy}{Av_{th}\int_{-\infty}^{\infty} e^{-y^2}dy}\\ &=\frac{\frac{1}{2}mAv_{th}^3 \frac{1}{2}}{A v_{th}}=\frac{1}{4}mv_{th}^2=\frac{1}{2}K_BT\end{aligned}\]

Thus the average kinetic energy is \(\frac{1}{2}K_BT\).
In three dimensions,

\[\begin{aligned} f(u,v,w)=n\left( \frac{m}{2\pi K_BT } \right)^{\frac{3}{2}}e^{-\frac{\frac{1}{2}m\left( u^2 + v^2 + w^2 \right)}{K_BT}}\end{aligned}\]

Using similar calculation we get

\[\begin{aligned} E_{av}=\frac{3}{2}KT\end{aligned}\]

The general result is that \(E_{av}\) equals \(\frac{1}{2}K_BT\) per degree of freedom.

\(\bullet\) Since \(T\) and \(E_{av}\) are so closely related, it is customary in plasma physics to give temperatures in units of energy.

\(\bullet\) To avoid confusion, it is not \(E_{av}\) but the energy corresponding to \(KT\) that is used to denote the temperature.

\(\bullet\) For \(KT=1eV= 1.6\times 10^{-19}[J]\)

\[\begin{aligned} T=\frac{1.6 \times 10^{-19}}{1.38 \times 10^{-23}}=11600\end{aligned}\]

Thus the conversion factor is

\[\begin{aligned} 1eV = 11,600 ^{\circ}K \end{aligned}\]

\(\bullet\) By a \(2eV\) plasma we mean that \(KT=2eV\), or \(E_{av}=3eV\) in three dimensions.

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[플라즈마 물리][Plasma Physics]CH4 Waves In Plasma - General ellipse equation angle derivation (tilted ellipse)

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Problem
Derive equation below given from the lecture note.

\[\begin{aligned} \tan(2\theta) = \frac{2E_{0x}E_{0y}}{E_{0x}^2 - E_{oy}^2}\cos \delta\end{aligned}\]

Answer
Figure below shows the equation of an ellipse tilted at an angle of \(\theta\) to the \(E_x\) axis.

Consider the transformation between the primed and unprimed coordinates.

\[\begin{aligned} E_x^{\prime} = E_x \cos \theta - E_y \sin \theta\\ E_y^{\prime} = E_x \sin \theta + E_y \cos \theta\end{aligned}\]

\[\begin{aligned} \left( \frac{E_x^{\prime}}{E_{0x}} \right)^2 + \left( \frac{E_y^{\prime}}{E_{0y}} \right)^2 - 2 \left( \frac{E_x^{\prime}}{E_{0x}} \right) \left( \frac{E_y^{\prime}}{E_{0y}} \right) \cos \delta = \sin^2 \delta\end{aligned}\]

\[E_{oy}^2 E_x^{\prime} + E_{0x}^2 E_y^{\prime} - 2 E_{0x}E_{oy} E_{x}^{\prime}E_y^{\prime}\cos \delta = \sin^2 \delta E_{ox}^2 E_{oy}^2\]

By sbustitution and rearranging it,

\[\begin{aligned} &E_x^2 [ E_oy^2 \cos^2 \theta + E_{0x}^2 \sin^2 \theta - 2 \cos \theta \sin \theta E_{0x} E_{0y} \cos \delta ] +\\ &E_y^2 [ E_oy^2 \sin^2 \theta + E_{0x}^2 \cos^2 \theta + 2 \cos \theta \sin \theta E_{0x} E_{0y} \cos \delta ] \\ &E_x E_y [ \boxed{-2 \cos\theta \sin \theta E_{0y}^2 + 2 \cos\theta \sin\theta E_{0x}^2 - 2 (\cos^2 \theta - \sin^2\theta) E_{0x}E_{0y}\cos \delta}] \\ &= E_{0x}^2 E_{0y}^2 \sin^2 \delta\end{aligned}\]

\(E_xE_y\) term must be zero in \(x^{\prime}\) and\(y^{\prime}\)coordinate system.

\[\begin{aligned} \sin(2\theta) (E_{0x}^2 - E_{0y}^2 ) = 2 \cos(2\theta) E_{0x} E_{0y} \cos \delta\end{aligned}\]

\[\begin{aligned} \tan(2\theta) = \frac{2E_{0x}E_{0y}}{E_{0x}^2 - E_{oy}^2}\cos \delta\end{aligned}\]

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[플라즈마 물리][Plasma Physics]Introduction

Introduction to Plasma

\(\bullet\) 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.

Plasma State

\(\bullet\) 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.

Definition of Plasma

\(\bullet\) 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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[플라즈마 물리][Plasma Physics]Debye Shielding 디바이 차폐

Example Problem: Debye Shielding
Consider a positive point charge immersed in a plasma as we discussed in the class. Show that the net charge in the Debye shielding cloud exactly cancels the test charge. Assume that the ions are fixed and that \(e\phi << KT_e\). Note that we assumed that ion distribution is similar to that of the electrons in the class.

Answer:

While deriving, we assume potential to be

\[\begin{aligned} \frac{e\phi}{KT_e} << 1 \quad \quad \quad n_i \simeq n_e\\ \lambda_D = \sqrt{\frac{\epsilon KT}{ne^2}}\\ \phi(r) = \frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}}\end{aligned}\]

Assume \(\frac{m_i}{m_e} \rightarrow \infty\) so that ions are fixed; \(n_i(r)=n\). Further assume electrons obey the Boltzmann distribution.

\[\begin{aligned} f(u) &= A e^{\frac{\frac{-1}{2}mv^2 + q\phi}{KT_e}}\\ n_e(r) &= ne^{\frac{e\phi(r)}{K_B T}}\end{aligned}\]

\[\begin{aligned} \rho &= -e(n_i - n_e) \\ &=-e \left(n - ne^{\frac{e\phi(r)}{K_BT}} \right)\\ &= ne \left(e^{\frac{e\phi(r)}{K_BT}} - 1 \right)\\\end{aligned}\]

for \(\frac{e\phi}{KT_e} << 1\),

\[\begin{aligned} e^{\frac{e\phi(r)}{K_B T}} = 1 + \frac{1}{2}\frac{e\phi(r)}{K_B T} + ...\end{aligned}\]

\[\begin{aligned} \rho &= ne(1+\frac{e\phi(r)}{K_B T} - 1) \\ &=\frac{ne^2 \phi(r)}{K_B T} \\ &= \frac{\epsilon_0}{\lambda_D^2}\phi(r)\end{aligned}\]

\[\begin{aligned} \int \rho dv &= Q \\ &= \int_V \frac{\epsilon_0}{\lambda_D^2}\frac{e}{4\pi\epsilon_0 r }e^{-\frac{r}{\lambda_D}} r^2 \sin\theta dr d\theta d\phi\\ &= \int \frac{4\pi \epsilon_0}{\lambda_D^2}\frac{e}{4\pi \epsilon_0} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} \int_{0}^{\infty} r e^{-\frac{r}{\lambda_D}}dr \\ &= \frac{e}{\lambda_D^2} (\lambda_D^2 e^{-\frac{r}{\lambda_D}}-\lambda_D r e^{-\frac{r}{\lambda_D}}) =0\end{aligned}\]

On the last step, when \(r = \lambda_D\), charge becomes zero. Hence, the Debye sheilding cloud exactly cancels the test charge.

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[플라즈마 물리]F.F. Chen Problem 1.5 Solution Debye Shielding

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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[플라즈마 물리]Normalization Constant & Average Kinetic Energy

Problem The 3D Maxwellian distribution is given by
\[f(\vec{v}) = A_3 e^{\left( -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} \right)}\]
with \[n = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z\]
(a) Show that the normalization constant is given by \[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
(b) Show that the average kinetic energy is \[E_{av} = \frac{3}{2}KT\]
Answer (a)

Notice that
\[\begin{aligned} \label{eq_gauss} \int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\end{aligned}\]
\[\begin{aligned} n &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(\vec{v})dv_x dv_y dv_z \nonumber \\ &= A_3 \int_{-\infty}^{\infty} e^{ -\frac{m(v_x^2)}{2KT}}dv_x \int_{-\infty}^{\infty} e^{ -\frac{m(v_y^2)}{2KT}}dv_y \int_{-\infty}^{\infty} e^{ -\frac{m(v_z^2)}{2KT}}dv_z \nonumber \\ &= A_3 \left( \sqrt{\frac{ 2 K T \pi}{m}} \right)^{3}\end{aligned}\]
Hence,
\[A_3 = n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}}\]
Answer (b)

Average energy can be calculated as
\[\begin{aligned} E_{av} &= \frac{\int_{-\infty}^{\infty} \frac{1}{2}mv^2 f(\vec{v})dv}{\int_{-\infty}^{\infty}f(\vec{v})dv} \nonumber \\ &=\frac{m}{2n}\int_{-\infty}^{\infty} (v_x^2 + v_y^2 + v_z^2) A_3 e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv\end{aligned}\]
The denominator is \(n\) as we calculated in previous problem. Notice that, \[\int x^2 e^{-ax^2}dx = \frac{1}{2}\sqrt{\frac{\pi}{a^3}}\]
If we calculate the first term of equation,
\[\begin{aligned} E_{av_x} &= \frac{m}{2n} \int_{-\infty}^{\infty} A_3(v_x^2 ) e^{ -\frac{m(v_x^2 + v_y^2 + v_z^2)}{2KT} }dv_x \nonumber \\ &=\frac{m}{2n}n \left( \frac{m}{2\pi KT} \right)^{\frac{3}{2}} \frac{1}{2} \left( \frac{2 KT}{m} \right)^{\frac{3}{2}}\sqrt{\pi}\frac{2KT \pi}{m} \nonumber \\ &=\frac{1}{2}KT\end{aligned}\]
Hence, if we calculate \(v_y\) and \(v_z\) components, final result becomes \[E_{av} = \frac{3}{2}KT\]
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