Question

(a) A cold homogeneous plasma supports oscillations at the plasma frequency $\omega = \omega_p$. To first order, the frequency of this oscillation is independent of the wavenumber. However, when pressure is considered, the situation changes. Here you are asked to find and explore the $\omega - k$ relationship of such oscillations. (i) The equations governing the electron fluid are $\frac{\partial n}{\partial t} + \nabla \cdot (nv) = 0$ $\m_e n \frac{dv}{dt} = -enE - \nabla p$ $\p n^{-\gamma} = p_0 n_0^{-\gamma}$ $\nabla \cdot \epsilon_0 E = -e(n - n_0),$ where the 0 subscript marks quantities of the homogeneous equilibrium. Considering a plasma in a stationary equilibrium, with no steady-state electric field, derive a set of linear equations sufficient to solve for the perturbed quantities. (ii) By assuming plane-wave solutions to the set of equations you derived above, show that the relation between $\omega$ and $k$ that permits non-trivial solutions is $\omega^2 = \omega_{pe}^2 + k^2 c_s^2$, where $\omega_{pe}$ is the plasma frequency and $c_s$ is the electron sound speed. 

          (a) A cold homogeneous plasma supports oscillations at the plasma frequency
$\omega = \omega_p$. To first order, the frequency of this oscillation is independent of the
wavenumber. However, when pressure is considered, the situation changes. Here
you are asked to find and explore the $\omega - k$ relationship of such oscillations.
(i) The equations governing the electron fluid are
$\frac{\partial n}{\partial t} + \nabla \cdot (nv) = 0$
$\m_e n \frac{dv}{dt} = -enE - \nabla p$
$\p n^{-\gamma} = p_0 n_0^{-\gamma}$
$\nabla \cdot \epsilon_0 E = -e(n - n_0),$
where the 0 subscript marks quantities of the homogeneous equilibrium.
Considering a plasma in a stationary equilibrium, with no steady-state
electric field, derive a set of linear equations sufficient to solve for the
perturbed quantities.
(ii) By assuming plane-wave solutions to the set of equations you derived above,
show that the relation between $\omega$ and $k$ that permits non-trivial solutions is
$\omega^2 = \omega_{pe}^2 + k^2 c_s^2$,
where $\omega_{pe}$ is the plasma frequency and $c_s$ is the electron sound speed.
Show more…
(a) A cold homogeneous plasma supports oscillations at the plasma frequency ω =. To first order, the frequency of this oscillation is independent of the wavenumber. However, when pressure is considered, the situation changes. Here you are asked to find and explore the ω - k relationship of such oscillations. (i) The equations governing the electron fluid are (∂ n)/(∂ t) + ∇· (nv) = 0 n (dv)/(dt) = -enE - ∇ p n^-γ = p0 n0^-γ ∇·ϵ0 E = -e(n - n0), where the 0 subscript marks quantities of the homogeneous equilibrium. Considering a plasma in a stationary equilibrium, with no steady-state electric field, derive a set of linear equations sufficient to solve for the perturbed quantities. (ii) By assuming plane-wave solutions to the set of equations you derived above, show that the relation between ω and k that permits non-trivial solutions is ω^2 = ωpe^2 + k^2 cs^2, where ωpe is the plasma frequency and cs is the electron sound speed.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Please show the mathematical procedure. a) A cold homogeneous plasma supports oscillations at the plasma frequency ωp. To first order, the frequency of this oscillation is independent of the wavenumber. However, when pressure is considered, the situation changes. Here you are asked to find and explore the ω-k relationship of such oscillations. (i) The equations governing the electron fluid are: ∂n/∂t + V · (∇n) = 0, m∂v/∂t - e∇φ - V · (∇p) = 0, ∂p/∂t + V · (∇p) = 0, ∇ · E = -e(n - n0). where the 0 subscript marks quantities of the homogeneous equilibrium. Considering a plasma in a stationary equilibrium, with no steady-state electric field, derive a set of linear equations sufficient to solve for the perturbed quantities. (ii) By assuming plane-wave solutions to the set of equations you derived above, show that the relation between ω and k that permits non-trivial solutions is: ω^2 = ωp^2 + k^2c^2, where ωp is the plasma frequency and cs is the electron sound speed.
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Transcript

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00:01 This question equilibrium vibration frequency of iodine molecule equilibrium vibration vibration vibration of iodine molecule it is given iodine molecule it is given 215 cm inverse so and hormonicity of constant and hormonicity constant value also that is x it is given as 0.
00:34 0 .003 and they have asked that at 300 kelvin temperature what is the intensity of hot band that is a v is equal to 1 2v is equal to 2 transition so at this particular transition this transition it is relative to the fundamental so it is relative to fundamental the value that is the fundamental that is v is equal to 0 to v is equal to 1 so now we have to find out so here what is given v value 215 cm inverse and x value it is given 0 .0 .03 and temperature it is given 300 kelvin we have to find out the hot band intensity so the hot band intensity is calculated by means of the formula delta e that is is v is equal to 1 -2 that is equal to 2115 into 1 minus 4 into 0 .003 so whenever we simplify this we'll get the value that is 1 minus 0 .0 .01 that is equal to 2112 .42 this is the intensity of the hot pan value then in this same question they have asked that here how many normal modes of vibration how how many normal modes of vibration that is possible for the following molecules.
02:12 They have given the following molecules, hbr, ocs, that is linear, and so2, that is, bent structure, and bcl3, and c -h -3, c -h -3 -1, c -h -h -c -h -4, c -h -3, i, c -6h -6.
02:33 For these molecules we have to find out normal modes of vibration so how many normal modes of vibration we have to find out so the compound here it is hbr and it is 3n minus 5 that is normal modes of vibration how to calculate normal modes of normal modes that is 3 into 2 4 2 minus 5 that is equal to 1 then for oxygen so for oxygen also formula it is 3n minus 5 3 into 2 minus 5 that is equal to 1 and for ocs ocs that is 3 n minus 5 that is 3 into 3 minus 5 that is equal to 4 then for for s .o 2 it is 3 n minus 6 that is 3 into 3 minus 6 that is equal to 3 then for bcl 3 then for b c l 3, bent structure 3n minus 6.
03:42 3n minus 6, that is 3 into 3n minus 6, that is 3 into 4 minus 6...
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