University:
Massachusetts Institute of Technology
Course:
22.611J | Introduction to Plasma Physics I
Academic year:

2024
Views:

458

Pages:

7
Author:

Jamar Russell

22.611J, 6.65J, 8.6137 PS#5 Solution Key The polarization drift is just: J_p = (m/e) * (ne/B²) * (E × B) = (m/eB²) * (E × B) So, J_p = (m/eB²) * (E × B) Discussion: Polarization effects are in MHD. (We just derived it!) There's no conflict with Ohm's law (E + v × B = 0) because the center of mass velocity of the fluid can be zero while the current is finite. Therefore, all the polarization effects in MHD are not affected. In MHD, J ≠ n₁v₁. The ions & electrons can still form independent currents, as long as these currents are reflected in J. 1 Cont. In other words, MHD gives one order higher J than Ohm's law. The dominant effect on J is E × B. But 2nd order effects, like the polarization drift, end up in J. 2) Wave Packets & Group Velocity First, show that: ψ(x,t) = Re[(1/2π) * ∫ dk * e^(ikx) * e^(-σ²(k - k₀)²/2)] We can get ψ(x,t) from (2) by: Let k' = k - k₀ + iσ dk' = dk Then, ψ(x,t) = Re[(1/2π) * ∫ dk' * e^(ik'x + ik₀x) * e^(-σ²(k' - iσ)²/2)] = Re[(1/2π) * e^(ik₀x) * ∫ dk' * e^(-σ²(k' - iσ)²/2 + ik'x)] Complete the square: = Re[(1/2π) * e^(ik₀x) * ∫ dk' * e^(-σ²(k' - iσ + ix/2σ)² + σ²/2 - ix²/4σ²)] Since k' = k - k₀ + iσ, then k'x = (k - k₀ + iσ)x = kx - k₀x + iσx Y = iσx/2σ² Then, ψ(x,t) = Re[(1/2π) * e^(ik₀x) * e^(σ²/2 - ix²/4σ²) * ∫ dk' * e^(-σ²(k' - Y)²)] Let N = k' - Y dN = dk' Then, ψ(x,t) = Re[(1/2π) * e^(ik₀x) * e^(σ²/2 - ix²/4σ²) * ∫ dN * e^(-σ²N²)] ψ(x,t) = Re[(1/2π) * e^(ik₀x) * e^(σ²/2 - ix²/4σ²) * √(π/σ²) * erf(∞) - erf(-∞)] ψ(x,t) = Re[(1/2π) * e^(ik₀x) * e^(σ²/2 - ix²/4σ²) * √(π/σ²) * 2] ψ(x,t) = Re[e^(ik₀x) * e^(σ²/2 - ix²/4σ²)] 2) Cont. Approximating ω(k) ≈ ω(k₀) + (∂ω/∂k)₀(k - k₀) in: A(x) = Re[(1/2π) * ∫ dk * exp(ikx - iw(k)t) * exp(-σ²(k - k₀)²)] A(x) = Re[(1/2π) * ∫ dk * e^(ikx - iw(k₀)t - i(∂ω/∂k)₀(k - k₀)t) * exp(-σ²(k - k₀)²)] A(x) = Re[(1/2π) * e^(-iw(k₀)t + i(∂ω/∂k)₀t) * ∫ dk * e^(ikx - i(∂ω/∂k)₀(k - k₀)t) * exp(-σ²(k - k₀)²)] Let k' = k - k₀ dk' = dk A(x) = Re[(1/2π) * e^(-iw(k₀)t + i(∂ω/∂k)₀t) * ∫ dk' * e^(ik'x - i(∂ω/∂k)₀k't) * exp(-σ²(k')²)] Now, e^(ik'x - i(∂ω/∂k)₀k't) factor is exactly the same form as A(x, t=0). In other words, exactly the same profile as A(x, t=0), but moving at (∂ω/∂k)₀. Which is the group velocity! 2) Extra Credit If we have: ω(k) = ω(k₀) + (∂ω/∂k)₀(k - k₀) + (1/2)(∂²ω/∂k²)₀(k - k₀)² Then since: A(x,t) = Re[(1/2π) * ∫ dk * exp(ikx - iw(k)t) * exp(-σ²(k - k₀)²)] A(x,t) = Re[(1/2π) * ∫ dk * exp(ikx - iw(k₀)t - i(∂ω/∂k)₀(k - k₀)t - (i/2)(∂²ω/∂k²)₀(k - k₀)²t) * exp(-σ²(k - k₀)²)] → which is the same as before, except now the wave packet has a dispersion factor added to σ²! → Hence, with increasing t: (σ² + (i/2)(∂²ω/∂k²)₀t) becomes longer and larger, increasing the packet size.

22.611J Problem Set 5 - Solution Key

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