Solution for Problem Set 19-20

Transcription

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2 Solution for Problem Set 19-0 compiled by Dan Grin and Nate Bode) April 16, 009 A 19.4 Ion Acoustic Waves [by Xinkai Wu 00] a) The derivation of these equations is trivial, so we omit it here. b) Write the proton density as n = n 0 +δn and substitute it into the equations of part a), keeping only terms linear in δn, u, and Φ. We find δn t + n u 0 z = 0 1) u t = e Φ m p z ) Φ z = e ɛ 0 δn n 0e k B T e Φ ). 3) Plugging the plane-wave solution where δn, u, and Φ are all of the form exp[ikz ωt)] into the above linearized equations, we find iωδn + n 0 iku = 0 4) iωu = e ikφ 5) m p k Φ = e δn n ) 0e Φ. 6) ɛ 0 k B T e For the above algebraic equation to possess a solution, the determinant of its coefficient matrix must vanish, which gives ω k n 0e ) + n 0e k = 0. 7) ɛ 0 k B T e ɛ 0 m p Solving this yields the dispersion relation ) 1/ ) with λ D = ɛ0k BT e n 0e and ωpp = n0e 1/. ɛ 0m p ω = ω pp 1 + 1/k λ D) 1/, 8) 1

3 For long-wavelength, kλ D << 1, one finds which agrees with eq. 0.36). ) 1/ kb T e ω ω pp kλ D = k, 9) m p 19.6 Dispersion and Faraday rotation of pulsar pulses [by Alexei Dvoretskii 000] a) The pulses are limited in time, so they must be composed of a range of frequencies. Such a wave packet will propagate at the group velocity. For high frequencies, v g = dω ) dk = c 1 ω p ω c 1 ω p ω. 10) The difference in propagation time is given by [ 1 t = dx v g ω H ) 1 ]. v g ω L ) Simplifying and assuming ω H ω L ω p and constant, we get t = L c Using n = m 3 we get ω p = 10 4 s 1, so indeed ω p ω L ω H. The distance to the pulsar is then L m. b) Consider the Faraday rotation and ω p ω L dχ dx = ω pω c ω c. The overall rotation angle is then given by χ = 1 ω ω c pω c dx. χ t = e ω p B m ω p dx = e nb = e m ndx m < B >. 11) From here the interstellar magnetic field can be estimated to be < B > G.

4 B 19.5 Ion Acoustic Solitons [by Keith Matthews 005] a) You ll note that the u coefficient k B T e /m p ) 1/ is the characteristic velocity of the system, so let s name it v c. Also define α k B T e /e. Then u = v c ɛu 1 + ɛ u +...) Φ = αɛφ 1 + ɛ Φ +...). 1) For an arbitrary field ψ, ψ t = α 1/ ψ ɛ λ D + 3/ ψ ω pp ɛ τ ψ z = ω pp 1/ ψ ɛ. 13) λ D i) From the continuity eqn at leading order ɛ 3/ we find n 1 + u 1 = 0, 14) and at ɛ 5/ order we have u + n 1u 1 ) = 0. 15) ii) To leading order ɛ 3/ the equation of motion produces However eα m pv c = 1 so we have u 1 eα Φ 1 m p v c = 0. We note that λ D ω p p = v c so to ɛ 5/ we have u 1 Φ 1 = 0. 16) u + u 1 τ + 1 u 1 ) = Φ.η 17) iii) en0λd αɛ 0 = 1 so at order ɛ and ɛ Poisson s eqn gives n 1 = Φ 1 18) Φ 1 = 1 n Φ 1 Φ 1 ). 19) 3

5 iv) From Eqs. 14), 16) and 18) we find Taking n 1 = u 1 = Φ 1. 0) of Eq. 19) and invoking Eq. 0) gives 3 n 1 3 = 1 n Φ 1 n 1 ) We find that we can take care of the n 17) and again invoking Eq. 0). Φ n Φ = n 1 τ + 3 Substituting Eq. ) into 1) gives us 3 n 1 3 = 1 n 1 τ + n 1 ). 1) term by adding Eqs. 15) and n 1 ). ) ) n 1. 3) which is just what the doctor ordered. Since n 1 = u 1 = Φ 1 any of the three can solve this KdV equation Exploration of Modes in CMA Diagram [by Kip Thorne 005] a) EM waves in a cold unmagnetized plasma can only propagate when their frequency exceeds the plasma frequency ω p, as can be seen from their dispersion relation ω = ωp + c k. Thus ω = ω p represents a cut-off. There is no turn-on. The unmagnetized condition places us on the x-axis, and the lower limit of the frequency places us to the left of unity. These waves correspond to the diagram in the lower left corner with the modification that with no magnetic field the distinction between X and O dissolves, and R and L modes have the same phase velocity, so the are represented by the same circle. b) Denote REM: Right hand polarized electromagnetic waves LEM: Left hand polarized electromagnetic waves RW: Right hand polarized Whistler waves LW: Left hand polarized Whistler waves. From Fig we can deduce the following relation between V φ = c n and ω. As we move from the lower left to the upper right on the CMA diagram, ω decreases with B and n fixed. So we sequentially encounter ) ) LEM LW LEM) RW ). REM RW 4

6 Figure 1: phase velocity vs. frequency for parallel propagating modes The boundaries corresponding to ɛ L = 0 and ɛ R = 0 are specified in the diagram and correspond to the mode changes above. c) Denote: O: ordinary mode XL: extraordinary lower mode XH: extraordinary hybrid mode XU: extraordinary upper mode. From Fig we can extract the following relation between the phase velocity and the frequency. As ω decreases with B and n fixed, we observe the following pattern ) ) XL XH XH) XL). O O The boundary for these regions are described by ɛ 1 = 0, ɛ 3 = 0, ɛ 1 = 0 respectively. Combined with the previous result b), we obtain the pattern shown in the CMA diagram. On the upper right in CMA one can see that RW becomes XL as we move from parallel to perpendicular mode and LW ceases to exist at some θ < π/. Similar phenomena can be read off from the CMA diagram. C 5

7 Figure : phase velocity vs. frequency for perpendicular propagating modes 19.8 Reflection of Short Waves by the Ionosphere [by Keith Matthews 005] The brute force approach: For adiabatic spatial variations in the index of refraction, Fermat s principle chooses rays of extremal time. This gives simple differential equations Eq. 6.4)) which we express as r component: θ component: d r ds + dñ/dr ñ ) [ dr ) ] ds 1 = 0 ñ dθ ds = C a constant. 4) We choose polar coordinates because, as we shall see, the maximum range is of the order of the radius of the earth r e. s parametrizes the path length. We are told that the electron density is exponential in altitude: n e = n 0 e y/y0. where y is the altitude r r e. From the values given in the problem: y 0 = 1.71 km and n 0 = 10 7 /m 3. From Eq ): ñ = 1 ωpe ω which we write as ñ 1 ηn e where η e m eɛ 0ω. ηn 0 = and r e = km.) Note that ñ = 1 y 0 ñ 1). I define ψ 0 at the point of transmission as the angle between the vertical 6

8 and the outgoing ray. As a result the initial conditions are cos ψ 0 = ) dr ds ) 0, and because sin ψ 0 = r 0 dθ ds ) 0, C = ñ 0 r 0 sin ψ 0. r 0 = r e and I set θ 0 = 0. Numerically integrate with s as the independent variable until rs f ) = r e. I used Mathematica.) The results give r and θ as a function of s, so you have to calculate the range R = r e θs f ). By searching around I found the maximum range to be 6677 km for ψ 0 = 90 deg. It is vital that this ray does not pass through the maximum altitude of y max = 00 km. I found that y top = 5.7 km. 0.1 Two-fluid Equation of Motion [by Xinkai Wu 00] We ll first write things in components and in the end convert back to vector/tensor notation. The Vlasov equation reads f s t + f vj s x j + f aj s = 0. 5) vj We ll make use of the following definitions n s = f s dv v n s u i s = f s v i dv v Ps ik = m s f s v i u i s)v k u k s)dv v = m s f s v i v k dv v m s n s u i s uk s. 6) Multiplying the Vlasov eq. by v k, integrating over velocity space, using integration by parts at various places and the fact that aj v = 0, also using the explicit j expression Eq. 0.3) for the acceleration due to external EM field, we get n s u k s) t + ) P jk s x j m + n su j su k s n sq s E k + u s B) k ) = 0. 7) m s Now using the continuity equation, n s t = x j n su j s), 8) one immediately sees that Eq. 7) reduces to Eq. 0.11) after converting to vector notation. D 7

9 0.4 Landau Contour Deduced Using Laplace Transforms [by Xinkai Wu 000] a) This part is nothing but a definition of Laplace transformation. b) The z-dependence is e ikz, giving / z ik. Also integration by parts gives 0 dte pt F s1 / t = F s1 v, 0) + p 0 dte pt F s1 v, t) 9) Noticing the above facts, we get by Laplace transforming the Vlasov equation: 0 = F s1 v, 0) + p F s1 v, p) + vik F s1 v, p) + q s /m s )F s0ẽp) 30) where s = p, e. Laplace transforming E = ρ/ɛ 0 gives us a second equation: ikẽp) = s q s /ɛ 0 ) dv[f s0 v)/p + F s1 v, p)] = s q s /ɛ 0 ) dv F s1 v, p). 31) Where to get the last equality we ve used the fact that the unperturbed charge density is zero, i.e. the contribution from F s0 v) vanishes. Combining these two equations we easily get Eq. 0.41). c) Setting ip = ω, and plugging Eq. 0.41) into Eq. 0.4), we immediately get Eq. 0.6) without that overall minus sign. E 0.5 Ion Acoustic Dispersion Relation [by Xinkai Wu 000] Recall the definitions of Debye length and plasma frequency for species s: ) ne 1/ ω ps = ɛ 0 m s ) 1/ ɛ0 k B T s λ Ds = ne, 3) and the Maxwellian distribution is ) 1/ ms F s v) = n exp [ m sv ]. 33) πk B T s k B T s 8

10 Now consider the integral For ωr k >> k BT s m s going to use for I p. For ωr k I s ω r k ) P F sv) dv. 34) v ω r /k, I s nk ω see eq. 1.37)), and this is the formula we are r << k BT s m s, I s nms k BT s as can be easily seen by ignoring the ω r /k in the denominator in the integral), and this is the formula we ll use for I e. In our problem, the total F v) is given by F = F e + me m p F p, and thus the total Iω r /k) is given by I = I e + me m p I p. Now that we have the explicit approximate) expressions for F and I, substitution into Eqs. 0.34) and 0.35) yields the desired expressions for ω r and ω i, after approximating F e ω r /k) as n me πk BT e ) 1/ in the numerator of 0.35), which is justified by our assumption ωr k << k BT e m e ). 0.8 Range of Unstable Wave Numbers [by Jeff Atwell] For instability, we need a ζ with ζ i > 0 such that k = Zζ) is real and positive. The corresponding ω = kζ specifies the mode. ζ with ζ i > 0 correspond to points in the interior of the closed curve in the Z plane see Fig. 0.4). So the intersection of the positive Z r axis and the interior of the closed curve gives the range of wave numbers that we are looking for. The shape of this closed curve depends on the distribution function F v). We are told our distribution function has two maxima, v 1 and v, and one minimum, v min. This means that the closed curve in the Z plane crosses the Z r axis three times, twice moving downward, and once moving upward. The curve in Fig. 0.4 happens to be an example of this situation.) Suppose v 1 < v. Then in the case of the closed curve shown in Fig 0.4, we may write the range of wave numbers which have at least one unstable mode as Zv ) < k < Zv min ) where Z is evaluated using Eq. 0.47)). For different shapes of the closed curve i.e. for different combinations of Zv 1 ), Zv min ), and Zv ) being positive and negative) the wave number range will be a bit different, but similar. The steps in going from Eq. 0.47) to Eq. 0.49) work for maxima of F v), in addition to minima. This means that in the case shown in Fig. 0.4 we can instead write the minimum wave number with an unstable mode as for example. e kmin = Zv ) = m e ɛ 0 + [F v) F v )] v v ) dv, 9

11 Solution for Problem Set 19-0 compiled by Dan Grin and Nate Bode) April 16, 009 A 19.4 Ion Acoustic Waves [by Xinkai Wu 00] a) The derivation of these equations is trivial, so we omit it here. b) Write the proton density as n = n 0 +δn and substitute it into the equations of part a), keeping only terms linear in δn, u, and Φ. We find δn t + n u 0 z = 0 1) u t = e Φ m p z ) Φ z = e ɛ 0 δn n 0e k B T e Φ ). 3) Plugging the plane-wave solution where δn, u, and Φ are all of the form exp[ikz ωt)] into the above linearized equations, we find iωδn + n 0 iku = 0 4) iωu = e ikφ 5) m p k Φ = e δn n ) 0e Φ. 6) ɛ 0 k B T e For the above algebraic equation to possess a solution, the determinant of its coefficient matrix must vanish, which gives ω k n 0e ) + n 0e k = 0. 7) ɛ 0 k B T e ɛ 0 m p Solving this yields the dispersion relation ) 1/ ) with λ D = ɛ0k BT e n 0e and ωpp = n0e 1/. ɛ 0m p ω = ω pp 1 + 1/k λ D) 1/, 8) 1

12 For long-wavelength, kλ D << 1, one finds which agrees with eq. 0.36). ) 1/ kb T e ω ω pp kλ D = k, 9) m p 19.6 Dispersion and Faraday rotation of pulsar pulses [by Alexei Dvoretskii 000] a) The pulses are limited in time, so they must be composed of a range of frequencies. Such a wave packet will propagate at the group velocity. For high frequencies, v g = dω ) dk = c 1 ω p ω c 1 ω p ω. 10) The difference in propagation time is given by [ 1 t = dx v g ω H ) 1 ]. v g ω L ) Simplifying and assuming ω H ω L ω p and constant, we get t = L c Using n = m 3 we get ω p = 10 4 s 1, so indeed ω p ω L ω H. The distance to the pulsar is then L m. b) Consider the Faraday rotation and ω p ω L dχ dx = ω pω c ω c. The overall rotation angle is then given by χ = 1 ω ω c pω c dx. χ t = e ω p B m ω p dx = e nb = e m ndx m < B >. 11) From here the interstellar magnetic field can be estimated to be < B > G.

13 B 19.5 Ion Acoustic Solitons [by Keith Matthews 005] a) You ll note that the u coefficient k B T e /m p ) 1/ is the characteristic velocity of the system, so let s name it v c. Also define α k B T e /e. Then u = v c ɛu 1 + ɛ u +...) Φ = αɛφ 1 + ɛ Φ +...). 1) For an arbitrary field ψ, ψ t = α 1/ ψ ɛ λ D + 3/ ψ ω pp ɛ τ ψ z = ω pp 1/ ψ ɛ. 13) λ D i) From the continuity eqn at leading order ɛ 3/ we find n 1 + u 1 = 0, 14) and at ɛ 5/ order we have u + n 1u 1 ) = 0. 15) ii) To leading order ɛ 3/ the equation of motion produces However eα m pv c = 1 so we have u 1 eα Φ 1 m p v c = 0. We note that λ D ω p p = v c so to ɛ 5/ we have u 1 Φ 1 = 0. 16) u + u 1 τ + 1 u 1 ) = Φ.η 17) iii) en0λd αɛ 0 = 1 so at order ɛ and ɛ Poisson s eqn gives n 1 = Φ 1 18) Φ 1 = 1 n Φ 1 Φ 1 ). 19) 3

14 iv) From Eqs. 14), 16) and 18) we find Taking n 1 = u 1 = Φ 1. 0) of Eq. 19) and invoking Eq. 0) gives 3 n 1 3 = 1 n Φ 1 n 1 ) We find that we can take care of the n 17) and again invoking Eq. 0). Φ n Φ = n 1 τ + 3 Substituting Eq. ) into 1) gives us 3 n 1 3 = 1 n 1 τ + n 1 ). 1) term by adding Eqs. 15) and n 1 ). ) ) n 1. 3) which is just what the doctor ordered. Since n 1 = u 1 = Φ 1 any of the three can solve this KdV equation Exploration of Modes in CMA Diagram [by Kip Thorne 005] a) EM waves in a cold unmagnetized plasma can only propagate when their frequency exceeds the plasma frequency ω p, as can be seen from their dispersion relation ω = ωp + c k. Thus ω = ω p represents a cut-off. There is no turn-on. The unmagnetized condition places us on the x-axis, and the lower limit of the frequency places us to the left of unity. These waves correspond to the diagram in the lower left corner with the modification that with no magnetic field the distinction between X and O dissolves, and R and L modes have the same phase velocity, so the are represented by the same circle. b) Denote REM: Right hand polarized electromagnetic waves LEM: Left hand polarized electromagnetic waves RW: Right hand polarized Whistler waves LW: Left hand polarized Whistler waves. From Fig we can deduce the following relation between V φ = c n and ω. As we move from the lower left to the upper right on the CMA diagram, ω decreases with B and n fixed. So we sequentially encounter ) ) LEM LW LEM) RW ). REM RW 4

15 Figure 1: phase velocity vs. frequency for parallel propagating modes The boundaries corresponding to ɛ L = 0 and ɛ R = 0 are specified in the diagram and correspond to the mode changes above. c) Denote: O: ordinary mode XL: extraordinary lower mode XH: extraordinary hybrid mode XU: extraordinary upper mode. From Fig we can extract the following relation between the phase velocity and the frequency. As ω decreases with B and n fixed, we observe the following pattern ) ) XL XH XH) XL). O O The boundary for these regions are described by ɛ 1 = 0, ɛ 3 = 0, ɛ 1 = 0 respectively. Combined with the previous result b), we obtain the pattern shown in the CMA diagram. On the upper right in CMA one can see that RW becomes XL as we move from parallel to perpendicular mode and LW ceases to exist at some θ < π/. Similar phenomena can be read off from the CMA diagram. C 5

16 Figure : phase velocity vs. frequency for perpendicular propagating modes 19.8 Reflection of Short Waves by the Ionosphere [by Keith Matthews 005] The brute force approach: For adiabatic spatial variations in the index of refraction, Fermat s principle chooses rays of extremal time. This gives simple differential equations Eq. 6.4)) which we express as r component: θ component: d r ds + dñ/dr ñ ) [ dr ) ] ds 1 = 0 ñ dθ ds = C a constant. 4) We choose polar coordinates because, as we shall see, the maximum range is of the order of the radius of the earth r e. s parametrizes the path length. We are told that the electron density is exponential in altitude: n e = n 0 e y/y0. where y is the altitude r r e. From the values given in the problem: y 0 = 1.71 km and n 0 = 10 7 /m 3. From Eq ): ñ = 1 ωpe ω which we write as ñ 1 ηn e where η e m eɛ 0ω. ηn 0 = and r e = km.) Note that ñ = 1 y 0 ñ 1). I define ψ 0 at the point of transmission as the angle between the vertical 6

17 and the outgoing ray. As a result the initial conditions are cos ψ 0 = ) dr ds ) 0, and because sin ψ 0 = r 0 dθ ds ) 0, C = ñ 0 r 0 sin ψ 0. r 0 = r e and I set θ 0 = 0. Numerically integrate with s as the independent variable until rs f ) = r e. I used Mathematica.) The results give r and θ as a function of s, so you have to calculate the range R = r e θs f ). By searching around I found the maximum range to be 6677 km for ψ 0 = 90 deg. It is vital that this ray does not pass through the maximum altitude of y max = 00 km. I found that y top = 5.7 km. 0.1 Two-fluid Equation of Motion [by Xinkai Wu 00] We ll first write things in components and in the end convert back to vector/tensor notation. The Vlasov equation reads f s t + f vj s x j + f aj s = 0. 5) vj We ll make use of the following definitions n s = f s dv v n s u i s = f s v i dv v Ps ik = m s f s v i u i s)v k u k s)dv v = m s f s v i v k dv v m s n s u i s uk s. 6) Multiplying the Vlasov eq. by v k, integrating over velocity space, using integration by parts at various places and the fact that aj v = 0, also using the explicit j expression Eq. 0.3) for the acceleration due to external EM field, we get n s u k s) t + ) P jk s x j m + n su j su k s n sq s E k + u s B) k ) = 0. 7) m s Now using the continuity equation, n s t = x j n su j s), 8) one immediately sees that Eq. 7) reduces to Eq. 0.11) after converting to vector notation. D 7

18 0.4 Landau Contour Deduced Using Laplace Transforms [by Xinkai Wu 000] a) This part is nothing but a definition of Laplace transformation. b) The z-dependence is e ikz, giving / z ik. Also integration by parts gives 0 dte pt F s1 / t = F s1 v, 0) + p 0 dte pt F s1 v, t) 9) Noticing the above facts, we get by Laplace transforming the Vlasov equation: 0 = F s1 v, 0) + p F s1 v, p) + vik F s1 v, p) + q s /m s )F s0ẽp) 30) where s = p, e. Laplace transforming E = ρ/ɛ 0 gives us a second equation: ikẽp) = s q s /ɛ 0 ) dv[f s0 v)/p + F s1 v, p)] = s q s /ɛ 0 ) dv F s1 v, p). 31) Where to get the last equality we ve used the fact that the unperturbed charge density is zero, i.e. the contribution from F s0 v) vanishes. Combining these two equations we easily get Eq. 0.41). c) Setting ip = ω, and plugging Eq. 0.41) into Eq. 0.4), we immediately get Eq. 0.6) without that overall minus sign. E 0.5 Ion Acoustic Dispersion Relation [by Xinkai Wu 000] Recall the definitions of Debye length and plasma frequency for species s: ) ne 1/ ω ps = ɛ 0 m s ) 1/ ɛ0 k B T s λ Ds = ne, 3) and the Maxwellian distribution is ) 1/ ms F s v) = n exp [ m sv ]. 33) πk B T s k B T s 8

19 Now consider the integral For ωr k >> k BT s m s going to use for I p. For ωr k I s ω r k ) P F sv) dv. 34) v ω r /k, I s nk ω see eq. 1.37)), and this is the formula we are r << k BT s m s, I s nms k BT s as can be easily seen by ignoring the ω r /k in the denominator in the integral), and this is the formula we ll use for I e. In our problem, the total F v) is given by F = F e + me m p F p, and thus the total Iω r /k) is given by I = I e + me m p I p. Now that we have the explicit approximate) expressions for F and I, substitution into Eqs. 0.34) and 0.35) yields the desired expressions for ω r and ω i, after approximating F e ω r /k) as n me πk BT e ) 1/ in the numerator of 0.35), which is justified by our assumption ωr k << k BT e m e ). 0.8 Range of Unstable Wave Numbers [by Jeff Atwell] For instability, we need a ζ with ζ i > 0 such that k = Zζ) is real and positive. The corresponding ω = kζ specifies the mode. ζ with ζ i > 0 correspond to points in the interior of the closed curve in the Z plane see Fig. 0.4). So the intersection of the positive Z r axis and the interior of the closed curve gives the range of wave numbers that we are looking for. The shape of this closed curve depends on the distribution function F v). We are told our distribution function has two maxima, v 1 and v, and one minimum, v min. This means that the closed curve in the Z plane crosses the Z r axis three times, twice moving downward, and once moving upward. The curve in Fig. 0.4 happens to be an example of this situation.) Suppose v 1 < v. Then in the case of the closed curve shown in Fig 0.4, we may write the range of wave numbers which have at least one unstable mode as Zv ) < k < Zv min ) where Z is evaluated using Eq. 0.47)). For different shapes of the closed curve i.e. for different combinations of Zv 1 ), Zv min ), and Zv ) being positive and negative) the wave number range will be a bit different, but similar. The steps in going from Eq. 0.47) to Eq. 0.49) work for maxima of F v), in addition to minima. This means that in the case shown in Fig. 0.4 we can instead write the minimum wave number with an unstable mode as for example. e kmin = Zv ) = m e ɛ 0 + [F v) F v )] v v ) dv, 9