lectures on plasma physics- problems and solutions

UKAEA

EURATOM/UKAEA Fusion Association

Culham Science CentreAbingdonOxfordshireOX14 3DBUnited Kingdom

Telephone: +44 1235 464181Facsimile: +44 1235 463435

UKAEA FUS 482

EURATOM/UKAEA Fusion

Lectures on Plasma Physics:Problems and Solutions

A. Thyagaraja

December 2002

Cosmic Plasmas, Physics 418

Problem Set for Lecture 1: Elements

A. Thyagaraja

EURATOM/UKAEA Fusion Association, Culham Science

Centre, Abingdon, OX14 3DB, UK.

Note: The problems (in this set and those for Lectures 2,3) are designed to bring outkey points made in the lectures and clarify them through explicit examples. Hints fortheir solutions are provided in some cases. Problems which are \hard" are starred;they will be dealt with in the \problems class", at least in outline. Solutions to theproblems will be handed out separately. Some additional problems are also providedfor entertainment for those who wish to go deeper into the subject. They are optionalextras and will not be required as a part of this course.

1. Calculate in joules the energy acquired by a deuterium nucleus (composed of aproton and a neutron) when accelerated in a straight line by an electric �eld of 2kV/m over a distance of 1 m. Assuming it started from rest, calculate its speed at thispoint.

2. Find the spherically symmetric solution to Eq.(2) of the text which matches withthe Coulomb potential of a point charge e at the origin, e

4��0R, as R ! 0 and which

goes to zero at in�nity. (Hint: You may �nd it useful to use the fact that for functionsf(R) with, R2 = x2 + y2 + z2, r2f = 1

R2

ddR(R2 df

dR). Write, � = f(R) = g=R. Solve the

equation which results for g, and apply the boundary conditions stated).

3. A uniform, in�nite cylindrical plasma (radius a, cylinder oriented along the z-axis) carries a steady current of Ip MA. Calculate the magnetic �eld at any radiusr in magnitude and direction due to this current. You may assume that the currentdensity, j = ( Ip

�a2)z MA/m2, for r � a and vanishes for r > a. (Hint: Solve \Amp�ere's

Law", Eq.(3), for B, using the given values for �0; c). Calculate, using the results, themagnetic �eld at r = a = 1m, when Ip = 2MA. Generalise this result to obtain aformula for the magnetic �eld in vacuum for r > a due to an arbitrary, cylindricallysymmetric distribution of the current density within the plasma (ie., when jz is an

1

arbitrary function of r2 = x2 + y2, but the total current owing is still Ip).

4�. Find the general solution of the 1-dimensional, collisionless, \free-particle" kineticequation for the distribution function f(x; v; t),

@f

@t+ v

@f

@x= 0

Using this solution, or otherwise prove the following results. Assume that the domainin question is all of \velocity space" (ie., �1 < v < 1) and periodic position space,(ie., f(x; v; t) is a periodic function of x=L with period, 2�, and L is a �xed, \periodicitylength"):

1. The integrals, In(t) =R R

fndxdv are constants of the motion for any n for whichthe integrals exist at t = 0; ie., dIn

dt= 0. Deduce that the evolution of f according

to this equation conserves the total number of particles, N = I1.

2. Show that if initially f is nonnegative (ie., f(x; v; 0) � 0), the kinetic equationpreserves this positivity property for all times (both for t > 0 and for t < 0!).

3. Show that the integral, H(t) =R R

(f ln f)dxdv (called Boltzmann's �-spaceentropy) is also a constant of the motion.

4. Assume that at t = 0 f(x; v; 0) = C:n0(x) exp [�M(v � u0(x))2=2T ], where, C is

a \normalization constant",M represents the mass of the particles, and T (joules)represents the uniform \temperature" of the particles and n0(x); u0(x) are cer-tain periodic functions of x, where n0(x) � 0. First, evaluate C in terms ofM;T;N given the condition that

R+1�1

dvR �� f(x; v; 0)

dxL= N , assuming that the

function, n0(x) satis�es the condition,R �� n0(x)

dxL

= N . Next, interpret phys-ically the meaning of n0(x) and u0(x). Taking, n0(x) =

P1

k=�1 �k cos(kx=L),and u0(x) =

P1

k=�1 Uk cos(kx=L), where the Fourier coeÆcients are certain con-stants characteristic of the initial conditions, �nd, using the general solution forf , the expressions for the velocity moments, < f >=

R1

�1f(x; v; t)dv;< vf >=

R1

�1vf(x; v; t)dv;< v2f >=

R1

�1v2f(x; v; t)dv as functions of x; t.

5. Show that, as jtj ! 1, these velocity moments tend to uniform and constantvalues, in spite of the fact that f itself (considered as a function of x; t for �xed v)does not! (Hint: For this exercise, you will need to be able to evaluate \Gaussianintegrals" like

R1

�1xn exp(��x2)dx, where � > 0; n are constants. For n = 0,

the value of the integral is, I0(�) = (��)1=2. For odd n the integral vanishes by

2

symmetry. For even n the values can be found by di�erentiation of I0 repeatedlywith respect to �).

The �nal result that the velocity moments of the solutions of the free particlekinetic equations approach constant values as jtj ! 1 is called \phase mixing"and is a crucial (but not the only) element in an important phenomenon called\Landau" or \collisionless damping" which you will encounter many times.

5. Show that if at t = 0 the Maxwell equation, r:B = 0 is satis�ed, it will hold for allt according to Faraday's Law.

6. Use the equation of continuity, Eq.(22) to derive Eq.(25) from Eq.(24) explicitly.

7. Show that in ideal MHD plasmas, E:B = 0.

8. Derive the ideal MHD equations in a form where E; j do not appear explicitly.

9. Show that for any bounded volume V with boundary surface, S,RS pndS =

RV rpdV . (Hint: let a be an arbitrary constant vector. Apply Gauss' divergencetheorem to

RV a:rpdV and infer the result from the fact that a is arbitrary).

Additional (optional) problems for entertainment

and self-study

10. Solve the collisionless kinetic equation for particles of mass M moving in a conser-vative, external force-�eld, Fx =

dKdx

with potential, K(x), a function of x in in�nite xspace, assuming that the distribution function does not vary with time:

v@f

@x+

1

M

dK

dx

@f

@v= 0

(Hint: Consider the particle energy, E(x; v) = 12Mv2 � K(x) and observe that the

equation can be written as, @(E;f)@(x;v)

= 0. Infer that if you can �nd a particular solutionto the above equation, an arbitrary function of it is also a solution! The same result canalso be obtained by the method of characteristics which can be used more generally).If the system is known to be in thermodynamic equilibrium, determine the solutioncompletely in terms of the total number of particles and the temperature. Calculatethe number density as a function of x, assuming that K(x) = �K0x

2 (ie., we have

3

harmonically bound particles). If K(x) does not grow unboundedly at in�nity butgoes to a constant, is thermodynamic equilibrium strictly possible at any non-zerotemperature?

10�. Using the results of Problem 9, derive an equation for the time rate of changeof B2. Assuming the potential, K to depend only on position, obtain an equation forthe time rate of change of the kinetic energy per unit volume, (1=2)�mu

2. From thesetwo equations, obtain the local form of the law of conservation of energy for an idealisentropic plasma.

11�. An ideal plasma is con�ned within an azimuthally symmetric toroidal vessel withan arbitrary cross section and a perfectly conducting, rigid wall. The electric �eld onthe wall satis�es, Etangential = n� (E� n) = 0, where n is the unit outward surfacenormal at the wall. The plasma is assumed to be moving inside the vessel with arbitraryvelocities u which satisfy the boundary condition , u:n = 0. Given that the plasmadynamics within this vessel is described by Eqs.(29)-(34) (Nb. The plasma motionis not assumed to be azimuthally symmetric or time independent!), demonstrate thefollowing results.

1. The plasma total mass, Mp =RV �mdV is a constant of the motion.

2. The toroidal magnetic ux, �t =RB:e�dRdZ is a constant of the motion, where

the z-axis is the axis of symmetry of the toroidal vessel and cylindrical coordinates(R;Z; �) are used to locate an arbitrary point within the plasma. The \wall"of the plasma is represented by a closed contour (does not vary with time),(R;Z) = const, and the integral is taken over the area enclosed by this contourin the \poloidal plane" with � = 0. The unit vector e� = eZ � eR.

3. The global helicity of the magnetic �eld is de�ned by the integral (taken overthe plasma volume bounded by the surface, = const), IB(t) =

RV A:BdV ,

where B = r � A;E = �@A@t� r�. On the perfectly conducting wall, the

electrostatic potential � may be taken to be zero. Then the vector potential Amay be assumed to satisfy the Coulomb gauge, r:A = 0 and Atangential = 0 onthe wall. Show that IB is a constant of the motion, and that B:n = 0.

4. From the local energy conservation equation, derive the global energy conservationlaw for the system.

4


Solutions to the problems for Lecture 1

A. Thyagaraja



Note: The suggested solutions here are essentially outlines or\sketches". You shouldcheck your own solutions against the suggestions made here and/or work through themif you had diÆculties the �rst time.

1. Calculate in joules the energy acquired by a deuterium nucleus (composed of aproton and a neutron) when accelerated in a straight line by an electric �eld of 2kV/m over a distance of 1 m. Assuming it started from rest, calculate its speed at thispoint.

Solution: We neglect the energy radiated by the accelerating charge. The work doneby the �eld is clearly (since the charge on a deuterium nucleus is equal and oppositeto that of an electron) 2 keV= 3:2� 10�16J. The mass, mD ' 2mp = 3:35� 10�27kg.Hence, v2 = 9:6�1010(m/s)2. Hence, v ' 3:1�105m/s. The non relativistic expressionis adequate since this speed is only a thousandth of c.

2. Find the spherically symmetric solution to Eq.(2) of the text which matches withthe Coulomb potential of a point charge e at the origin, e

4��0R, as R ! 0 and which

goes to zero at in�nity. (Hint: You may �nd it useful to use the fact that for functionsf(R) with, R2 = x2 + y2 + z2, r2f = 1

R2

ddR(R2 df

dR). Write, � = f(R) = g=R. Solve the

equation which results for g, and apply the boundary conditions stated).

Solution: The spherically symmetric Debye equations is,

1

R2

d

dR(R2 d�

dR) = (

2�ne2

�0T)�

1

Putting, � = g=R, we see that g satis�es the equation,

d2g

dR2= (

2�ne2

�0T)g

The solution of this equation which goes to zero at in�nity is clearly, g = C: exp(�R=�Debye),where, �Debye = ( �0T

2�ne2)1=2, and C is a constant. Since we require the solution, � = g=R

to tend to the Coulomb potential at the origin, we �nd that, C = e4��0

.

3. A uniform, in�nite cylindrical plasma (radius a, cylinder oriented along the z-axis) carries a steady current of Ip MA. Calculate the magnetic �eld at any radiusr in magnitude and direction due to this current. You may assume that the currentdensity, j = ( Ip

�a2)z MA/m2, for r � a and vanishes for r > a. (Hint: Solve \Amp�ere's

Law", Eq.(3), for B, using the given values for �0; c). Calculate, using the results, themagnetic �eld at r = a = 1m, when Ip = 2MA. Generalise this result to obtain aformula for the magnetic �eld in vacuum for r > a due to an arbitrary, cylindricallysymmetric distribution of the current density within the plasma (ie., when jz is anarbitrary function of r2 = x2 + y2, but the total current owing is still Ip).

Solution: The only nontrivial component of Amp�ere's equation in cylindrical polarcoordinates (r; �; z) is:

c2�01

r

d

dr(rB�) = (

Ip�a2

); (r < a)

= 0; (r � a)

We must �nd a solution for which the �eld vanishes at in�nity and is non singular atr = 0. Clearly, we have, for r < a,

B�(r) = (Ip

2�a2c2�0)r

as can be veri�ed immediately by direct substitution. Evidently, in the region, r > a,the equation is satis�ed by, B� = C=r, where C is a constant to be determined. Thismust match smoothly at r = a (if not there will be a discontinuity there which willimply a \current sheet", which by the conditions of the problem does not exist!). Itfollows therefore that C = ( Ip

2�c2�0). In the general case of the current distribution within

the cylinder being an arbitrary function of radius, we see that only the integral withinthe cylinder is modi�ed and the interior �eld distribution is changed. The constant Cdepends only on the total current, Ip, and hence the vacuum �eld, outside the plasma

is B� = ( Ip2�rc2�0

). This formula gives the �eld in teslas in the numerical example.

2

4�. Find the general solution of the 1-dimensional, collisionless, \free-particle" kineticequation for the distribution function f(x; v; t),

@f

@t+ v

@f

@x= 0

Using this solution, or otherwise prove the following results. Assume that the domainin question is all of \velocity space" (ie., �1 < v < 1) and periodic position space,(ie., f(x; v; t) is a periodic function of x=L with period, 2�, and L is a �xed, \periodicitylength"):

1. The integrals, In(t) =R R

fndxdv are constants of the motion for any n for whichthe integrals exist at t = 0; ie., dIn

dt= 0. Deduce that the evolution of f according

to this equation conserves the total number of particles, N = I1.

2. Show that if initially f is nonnegative (ie., f(x; v; 0) � 0), the kinetic equationpreserves this positivity property for all times (both for t > 0 and for t < 0!).

3. Show that the integral, H(t) =R R

(f ln f)dxdv (called Boltzmann's �-spaceentropy) is also a constant of the motion.

4. Assume that at t = 0 f(x; v; 0) = C:n0(x) exp [�M(v � u0(x))2=2T ], where, C is

a \normalization constant",M represents the mass of the particles, and T (joules)represents the uniform \temperature" of the particles and n0(x); u0(x) are cer-tain periodic functions of x, where n0(x) � 0. First, evaluate C in terms ofM;T;N given the condition that

R+1�1

dvR �� f(x; v; 0)

dxL= N , assuming that the

function, n0(x) satis�es the condition,R �� n0(x)

dxL

= N . Next, interpret phys-ically the meaning of n0(x) and u0(x). Taking, n0(x) =

P1

k=�1 �k cos(kx=L),and u0(x) =

P1

k=�1 Uk cos(kx=L), where the Fourier coeÆcients are certain con-stants characteristic of the initial conditions, �nd, using the general solution forf , the expressions for the velocity moments, < f >=

R1

�1f(x; v; t)dv;< vf >=R

1

�1vf(x; v; t)dv;< v2f >=

R1

�1v2f(x; v; t)dv as functions of x; t.

5. Show that, as jtj ! 1, these velocity moments tend to uniform and constantvalues, in spite of the fact that f itself (considered as a function of x; t for �xed v)does not! (Hint: For this exercise, you will need to be able to evaluate \Gaussianintegrals" like

R1

�1xn exp(��x2)dx, where � > 0; n are constants. For n = 0,

the value of the integral is, I0(�) = (��)1=2. For odd n the integral vanishes by

symmetry. For even n the values can be found by di�erentiation of I0 repeatedlywith respect to �).

3

The �nal result that the velocity moments of the solutions of the free particlekinetic equations approach constant values as jtj ! 1 is called \phase mixing"and is a crucial (but not the only) element in an important phenomenon called\Landau" or \collisionless damping" which you will encounter many times.

Solution: Either by the method of characteristics or just by inspection, we see thatthe general solution to the given �rst order, linear pde. is, f(x; v; t) = F (x � vt; v),where, F (�; �) is an \arbitrary" function of its arguments, � = x � vt; � = v. Fromthe given conditions, we see that F must be a periodic function of �=L with period 2�,and at t = 0, must be nonnegative for values of its arguments (since, by de�nition, itrepresents a distribution function or a probability).

1. We can prove this in two ways: without even solving the equation, if we multiplyit by fn�1 and integrate over the x; v ranges stated, we see that,

dIndt

=Zx;v

@fn

@t

= �Zx;v

@vfn

@xdxdv

= 0

The last following from the periodicity of f with respect to the space variable,x. Alternatively, In =

Rx;v F

n(x � vt; v)dxdv =R�;v F

n(�; v)d�dv. Clearly aperiodic function integrated over its period can be \shifted" by an arbitraryamount without altering the value of the integral! Hence the time variable simplydisappears from In and the result follows! Since, for n = 1, the integral equalsthe total number N of particles, the evolution equation clearly conserves this.

2. Nothing in the general solution says we have to take t > 0. Since the initialconditions dictate that F (�; �) � 0, we must have that, f(x; v; t) � F (x�vt; v) �0 for all time.

3. Observe that if f(x; v; t) is a solution of the kinetic equation, and G(z) is anarbitrary (suÆciently regular, of course) function of its argument, then, G(F ) =g(x; v; t) is also a solution! This is because,

@G

@t+ v

@G

@x= G0

"@f

@t+ v

@f

@x

#

= 0

4

where G0 is the derivative with respect to its argument and we have used the\chain rule" of calculus. It follows that G = f ln f is also a solution, and fromthe preceding result, the constancy in time of Boltzmann's H follows.

4. From the given relation for the \Gaussian" integrals, we have,

Z +1

�1

exph�M(v � u0(x))

2=2Tidv = (

2�T

M)1=2

irrespective of the function, u0(x). Since we are given the conditions that n0(x)is periodic in x=L with period 2� and,

R �� n0(x)

dxL

= N , it is clear that C =(2�TM

)�1=2, in order to satisfy the normalization requirement. It is useful to thinkof the position coordinate, x=L as an \angle", �, which varies from �� to �.It is then easily seen that n0(�) is the number density (ie., number of particlesper radian) of the system at t = 0. From the symmetry properties of Gaussianintegrals, we know that,

Z +1

�1

exph�M(v � u0(x))

2=2Ti(v � u0(x))dv = 0

It is clear therefore that,

Z +1

�1

f(x; v; 0)dv = n0(x)Z +1

�1

f(x; v; 0)vdv = n0(x)u0(x)

It is then clear that u0(x) is the average \ uid velocity" of the particles at thelocation x at t = 0.

5. We know from the general solution that f(x; v; t) = C:n0(x�vt) exp [�M(v � u0(x� vt))2=2T ],where,it is clear that C = (2�T

M)�1=2. This solution is still periodic in � = x=L

as before, but its v dependence is extremely complicated, as well as the timedependence! We have the following relations for the lowest velocity moment:

Z +1

�1

f(x; v; t)dv = CZ +1

�1

1Xk=�1

�k cos(kx=L� kvt=L) exph�M(v � u0(x� vt))2=2T

idv

The general evaluation of such series is indeed hopeless, but we are only interestedin the behaviour of these averages in the limit, jtj ! 1. Furthermore, to illustratethe points with the minimum of algebraic complexity, we consider the special casewhen ju0(x)j � (2T

M)1=2. This simply says that the ow velocity is initially small

compared to the \random" or \thermal" velocity. With this approximation, we

5

may expand the local Maxwellian about the zero mean ow one and use theFourier representation. We must discuss the long time behaviour of series like,

< f > =Z +1

�1

1Xk=�1

�k(x) exp(�ikvt=L) exph�Mv2=2T

idv

where �k(x) can be calculated from the given expressions for n0(x); u0(x). Theimportant thing is the form of this series. Now consider the integrals,

Jk(t) =Z +1

�1

exp(�ikvt=L) exph�Mv2=2T

idv

For k 6= 0, it is easily seen, either by explicit evaluation, or by the generalRiemann-Lebesgue lemma of Fourier integrals, that as jtj ! 1, Jk(t) ! 0.For, setting Vth = (2T

M)1=2, we see that,

Jk(t) = Vth

Z +1

�1

exph��2 � i(kVtht=L)�)

id�

= (�)1=2Vth exph�(kVtht=2L)

2i

from the known properties of Gaussian (complex!) integrals (consult almost anybook on applied mathematics, or even the Feynman Lectures!). Hence, in thelimit as jtj ! 1, all the terms of the series for < f >, except the k = 0 termtend very rapidly to zero! Hence < f >! �0 = N

2�. The same proof can be

clearly extended, almost with no changes to all the velocity moments. Thus, allof such moments \relax" to their spatial average values which would be the limitof thermodynamic equlibrium!

The interesting thing is that this \approach" to equilibrium by the moments ofthe distribution function takes place reversibly in time, and in the presence ofan in�nity of constants of the motion! Furthermore, it is very easy to verifythat f(x; v; t) itself does not show this \Landau damping" behaviour, nor is anyparticular \wave particle interaction" is necessary. The simplest of all possiblekineticequations thus exhibits \phase mixing" and demonstrates a very genericway in which time reversible systems can show apparently irrecersible approachto equilibrium of statistical averages over an in�nite velocity space under certainconditions (which are met in our example).

5. Show that if at t = 0 the Maxwell equation, r:B = 0 is satis�ed, it will hold for allt according to Faraday's Law.

6

Solution: Take the divergence of Faraday's Law and obtain, @@t(r:B) = 0. The result

follows.

6. Use the equation of continuity, Eq.(22) to derive Eq.(25) from Eq.(24) explicitly.

Solution: Expand the terms using the product rule of di�erentiation.

7. Show that in ideal MHD plasmas, E:B = 0.

Solution: In ideal MHD, E = �v �B. Taking dot product with B gives the result.

8. Derive the ideal MHD equations in a form where E; j do not appear explicitly.

Solution: Trivial!

9. Show that for any bounded volume V with boundary surface, S,RS pndS =R

V rpdV . (Hint: let a be an arbitrary constant vector. Apply Gauss' divergencetheorem to

RV a:rpdV and infer the result from the fact that a is arbitrary).

Solution: Use the hint!

7


Problem Set for Lecture 2: Particle orbit theory

A. Thyagaraja



Note: The problems are designed to bring out key points made in the lectures and

clarify them through explicit examples. Hints for their solutions are provided in some

cases. Problems which are \hard" are starred; they will be dealt with in the \problems

class", at least in outline. Solutions to the problems will be handed out separately.

Some additional problems are also provided for entertainment for those who wish to go

deeper into the subject. They are optional extras and will not be required as a part of

this course.

1. Using the nonrelativistic Lagrangian given in Eq.(6), show that Lagrange's equationsare the same as Newton's equations in Cartesian coordinates.

2. Cylindrical coordinates, (R; �; Z) are related to Cartesian coordinates through thetransformation laws, x = R cos �; y = R sin�; z = Z. Show that the velocity compo-nents, VR; V�; VZ of a particle with coordinates, (R(t); �(t); Z(t)) can be expressed interms of dR

dt; d�dt; dZdt

according to the formula,

V =dR

dt

=dR

dteR +R

d�

dte� +

dZ

dteZ

where eR;�;Z are, respectively the unit vectors in the corresponding directions. Incylindrical coordinates, the vector potential, A = AReR + A�e� + AzeZ and B =

r�A = ( 1

R@AZ@��

@A�@Z

)eR + (@AR@Z�

@AZ@R

)e� + ( 1

R@@R(RA�)�

1

R@AR@�

)eZ . If � representsthe scalar potential, obtain the nonrelativistic Lagrangians for a charged particle withmassM and charge e moving in the �elds due to these potentials. (Hint: R = x(t)ex+y(t)ey + Z(t)eZ ;; express the unit vectors, eR(t); e�(t) in terms of ex; ey. Write downthe expression for the kinetic energy, T = 1

2MV:V, using the results of Pb. 1 and the

1

nonrelativistic Lagrangian in the Notes (see Eq.(6))).

3�. Using the nonrelativistic Lagrangian of the previous problem and Lagrange's equa-tions, obtain the forms for the Newton's equations of motion of a charged particle incylindrical coordinates. If the potentials of Problem 3 are independent of �, the az-imuthal angle, obtain the corresponding constant of motion. This constant is calledthe \canonical angular momentum" of the particle. Is the ordinary angular momentum

of the particle about the z-axis a constant of the motion in the presence of a magnetic�eld, even if the Lagrangian has azimuthal symmetry? If, in addition, the potentialsare also independent of t, what can you say about the energy, H?

4. Estimate the gyroradius rL of an alpha particle (Helium 4 nucleus) with Massnumber A = 4, atomic number, Z = 2 and energy 4 MeV in the Earth's magnetic �eld(approx. 10�4T). Given that the typical scale length of the �eld is of the order of 1000km, calculate �� for this particle. Look up the data for the Sun's average �eld near itssurface and for Jupiter and calculate the corresponding numbers.

5. Imagine a charged particle moving with a speed vk along a circular B-�eld line,the radius of curvature Rc is assumed much larger than the Larmor radius, rL of theparticle in the �eld of magnitude B. Treat the centrifugal force on the particle,Mv2k=Rc

as an \external" force, f? (cf. Eq.(21)) and work out the resultant drift. Compare yourresult in magnitude and direction with the last term of Eq.(50).


and self-study

6. Taking b = ez and B = Bez;E = Eez, where B;E are constants, integrate therelativistic equations,

dp?dt

= (eB=M)(p? � ez)

dpkdt

= eEez

subject to the conditions, pk(0) = 0;p?(0) = p0ex. Determine the time variation ofM . Obtain the expressions for the velocities, v = p=M and solve for the trajectory ofthe system, given that the particle starts from (x0; 0; 0) at t = 0. Determine the timevariation of the Larmor frequency and the Larmor radius in the laboratory frame.(Hint:

2

it is advantageous to introduce the new independent \gyrophase" variable, through,d�dt= �eB=M(t) and write the equations in component form in Cartesian coordinates).

7.� Solve the equations,

dp

dt= eEex + (eB=M)p� ez (1)

together with the trajectory equations, dxdt

= p

Min the relativistic case in complete

generality, assuming only that E;B are constants and M = (m2 + p2=c2)1=2, where mis the rest mass and e is the charge of the particle. In particular, consider arbitraryinitial data and all possible values of E=B. What happens to the \E�B" drift of acharged particle when E=B > c?

8.� A uniform B �eld along the z-axis is represented by the vector potential, A =(BR

2

2)e�. Consider the plane non relativistic motion (in the plane, Z = 0) of a charged

particle of mass M and charge e under the action of a central gravitational potential,G�=R where � is the mass of the central attracting body (neglect the reaction on thisbody which can be assumed to be at rest at the origin). Set up the two-dimensionalequations of motion and solve them after obtaining the constants of the motion. Inparticular, obtain the period and the trajectory of the particle's motion in the plane.

9�. Reconsider the previous problem, assuming that dZdt

= 0 for all time, but the motionis relativistic (ignoring any radiation due to accelerations) due to the very large valueof the gravitational attraction, what happens to the motion? What happens if thecentral force is a repulsive, electrostatic one?

10. Derive Eq.(43) of the text for the nonrelativistic averaged \drift" Lagrangian indetail.

11. Using the previous result and the suggestion made in the text, derive Eq.(26) ofthe text. Similarly show in detail that Eq.(48) follows by varying the drift Lagrangianand taking the parallel component.

12. Justify the statement made in the text that the terms neglected in deriving Eq.(50)are indeed smaller than the retained terms by at least ��.

13�. A \wiggler" magnetic �eld can be speci�ed approximately in cylindrical coordi-nates thus:

BR = �1

R

@

@Z

3

BZ =1

R

@

@RB� = 0

where the function (R;Z) =hB0 +B1 cos(

ZL)i(R

2

2), where B0 � B1 are constants

and L is a scale length, much larger than the Larmor radius rL of a particle with mass,M and charge e moving nonrelativistically. Show that this �eld is indeed divergencefree and determine the components of the vector potential in terms of .

Assume that there is no electric �eld in the system. Integrate the equation Eq.(48) and�nd the conditions on the initial values of vk; v? and location for the particles to be\trapped" in the minima of the �eld and for them to be \passing" (ie., be \untrapped").Assume that the probability of �nding a particle with a given set of initial velocitiesis a Maxwellian at a constant temperature, calculate, at a radius R0 and location, Z0,the probability of being trapped or passing. Apart from � and kinetic energy, whatother constants of motion exist in this case, and why?

14�. A purely azimuthal �eld in a vacuum is of the form, B = B�e� = B0(R0

R)e� =

�@AZ@R

e�. It is regular for R > 0 and R0 is a \reference radius" where the value of the�eld is B0. Assuming there are no electric �elds and the magnetic �eld is stationary,solve the exact relativistic (and nonrelativistic) equations of motion by �nding the threeconstants of the motion. Assume that at t = 0 the particle is at R = Rin; � = 0; z = 0and v = vR(0)eR + v�(0)e� + vZ(0)eZ. Obtain the geometrical characteristics of theorbit. Generalize the solution to the case when there is an electrostatic potential, �(R).

15�. Reconsider the above problem non relativistically, using the \drift" equationsderived for the guiding centre motion in leading order of ��. Calculate the drifts ofcharged particles (positive and negative) in such a �eld, amd work out the guidingcentre orbit. How does this compare with the exact solution in the nonrelativistic caseobtained in the problem above?

16. Carry out the gyro averaging procedure directly on the nonrelativistic equationsof motion and derive the drift equations for the guiding centres. (Hint: Consult theliterature, particularly, Francis Chen's book, Ch. 2).

17. Derive the expressions for V = dRdt

for a particle in terms of spherical coordinates,r; �; � (where, x = r sin � cos�; y = r sin � sin�; z = r cos �), and using it, write downthe nonrelativistic Lagrangian for a charged particle of mass M and charge e in thesecoordinates. Obtain the equations of motion from it in the presence of general E;B�elds (expressed as usual in terms of an electric potential, � and vector potential A in

4

spherical coordinates) and an external gravitational force �eld f =MrK.

18�. Discuss, using the nonrelativistic, Newtonian guiding centre equations, the motionof a charged particle in the dipole magnetic �eld of a star of mass M�. Find all therelevant constants of the motion of the exact system and reduce the equations as muchas possible towards obtaining a numerical solution (the latter is not expected!). Youmay assume that the dipole is oriented along the spin axis of the star and the initialposition of the particle is in the equatorial plane, well away from the \edge" of thestar.

5



A. Thyagaraja




1. Using the nonrelativistic Lagrangian given in Eq.(6), show that Lagrange's equationsare the same as Newton's equations in Cartesian coordinates.

Solution: The Lagrangian is,

Lnr = (1=2)Mv2� e�(r; t) + eA(r; t):v

Let r = (x1; x2; x3);v = (V1; V2; V3) = ( _x1; _x2; _x3);A = (A1; A2; A3). Clearly, @Lnr

@V1=

MV1 + eA1,@Lnr

@x1= �e @�

@x1+ e@A1

@x1V1 + e@A2

@x1V2 + e@A3

@x1V3.

Substituting in Lagrange's equation, ddt(@Lnr

@V1) = @Lnr

@x1, and using the facts that dA

dt=

@A@t+V1

@A@x1

+V2@A@x2

+V3@A@x3

;E = �@A@t�r�;B = r�A, we obtain Newton's equations.

2. Cylindrical coordinates, (R; �; Z) are related to Cartesian coordinates through thetransformation laws, x = R cos �; y = R sin�; z = Z. Show that the velocity compo-nents, VR; V�; VZ of a particle with coordinates, (R(t); �(t); Z(t)) can be expressed interms of dR

dt; d�dt; dZdt

according to the formula,

V =dR

dt

=dR

dteR +R

d�

dte� +

dZ

dteZ

where eR;�;Z are, respectively the unit vectors in the corresponding directions. Incylindrical coordinates, the vector potential, A = AReR + A�e� + AzeZ and B =

1

r�A = ( 1

R@AZ@��

@A�@Z

)eR + (@AR@Z�

@AZ@R

)e� + ( 1

R@@R(RA�)�

1

R@AR@�

)eZ . If � representsthe scalar potential, obtain the nonrelativistic Lagrangians for a charged particle withmassM and charge e moving in the �elds due to these potentials. (Hint: R = x(t)ex+y(t)ey + Z(t)eZ ;; express the unit vectors, eR(t); e�(t) in terms of ex; ey. Write downthe expression for the kinetic energy, T = 1

2MV:V, using the results of Pb. 1 and the

nonrelativistic Lagrangian in the Notes (see Eq.(6))).

Solution: R = R cos�ex + R sin�ey + Zez. Di�erentiate w.r.t t, bearing in mindthat the basis vectors ex;y;z are constants. We obtain, V = ( _R cos� � R sin� _�)ex +( _R sin�+R cos� _�)ey + _ZeZ = dR

dteR +R d�

dte� +

dZdteZ, from the de�nitions of the unit

vectors in the cylindrical system.

Using these results, we see that, Lnr( _R; _�; _Z;R; �; Z; t) can be written as,

Lnr =1

2M

h_R2 +R2 _�2 + _Z2

i� e� + e _RAR + eR _�A� + e _ZAZ

3�. Using the nonrelativistic Lagrangian of the previous problem and Lagrange's equa-tions, obtain the forms for the Newton's equations of motion of a charged particle incylindrical coordinates. If the potentials of Problem 3 are independent of �, the az-imuthal angle, obtain the corresponding constant of motion. This constant is calledthe \canonical angular momentum" of the particle. Is the ordinary angular momentumof the particle about the z-axis a constant of the motion in the presence of a magnetic�eld, even if the Lagrangian has azimuthal symmetry? If, in addition, the potentialsare also independent of t, what can you say about the energy, H?

Solution: Lagrange's equations give the equations of motion in cylindrical coordinates.The method is exactly similar to that used in Pb. 1 and will not be repeated. Notehowever that they automatically include the \centrifugal" and Coriolis terms in theradial and azimuthal directions.

If � and A are independent of the azimuthal angle, �, the cyclicity theorem saysthat the \conjugate canonical momentum", P� =

@Lnr

@ _�is a constant of the motion. We

see easily that, P� = MR2 _� + eRA�. If A� is identically zero, we would indeed have,P� = MR2 _� = constant, and this implies the constancy of the mechanical angularmomentum. If A� is a general function of R;Z; t, however, the mechanical angularmomentum is not constant! If � and A are independent of t, the Notes give the proofthat the energy, H = T + e� is also a constant of the motion.

2

4. Estimate the gyroradius rL of an alpha particle (Helium 4 nucleus) with Massnumber A = 4, atomic number, Z = 2 and energy 4 MeV in the Earth's magnetic �eld(approx. 10�4T). Given that the typical scale length of the �eld is of the order of 1000km, calculate �� for this particle. Look up the data for the Sun's average �eld near itssurface and for Jupiter and calculate the corresponding numbers.

Solution: The rest mass of the alpha is, Amp = 4 � 1:67 � 10�27kg. The charge isZ:e = 2 � 1:6 � 10�19C. Now, 4 MeV corresponds to 6:4 � 10�13J. Since this is lowcompared to the rest energy, we may use nonrelativistic formulae. We may assume thatthe \perpendicular" energy, E?, of the particle is equal to its \parallel" energy, and takethe former to be 2MeV. Thus, c? = (2E?=M)1=2 = (4�1:6�10�13=4�1:67�10�27)1=2 '107m/s. The gyro frequency, c = ZeB=Amp = 10�4(2 � 1:6 � 10�19)=(4 � 1:67 �10�27) = 5000rads/s. Then, rL = c?=c = 2km. Hence, �� ' 2 � 10�3, and theparticle is indeed \stuck" to the �eld line, in comparison with the length scale overwhich the �eld varies. I leave the reference work on the Sun and Jupiter to you.

5. Imagine a charged particle moving with a speed vk along a circular B-�eld line,the radius of curvature Rc is assumed much larger than the Larmor radius, rL of theparticle in the �eld of magnitude B. Treat the centrifugal force on the particle,Mv2k=Rc

as an \external" force, f? (cf. Eq.(21)) and work out the resultant drift. Compare yourresult in magnitude and direction with the last term of Eq.(50).

Solution: The centrifugal force f = (Mv2k=Rc)eR. From the drift formula, we see that

f � b=eB = (Mv2k=eBRc)eZ = (v2k=cRc)eZ = (v2k=c)b� k, since, k = �( 1

Rc)eR, for

a circle with centre at the origin and radius Rc, at any point where the radius vectoris along eR(�). Furthermore, b is the unit tangent vector, = e�(�), and we obviouslyhave, eZ = b� k. This proves the result required. You should draw a diagram toillustrate the relevant vectors in both magnitude and direction. Note incidentally thatthis drift depends only upon the parallel kinetic energy of the particle and its charge.

3


Problem Set for Lecture 3: Applications of the uid

description

A. Thyagaraja









this course.

1. Show that Eqs.(1-3) imply that the internal energy per unit volume, e = p

�1, of a

perfect gas satis�es the equation,

@e

@t+r:(ue) + pr:u = 0

If m is the mass of a molecule, the number density of the gas is related to the massdensity �m through, n = �m

m. Given that in a perfect gas, with temperature measured

in joules, p = nT , �nd the equation satis�ed by the temperature T .

2. Find the equation satis�ed by the entropy per particle, s = � ln p +

�1lnT of a

perfect gas. Deduce the conservation equation satis�ed by the entropy per unit volume,� = ns = �ms=m.

3. Show that Eq.(11) follows from Eq.(10).

4. Show that Eq.(13) follows from Eq.(9) and the equations of continuity, Eq.(1).

5. In deriving the wave equation for small amplitude sound waves, Eq.(35), we used

1

the isentropic relation between density and pressure. Newton assumed a perfect gasand related pressure and density by treating the temperature (rather than the entropy,as we and Laplace did) as a constant. Derive the analogue of Eq.(35) with Newton'sisothermal relation. How could you experimentally distinguish between the theory ofsound waves due to Laplace (isentropic) and Newton (isothermal)?

6. If a source emits sound with a constant wavelength in vessels �lled with hydrogen andhelium respectively at the same temperature, where would one hear a higher frequencyand why?

7. Show from Eq.(30) that W:rH = 0, in steady ideal ow. In Eq.(39) show why itis permissible to set H(t) = 0.

8. Consider a steady, incompressible ideal ow through a convergent-divergent \nozzle"formed by rotating a parabola (y = (ymax�ymin)(

xL)2+ymin) about the x-axis. Apply the

mass conservation and the Bernoulli equations to obtain the variation of the pressurealong the symmetry axis. Assume that the \inlet" is located at x = �L and thecentreline velocity there is u0. Calculate the di�erence in pressure between the inletand the point where the pressure is a minimum (where is this point?).

9. We have derived the sound wave equation in a gas at rest. If the gas were movingwith a uniform and constant velocity, u0 with respect to an observer, what is the equa-tion of satis�ed by the density uctuations in that frame? Derive the famous \Dopplershift" formula from the corresponding dispersion equation (analogue of Eq.(36)).(Hint:Observe that �m = �0; p = p0;u = u0 is an exact solution of the Euler equations.Discuss \small oscillations" about this state as has been done in the Notes about thestate of \rest" where, u0 = 0.)


and self-study

10. In deriving Eq.(29) (\Kelvin's circulation theorem"), justify the statement that allthe terms on the right vanish. If p is not a function of �m, is the theorem true?

11. Show that, in incompressible ow, the pressure can always be \eliminated" bysolving the Poisson equation for it in terms of the velocity and vorticity.,

r2~p = ��0r:(W� u)� �0r

2(u2=2)

2

If we further assume that the ow is irrotational, what does the resultant equationsimplify to and how is it related to the Bernoulli relation, Eq.(52)?

12. Show that in steady, two dimensional, incompressible, but rotational ow, thevorticity ~! is a function of the stream function, .(Hint: If f; g are two functions of(x; y) such that one is a function of the other, it is necessary and suÆcient that, the

Jacobian, @(f;g)@(x;y)

= 0, identically).

13.� (\Water waves"): Consider water �lling an in�nite (in the x; y plane) lake ofuniform depth �H, with uniform gravity acting downwards (ie., the gravitational ac-celeration is, �gz). At rest, the water surface is taken to be the z = 0 plane, abovewhich we assume there is a gas at constant pressure, p1.

1. Solve for the hydrostatic pressure equilibrium within the lake, assuming the waterdensity to be uniform and constant at �0.

2. Next consider \small" oscillations of the water surface, given by the equation,z = �(x; y; t). Assuming the ow due to the water wave to be small amplitude and ir-rotational, discuss the wave motion of the surface and the velocity �elds. Assume thatthe \free surface", F (x; y; z; t) � z�h(x; y; t) is a \material surface" (this is called thekinematic boundary condition) and the pressure ~p = p1 on this surface as the boundaryconditions at the free surface. The bottom of the lake is assumed to be \impenetrable"(ie., u:z = 0). Periodic boundary conditions may be assumed in the x and y direc-tions. Thus you may assume that all quantities vary like, F (z) exp [ikxx + ikyy � i!t],where kx; ky are \wave numbers" in the respective directions and ! is the frequency.Determine, in particular, the function F for the perturbed velocity potential � and thepressure, and the dispersion relation which connects! with kx; ky.

3. What is the primary di�erence between this \water wave" and the sound wave asregards the propagation speed of the disturbances of a given wave number?

14�. This problem is typical of the so-called \exterior ow problem" in ideal incom-pressible, irrotational ow theory and illustrates the utility of the velocity potential:Let x = (x; y; z).

1. Show that the velocity potential de�ned by, �0 = U:x is a possible, 3-d exactsolution of the incompressible, irrotational hydrodynamic equations of motion, whereU(t) is a spatially uniform but time-dependent vector.(Hint. Verify that, r2(�0) = 0).

3

2. Calculate the pressure at every point of the ow and interpret the solution physically.

3. Show also that �1 =U:xr3

; r2 = x2 + y2 + z2 is a solution, as is the sum, �a;U(x) =

�0+(a3

2)�1, where a is a positive constant.(Hint: 1=r is a solution of Laplace's equation,

as are its spatial derivatives. The solution is simpli�ed if at any instant t, you takeU = U(t)ez and use spherical polar coordinates, r; �; �, noting that only r; � enter theproblem. The time is merely a parameter in this problem!)

4. Show that n:r�a;U = 0 on the spherical surface, r = a, where n is the unit normalto the surface. Interpret the solution as the ow past a solid sphere of radius a (in theregion outside the sphere) which becomes uniform and equal to U at in�nity. (Hint:calculate the uid velocity on the surface of the sphere and at in�nity from the velocitypotential).

5. When U is independent of time, work out the pressure everywhere in the uidand by integrating it over the surface of the sphere, and show that the total forceexerted by the uid on the sphere vanishes. In particular, show that the sphere doesnot experience a \drag" force parallel to the ow direction at in�nity. This is a specialcase of an important theorem of steady, ideal, incompressible, irrotational ow knownas D'Alembert's Theorem (sometimes mistakenly called \D'Alembert's paradox").

4



A. Thyagaraja




1. Show that Eqs.(1-3) imply that the internal energy per unit volume, e = p

�1, of a

perfect gas satis�es the equation,

@e

@t+r:(ue) + pr:u = 0

If m is the mass of a molecule, the number density of the gas is related to the massdensity �m through, n = �m

m. Given that in a perfect gas, with temperature measured

in joules, p = nT , �nd the equation satis�ed by the temperature T .

Solution: The continuity equation, upon expansion and using the fact that �m = m:ngives,

@n

@t+ u:rn + nr:u = 0

Multiply this equation by n �1 to get,

@n

@t+ u:rn + n r:u =

@n

@t+r:(un ) + ( � 1)n r:u

= 0 (1)

Now, p = p�(n=n�) . The required equation for e = p

�1follows upon multiplying by

the constant, p�(n�)� and using the de�nitions. Substituting, e = nT=( � 1) in theequation for e (just derived), we �nd that T satis�es,

@T

@t+ u:rT + ( � 1)Tr:u = 0

1

2. Find the equation satis�ed by the entropy per particle, s = � ln p +

�1lnT of a

perfect gas. Deduce the conservation equation satis�ed by the entropy per unit volume,� = ns = �ms=m.

Solution: From the de�nitions, and the preceding results, observe that,

Ds

Dt=

@s

@t+ u:rs

= �1

p

Dp

Dt+

� 1

1

T

DT

Dt

= �( � 1)r:u + r:u

= r:u

Since, we have from the continuity equation, the relation, DnDt

+ nr:u = 0, it followsthat the entropy per unit volume, � = ns satis�es,

D�

Dt= 0

We see that in an ideal (ie., isentropic) uid, the material derivative of the entropy perunit volume vanishes. This is in fact what is meant by isentropic ow. Note that �need not be uniform in space or constant in time.

3. Show that Eq.(11) follows from Eq.(10).

Solution: It is easy to verify that Eq.(11) reduces to Eq.(10). Firstly, since K is onlya function of position, using the equation of continuity and preceding results, we getthe following:

@�mK

@t+r:(�mKu) = �mu:rK

@

@t(

p

� 1) +r:(u

p

� 1) = �pr:u

2

We expand the LHS of Eq.(11):

@

@t

"(�mu

2

2) + (

p

� 1)� (�mK)

#=

@�m@t

u2

2+ �m

@

@t(u2

2)�r:

"(

p

� 1)u

#� pr:u

+r:(�mKu)� �mu:rK

= �r:

"u(

1

� 1p�K�m)

#�r:(�m

u2

2u) + �mu:r(

u2

2)

+�m@

@t(u2

2)�r:(pu) + u:rp� �mu:rK

= �r:

"u(�mu

2

2+

� 1p�K�m)

#

+�m@

@t(u2

2) + �mu:r(

u2

2) + u:rp� �mu:rK

By virtue of Eq.(8) (it is equivalent to Eq.(10)!), the terms in the last row cancel, andwe have arrived at Eq.(11).

4. Show that Eq.(13) follows from Eq.(9) and the equations of continuity, Eq.(1).

Solution: We have seen in the Notes that Eq.(9) can be expanded to,

@W

@t+ u:rW = �Wr:u+W:ru

From the equation of continuity, we �nd that 1

�msatis�es the equation,

@

@t(1

�m) + u:r(

1

�m) = (

1

�m)r:u

It follows readily that W� =W=�m satis�es Eq.(13).

5. In deriving the wave equation for small amplitude sound waves, Eq.(35), we usedthe isentropic relation between density and pressure. Newton assumed a perfect gasand related pressure and density by treating the temperature (rather than the entropy,as we and Laplace did) as a constant. Derive the analogue of Eq.(35) with Newton'sisothermal relation. How could you experimentally distinguish between the theory ofsound waves due to Laplace (isentropic) and Newton (isothermal)?

Solution: The linearized isentropic relation says, ~p

p0= ~�

�0. If we take the isothermal

assumption of Newton, p = �(T0=m), where T0 is the temperature (in joules). Then,

3

linearization yields, ~p

p0= ~�

�0. In e�ect, we will getNewton's theory, simply by replacing

= 1 in all the formulae! We get the same wave equation for small disturbances, butC2

s = p0=�0. It is now obvious how we distinguish between the two theories. Wesimply measure the speed of sound at �xed ambient conditions. Since C2

s (Laplace) = C2

s (Newton) we have a clear experimental verdict. Indeed, Laplace modi�ed Newton'stheory precisely because Newton's prediction of the speed of sound was considerablylower than the actual value!

6. If a source emits sound with a constant wavelength in vessels �lled with hydrogen andhelium respectively at the same temperature, where would one hear a higher frequencyand why?

Solution: Hydrogen being the lighter gas would have a larger sound speed than Heliumat the same temperature. Since ! = Csk, at given wave length, the frequency wouldbe higher in Hydrogen.

7. Show from Eq.(30) that W:rH = 0, in steady ideal ow. In Eq.(39) show why itis permissible to set H(t) = 0.

Solution: Taking the dot product of Eq.(30) with W gives the �rst result. Sinceu = r�, we may add an arbitrary function of t to � without altering any physics!Hence we can simply \absorb" an arbitrary function H(t) into the de�nition of thevelocity potential �. This new � will give the same velocities as the old one and willhave H = 0, identically.

8. Consider a steady, incompressible ideal ow through a convergent-divergent \nozzle"formed by rotating a parabola (y = (ymax�ymin)(

x

L)2+ymin) about the x-axis. Apply the

mass conservation and the Bernoulli equations to obtain the variation of the pressurealong the symmetry axis. Assume that the \inlet" is located at x = �L and thecentreline velocity there is u0. Calculate the di�erence in pressure between the inletand the point where the pressure is a minimum (where is this point?).

Solution: Since the ow is stated to be steady and incompressible, we may assumethe density to be �m. Let all the inlet quantities be denoted by the subscript 0 andthe \throat" (y = ymin) quantities, by the subscript, \t". Thus the inlet area isA0 = �y2

max; At = �y2

min. The steady mass continuity relation reads,

�m�u0A0 = �m�utAt

4

The steady, incompressible Bernoulli relation is,

�m�u20

2+ �p0 = �m

�u2t2+ �pt

It follows that,

�p0 � �pt(12�m �u02)

= (�ut�u0)2 � 1

= (A0

At

)2 � 1

= (ymax

ymin

)4 � 1

It is evident that �p0 > �pt. It is evident that the pressure at the throat is lower than atany other point, and correspondingly the velocity is the greatest there. Note that thissolution is only valid if the velocity at the throat, �utis very much smaller than the speedof sound there (this is the condition for incompressible ow). It is a problem for the\advanced" student to consider the same general situation when the ow isnot assumedto be incompressible, but still under steady, continuous and isentropic conditions (thegeneral case of 1-d compressible \nozzle ow").

9. We have derived the sound wave equation in a gas at rest. If the gas were movingwith a uniform and constant velocity, u0 with respect to an observer, what is the equa-tion of satis�ed by the density uctuations in that frame? Derive the famous \Dopplershift" formula from the corresponding dispersion equation (analogue of Eq.(36)).(Hint:Observe that �m = �0; p = p0;u = u0 is an exact solution of the Euler equations.Discuss \small oscillations" about this state as has been done in the Notes about thestate of \rest" where, u0 = 0.)

Solution: Firstly, note that u0; �0; p0 do, indeed, satisfy the compressible, steadyEuler equations. Next, we \linearize" the system, exactly as before, substituting,�m = �0 + ~�; p = p0 + ~p;u = u0 + ~u in the full equations of motion and get (upondropping terms quadratic in the disturbances)

@~�

@t+ u0:r~� + �0r:~u = 0

@~u

@t+ u0:r~u = �(

1

�0)r~p

~p

p0= (

~�

�0)

5

It is easily seen that the relative density uctuation, � = ~�

�0satis�es the modi�ed wave

equation,

(@

@t+ u0:r)

2� = C2

sr2�

in place of Eq.(35). As before, we try the \harmonic plane wave", � ' exp(ik:x� i!t).We �nd that the dispersion relation relating k; ! to be,

(! � u0:k)2 = C2

sk2

The frequency is no longer given by, ! = �Csjkj, as in the case of sound wavespropagating in \still" air, but is in fact, ! = u0:k�Csjkj. Suppose the source of soundmoves \with the uid", ie., with the velocity, u0. To a stationary observer, alignedalong the direction of this velocity, sound will be appearing to come towards him withspeed, Cs + V and as the source recedes away from him, the wave will seem to betravelling at speed, Cs � V . Since the wave length remains unchanged, he will hearthe higher pitched Doppler \up shift" on approach of the source towards him and the\down shift" as the source moves past and away from him. Note that the D'Alembert

wave equation for sound is not form invariant under Galilean transformations. Thissays that sound speed depends upon the relative motion of the observer, the \medium"(ie., the gas), and the source in general and is not absolute like the speed of light whichis invariant for all observers and requires time and space to be relative for relativelymoving observers.

This simple analysis can be generalized to situations when u0 is nonuniform (ie varies inspace) but still satis�es the equations of motion. In this \sheared ow" case, the waveequation is more complex (involving equations with spatially variable coeÆcients asopposed to Eq.(35) with its constant coeÆcients), as it also is when �0; p0 are spatiallyvariable, steady exact solutions of the Euler equations. Such modi�ed wave propa-gation problems in strati�ed/moving gases are of great importance in meteorology,astro/geophysics and in engineering.

6


Problem Set for Lecture 4: Collisions, two- uid

theory and qualitative ideas of plasma turbulence

A. Thyagaraja









this course.

1. Consider the 1-dimensional di�usion equation,

@n

@t= D

@2n

@x2

where, �1 < x < +1. At t = 0, n(x; 0) = n0(x), where n0 is an even, positivefunction of x which tends to zero like an exponential as x ! 1. Show, withoutsolving the equation that,

1.R+1�1

n(x; t)dx = I0(t) is a constant equal to I0(0) =R+1�1

n0(x)dx

2. < (�x)2 >�

R+1

�1n(x;t)x2dx

I0satis�es the equation, d<(�x)2>

dt= D, and integrate

this equation.

3. Interpret p(x; t) = n(x; t)=I0 as a \probablility distribution" and work out thephysical meaning of the above two results.

2. Consider heat conduction in a copper bar (cf. Eqs.(4,5) of the lectures) to proceedaccording to Fourier's hypothesis. Since the volume doe not change, the change in

1

the entropy, TdS = CV dT , according to thermodynamics. Since entropy- ux can bede�ned as, q

T, show that heat conduction equation, Eq.(5) implies that S continually

increases within a bounded volume � of copper whose boundaries are \adiabatic" (ie.,q:n = 0 on �, the bounding surface of �, where, n is the unit normal to �), providedKT > 0 everywhere. If you start with an arbitrary initial distribution of temperaturewithin �, determine the temperature distribution within the region � as t ! 1.How does this asymptotic distribution depend on the thermal di�usivity �, on theinitial distribution, T (r; 0)? Explain how Fourier's Law of heat conduction embodiesthe Second Law of Thermodynamics on the basis of this example.

3. Consider the two- uid equilibrium obtained in the Notes (Eqs.(13-19)). If Sp =( ra)S�, for 0 � r � a and S� is a constant (units: electrons. m�3.s�1), obtain ne(r)

and Bz(r), assuming that �f is given by Braginskii's formula mentioned in the text.Te; B0;� are to be treated as known (constant) quantities.

4. Find the condition on Ln; k�; �s for the ordering, !� � ci to be valid.

5. Show that the solution to Eq.(31) of the Notes is given by, e~�Te

= F (��!�t; r; z), wheree~�Te(�; r; z; 0) = F (�; r; z), at t = 0 and F is periodic in �, but may be an \arbitrary"

function of its other arguments.

6�. Show that if ne0 = n� exp [�(r=Ln)2], where n�; Ln are constants, !� is a constant.

Solve Eq.(40) of the text for A and determine the eigenvalues ! and eigenfunctions.(Hint: a knowledge of Bessel functions is needed for this problem!)

7. Derive Eq.(46) of the Notes in detail.

8. Taking �ei to be given by Braginskii's formula for �f and estimate how large k?kz

must be to get !im = 0:01!re. Assume that Te ' 1KeV, B ' 1T, Ln ' 1m, �sk? = 0:1.These are typical \tokamak" values. Take � = 20.

2



A. Thyagaraja




Some important corrections to the Lecture Notes for Lecture 11 and the

problem set distributed. In the lecture Notes, a formula was given for the Bragin-skii collision frequency, �f Unfortunately, the formula quoted was in Gaussian units!The correct SI formula is the following:

�f =21=2ne�e

4

12�3=2�20m1=2e T

3=2e

The notation and the symbols are exactly as in the lecture Notes.

Corrections to the problems: Problem 1 needs a correction in Part 2, as indicated below.In the last problem, the number density, ne was not speci�ed. It should be taken as5� 1019m�3.

1. Consider the 1-dimensional di�usion equation,

@n

@t= D

@2n

@x2

where, �1 < x < +1. At t = 0, n(x; 0) = n0(x), where n0 is an even, positivefunction of x which tends to zero like an exponential as x ! 1. Show, withoutsolving the equation that,

1.R+1�1

n(x; t)dx = I0(t) is a constant equal to I0(0) =R+1�1

n0(x)dx

1

2. < (�x)2 >�

R +1

�1n(x;t)x2dx

I0satis�es the equation, d<(�x)2>

dt= 2D (this was wrongly

given in the distributed sheet!), and integrate this equation.

3. Interpret p(x; t) = n(x; t)=I0 as a \probablility distribution" and work out thephysical meaning of the above two results.

Solution: 1. Integrate the equation from �1 to 1 with respect to x. From theboundary conditions at in�nity, the rhs vanishes. Hence, we �nd, dI0

dt= 0. Evaluating

it at t = 0 gives the required result.

2. Multiply the di�usion equation by x2 and integrate w.r.t x over the whole domain.We obtain,

d

dt(Z +1

�1

n(x; t)x2dx) = DZ +1

�1

x2@2n

@x2dx

= �2DZ +1

�1

x@n

@xdx

= 2DZ +1

�1

ndx

= 2DI0(0)

where we use integration by parts (twice!) and the b.c's. Note that we also use theresult obtained above for the constancy of I0, incidentally correcting a typo in theproblem. The rhs should be 2 � DI0, not DI0 as the problem originally stated! Theintegration is straight forward: < (�x)2 > (t) = 2Dt+ < (�x)2 > (0). This showsthat the mean-square displacement is linearly proportional to the time, a typical resultof random walk theory.

3. It is evident that p(x; t) = n=I0 is a positive probability density, as it is clearly truethat

R +1�1

p(x; t)dx = 1. Thus p(x; t)dx is the probability of �nding a particle in theinterval, x; x+dx at time t. Obviously, < (�x)2 > (t) is the mean square displacementof the random walking particle (\drunkard"!) at time t. We have shown that thisincreases linearly with time, although < �x >= 0 for all time. Thus, even though thedrunkard may have had a zero probability of being outside a �nite interval initially, ast!1, there will be a small likelihood of �nding him arbitrarily far from the startingpoint.

2. Consider heat conduction in a copper bar (cf. Eqs.(4,5) of the lectures) to proceedaccording to Fourier's hypothesis. Since the volume doe not change, the change in

2

the entropy, TdS = CV dT , according to thermodynamics. Since entropy- ux can bede�ned as, q

T, show that heat conduction equation, Eq.(5) implies that S continually

increases within a bounded volume � of copper whose boundaries are \adiabatic" (ie.,q:n = 0 on �, the bounding surface of �, where, n is the unit normal to �), providedKT > 0 everywhere. If you start with an arbitrary initial distribution of temperaturewithin �, determine the temperature distribution within the region � as t ! 1.How does this asymptotic distribution depend on the thermal di�usivity �, on theinitial distribution, T (r; 0)? Explain how Fourier's Law of heat conduction embodiesthe Second Law of Thermodynamics on the basis of this example.

Solution: We convert the equation for T into one for S!

CV@T

@t= �r:q

T@S

@t= �r:

�Tq

T

�

= �Tr:(q

T)�rT:

q

T

We divide this equation by T and use Fourier's Law, q = �KTrT , to get,

@S

@t+r:(

q

T) = KT (

rT

T)2

Next we integrate this equation over the domain � and make use of the divergencetheorem and the b.c., to obtain,

d

dt(Z�SdV ) + (

Z�r:

q

TdV ) =

Z�KT (

rT

T)2dV

d

dt(Z�SdV ) +

Z�

q

T:ndA =

Z�KT (

rT

T)2dV

d

dt(Z�SdV ) =

Z�KT (

rT

T)2dV

where the surface integral vanishes from the \adiabatic wall condition". For KT >0 everywhere, the rhs is evidently positive (unless rT � 0 everywhere within �!).Hence the total entropy of the system,

R� SdV increases monotonically during the

process of thermal conduction. Let us observe that as in problem 1, Fourier's heatconduction equation in � with the adiabatic b.c conserves the total internal energy.Thus integrating the basic equauation over � and making use of the divergence theoremand the b.c, we �nd that E(t) = CV

R� TdV is a constant. Since it is plain that in the

\�nal state" as t tends to in�nity, the entropy is a maximum if and only if T is uniform

3

throughout �, we see that this uniform temperature must be equal to, T�nal =E(0)

V (�)CV,

where V (�) is the volume of the domain �. This does not depend on KT ! In fact itonly depends upon the total initial internal energy, E(0), the speci�c heat, CV and thevolume of the domain.

Fourier's Law clearly implies that a nonequilibrium temperature distribution in anadiabatic enclosure (ie., a thermally \insulated" system) will evolve in such a manneras to continually and irreversibly increase the entropy, which reaches its maximumvalue in the thermodynamic equilibrium state when the temperature becomes uniformeverywhere. This evolution must of course be consistent with the First Law of Ther-modynamics, ie., the Law of Conservation of Energy, and it indeed is, as shown byour simple calculations. The thermal conductivity must be everywhere positive, butits size is irrelevant as far as qualitative consistency with Thermodynamic Laws areconcerned.

3. Consider the two- uid equilibrium obtained in the Notes (Eqs.(13-19)). If Sp =( ra)S�, for 0 � r � a and S� is a constant (units: electrons. m�3.s�1), obtain ne(r)

and Bz(r), assuming that �f is given by Braginskii's formula mentioned in the text.Te; B0;� are to be treated as known (constant) quantities.

Solution: Consider Eq.(19):

1

r

d

dr(rD?e

dnedr

) + (r

a)S� = 0

multiplying by r and integrating once from the origin, we �nd, quite generally,D?ednedr

=

� r2

3aS�, where we use the fact the particle ux must be zero at the origin (ie., the axis).

Now, we are given that �f '21=2ne�e4

12�3=2�20m1=2e T

3=2e

. Note that all quantities except ne are con-

stant. We may therefore write, D?e = D� ne(r)ne(0)

, D� = 12�2e�0, where, �0 =

21=2ne(0)�e4

12�3=2�20m1=2e T

3=2e

.

We then �nd that ne(r) satis�es the equation,

nednedr

= �r2

3a

ne(0)S�

D�

This equation is readily integrated to give, n2e = C � 2r3

9ane(0)S�

D�. We can obtain

the constant C from the condition, ne(a) = 0. In fact, we get, C = 2a2ne(0)S�

9D�.

Since, at r = 0; ne = n0(0), by de�nition, we have, 9D� = 2a2S�. This then gives,

ne(0) = (4a2S�

9�2e)(

12�3=2�20m1=2e T

3=2e

21=2�e4). The density pro�le can be written more transpar-

ently in the form, ne(r=a) = ne(0)h1� ( r

a)3i. It follows from Eq.(15) that, Bz(r) =

4

B0

h1� 2�0Tene

B20

i1=2. It is an interesting problem to �gure out how big S� can be to have

a consistent solution, and what happens if this is exceeded.

4. Find the condition on Ln; k�; �s for the ordering, !� � ci to be valid.

Solution: From Eq.(33) of the Notes, we have, !�ci

= ( CsciLn

)(k��s) = �sLn(k��s) =

��k��s. Hence the condition is �� 1, for k��s ' 1.

5. Show that the solution to Eq.(31) of the Notes is given by, e~�Te

= F (��!�t; r; z), wheree~�Te(�; r; z; 0) = F (�; r; z), at t = 0 and F is periodic in �, but may be an \arbitrary"

function of its other arguments.

Solution: Simply substituting the given functional form into Eq.(31), we get the result!It is evident that at t = 0, we match the initial condition. The periodicity of F wrt. �is a simple consequence of the single-valuedness of the electrostatic potential.

6�. Show that if ne0 = n� exp [�(r=Ln)2], where n�; Ln are constants, !� is a constant.

Solve Eq.(40) of the text for A and determine the eigenvalues ! and eigenfunctions.(Hint: a knowledge of Bessel functions is needed for this problem!)

Solution: Substituting for ne0 and its derivative in Eq.(32), we see immediately that!� for this pro�le is independent of r and satis�es, !� =

2TeeB

mLn. We may plainly write

Eq.(40) in the form,

1

r

d

dr(rdA

dr)� (

m

r)2A = �KA

K =

h!�!� 1 + Csk2z

!2

i�2s

Suppose we set, z = rK1=2. The equation then becomes (with A = f(z)),

d2f

dz2+

1

z

df

dz+ (1�

m2

z2) = 0

We recognize this as Bessel's equation of integer order m in the variable z (see, forexample, Handbook of Mathematical Functions, eds. Abramowitz and Stegun, p. 358).It has two linearly independent solutions, Jm(z); Ym(z), of which only Jm(z) is regularat the origin. We require also that A(r = a) = 0. This implies that, Jm(aK

1=2) = 0.Now, the zeros of Bessel functions are well-documented (see above reference or books

5

on Bessel functions). They are all real and increasing numbers. We denote them

by �(m)1 ; �

(m)2 ; ::. We then obtain the result that aK1=2 = �

(m)1 , for example. The

\eigenvalue equation" determining the frequency, ! then becomes,"!�!� 1 +

Csk2z

!2

#=

(�(m)1 �s)

2

a2

(!

!�)2 �

1

1 +(�

(m)1 �s)2

a2

!�!�

1

1 +(�

(m)1 �s)2

a2

Csk2z

!2�

= 0

It is entirely reasonable to regard,�(m)1

aas a typical \perpendicular wavenumber" asso-

ciated with the zero,�(m)1 of Jm. Setting this to be, �

(m)1 =

�(m)1

a, we see that the above

eigenvalue relation for the \dimensionless frequency", !!�

becomes,

(!

!�)2 �

1

1 + (�(m)1 �s)2

(!�!)�

1

1 + (�(m)1 �s)2

(Csk

2z

!2�

) = 0

Note that this is virtually identical with the \local dispersion relation", Eq.(41) derivedin the text, upon dividing the latter by !2

�! Incidentally, we see that the \normal

modes" are labelled by the \azimuthal quantum number", m and the \radial quantumnumber", �(m)

k ; k = 1; 2; ::. This completes the exact solution of the drift wave poblemfor this \Gaussian" density pro�le.

7. Derive Eq.(46) of the Notes in detail.

Solution: Using Eq.(35) for ~u?i, and Eq.(45) for ~u?e, we �nd that the perturbedperpendicular current density,

~j? = ene0(~u?i � ~u?e)

= ene0

"(Cs�s)ez �r?(

e~�

Te)� �2sr?

@

@t(e~�

Te)� Cs�sez �r?

e~�

Te�Cs�sne0

ez �r?~n

#

= ene0

"��2sr?

@

@t(e~�

Te)

#� eCs�sez �r?~n

Note that the E�B drifts in both species cancel. This drift does not ever driveperpendicular currents in quasi neutral conditions, only ows. Now, Eq.(44) gives anexpression for the uctuating parallel current density:

~jz = ezene0(~uzi � ~uze)

= ezne0Teme�ei

@

@z(~n

ne0�e~�

Te)

6

We are almost done! Let us recall that \quasi-neutrality" requires that r:~j = 0. Thismeans, r:(~j?+~jz) = 0. It is elementary to note that the second term in the expressionfor the perpendicular current is divergence-free and substitution gives Eq.(46) of thetext. This completes the derivation.

It is an excellent exercise (left to the interested student!) to repeat the above analysisof drift waves carefully in the case when the diamagnetic variation of Bz with r is fullytaken into account. Some terms we have approximated away (can you identify which?)will enter this analysis.

8. Taking �ei to be given by Braginskii's formula for �ei '21=2ne�e4

12�3=2�20m1=2e T

3=2e

and estimate

how large k?kz

must be to get !im = 0:01!re. Assume that Te ' 1KeV, ne ' 5�1019m�3,B ' 1T, Ln ' 1m, �sk? = 0:1. These are typical \tokamak" values. Take � = 20.

Solution: Taking the ions to be protons, we know that me

mi= 5:4 � 10�4 and ci =

eB=mi = 4:8 � 10�19=1:67 � 10�27 ' 3 � 108rads/s, for B = 1T. Similarly, Cs =(Te=mi)

1=2 = (1:6 � 10�16=1:67 � 10�27)1=2 ' 3:2 � 105m/s. It follows that, �s =Cs=ci ' 10�3m. Taking �0 = 8:85 � 10�12 (SI) , � = 20, we calculate �ei '

21=2�5�1019�20�6:55�10�76

12�5:56�78:3�10�24�0:95�10�15�2�10�24= 9:3�10�55

9:9�10�60' 105s�1.

We approximate !re ' (Cs=Ln)(�sk?) ' 3:2� 104. Then,

!im!re

' (me

mi)(!re�ei!2ci

)(k?kz

)2

' (5:4� 10�4)(3:2� 104 � 105

32 � 1016)(k?kz

)2

' 1:9� 10�11(k?kz

)2 (1)

Thus we �nd that (k?kz)2 ' ( 1

1:9) � 109. This implies that jk?

kzj ' 2:3 � 104! Since

�sk? ' 0:1, kz ' 0:1=(2:3� 104�s) ' 4� 10�3m�1. This is, even compared to 1=Ln avery long wave length! This example illustrates the sort of relatively simple estimatesone can make in drift wave physics.

7

lectures on plasma physics- problems and solutions

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