Astronomy 9620a / Physics 9302a - 1st Problem List and Assignment

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1 Astronomy 9620a / Physics 9302a - st Problem List and Assignment Your solution to problems 4, 7, 3, and 6 has to be handed in on Thursday, Oct. 7, Prove the following relations (going from the left- to the right-hand side of the equality): Solution. ( c d) ( a c) ( b d) ( a d) ( b c) ψ a + ψ a ( a )b + ( b )a + a ( b) + b ( a) a( b) b( a) + ( b )a ( a )b. a) a b b) ψ a c) a b d) a b a) We use index calculus to get (Einstein summation convention assumed) ( a b) ( c d) ( ε ijk a j b k )( ε imn c m d n ) ε ijk ε imn a j b k c m d n ( δ jm δ kn δ jn δ km )a j b k c m d n (.) b) The second relation (with i x i ) a j b k c j d k a j b k c k d j ( b d) ( a d) b c a c. ( ψ a) i ε ijk j ( ψ a k ) ε ijk a k j ψ + ψ j a k [ ] i. ψ a + ψ a (.2) c) The third relation ( a b) i i ( a j b j ) b j i a j + a j i b j b j i a j + a j i b j + b j j a i + a j j b i b j j a i a j j b i a j j b i + b j j a i + a j i b j j b i + b j ( i a j j a i ) (.3) a j j b i + b j j a i + a j ε ijk ε mnk m b n + b j ε ijk ε mnk m a n a j j b i + b j j a i + ε ijk a j ε kmn m b n + ε ijk b j ε kmn m a n ( a )b + ( b )a + a ( b) + b ( a) i.

2 d) Finally, the fourth relation a b i ε ijk j ε kmn a m b n ( b n j a m + a m j b n ) δ im δ jn δ in δ jm b j j a i + a i j b j b i j a j a j j b i a( b) b( a) + ( b )a ( a )b i. (.4) 2. The time-average potential of a neutral hydrogen atom in its ground state is given by Φ q e αr r + αr 2, (2.) where q is the magnitude of the electronic charge, and α a 0 2 with a 0 the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically (for example, calculate the total charge). Solution. Let us first rewrite equation (2.) as follows Φ q e αr α 2 + r. (2.2) We now need to apply the Poisson equation to equation (2.2) remembering the following relation for the product of two functions 2 ( fg) fg ( g f + f g) g 2 f + 2 f g + f 2 g. (2.3) Setting f e αr g α 2 + r, (2.4) with 2

3 f f r αe αr e r 2 2 f r r 2 ( rf ) r r g g r r e 2 r 2 g 2 r 4πδ ( x). ( αr)e αr α 2 r αe αr (2.5) If we now apply these equations to the Poisson equation, we find 2 Φ ρ ε 0 q e αr α 2 + r α 2 r α + 2 α r 4πδ ( x) 2 q 4πδ x α 3 2 e αr. (2.6) Then, the charge distribution is ρ q δ x 2r 3 πa e a 0 0. (2.7) We see that the first term corresponds to the proton charge, while the second term is the square of the norm of the radial wave function of the electron in the ground state (multiplied by its charge). Integrating ρ over all space yields a total charge of zero, as expected. 3. A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and opposite charges are placed on the conductors, there will be a certain difference of potential between them. The ratio of the magnitude of the charge on one conductor to the magnitude of the potential difference is called the capacitance (in SI units it is measured in farads). Using Gauss Law, calculate the capacitance of a) two large, flat, conducting sheets of area A, separated by a small distance d ; ; b) two concentric conducting spheres with radii a, b b > a c) two concentric conducting cylinders of length L, large compared to their radii a, b b > a. Solution. 3

4 a) It is clear by the symmetry of the problem that if the sheets are large enough (we neglect the edge effects when d A 2 ), then the electric field between them will be oriented normal to their surface. If we now consider an imaginary closed, rectangular surface that straddles one conducting sheet, extending some distance ξ ( < d) away on either side, then using Gauss Law for this surface we find 2EΔa q Δa, (3.) ε 0 A with E and Δa the electric field at, and the area of, the imaginary Gauss surface inside the capacitor (the electric field in the conducting sheet is zero). Alternatively, we can write E q 2ε 0 A. (3.2) Obviously, the surface charge density of the other sheet will contribute equally to the electric field E c within the capacitor. It is therefore clear that this electric field is constant with E c and that the potential difference between the two plates is The capacitance is therefore q ε 0 A, (3.3) Φ E c d qd ε 0 A. (3.4) C q Φ ε 0A d. (3.5) b) We consider a closed spherical surface of radius r located in the space between the two conducting spheres (it is also concentric to the two conducting spheres). Since the outer sphere does not contribute to the electric filed within its volume, the electric field between the spheres is entirely due to the charge distribution on the inner sphere with E q r 2 e r, (3.6) where q is the total charge on a sphere. The potential difference between the two spheres that form the capacitor is therefore 4

5 Φ q a dr q b r 2 a b. (3.7) The capacitance is then C q Φ 4πε 0 a b. (3.8) c) Because the cylinders are very long compared to their radii, we are justified in neglecting edge effects. Then, we see from the symmetry of the problem that electric field will only have a radial component. Just as in the preceding case, the electric field in the space between the two conducting surfaces is entirely due to the charge located on the smaller surface. Considering a cylindrical surface of radius r extending within the space between the two conducting cylinders and using Gauss Law, we find that the electric field in that space is due to the charge q located on the surface of the smaller cylinder is E q 2πε 0 rl. (3.9) The potential difference between the two conducting cylinders is therefore Φ q 2πε 0 L b a dr r q 2πε 0 L ln b a. (3.0) And the capacitance is C q Φ 2πε L 0 ln b. (3.) a 4. a) Find the electric field a distance z along the symmetry axis of a uniformly charged ring of total charge Q and radius R. b) Use the field for a ring calculated in a) to find the electric field a distance z along the symmetry axis of a uniformly charged disk of total charge Q and radius R. c) Use the results of b) to find the electric field due to the charged disk at z 0 and z R. For the latter, would it have been possible to derive this result based simply on physical arguments? Explain your answer. 5

6 Solution. a) The field from the elemental charge dq ( Q 2π R)Rdϕ ( Q 2π )dϕ, where dϕ is an infinitesimal angle in the plane of the ring is given by de dq r 2 e r Qdϕ e r, 8π 2 ε 0 R 2 + z 2 (4.) where e r is the unit vector linking the elemental charge to the point where the field is evaluated. Since that location is on the symmetry axis, all the elemental charges from the ring will contribute to yield a total field oriented along the z-axis, the field component in direction perpendicular to the z-axis will cancel out. We then have 2π E z e z de 0 Qcos( θ ) R 2 + z 2 Qz R 2 + z 2 3 2, (4.2) where e z e r cos( θ ). b) The disk can be thought of as being made of rings of radius ρ and thickness dρ, with 0 ρ R. We then write E z ( z) 0 R Qz 2πε 0 R 2 Q 2πε 0 R 2 Q 2πρ π R 2 0 R ρdρ ( ρ 2 + z 2 ) 3 2 z ( ρ 2 + z 2 ) 3 2 dρ z z 2 + R 2. (4.3) c) It is straightforward to calculate that 6

7 E z E z ( 0) Q 2πε 0 R 2 Q ( z R) + R2 2πε 0 R 2 z 2 Q 2πε 0 R 2 Q z 2. 2 R2 2z + 2 (4.4) This last equation could have be easily guessed from the fact that at z R the disk appears as a point charge Q, resulting in the calculated approximation for the electric field. 5. Electrostatic Screening and the Debye Length. Let s assume that we have a system consisting of distributions (i.e., number densities in m 3 ) n e and n i of electrons and positive ions, respectively. The distributions are initially in equilibrium with homogeneous densities n e n i (independent of position), such that the electrostatic potential Φ is zero everywhere. We now introduce a charge Q that locally perturbs the distributions of electrons and ions, and destroys their homogeneity in its vicinity. If we assume that far away from the charge both densities still have the same value n, then we can expect from thermodynamic equilibrium considerations that the following relations will hold in the neighborhood of Q n i ne qφ κ B T qφ κ n e ne T B, (5.) where q, κ B, and T are the charge of the ions ( q for the electrons), the Boltzmann constant, and the temperature of the system, respectively. a) Use equations (5.) and the Poisson equation to show that the electrostatic potential around Q is given by Φ( r) where r is the distance (radius) away from Q and r Q r e λ D, (5.2) 7

8 λ D ε 0κ B T 2nq 2 2 (5.3) is the so-called Debye length. To obtain equation (5.2) you can safely assume that qφ ( κ B T ) is a small quantity, such that it is appropriate to linearize the exponentials of equations (5.). Also, assume that the electrostatic potential is zero at infinity. b) Using just a few sentences, explain the physical implication of equation (5.2). That is, what is the spatial extent over which the presence of Q is felt? Over what length scale can this plasma be considered neutral? [Hints: Poisson s equation is 2 Φ x ρ( x) and the Laplacian operator in spherical coordinates is given by 2 2 Φ r r 2 ( rφ) + r 2 sin θ θ sin θ Also, the following second order differential equation ε 0, (5.4) θ + r 2 sin 2 θ Φ 2 Φ ϕ 2. (5.5) has for solution with R a constant.] Solution. ψ r 2 r 2 [ ψ ] ψ R, (5.6) 2 Ae r r R + Be R + C, (5.7) The Laplace equation for this system in the vicinity of, but not at the position of, Q is 2 Φ q ε 0 ( n i n e ) qn ε 0 e qφ κ B T e qφ κ B T (5.8) 2nq2 ε 0 κ B T Φ, 8

9 where we have used e ±qφ ( κ BT ) ± qφ ( κ B T ) for the last step. Using the equations (5.5) and (5.8), and the fact that the system is spherically symmetric, we can write or, if we define a new function ψ ( r) rφ( r), Equation (5.0) is easily solved to yield 2 r r rφ( r) Φ r, (5.9) 2 2 λ D 2 ψ r 2 ψ λ D 2. (5.0) ψ ( r) Φ r r Ae r r λ D + Be r λ D + C, (5.) where A, B, and C are some constants. However, we require that the electrostatic potential tends towards zero as r, and that lim Φ( r) r 0 then B C 0 and A Q ( ), which finally leaves Φ( r) Q r, (5.2) r Q r e λ D. (5.3) b) Equation (5.3) implies that the presence of the charge Q introduced in the system will only affect the electron and ion distributions within a volume of radius approximately equal to the Debye length. It is seen that beyond r λ D the potential caused by the charge becomes negligible. It follows that a plasma that is globally neutral can be also considered neutral everywhere when the relevant physical quantities are averaged on scales large compared to the Debye length. 6. Consider a hollow sphere of radius R, on which the potential V ( ϕ) V cos( ϕ) (6.) 9

10 is applied (ϕ is the usual azimuth angle for spherical coordinates). Find the potential Φ x everywhere inside the sphere. [Hints: To ensure that the potential is unambiguously defined everywhere, you can assume that there is a tiny hole at both poles on the sphere. Use the most general (and physically plausible) expansion for the potential. Don t forget that any function can be expressed as a series of functions belonging to a complete set of basis functions; choose the right basis. You may use the table of spherical harmonics given at the end, but limit yourself to l 3.] Solution. Using spherical coordinates, and spherical harmonics, we can expand the potential Φ x as follows A lm r l Y lm ( θ,ϕ) Φ r,θ,ϕ m l 0 l m, (6.2) where we omitted any terms in r ( l +) since the potential must be finite at the origin. On the surface of the sphere, the potential must satisfy the boundary condition A lm R l Y lm ( θ,ϕ) Φ R,θ,ϕ m l 0 l m V cos( ϕ). (6.3) It follows from equation (6.3) that the coefficients A lm R l are simply given by However, because Y lm θ,ϕ unless m ±. Therefore, A lm R l 2π π * V dϕ cos( ϕ)y lm ( θ,ϕ)sin( θ) dθ. (6.4) 0 P lm θ 0 e imϕ, and cos ϕ e ±iϕ, then A lm R l will be zero A l,± πv R l πv R l π 0 * P l,± ( θ)sin θ dθ π P l,± ( θ)sin( θ) dθ, 0 (6.5) where P l,± θ * P l,± ( θ) can be obtained from the table at the end. More precisely, 0

11 P,± ( θ) 3 8π sin θ P 2,± ( θ) 5 8π sin( θ)cos( θ) P 3,± ( θ) 2 64π sin θ ( 5cos2 θ). (6.6) Upon inserting the different functions in equation (6.5), it is easy to see that terms in l 2 will vanish since their integrand is odd. Now, from we find π dθ sin 2 θ π 0 2 π sin 2 ( θ )cos 2 ( θ) dθ π (6.7) 8, 0 A,± V π 2 R 2 A 3,± V R 3 π π 2 64π, (6.8) and Vπ 2 Φ r,θ,ϕ r 8π R Y, θ,ϕ 2 r 3 64π R Y 3, 3 θ,ϕ Y ( θ,ϕ) Y 3 ( θ,ϕ). (6.9) Rearranging the different terms, we finally get Φ( r,θ,ϕ) V 3π 8 sin( θ)cos( ϕ) r + 7 r R 32 R 3 5 cos 2 θ. (6.0) 7. An arbitrary function Φ that is a solution to the Laplace equation can always be expressed in cylindrical coordinates using a Bessel-Fourier expansion with

12 A mn J m ( k mn ρ) + B mn N m ( k mn ρ) Φ ρ,φ, z m0 n { C mn cos k mn z E m cos mφ + D mn sin( k mn z) + F m sin( mφ ) } (7.) and N m ( x) are the Bessel functions of, respectively, the first and second where J m x kind of order m, and it is implicit that F 0 0 and E 0 agrees with the usual definition for the corresponding Fourier coefficient (see Equation (2.2) of the Lecture Notes). The functions behaviors at the origin are such J m 0 N m 0 is finite and well behaved whereas is not. The validity of equation (7.) stems, in part, from the fact that the set of can be Bessel functions is complete and forms a basis, such that any function f ρ expanded over a given interval for ρ of length a with where ρ α mn J m k mn f ρ ρ a + β mn N m k mn a, (7.2) m0 n α mn β mn 2 a 2 2 J m+ 2 a 2 2 N m+ k mn k mn ρ J m k mn ρ f ρ ρ f ρ ρ N m k mn a dρ a dρ (7.3) and the integrals are performed over the aforementioned interval. Now consider a hollow cylindrical tube of radius a, with its axis of symmetry coincident with the z-axis, and its ends located at z 0 and z L. The potential on its end faces is zero, while the potential on the cylinder is given as V ( φ, z) V for π 2 < φ < π 2 V for π 2 < φ < 3π 2. (7.4) Find the potential anywhere inside the cylinder. Solution. Since we require the potential to be finite at the origin ρ 0, equation (7.) reduces to 2

13 A mn J m ( k mn ρ) C mn cos( k mn z) + D mn sin( k mn z) Φ ρ,φ,z m0 n (7.5) E m cos mφ. + F m sin( mφ ) Applying the boundary conditions specified for the end faces we also find that C mn 0. Furthermore, the symmetry of the potential V ( φ, z) on the cylinder implies that F m 0. We then have A mn J m k mn ρ Φ ρ,φ, z sin( k mn z)cos( mφ ), (7.6) m0 n where the Bessel-Fourier coefficient A mn has been accordingly redefined, and A mn J m k mn a V φ,z sin( k mn z)cos( mφ ). (7.7) m0 n Clearly, for the potential to vanish at the end faces it must be that k mn nπ L, n, 2, 3, (7.8) and k mn is independent of m. It follows from the definition for Fourier coefficients that when m 0 A mn J m ( k mn a) 2 2π L dφ cos( mφ ) dzv ( φ,z) π L sin( k mn z) 0 0 2V π 2 3π 2 L cos( mφ )dφ π L cos( mφ )dφ π 2 sin k π 2 ( mn z)dz, 0 (7.9) which further reduces to 8V mnπ A mn J m ( k mn a) 2 8V mnπ 2 It is also understood that A 0n J 0 k 0n a A 0n 0. We therefore find that n n, m,5,9, (7.0) m 3, 7,, 0, which according to equation (7.8) means that 3

14 Φ( ρ,φ,z) 8V π 2 8V π 2 n sin nπ z L J m nπ ρ L mn nπ a L cos( mφ ), m,5,9, m n m n n mn J m J m J m ( nπ ρ L) nπ a L sin nπ z L cos mφ m 3,7,, (7.) for 0 ρ a and 0 z L. the values of moments of the first non-vanishing multipole are independent of the origin of 8. Prove the following theorem: For an arbitrary charge distribution ρ x the 2l + the coordinate axes, but the value of all higher multipole moments do in general depend on the choice of origin. (The different moments q lm for fixed l depend, of course, on the orientation of the axes.) Solution. For a given l and m the corresponding multipole moment is given by * q lm Y lm ( θ,ϕ)r l ρ( x)d 3 x. (8.) Before we can prove the theorem, we first have to show that the spherical harmonics obey the following relation Y lm ( θ,ϕ)r l C a,b,c x a y b z c, (8.2) a+b+cl where C a,b,c is some constant. To do so we consider a function Φ l we can express with the two following series ( x) of order l, which Φ l l ( x) A lm r l Y lm ( θ,ϕ) B a,b,c x a y b z c, (8.3) m l a+b+cl with the constraint that a, b, and c are integer numbers. If we next consider one specific term of the first series, say for m m, we can write A l m r l Y l m θ,ϕ Φ l ( x) A lm r l Y lm ( θ,ϕ). (8.4) m m It is obvious, however, that the right hand side of equation (8.4) is just another function of the coordinates, say x Φ l, that can also be expanded as follows 4

15 A l m r l Y l m θ,ϕ ( x) Φ l B a,b,c x a y b z c, a+b+cl (8.5) which proves equation (8.2). We are now ready to proceed, and we replace equation (8.) with q lm C a,b,c x a y b z c ρ( x)d 3 x. (8.6) a+b+cl If we now change the origin of the coordinate axes and make the following substitution x x x 0 y y y 0 z z z 0, (8.7) then equation (8.6) becomes a ( ) b ( ) c ρ q lm C a,b,c x + x 0 y + y 0 z + z 0 ( x )d 3 x. (8.8) a+b+cl Now each term within parentheses in equation (8.8) can be transformed with the binomial expansion as follows ( x + x 0 ) a a a n x a n n x 0, (8.9) n0 with similar relations for the terms in y and z. Inserting equation (8.9) in equation (8.8) we have q lm a b c D n, p,r x a,b,c a n y b p z c r ρ ( x )d 3 x, (8.0) a+b+cl n0 p0 r 0 n, p,r with D a,b,c a new constant. Now, it should be clear that for every combination of n, p, and r with n + p + r > 0 corresponds a value l such that l ( a n) + ( b p) + ( c r) ( a + b + c) (n + p + r) l (n + p + r) < l. (8.) If l is for the first non-vanishing multipole (i.e., q l m 0 for l < l ), then x a n y b p z c r ρ ( x )d 3 x 0, for l l (n + p + r) < l, (8.2) 5

16 since we can always write l x e y f z g D e, f,g r l Y l m ( θ, ϕ ), for e + f + g l, (8.3) m l while it is assumed here that * l m Y l m ( θ, ϕ ) r l ρ ( x )d 3 x 0, for l < l. (8.4) q Inserting equation (8.2) into equation (8.0) we are left with q lm C a,b,c x a y b z c ρ ( x )d 3 x a+b+cl * Y lm ( θ, ϕ ) r l ρ ( x )d 3 x * Y lm ( θ,ϕ)r l ρ( x)d 3 x, (8.5) and q lm is therefore independent of the origin of the coordinates axes. Finally, if we consider the case for l l +, because (contrary to the case where l < l ) we have x a y b z c ρ ( x )d 3 x 0, when a + b + c l < l, (8.6) and we cannot simplify equation (8.0) as before, implying that equation (8.8) applies. The value of these multipole moments (i.e., for l > l ) depends on the choice of origin. Obviously, the same is true in general for all values greater than l. 9. Given that the potential Φ x with the following relation due to a charge distribution ρ ( x ) can be evaluated Φ( x) ρ x x x d 3 x, (9.) expand x x with a Taylor series (see equation (.84) of the lecture notes) to show that Φ( x) q r + p x + r 3 2 Q ij x i x j + r 5, (9.2) where a summation on a repeated index is assumed, and 6

17 q ρ x d 3 x p x ρ ( x ) d 3 x Q ij ( 3 x i x r 2 δ )ρ( x ) d 3 x. j ij (9.3) Solution. Using a Taylor series expansion for x x we can write x x r x x x x with r x. We, therefore, need the following derivatives ( x ) 2 x x x 0, (9.4) x i x x x 0 x i x j x j x x i i x ( x j x j ) x 3 x 0 x i r 3, x 0 (9.5) and x x 2 x i x j x 0 x i x i x j x x 3 δ ij r x ix j r 5. x 0 (9.6) Now, from equation (9.6) and the last term in the right-hand side of equation (9.4), we have (summations on repeated indices implied) r 5 3x i x j δ ij r 2 d 3 x i x j ρ x x 3x r 5 i x j δ ij δ mn x m x n d 3 x i x j ρ x d 3 x r x ix 5 j 3 x i x j ρ x x δ mn x m x n x j x j ρ x ( r 2 ) j ρ x r x ix 5 j 3 x i x δ j ij r 5 Q ij x i x j, d 3 x d 3 x (9.7) 7

18 where the quadrupole moment Q ij is defined in the last of equations (9.3). Inserting equations (9.5) and (9.7) in equation (9.), we find Φ( x) q r + x r x ρ ( x ) d 3 x Q x i x j ij r 5 q r + x p + (9.8) r 3 2 Q x i x j ij r 5. The dipole moment p is defined in the second of equations (9.3). 0. A perfectly conducting object has a hollow cavity in its interior. If a point charge is introduced in the cavity, what is the total charge induced on the surface of the cavity. Solution. The electric field inside a perfect conductor is zero. So, if we choose an imaginary surface S that encloses the hollow cavity but is located within the conductor, then we must have (with n is the unit vector normal to S ) E n da 0. (0.) S This implies, from equation (.53) of the lecture notes, that the total charge enclosed by S must be zero. Therefore, if a point charge q is introduced anywhere within the cavity, then a charge q must be induced on its surface to keep the total charge inside S to zero.. An electric dipole is pointing in the z direction and is placed at the origin of the coordinate system. Find the value of the electric field at any point. Solution. We will use the usual definitions for the spherical coordinates r, θ, and ϕ x r sin( θ)cos ϕ y r sin( θ)sin( ϕ) z r cos( θ). (.) The potential due to the dipole, given its orientation and that it is located at the origin, is Φ( x) x p r 3 Staying with spherical coordinates pz r 3 pcos( θ). (.2) r 2 8

19 E r Φ r 2 p cos θ r 3 E θ r E ϕ Φ θ Φ r sin( θ) ϕ 0. r 3 psin θ (.3) 2. We know that the force F acting on a charge q is given by F qe, where E is the electric field at the position of the charge. a) Now, consider a charge distribution ρ x the total force acting on it. and write down the general expression for b) We choose a position x 0 as the origin for calculating the different multipole moments of the distribution. Show that the force acting on the electric dipole moment is F d ( p )E, (2.) where p is the dipole moment, and it is understood that the gradient is evaluated at x 0. c) We also know that the electrostatic energy of the dipole moment is U d p E. Starting from equation (2.) show that the total force on the dipole can also be expressed as F d U. That is, F d ( p E). (2.2) [Hint. You may need the following relations b ( a) a ( b) a( b) b( a) + ( b )a ( a )b ( a )b + ( b )a + a ( b) + b a a b a b a b. (2.3) ] Solution. a) The total force acting on the charge distribution is F ρ( x)e( x) d 3 x. (2.4) 9

20 b) Let s expand the electric field with a Taylor series about x 0, then E( x) E( x 0 ) + ( x x 0 ) x0 Inserting equation (2.5) into equation (2.4) yields E ( x ). (2.5) E( x 0 ) + ( x x 0 ) F ρ x E d 3 x E( x 0 ) ρ( x) d 3 x + { ρ( x) ( x x 0 ) d 3 x }E qe( x 0 ) + ( p )E, (2.6) where it is understood that the gradient is evaluated at x 0, q is the total charge, and p is the dipole moment at x 0,. That is, Then the force acting on the dipole moment is c) Using the last of equations (2.3), we have ( x x 0 ) d 3 x p ρ x. (2.7) F d ( p )E. (2.8) ( p E) ( p )E + ( E )p + p ( E) + E ( p). (2.9) But since the dipole moment is independent of x, then i p j 0 and the second and fourth terms on the right-hand side of equation (2.9) vanish. Moreover, since E 0 in electrostatics, then we are left with ( p E) ( p )E. (2.0) 3. We know that the potential energy of a dipole p subjected to an electric field E is given by W p E, where the electric is evaluated at the position occupied by the dipole. Assume that the dipole changes by an amount dp following a rotation dθ caused by the torque resulting from the electric field. a) Consider the corresponding change in potential energy dw stemming from the previous equation and compare it to the general relation dw τ dθ, which links it to the applied torque τ, and derive the equation for the torque acting on the dipole. qe( x), consider and write down the general expression for the total torque b) Given that the force F acting on a charge q is expressed with F x a charge distribution ρ x 20

21 acting on it. Show that the lowest order term for the torque, when expanding the electric in a Taylor series, is the same as that obtained in a). c) Determine the torque acting on a dipole p 2 due to the presence of another dipole p located some distance r away. d) Find the torque on p due to p 2. e) Determine the total torque on the system composed of p and p 2 (make sure to consider the contributions from the interaction force each dipole applies on the other). Solution. a) Using a simple differential we write dw E dp τ dθ. (3.) Since, like any vector field, the dipole p transforms as dp dθ p under a rotation dθ, we then have dw E dθ p dθ ( p E), (3.2) which under comparison with the last of equations (3.) yields τ p E. (3.3) b) Generalizing to a charge density we write for the torque acting on it τ x E( x)ρ( x)d 3 x, (3.4) where we recognize in E( x)ρ( x)d 3 x the electric force acting on the elemental charge contained with the volume d 3 x. Assuming the dipole to be located at x 0 we expand the electric field with E( x) E( x 0 ) + x x 0 which upon keeping only the leading term in the expansion yields ( ) x0 E ( x ), (3.5) τ x E( x)ρ( x)d 3 x xρ( x)d 3 x E x 0 p E( x 0 ). (3.6) 2

22 This is the same expression as derived in equation (3.3). c) We know from equation (2.) of the Lecture Notes that the electric field due to a dipole p some distance r away from it is E p ( r) 3n n p, (3.7) r 3 with n r r. Therefore the torque on p 2 stemming from this field is τ 2 p 2 E ( r) ( n p ) p 2 p 3 p 2 n r 3. (3.8) d) To determine the torque τ on p due to p 2 we simply need to make the changes p p 2 and n n in equation (3.8). This results in ( n p 2 ) + p 2 p τ 3 p n r 3. (3.9) e) To evaluate the total torque on the system we may be tempted to simply add together the two torques previously calculated to get τ + τ 2 3 r 3 ( p n) ( n p 2 ) + ( p 2 n) ( n p ). (3.0) But this result cannot be complete since the conservation of angular momentum requires that the total torque must be zero (i.e., there is no external force or torque acting on the system composed of the two dipoles). We must also consider the force of interaction between the two torques. For example, we know from equation (2.0) that the force F 2 on p 2 due to p is F 2 ( p 2 E ) ( p 2 )E. Applying the needed derivatives to equation (3.7) yields p F 2 3 n p 2 r 3n n p 4 3 n p + p r 3 2 n( n p 2 ) r + 3n p p 2 ( n p )( n p 2 ) r 3 r ( n p 4 )p 2 + ( n p 2 )p + ( p p 2 )n 5( n p )( n p 2 )n, (3.) 22

23 where we have used n r 0, n θ e θ, and n ϕ sin( θ )e ϕ (see equations (2.95) in the Lecture Notes when n e r ). Since according to Newton s Third Law F 2 F 2 (it is straightforward to verify this by changing n n in equation (3.)) we find that the torque due to this force of interaction, calculated about the mid-point between the two dipoles, is τ 2 r 2 ( F F 2 2 ) r F 2 3 r ( p 3 n) ( n p 2 ) + ( p 2 n) ( n p ). (3.2) Adding equations (3.0) and (3.2) we find that the total torque on the system is zero, as expected. 4. Two dipoles lie in the ( x,z)-plane. p is placed at the origin and is pointing along the z-axis ; p 2 is placed at the point ( x, z) and makes an angle α with p. Calculate the force on p 2 in the general case (limit yourself to the x and z components of the force). Solution. If E is the electric due to the first dipole p at the position of the second dipole p 2, then the force acting on p 2 is F ( p 2 )E. (4.) (You should try to prove this.) Using equations (.3) with ϕ 0 (to work in the x,z -plane ), and transforming to Cartesian coordinates, we find 23

24 E x 3p xz ( x 2 + z 2 ) E z p 2z 2 x 2. x 2 + z 2 (4.2) From these equations we can evaluate the following derivatives 7 2 ( x 2 + z 2 ) 7 2 E x x 3z z 2 4x 2 x 2 + z 2 E z x 3x x 2 4z 2 p, p, 7 2 ( x 2 + z 2 ) 7 2 E x z 3x x 2 4z 2 x 2 + z 2 E z z 3z 3x 2 2z 2 p p (4.3) and the force components are F x p p 2 F z p p 2 sin α sin α ( x 2 + z 2 ) 7 2 3z z2 4x 2 ( x 2 + z 2 ) 7 2 3x x2 4z 2 + cos α + cos α ( x 2 + z 2 ) 7 2 3x x2 4z 2 3z 3x2 2z 2 ( x 2 + z 2 ) 7 2. (4.4) 5. Find the B field along the symmetry axis of a circular loop of radius a carrying a current I. Solution. We will use the cylindrical coordinates r, θ, and z. The Biot and Savard law states that db µ 0 I ( dl x), (5.) 4π x 3 where x ze z ae r is the vector going from a given point on the loop to the observation point on the symmetry axis, and dl adθe θ. Therefore, db µ 0 4π µ 0 4π I x 3 I x 3 a dθe θ ze z ae r ( az dθe r + a 2 dθe z ). (5.2) 24

25 If we integrate over θ from 0 to 2π, we find that the first term vanishes since e r cos( θ)e x + sin( θ)e y, and finally B µ 0 Ia z 2 + a 2 e z. (5.3) 6. A small current loop of radius R lies in the ( x, y)-plane. A current J passes through the loop. a) Find the magnetic dipole moment m of the loop. b) Find the asymptotic (i.e., r R ) magnetic induction B from the magnetic potential due to m. c) Write the equation of motion for a particle of mass M and charge q in the asymptotic B field of the loop. Show that the particle stays in the ( x, y)-plane if its original velocity is restricted to that plane. Solution. a) The magnetic moment is located in a plane; therefore, it is simply expressed as the product of the current and the area of the loop. That is, m Jπ R 2 e z. (6.) b) We know from equation (3.48) of the lecture notes that the magnetic field away from the magnetic dipole is given by B( x) µ 0 m x 3 3n n m 4π, (6.2) with n x x x r. If the vector x makes an angle θ with the z-axis, then n m m cos( θ)e z, and m 4π r 3 B( x) µ 0 3cos( θ)e r e z. (6.3) However, we know that Cartesian and spherical coordinates are related in such a way that (see Problem 8) e z cos( θ)e r sin( θ)e θ, and B( x) µ 0 m 4π r 2cos( θ)e 3 r + sin( θ)e θ. (6.4) 25

26 c) Using the equation of the Lorentz force for a charged particle subjected to a magnetic induction, we can write M dv dt qv B. (6.5) Breaking up the velocity of the particle into two components, one parallel and another perpendicular to B (i.e., v v e + v e ), then equation (6.5) transforms to d 2 r dt 0 2 d 2 r qb dt 2 M v, (6.6) with B µ 0 m + 3cos 2 ( θ) 4π r 2 3. But if the particle has an initial motion restricted to the ( x, y)-plane (i.e., z 0 v z ( 0) 0, and θ π 2 ), then B µ 0 m e 4π ( x 2 + y 2 ) 3 z, (6.7) 2 and e e z, and e is in the ( x, y)-plane. The solution to the first of equations (6.6) is easily shown to be z z 0 + v z ( 0)t z 0 0. (6.8) The particle s motion will therefore remain in the ( x, y)-plane. 7. Magnetostatic problems can be treated in a way similar to electrostatic ones. Show that for the calculation of the fields, a material of magnetization M( x) can be replaced by a volume polarization charge density ρ M M and a surface polarization charge densityσ M n M. Solution. Consider Ampère s law in cases when a magnetization M is present B µ 0 [ J + M], (7.) and let us define an equivalent magnetization M J that is due to the current J. That is, we have 26

27 J M J. (7.2) Furthermore, we introduce the total effective magnetization M T M J + M. Limiting ourselves to cases where µ 0 is constant, we can transform equation (7.) to B M T µ 0 0, (7.3) or, equivalently, if we define the magnetic field H as H µ 0 B M T, (7.4) then we have H 0. (7.5) Turning our attention to the equation for the divergence of the magnetic induction, we can write B µ 0 ( H + M T ) 0, (7.6) or alternatively H M T. (7.7) If we further define a volume polarization charge density ρ M M T, equation (7.7) becomes H ρ M. (7.8) Equations (7.5) and (7.8) are similar in form to the equations of electrostatics E 0 E ρ T ε 0, (7.9) where ρ T ρ P is the total effective charge that takes into account the polarization of the medium. So, in analogy to electrostatics we can postulate that for magnetostatics H Φ M, (7.0) 27

28 with Φ M the magnetic scalar potential. Finally, again by analogy with electrostatics, if we assume that our medium of effective magnetization M T has an outer bound beyond which the magnetization suddenly falls to zero, then, considering an infinitesimally small pillbox straddling the boundary, we obtain the following boundary condition (see equation (.65)) n H σ M, (7.) where σ M is the effective magnetic surface-charge density, and n the outwardly directed normal. But since the total effective magnetic charge q M inside the pillbox is given by and q M ρ M d 3 x σ M da V, (7.2) S d 3 x σ M da ρ M V S V M T d 3 x M T nda, S (7.3) then σ M n M T. (7.4) Although our treatment was limited to media of constant permeability, it can also be extended to ferromagnetic material, since in this case ρ J When making a change from a system of Cartesian coordinates x, y, z system α, β,γ we find that the volume integral is expressed as to another dxdydz J dαdβdγ, (8.) where J is the determinant of the Jacobian matrix that expresses the transformation of infinitesimal changes between the two systems of coordinates. More precisely, we have dx dy dz J dα dβ. (8.2) dγ 28

29 We assume that the α,β,γ with the usual definitions system is actually a spherical coordinate system ( r,θ,ϕ), x r sin( θ)cos ϕ y r sin( θ)sin( ϕ) z r cos( θ). (8.3) a) Write down equations for dx, dy, and dz as functions of r, θ, ϕ, dr, dθ, and dϕ, and using equations (8.2) and (8.) show that dxdydz r 2 sin( θ)drdθdϕ. (8.4) and ( e r,e θ,e ϕ ). b) Consider the two corresponding sets of unit basis vectors e x,e y,e z Express the vectors of the Cartesian basis as functions of the spherical basis, and write down the transformation matrix equation that links them as ( e x,e y,e z ) e r,e θ,e ( ϕ ) T. (8.5) Give an expression for T. c) Just as the infinitesimal vector dr can be written as dx dy dr e x,e y,e z, (8.6) dz or dr dx e x + dye y + dz e z, use the matrices J and T of a) and b) to transform equation (8.6) to express dr as M dr e r,e θ,e ϕ dr dθ, (8.7) dϕ or dr ae r + be θ + ce ϕ. Give expressions for M, a, b, and c. d) An equivalent way of expressing the infinitesimal Cartesian volume element of equation (8.) is to use the components of the vector dr dx e x + dye y + dz e z to 29

30 calculate dx dydz dz e z dx e x dye y. Similarly, use dr ae r + b e θ + ce ϕ to verify that the infinitesimal spherical volume element equals r 2 sin( θ)dr dθ dϕ. e) Finally, invert equation (8.5) to obtain ( e r,e θ,e ϕ ) e x,e y,e ( z ) T, (8.8) then starting with r r e r verify that dr ae r + be θ + ce ϕ. Solution. a) From equation (8.3) we calculate dx dr sin( θ)cos( ϕ) + r cos( θ)cos( ϕ)dθ r sin( θ)sin( ϕ)dϕ dy dr sin( θ)sin( ϕ) + r cos( θ)sin( ϕ)dθ + r sin( θ)cos( ϕ)dϕ dz dr cos( θ) r sin( θ)dθ. (8.9) We then have dx sin( θ)cos ϕ dy sin( θ)sin ϕ dz cos θ r cos( θ)cos( ϕ) r sin θ r cos( θ)sin( ϕ) r sin θ r sin( θ) 0 sin( ϕ) dr cos( ϕ) dθ, (8.0) dϕ where the matrix in bracket is J, the Jacobian matrix. Using the third row as the anchor, we can calculate its determinant as J cos θ r sin θ { r 2 sin( θ)cos( θ) cos 2 ( ϕ) + sin 2 ( ϕ) } r sin 2 ( θ) sin 2 ( ϕ) + cos 2 ( ϕ) r 2 sin( θ). { } Using equation (8.), we find that dxdydz r 2 sin( θ)drdθdϕ b) We can also write for the basis vectors. (8.) e x sin( θ)cos( ϕ)e r + cos( θ)cos( ϕ)e θ sin( ϕ)e ϕ e y sin( θ)sin( ϕ)e r + cos( θ)sin( ϕ)e θ + cos( ϕ)e ϕ e z cos( θ)e r sin( θ)e θ. (8.2) 30

31 We then have ( e x,e y,e z ) ( e r,e θ,e ϕ ) sin θ cos θ cos ϕ cos ϕ sin ϕ sin( θ)sin( ϕ) cos( θ) cos( θ)sin( ϕ) sin( θ) cos( ϕ) 0. (8.3) c) Inserting equations (8.2) and (8.5) into equation (8.6) we get where T dr e r,e θ,e ϕ e r,e θ,e ϕ M J dr dθ, dϕ dr dθ dϕ (8.4) [ M ] [ T ][ J ] r 0. (8.5) 0 0 r sin( θ) Thus we have r 0 dr e r,e θ,e ϕ dr dθ. (8.6) 0 0 r sin( θ) dϕ Using equation (8.6), we can write and we have a dr, b r dθ, and c r sin( θ)dϕ. dr dr e r + r dθ e θ + r sin( θ)dϕ e ϕ, (8.7) d) With equation (8.7) we can calculate the infinitesimal spherical volume element with 3

32 dr e r r 2 sin θ dr e r r dθ e θ r sin( θ)dϕ e ϕ dθ dϕ e r r 2 sin( θ)dr dθ dϕ. (8.8) e) It can easily be verified that the matrix T is orthonormal, so T T T (T T is the transpose of T ), and We can now write ( e r,e θ,e ϕ ) ( e x,e y,e z ) sin θ sin θ cos ϕ sin ϕ cos θ cos( θ)cos( ϕ) sin( ϕ) cos( θ)sin( ϕ) cos( ϕ) sin( θ) 0. (8.9) r e r r sin( θ)cos( ϕ)e x + sin( θ)sin( ϕ)e y + cos( θ)e z, (8.20) from which we calculate the differential dr r r r dr + dθ + r θ ϕ dϕ e r + r e r r dr + r e r θ dθ + r e r ϕ dϕ dr e r +r dθ cos( θ)cos( ϕ)e x + cos( θ)sin( ϕ)e y sin( θ)e z +r dϕ ( sin( θ)sin( ϕ)e x + sin( θ)cos( ϕ)e y ) dr e r + r dθ e θ + r sin( θ)dϕ e ϕ, (8.2) where we used equation (8.9) for the last step. We, therefore, get the same result than was obtained in a different manner in d) (cf., equation (8.7)) 9. Consider the gradient vector operator in Cartesian coordinates e x x + e y y + e z z, (9.) where we used the following short hand notation for the partial derivatives x x, etc. We are interested in investigating the form that this operator takes when we make the change from a Cartesian to a spherical coordinate system. We use the usual definitions 32

33 x r sin( θ)cos ϕ y r sin( θ)sin( ϕ) z r cos( θ), (9.2) or, alternatively, r x 2 + y 2 + z 2 tan( θ) x2 + y 2 z (9.3) tan( ϕ) y x. a) Using the chain rule we can write x r x r + θ x θ + ϕ x ϕ. (9.4) Write similar equations for y and z, and then write the matrix equation relating the components of the gradient operator from the two coordinate systems, that is x y z S r θ ϕ. (9.5) (You have to express the matrix S using the different partial derivatives as its components, but do not evaluate the derivatives yet.) b) Using, when needed, the following equation for the derivative of a tangent function or, alternatively, use equations (9.3) to show that d du tan α ( u) cos 2 α dα du cos2 α d du tan α ( u) dα du, (9.6), (9.7) 33

34 sin( θ)cos( ϕ) S sin( θ)sin( ϕ) r cos θ cos ϕ r cos θ sin ϕ r sin( θ) sin( ϕ) r sin( θ) cos( ϕ) cos( θ) r sin( θ) 0 (Here is an example as how to use equation (9.7). (9.8) ϕ y ( cos2 ϕ) y tan( ϕ) y ( cos2 ϕ) y x ( cos2 ϕ) x cos 2 ϕ r sin( θ)cos ϕ cos( ϕ) r sin( θ), (9.9) where the first of equations (9.2) was used for the second line.) c) Just as equation (9.) can be written as x y e x,e y,e z z, (9.0) use equation (9.5) and (9.8), along with the following result relating the two unit bases obtained in part b) of Problem 8 ( e r,e θ,e ϕ ) sin( θ)cos ϕ cos( θ)cos ϕ sin ϕ ( e x,e y,e z ) e r,e θ,e ( ϕ ) T sin( θ)sin( ϕ) cos( θ) cos( θ)sin( ϕ) sin( θ) cos( ϕ) 0 (9.) to prove that e r r + e θ r θ + e ϕ r sin θ ϕ. (9.2) d) Given a vector A A x e x + A y e y + A z e z A r e r + A θ e θ + A ϕ e ϕ, (9.3) 34

35 we would like to find an expression for its divergence. To do so start with A e r r + e θ r θ + e ϕ r sin θ ϕ ( A re r + A θ e θ + A ϕ e ϕ ), (9.4) and use the inverse of equation (9.) (which was calculated in part e) of Problem 8) to prove that Solution. A r A r + 2 A r r + r A + cos θ θ θ r sin θ r 2 r ( r 2 A r ) + a) Using the chain rule we have r sin θ A + θ r sin( θ) A ϕ ϕ + θ A θ sin( θ) x r x r + θ x θ + ϕ x ϕ y r y r + θ y θ + ϕ y ϕ z r z r + θ z θ + ϕ z ϕ. r sin( θ) A. ϕ ϕ (9.5) (9.6) The system of equations (9.6) can be written in a matrix form as x y z S r θ ϕ, (9.7) with r x r S y r z b) From equations (9.3) we calculate θ x θ y θ z ϕ x ϕ y. (9.8) ϕ z 35

36 r x r y r z x x 2 + y 2 + z 2 y x 2 + y 2 + z 2 z x 2 + y 2 + z 2 sin( θ)cos( ϕ) sin( θ)sin( ϕ) cos( θ), (9.9) and by further using equation (9.7) θ x cos2 θ θ y cos2 θ x z x 2 + y 2 r cos( θ)cos ϕ y z x 2 + y 2 r cos( θ)sin ϕ θ z x ( cos2 θ) 2 + y 2 z 2 ϕ x ( cos2 ϕ) y x 2 r sin θ ϕ y cos2 ϕ ϕ z 0. r sin( θ) sin( ϕ) x r sin( θ) cos( ϕ) (9.20) Inserting equations (9.9) and (9.20) into equation (9.8) we get c) We know that sin( θ)cos( ϕ) S sin( θ)sin( ϕ) r cos θ cos ϕ r cos θ sin ϕ r sin( θ) sin( ϕ) r sin( θ) cos( ϕ) cos( θ) r sin( θ) 0 x y e x,e y,e z z. (9.2), (9.22) and upon using equations (9.) and (9.7) we can express the gradient operator as 36

37 N e r,e θ,e ϕ r θ ϕ, (9.23) with N TS. (9.24) Inserting equations (9.) and (9.2) for the matricest and S, respectively, into equation (9.24) we get and 0 0 N 0 0 r 0 0 r sin θ e r r + e θ r θ + e ϕ c) Starting with part e) of Problem 8 we know that ( e r,e θ,e ϕ ) ( e x,e y,e z ) sin θ sin θ cos ϕ sin ϕ cos θ, (9.25) r sin θ ϕ. (9.26) cos( θ)cos( ϕ) sin( ϕ) cos( θ)sin( ϕ) cos( ϕ) sin( θ) 0 from which we can easily calculate the following derivatives, (9.27) r e r r e θ r e ϕ 0 θ e r cos( θ)cos( ϕ)e x + cos( θ)sin( ϕ)e y sin( θ)e z e θ θ e θ sin( θ)cos( ϕ)e x sin( θ)sin( ϕ)e y cos( θ)e z e r θ e ϕ 0 ϕ e r sin( θ)sin( ϕ)e x + sin( θ)cos( ϕ)e y sin( θ)e ϕ ϕ e θ cos( θ)sin( ϕ)e x + cos( θ)cos( ϕ)e y cos( θ)e ϕ ϕ e ϕ cos( ϕ)e x sin( ϕ)e y sin( θ)e r cos( θ)e θ. (9.28) 37

38 Using equation (9.4) A e r r + e θ r θ + e ϕ r sin θ ( r A r ) + ( r A r + θ A θ ) + ϕ r sin θ i A re r + A θ e θ + A ϕ e ϕ, A r sin( θ) + A θ cos( θ) + ϕ A ϕ (9.29) and finally A r A r + 2 r A r + r θ A θ + r 2 r ( r 2 A r ) + r sin θ r sin( θ) A θ cos( θ) + + θ A θ sin( θ) r sin θ ϕ A ϕ r sin( θ) ϕ A ϕ. (9.30) 38

39 Table of spherical harmonics Y 00 4π Y 0 Y,± Y π cos θ 3 8π sin( θ)e±iϕ 5 6π 3cos2 θ Y 2,± 5 8π sin( θ)cos( θ)e±iϕ Y 2,±2 Y 30 Y 3,± 5 ( 32π sin2 θ)e ±i2ϕ 7 6π 5cos3 θ 2 64π sin θ 3cos( θ) ( 5cos2 θ) Y 3,±2 05 ( 32π sin2 θ)cos( θ)e ±i2ϕ Y 3,±3 35 ( 64π sin3 θ)e ±i 3ϕ e ±iϕ 39