Electromagnetism HW 1 math review

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1 Electromagnetism HW math review Problems -5 due Mon 7th Sep, 6- due Mon 4th Sep Exercise. The Levi-Civita symbol, ɛ ijk, also known as the completely antisymmetric rank-3 tensor, has the following properties: (a) ɛ 3 = ɛ 3 = ɛ 3 = + (b) ɛ 3 = ɛ 3 = ɛ 3 = (c) all other entries are zero. It follows that ɛ jik = ɛ ijk and similar antisymmetric relations hold.. Using a notation in which repeated indices are summed over, show that (a) ɛ ijk ɛ ilm = δ jl δ km δ jm δ kl, (b) ɛ ijk ɛ ijm = δ km, and (c) ɛ ijk ɛ ijk = 6, where the Kronecker delta, δ ij, is equal to + if i = j and zero if i j.. Show that C i = ɛ ijk A j B k has the same components as the vector C = A B.3 Using properties of the Levi-Civita symbol, prove that (a) A B C = ( A C) B ( A B) C, (b) ( A B) ( C D) = ( A C)( B D) ( A D)( B C), (c) ( A B) ( C D) = ( A C D) B ( B C D) A, (d) ( g) =, (e) ( g) = ( g) g, (f) (f g) = f g + g f, (g) (f g) = f g g f, (h) ( f g) = g ( f) f ( g), (i) ( f g) = f( g) g( f) ( f ) g + ( g ) f, and (j) ( g r) = g + r r g r( g), where r = xˆx + yŷ + zẑ

2 Exercise. The determinant of the Jacobian matrix, J, relates volume elements when changing variables in an integral. For example if we change from variables x = (x, x... x N ) to variables y = (y, y... y N ), the volume elements are related by d N x = J( x, y) d N y = x x y x x y x N y y... y.... x N y... x y N x y N x N y N d N y. Starting from the volume element in cartesian coordinates, dx dy dz, use the Jacobian to show that the volume element is (a) ρ dρ dφ dz in cylindrical coordinates, and (b) r sin θ dr dθ dφ in spherical coordinates. Exercise 3. Starting from the divergence theorem, d 3 r F = ds F, show that (a) V d3 r φ = S d S φ for a scalar field φ( r) [Hint: choose F = c φ( r) with a constant vector c ], (b) V d3 r A = S d S A for a vector field A( r) [Hint: choose F = A( r) c with a constant vector c ], (c) V d3 r ( φ ψ ψ φ ) = S d S (φ ψ ψ φ ), V (d) for a closed surface, S, enclosing a volume, V, S d S =, and S d S r = 3V. S

3 Exercise 4. Show that a vector field with zero curl can be expressed as the gradient of a scalar field, F ( r) = Φ( r). Show that the line integral, r f r i d l F, of a curl-less vector field, F, between two points r = r i and r = r f is independent of the path taken between the two points. For the particular field show that (a) F =, F = yˆx + xŷ x + y, (b) the line integral between (x =, y = ) and (x =, y = ) has value π for each of the paths, A, B, C, shown in the figure. y C (,) B A (-,) (,) x (c) Find a scalar field, Φ(x, y), such that F = Φ. [Hint: integrate F d l from (, ) to (x, y) choosing a path that makes the integrals as simple as possible ] 3

4 Exercise 5. Evaluate d S F where S F = (y x )ˆx + (xy y)ŷ + 3zẑ for the entire surface of the tin can bounded by the cylinder x + y = 6, z = 3, z = 3, (a) by explicitly computing the surface integral, and (b) by using the divergence theorem. Exercise 6. Consider the following vector field, expressed in cylindrical coordinates, W (ρ, φ, z) = α ρ ˆφ. Show that the z-component of the curl is zero everywhere except at the origin and ( W ) z = α ρ δ(ρ) Exercise 7. In lectures we showed, using the divergence theorem, that r = 4πδ( r). Let s explore another way of deriving the same result: (a) Show that the function Λ(r, a) = r +a is Λ(r, a) = 3a (r +a ) 5/, and plot Λ(r, a) versus r for a range of values of a decreasing toward zero. (b) Show that the integral over all space of Λ(r, a) is of value 4π. (c) Explain how we can conclude that lim a Λ(r, a) = 4πδ( r) Exercise 8. The Helmholtz theorem for a divergence-less and curl-less vector field: Suppose a vector field satisfies E = and E = everywhere in a volume, V, bounded by a surface, S. Use a derivation like the one we used in lecture to show that E( r) can be found everywhere in V if its value is known at all points on the surface, S. Notice that your result shows that if E is zero everywhere on the boundary, then E is also zero everywhere inside the volume this will be realized when we consider an empty cavity in a perfect conductor. 4

5 Exercise 9. The Legendre differential equation is ( x ) d y dx xdy + l(l + )y =. dx One set of solutions to this equation, when l takes positive integer values, are known as the Legendre polynomials, y(x) = P l (x). They are polynomials of order l that are normalized such that P l () =. The first few are P (x) =, P (x) = x, P (x) = (3x ), P 3 (x) = (5x3 3x), Sketch graphs of P (x), P (x), P (x), P 3 (x) in the range x. 9. We can prove that the solutions to Legendre s equation also satisfy Rodrigues formula, P l (x) = l l! d l dx l (x ) l. (a) Check that you can obtain P (x)... P 3 (x) using Rodrigues formula. (b) Using Rodrigues formula show that + dx xm P l (x) = if m < l. 9.3 The generating function for Legendre polynomials is Φ(x, h) = = h l P l (x). xh + h (a) Show that the functions P l (x) in the sum here do indeed satisfy Legendre s equation and have the property P l () =. [Hint: consider Φ, Φ and h (hφ) ] x x h (b) Prove the identity xp l (x) P l (x) = lp l(x). [Hint: consider Φ x l= and Φ h ] 9.4 The Legendre polynomials form a complete orthogonal basis with the property that + dxp l (x)p m (x) = δ ml N l (a) Show that N l = in 9.(b). l+ using the identity derived in 9.3(b) and the result obtained (b) Prove that the P l (x) are complete on the interval x by showing that l+ P l(x)p l (x ) is a representation of the Dirac delta function, D(x, x ) = δ(x x ). D(x, x ) = l= [ Hint: Start from an arbitrary function f(x) expanded as a infinite superposition of P l (x) and then show that + dx D(x, x )f(x) = f(x ) ] 5

6 Exercise. The Bessel differential equation is x d y dx + xdy dx + (x p )y =. If we restrict ourselves to x and integer values of p, the linearly independent solutions to this equation are known as Bessel functions, J p (x), and Neumann functions, N p (x). The first few such functions are plotted below. Note that the Neumann functions diverge at the origin.. Bessel functions are orthogonal, but in a way that might look strange: dx x J p (a n x) J p (a m x) = δ n,m N n In this equation, a n and a m are positions of zeros of the Bessel function, i.e. J p (a n ) =. Prove the orthogonality expression by taking the following steps: (a) Show that J p (ax) satisfies the equation x d ( x d ) dx dx J p(ax) + (a x p )J p (ax) =. 6

7 (b) Show that J p (bx) d dx ( x d dx J p(ax) ) J p (ax) d ( x d ) dx dx J p(bx) + (a b ) x J p (ax) J p (bx) =, and integrate this expression in the case that a and b are chosen to be two different zeroes of the Bessel function J p (x) to obtain the orthogonality relation. (c) Finally, show that N n = ( J p (a n ) ) [ Hint: you might want to consider the proof in (b) but only assuming that b is a zero of J p and not a, and be careful taking the limit a b ].. Bessel functions get used when we solve problems with cylindrical symmetry. Consider the following functions of cylindrical coordinates, ρ, z (and independent of φ), f (ρ, z) = ρ + z, f (ρ, z) = dk A(k)J (kρ)e k z. (a) Show that these functions both satisfy Laplace s equation f =. (b) Assuming that f = f, by considering ρ = show that A(k) = and thus that, ρ + z = dk J (kρ)e k z, which is known as a Fourier-Bessel representation of ρ +z. Exercise. Functions can often usefully be expanded as superpositions of orthogonal basis functions, for example as Fourier series and transforms.. Show that the function f(x) = x on the interval π < x < π representation π 4 π odd n cos nx n has a Fourier series. Show that the function f(x) = {, < x <, x > has a Fourier representation f(x) = π dk sin k k cos kx. Using the result that sin y dy = π, explicitly compute the above integral in the cases y x < and x > and check that you get and respectively. 7