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• Ravi Roy

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Physics 506 : Problem set 5 Due : October 13, 2014

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Problem 1

Let us consider the circular cylindrical coordinate system (ρ, ϕ, z) with the corresponding unit vector (ˆ eρ , eˆϕ , eˆz ): a) Compute divergence and curl of the unit vector in the direction of increasing radial coordinate eˆρ . Sol: It is given by ∂(hϕ hz ) 1 ∂ρ 1 1 = = , hρ hϕ hz ∂ρ ρ ∂ρ ρ     1 1 ∂hρ 1 ∂hρ 1 ∂1 1 ∂1 ∇ × eˆρ = eˆϕ − eˆz = eˆϕ − eˆz =0. hρ hz ∂z hϕ ∂ϕ hρ ∂z ρ ∂ϕ

∇ · eˆρ =

b) Rewrite the circular cylindrical unit vectors into their Cartesian components. Sol: They are given by eˆρ = cos ϕ x ˆ + sin ϕ yˆ ,

eˆϕ = − sin ϕ x ˆ + cos ϕ yˆ ,

eˆz = zˆ .

c) Using b) compute all the first derivatives of the circular cylindrical unit vectors with respect to the circular cylindrical coordinates. Sol: The non-vanishing ones are ∂ˆ eρ = eˆϕ , ∂ϕ

∂ˆ eϕ = −ˆ eρ , ∂ϕ

and all the others vanish. d) Write a radial vector ~r from the origin to a general point in space in terms of the coordinates and their unit vectors. And compute ∇ · ~r and ∇ × ~r. Sol: It is ~r = eˆρ ρ + eˆz z. And     ∂(rρ hϕ hz ) ∂(rϕ hρ hz ) ∂(rz hϕ hρ ) ∂z 1 1 ∂ρ2 + + +ρ ∇ · ~r = = =3, hρ hϕ hz ∂ρ ∂ϕ ∂z ρ ∂ρ ∂z     hρ eˆρ hϕ eˆϕ hz eˆz eˆρ ρˆ eϕ eˆz 1 1  ∂ρ ∂ϕ ∂z  =  ∂ρ ∂ϕ ∂z  = 0 . ∇ · ~r = hρ hϕ hz ρ hρ rρ hϕ rϕ hz rz ρ 0 z e) Find the circular cylindrical components of the velocity and acceleration of a moving particle, by using the radial vector ~r(t) with time dependence on all the coordinate ρ(t), ϕ(t), z(t). Sol: Using ~r(t) = eˆρ (t)ρ(t) + eˆz z(t) = (cos ϕ(t) x ˆ + sin ϕ(t) yˆ)ρ(t) + zˆz(t), we can show ~v (t) = eˆρ (t)ρ(t) ˙ + eˆϕ (t)ρ(t)ϕ(t) ˙ + eˆz z(t) ˙ ,
2 ~a(t) = eˆρ (t) ρ¨(t) − ρ(t)(ϕ(t)) ˙ + eˆϕ (t) (ρ(t)ϕ(t) ¨ + 2ρ(t) ˙ ϕ(t)) ˙ + eˆz z¨(t) . f) Parity is the operation that reflects the coordinate through the origin in 3-dimensions (not true for other dimensions). In Cartesian coordinate, this operation gives P : (x, y, z) → (−x, −y, −z). Write the parity operation for the circular cylindrical components. If some physical quantities are invariant under the parity operation, it is called even, otherwise called odd. What are the properties of the circular cylindrical basis vectors under the parity? (The Cartesian unit vectors remain constant under the parity. ) Sol: Under the parity, we have P : (ρ, ϕ, z) → (ρ, ϕ ± π, −z). The unit vectors eˆρ , eˆϕ are parity odd, while the unit vector eˆz is even.

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Problem 2

Let us consider the quantum mechanical orbital angular momentum operator ~ = −i ~r × ∇ ~ . L ~ in terms of spherical polar coordinate (r, θ, ϕ) and the corresponding unit vector (ˆ a) Write L er , eˆθ , eˆϕ ). Sol:   ~ = −i ~r × ∇ = i eˆθ 1 ∂ − eˆϕ ∂ . L sin θ ∂ϕ ∂θ b) Determine the explicit form Lx , Ly , Lz , Lx +iLy , Lx −iLy in spherical polar coordinates by resolving the unit vectors in terms of the unit vectors in Cartesian coordinates. Sol: Using the decomposition of the unit vectors in Cartesian coordinates, one can show     ∂ ∂ ∂ ∂ ∂ + sin ϕ , Ly = i cot θ sin ϕ − cos ϕ , Lz = −i , Lx = i cot θ cos ϕ ∂ϕ ∂θ ∂ϕ ∂θ ∂ϕ   ∂ ∂ ∓i . Lx ± Ly = ie±iϕ cot θ ∂ϕ ∂θ c) Determine the L2 . Sol: ~ 2 = −~r2 ∇2 + ∂ L ∂r



∂ r ∂r 2



1 ∂ =− sin θ ∂θ



∂ sin θ ∂θ

 −

1 ∂2 . sin θ2 ∂ϕ2

~ L ~ = iL ~ in spherical polar coordinate. This is equivalent to verify the familiar commutation d) Verify L× relation [Li , Lj ] = iijk Lk , where i, j, k = x, y, z. Sol:     1 ∂ ∂ ~ ×L ~ = − eˆθ 1 ∂ − eˆϕ ∂ × eˆθ 1 ∂ − eˆϕ ∂ = −ˆ ~ . L eθ + eˆϕ = iL sin θ ∂ϕ ∂θ sin θ ∂ϕ ∂θ sin θ ∂ϕ ∂θ where we use the fact that the unit vector is not constant. For example, we have     1 ∂ ∂ 1 ∂ 1 ∂ 1 ∂ ∂ 1 ∂ ∂ 1 − eˆθ × eˆθ × ( eˆθ ) + eˆθ ( ) + eˆθ = −ˆ eθ sin θ ∂ϕ sin θ ∂ϕ sin θ ∂ϕ sin θ ∂ϕ ∂ϕ sin θ ∂ϕ sin θ ∂ϕ ∂ϕ   1 ∂ 1 ∂ ∂ 1 cos θ ∂ × cos θˆ eϕ + 0 + eˆθ , = −ˆ eθ = −ˆ er 2 sin θ sin θ ∂ϕ sin θ ∂ϕ ∂ϕ sin θ ∂ϕ where we use eˆθ × eˆϕ =  eˆr and eˆθ × eˆθ = 0. Now this term actually cancels another term comes ∂ 1 ∂ ∂ ∂ cos θ ∂ from eˆϕ ∂θ × eˆθ sin θ ∂ϕ = · · · + eˆϕ × eˆθ ∂θ ( sin1 θ ) ∂ϕ = · · · + eˆr sin 2 θ ∂ϕ . All the other terms can be computed accordingly. e) Verify the operator identities ~ ~ = eˆr ∂ − i ~r × L , ∇ ∂r r2

  ∂ ~ ~ . ~r∇ − ∇ 1 + r = i∇ × L ∂r 2

Sol: The first identity comes from the following manipulation ~ = −i~r × (~r × ∇) = −iijk klm rj rl ∇m = −i~r(~r × ∇) + ir2 ∇ ~ = −iˆ ~ , ~r × L er r2 ∂r + ir2 ∇ while the second identity comes from

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Problem 3

The u−, v−, z− coordinate system frequently used in electrostatics and hydrodynamics is orthogonal and is defined by xy = u , x2 − y 2 = v , z = z . a) Describe the nature of each of the three families of coordinate surfaces, and draw the system in the xy-plane showing the intersection of surfaces of constant u and constant v with the xy-plane. Use the Mathematica to draw the picture. b)Indicate the directions of the unit vectors eˆu and eˆv in all four quadrants. c) Is the u−, v−, z− coordinate system right-handed eˆu × eˆv = +ˆ ez or left-handed eˆu × eˆv = −ˆ ez ? Sol: The following picture represents all four quadrants of the xy plane. The solid lines, including x = 0, y = 0 coordinate lines for u = 0, are the lines for u = const. The dashed lines are the constant v lines. The straight dashed lines are for v = x2 − y 2 = 0. For all the quadrants, constant u solid lines are for the increasing v along with the flow of the arrow and thus the direction of eˆv , while the constant v dashed lines are for increasing u along the flow of the arrow and thus the direction of eˆu . This tells us that in the first quadrant x > 0, y > 0, the (u, v, z) coordinate is left-handed. And all the other quadrants have the same handedness.

y

x

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Problem 4

Construct a Mathematica package file with extension ”InvMat.m” that takes a n × n matrix g and return its inverse. Demonstrate this with the following matrix.   a b c g= d e f  . g h i Sol: Mathematical file will be posted in separate file.