Mathematical Method For Physicists Ch. 1 & 2 Selected solutions Webber and Arfken

Physics 451

Fall 2004 Homework Assignment #3 — Solutions

Textbook problems: Ch. 1: 1.7.1, 1.8.11, 1.8.16, 1.9.12, 1.10.4, 1.12.9 Ch. 2: 2.4.8, 2.4.11 Chapter 1 1.7.1 For a particle moving in a circular orbit ~r = x ˆ r cos ωt + yˆ r sin ωt (a) evaluate ~r × ~r˙ Taking a time derivative of ~r, we obtain ~r˙ = −ˆ x rω sin ωt + yˆ rω cos ωt

(1)

Hence ~r × ~r˙ = (ˆ x r cos ωt + yˆ r sin ωt) × (−ˆ x rω sin ωt + yˆ rω cos ωt) = (ˆ x × yˆ)r2 ω cos2 ωt − (ˆ y×x ˆ)r2 ω sin2 ωt = zˆ r2 ω(sin2 ωt + cos2 ωt) = zˆ r2 ω (b) Show that ~¨r + ω 2~r = 0 The acceleration is the time derivative of (1) ~¨r = −ˆ x rω 2 cos ωt − yˆ rω 2 sin ωt = −ω 2 (ˆ x r cos ωt + yˆ r sin ωt) = −ω 2~r Hence ~¨r + ω 2~r = 0. This is of course the standard kinematics of uniform circular motion. 1.8.11 Verify the vector identity ~ × (A ~ × B) ~ = (B ~ · ∇) ~ A ~ − (A ~ · ∇) ~ B ~ − B( ~ ∇ ~ · A) ~ + A( ~ ∇ ~ · B) ~ ∇ This looks like a good time for the BAC–CAB rule. However, we have to be ~ has both derivative and vector properties. As a derivative, it careful since ∇ ~ and B. ~ Therefore, by the product rule of differentiation, we operates on both A can write ↓ ↓ ~ ~ ~ ~ ~ ~ ~ ~ ~ ∇ × (A × B) = ∇ × (A × B) + ∇ × (A × B)

where the arrows indicate where the derivative is acting. Now that we have ~ as a vector. Using the specified exactly where the derivative goes, we can treat ∇ BAC–CAB rule (once for each term) gives ↓ ↓ ↓ ↓ ~ ~ ~ ~ ~ · A) ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ∇ × (A × B) = A(∇ · B) − B(∇ · A) + A(∇ · B) − B(∇

(2)

The first and last terms on the right hand side are ‘backwards’. However, we can turn them around. For example ↓ ↓ ↓ ~ ~ ~ ~ ~ ~ ~ ~ A(∇ · B) = A(B · ∇) = (B · ∇)A With all the arrows in the right place [after flipping the first and last terms in (2)], we find simply ~ × (A ~ × B) ~ = (B ~ · ∇) ~ A ~ − B( ~ ∇ ~ · A) ~ + A( ~ ∇ ~ · B) ~ − (A ~ · ∇) ~ B ~ ∇ which is what we set out to prove. 1.8.16 An electric dipole of moment p~ is located at the origin. The dipole creates an electric potential at ~r given by p~ · ~r ψ(~r ) = 4πo r3 ~ = −∇ψ ~ at ~r. Find the electric field, E We first use the quotient rule to write ~ ~ = −∇ψ ~ =− 1 ∇ E 4π0



p~ · ~r r3

 =−

~ p · ~r ) − (~ ~ 3) 1 r3 ∇(~ p · ~r )∇(r 4π0 r6

Applying the chain rule to the second term in the numerator, we obtain 3~ ~ p · ~r ) − 3r2 (~ p · ~r )∇(r) ~ = − 1 r ∇(~ E 4π0 r6

We now evaluate the two separate gradients ∂ ∂xj ~ p · ~r ) = x ∇(~ ˆi (pj xj ) = ~xi pj =x ˆi pj δij = x ˆi pi = p~ ∂xi ∂xi and ∂ ~ =x ∇r ˆi ∂xi

1 x ˆi xi ~r x21 + x22 + x23 = x ˆi p 2 2x = = = rˆ i r r 2 x1 + x22 + x23

q

Hence

3 2 1 p~ − 3(~ p · ~r )ˆ r p · rˆ)ˆ r ~ = − 1 r p~ − 3r (~ =− E 6 3 4π0 r 4π0 r

Note that we have used the fact that p~ is a constant, although this was never stated in the problem. 1.9.12 Show that any solution of the equation ~ × (∇ ~ × A) ~ − k2 A ~=0 ∇ automatically satisfies the vector Helmholtz equation ~ + k2 A ~=0 ∇2 A and the solenoidal condition ~ ·A ~=0 ∇ We actually follow the hint and demonstrate the solenoidal condition first. Taking the divergence of the first equation, we find ~ ·∇ ~ × (∇ ~ × A) ~ − k2 ∇ ~ ·A ~=0 ∇ However, the divergence of a curl vanishes identically. Hence the first term is ~ ·A ~ = 0 or (upon dividing automatically equal to zero, and we are left with k 2 ∇ ~ ·A ~ = 0. by the constant k) ∇ We now return to the first equation and simplify the double curl using the BAC– ~ CAB rule (taking into account the fact that all derivatives must act on A) ~ × (∇ ~ × A) = ∇( ~ ∇ ~ · A) ~ − ∇2 A ~ ∇

(3)

As a result, the first equation becomes ~ ∇ ~ · A) ~ − ∇2 A ~ − k2 A ~=0 ∇( ~ ·A ~ = 0 for this problem. Thus (3) reduces However, we have shown above that ∇ to ~ + k2 A ~=0 ∇2 A which is what we wanted to show.

1.10.4 Evaluate

H

~r · d~r

We have evaluated this integral in class. For a line integral from point 1 to point 2, we have Z 2 Z 2 2 1 d(r2 ) = 12 r2 1 = 12 r22 − 21 r12 ~r · d~r = 2 1

1

However for a closed path, point H 1 and point 2 are the same. Thus the integral along a closed loop vanishes, ~r · d~r = 0. Note that this vanishing of the line integral around a closed loop is the sign of a conservative force. Alternatively, we can apply Stokes’ theorem I Z ~ × ~r · d~σ ∇ ~r · d~r = S

It is easy to see that ~r is curl-free. Hence the surface integral on the right hand side vanishes. 1.12.9 Prove that

I

~ · d~λ = − u∇v

I

~ · d~λ v ∇u

This is an application of Stokes’ theorem. Let us write I Z ~ ~ × (u∇v ~ + v ∇u) ~ · d~σ ~ ~ ∇ (u∇v + v ∇u) · dλ =

(4)

S

We now expand the curl using ~ × (u∇v) ~ = (∇u) ~ × (∇v) ~ + u∇ ~ × ∇v ~ = (∇u) ~ × (∇v) ~ ∇ where we have also used the fact that the curl of a gradient vanishes. Returning to (4), this indicates that I

~ + v ∇u) ~ · d~λ = (u∇v

Z

~ × (∇v) ~ + (∇v) ~ × (∇u)] ~ [(∇u) · d~σ = 0

S

where the vanishing of the right hand side is guaranteed by the antisymmetry of ~×B ~ = −B ~ × A. ~ the cross-product, A

Chapter 2 2.4.8 Find the circular cylindrical components of the velocity and acceleration of a moving particle We first explore the time derivatives of the cylindrical coordinate basis vectors. Since ρˆ = (cos ϕ, sin ϕ, 0),

ϕˆ = (− sin ϕ, cos ϕ, 0),

zˆ = (0, 0, 1)

their derivatives are ∂ ρˆ = (− sin ϕ, cos ϕ, 0) = ϕ, ˆ ∂ϕ

∂ ϕˆ = (− cos ϕ, − sin ϕ, 0) = −ˆ ρ ∂ϕ

Using the chain rule, this indicates that ∂ ρˆ ϕ˙ = ϕˆϕ, ˙ ρˆ˙ = ∂ϕ

∂ ϕˆ ϕˆ˙ = ϕ˙ = −ˆ ρϕ˙ ∂ϕ

(5)

Now, we note that the position vector is given by ~r = ρˆρ + zˆz So all we have to do to find the velocity is to take a time derivative ~v = ~r˙ = ρˆρ˙ + zˆz˙ + ρˆ˙ ρ + zˆ˙ z = ρˆρ˙ + zˆz˙ + ϕρ ˆ ϕ˙ Note that we have used the expression for ρˆ˙ in (5). Taking one more time derivative yields the acceleration ~a = ~v˙ = ρˆρ¨ + zˆz¨ + ϕ(ρ ˆ ϕ¨ + ρ˙ ϕ) ˙ + ρˆ˙ ρ˙ + zˆ˙ z˙ + ϕρ ˆ˙ ϕ˙ = ρˆρ¨ + zˆz¨ + ϕ(ρ ˆ ϕ¨ + ρ˙ ϕ) ˙ + ϕˆρ˙ ϕ˙ − ρˆρϕ˙ 2 = ρˆ(¨ ρ − ρϕ˙ 2 ) + zˆz¨ + ϕ(ρ ˆ ϕ¨ + 2ρ˙ ϕ) ˙ 2.4.11 For the flow of an incompressible viscous fluid the Navier-Stokes equations lead to ~ × (~v × (∇ ~ × ~v )) = η ∇2 (∇ ~ × ~v ) −∇ ρ0 Here η is the viscosity and ρ0 the density of the fluid. For axial flow in a cylindrical pipe we take the velocity ~v to be ~v = zˆv(ρ)

From Example 2.4.1 ~ × (~v × (∇ ~ × ~v )) = 0 ∇ for this choice of ~v . Show that ~ × ~v ) = 0 ∇2 (∇ leads to the differential equation 1 d ρ dρ



d2 v ρ 2 dρ

 −

1 dv =0 ρ2 dρ

and that this is satisfied by v = v0 + a2 ρ2 This problem is an exercise in applying the vector differential operators in cylin~ =∇ ~ × ~v drical coordinates. Let us first compute V ρˆ 1 ~ =∇ ~ × ~v = ∂ V ρ ∂ρ 0

ρϕˆ ∂ ∂ϕ 0

zˆ ∂ dv = −ϕˆ ∂z dρ v(ρ)

⇒

Vϕ = −

dv dρ

Note that, since v(ρ) is a function of a single variable, partial derivatives of v are the same as ordinary derivatives. Next we need to compute the vector Laplacian ~ × ~v ) = ∇2 V ~ . Using (2.35) in the textbook, and the fact that on the Vϕ ∇2 (∇ component is non-vanishing, we find 2 ∂Vϕ =0 ρ2 ∂ϕ   1 dv 1 dv 2~ 2 2 (∇ V )ϕ = ∇ (Vϕ ) − 2 Vϕ = −∇ + 2 ρ dρ ρ dρ ~ )z = 0 (∇2 V ~ )ρ = − (∇2 V

This indicates that only the ϕ component of the vector Laplacian gives a nontrivial equation. Finally, we evaluate the scalar Laplacian ∇2 (dv/dρ) to obtain 1 d (∇ V )ϕ = − ρ dρ 2~



d2 v ρ 2 dρ

 +

1 dv ρ2 dρ

(6)

Setting this equal to zero gives the equation that we were asked to prove. To prove that v = v0 + a2 ρ2 satisfies the (third order!) differential equation, all we have to do is substitute it in. However, it is more fun to go ahead and solve

the equation. First we notice that v only enters through its derivative f = dv/dρ. Substituting this into (6), we find 1 d ρ dρ



df ρ dρ

 −

1 f =0 ρ2

Expanding the derivatives in the first term yields 1 d2 f 1 df + − 2f = 0 2 dρ ρ dρ ρ Since this is a homogeneous equation, we may substitute in f = ρα to obtain the algebraic equation α(α − 1) + α − 1 = 0

⇒

α = ±1

This indicates that the general solution for f (ρ) is of the form f = 2aρ + bρ−1 where the factor of 2 is chosen for later convenience. Integrating f once to obtain v, we find Z v=

f dρ = v0 + aρ2 + b log ρ

which agrees with the given solution, except for the log term. However, now we can appeal to physical boundary conditions for fluid flow in the cylindrical pipe. The point ρ = 0 corresponds to the central axis of the pipe. At this point, the fluid velocity should not be infinite. Hence we must throw away the log, or in other words we must set b = 0, so that v = v0 + aρ2 . Incidentally, the fluid flow boundary conditions should be that the velocity vanishes at the wall of the pipe. If we let R be the radius of the pipe, this means that we can write the solution as   ρ2 v(ρ) = vmax 1 − 2 R where the maximum velocity vmax is for the fluid along the central axis (with the velocity going to zero quadratically as a function of the radius).