Weber & Arfken Mathematical Methods For Physicists Ch. 5 selected solutions

Physics 451

Fall 2004 Homework Assignment #5 — Solutions

Textbook problems: Ch. 5: 5.1.1, 5.1.2 Chapter 5 5.1.1 Show that

∞ X

1 1 = (2n − 1)(2n + 1) 2 n=1 We take the hint and use mathematical induction. First, we assume that sm =

m 2m + 1

(1)

In this case, the next partial sum becomes m 1 + 2m + 1 (2(m + 1) − 1)(2(m + 1) + 1) m 1 m(2m + 3) + 1 = + = 2m + 1 (2m + 1)(2m + 3) (2m + 1)(2m + 3) 2 (m + 1)(2m + 1) 2m + 3m + 1 = = (2m + 1)(2m + 3) (2m + 1)(2m + 3) (m + 1) = 2(m + 1) + 1

sm+1 = sm + am+1 =

which is of the correct form (1). Finally, by explicit computation, we see that s1 = 1/(1 · 3) = 1/3 = 1/(2 · 1 + 1), so that (1) is correct for s1 . Therefore, by induction, we conclude that the mth partial sum is exactly sm = m/(2m + 1). It is now simple to take the limit to obtain S = lim sm = lim m→∞

m→∞

m 1 = 2m + 1 2

Note that we could also have evaluated this sum by partial fraction expansion  ∞  X 1 1 1 = − (2n − 1)(2n + 1) 2(2n − 1) 2(2n + 1) n=1 n=1 ∞ X

Since this is a telescoping series, we have sm =

1 1 m − = 2(2 · 1 − 1) 2(2m + 1) 2m + 1

which agrees with (1). 5.1.2 Show that

∞ X

1 =1 n(n + 1) n=1 This problem may be solved in a similar manner. While there is no hint of for the partial sum, we may try a few terms s1 =

1 , 2

s2 = s1 +

1 2 = , 2·3 3

This suggests that sm =

s3 = s2 +

1 3 = 3·4 4

m m+1

(2)

We now prove this statement by induction. Starting from sm , we compute 1 m(m + 2) + 1 m + = m + 1 (m + 1)(m + 2) (m + 1)(m + 2) (m + 1)2 m+1 (m + 1) = = = (m + 1)(m + 2) m+2 (m + 1) + 1

sm+1 = sm + am+1 =

Therefore if (2) holds for m, it also holds for m + 1. Finally, since (2) is correct for s1 = 1/2, it must be true for all m by induction. Taking the limit yields S = lim sm = lim m→∞

m→∞

m =1 m+1

The partial fraction approach to this problem is to note that ∞ X

 ∞  X 1 1 1 = − n(n + 1) n n+1 n=1 n=1 Hence sm =

1 m 1 − = 1 m+1 m+1

which reproduces (2). Additional Problems 1. The metric for a three-dimensional hyperbolic (non-Euclidean) space can be written as dx2 + dy 2 + dz 2 ds2 = L2 z2

where L is a constant with dimensions of length. Calculate the non-vanishing Christoffel coefficients for this metric. We first note that the metric is given in matrix form as 

L2 /z 2 gij =  0 0

0 L2 /z 2 0

 0 0  2 L /z 2

L2 , z2

gzz =

so the non-zero components are gxx =

L2 , z2

gyy =

L2 z2

(3)

The covariant components of the Christoffel connection are obtained from the metric by Γijk = 21 (gij,k + gik,j − gjk,i ) where the comma denotes partial differentiation. According to (3), the only nonzero metric components have repeated indices. In addition, only the z-derivative is non-vanishing. Hence we conclude that the only non-vanishing Christoffel symbols must have two repeated indices combined with a z index. Recalling that Γijk is symmetric in the last two indices, we compute L2 Γzxx = = 3, z 2 L Γzyy = − 12 gyy,z = 3 , z L2 Γzzz = 12 gzz,z = − 3 z − 21 gxx,z

Γxzx = Γxxz =

1 2 gxx,z

Γyzy = Γyyz = 12 gyy,z

L2 =− 3 z L2 =− 3 z

Raising the first index using the inverse metric g ij = (z 2 /L2 )δ ij finally yields 1 , z 1 = , z 1 =− z

1 z 1 =− z

Γz xx =

Γx zx = Γx xz = −

Γz yy

Γy zy = Γy yz

Γz zz

(4)

2. The motion of a free particle moving along a path xi (t) in hyperbolic space is governed by the geodesic equation x ¨i (t) + Γi jk x˙ j (t)x˙ k (t) = 0

Taking (x1 , x2 , x3 ) to be (x, y, z), and using the Christoffel coefficients calculated above, show that the geodesic equation is given explicitly by 2 x ¨ − x˙ z˙ = 0 z 2 y¨ − y˙ z˙ = 0 z 1 2 z¨ + (x˙ + y˙ 2 − z˙ 2 ) = 0 z Using the Christoffel coefficients in (4), we compute the three components of the geodesic equation x ¨ + Γx xz x˙ z˙ + Γx zx z˙ x˙ = 0

⇒

2 x ¨ − x˙ z˙ = 0 z

(5)

2 (6) y¨ − y˙ z˙ = 0 z 1 z¨ + Γz xx x˙ x˙ + Γz yy y˙ y˙ + Γz zz z˙ z˙ = 0 ⇒ z¨ + (x˙ 2 + y˙ 2 − z˙ 2 ) = 0 (7) z The geodesic equation is important because it describes the motion of free particles in curved space. However, for this problem, all that is necessary is to show that it gives a system of coupled ordinary differential equations (5), (6), (7). y¨ + Γy yz y˙ z˙ + Γy zy z˙ y˙ = 0

⇒

3. Show that a solution to the geodesic equation of Problem 2 is given by x = x0 + R cos ϕ tanh(v0 t) y = y0 + R sin ϕ tanh(v0 t) z = R sech(v0 t) where x0 , y0 , R, ϕ and v0 are constants. Show that the path of the particle lies on a sphere of radius R centered at (x0 , y0 , 0) in the Cartesian coordinate space given by (x, y, z). Note that this demonstrates the non-Euclidean nature of hyperbolic space; in reality the ‘sphere’ is flat, while the space is curved. It should be a straightforward exercise to insert the x, y and z equations into (5), (6) and (7) to show that it is a solution. However it is actually more interesting to solve the equations directly. We start with the x equation, (5). If we are somewhat clever, we could rewrite (5) as x ¨ z˙ =2 x˙ z

⇒

d d log x˙ = 2 log z dt dt

Both sides of this may be integrated in time to get x˙ = ax z 2

(8)

where ax is a constant. It should be clear that the y equation, (6) can be worked on in similar manner to get y˙ = ay z 2 (9) Of course, we have not yet completely solved for x and y. But we are a step closer to the solution. Now, inserting (8) and (9) into the z equation, (7), we obtain z¨ + (a2x + a2y )z 3 −

z˙ 2 =0 z

This non-linear differential equation can be simplified by performing the substitution z(t) = 1/u(t). Noting that z˙ = −

u˙ , u2

z¨ = −

u ¨ u˙ 2 + 2 u2 u3

the z equation may be rewritten as u¨ u − u˙ 2 = (a2x + a2y ) While this equation is still non-linear, it is possible to obtain a general solution u(t) =

1q 2 ax + a2y cosh(v0 (t − t0 )) v0

where v0 and t0 are constants. Given the solution for z = 1/u, we now insert this back into (8) to obtain v02 ax ax x˙ = 2 = 2 sech2 (v0 (t − t0 )) 2 u ax + ay which may be integrated to yield x(t) = x0 +

v0 ax tanh(v0 (t − t0 )) + a2y

a2x

Similarly, for y, we integrate (9) to find y(t) = y0 +

v0 ay tanh(v0 (t − t0 )) + a2y

a2x

Note that the three (coupled) second order differential equations give rise to six constants of integration, (x0 , y0 , ax , ay , v0 , t0 ). The expressions may be simplified by defining v0 v0 ax = cos ϕ, ay = sin ϕ R R

in which case we see that x = x0 + R cos ϕ tanh(v0 (t − t0 )) y = y0 + R sin ϕ tanh(v0 (t − t0 )) z = R sech(v0 (t − t0 )) which is the answer we wanted to show, except that here we have retained an extra constant t0 related to the time translation invariance of the system. Finally, to show that the path of the particle lies in a sphere, all we need to demonstrate is that (x − x0 )2 + (y − y0 )2 + z 2 = R2 cos2 ϕ tanh2 (v0 t) + R2 sin2 ϕ tanh2 (v0 t) + R2 sech2 (v0 t) = R2 (tanh2 (v0 t) + sech2 (v0 t)) = R2 This is indeed the equation for a sphere, (x − x0 )2 + (y − y0 )2 + z 2 = R2 .