PC235 Winter 2013 Classical Mechanics
Assignment #1 Solutions #1 (10 points) ˆ. Find A vector field in cylindrical polar coordinates is A = 2ρρˆ + ρz ˆ φ + ρφz z A. Is this a conservative field? Why or why not? and
∇ ×
∇·A
Solution ˆ, where A ρ + Aφ ˆ φ + Az z Here, we have A = A = A ρ ˆ where A ρ = 2ρ, Aφ = ρz , and Az = ρφz . From the inside of the back cover of the text, we see that
∇·A
1 ∂ 1 ∂ ∂ (ρAρ ) + Aφ + Az ρ ∂ρ ρ ∂φ ∂z 1 1 = (4ρ (4ρ) + (0) + ρφ + ρφ ρ ρ = 4 + ρφ + ρφ..
=
(1) (2) (3)
Next, we have
∇×
1 ∂ ˆ A = ρ Az ρ ∂φ = (z
− ρ)ρˆ −
∂ ˆ ∂ Aρ Aφ + φ ∂z ∂z 2z zˆ. φ + 2z φz ˆ
−
−
1 ∂ ∂ (ρAφ ) Az + ˆz ∂ρ ρ ∂ρ
−
∂ Aρ ∂φ
(4) (5)
Since
∇ × A = 0, A is not a conservative field.
#2 (10 points) Consider Consider an object ob ject that is moving moving in a cylindrical cylindrical coordinate coordinate system. system. Its position at time t time t is given by r = ρ = ρ((t)ρˆ + z (t)zˆ. Find an expression for the three components of the object’s velocity v and acceleration a. Show all of your work. Solution This derivation follows very closely with the one for 2D polar coordinates (slides 19-21 of chapter 1 notes); the only difference is the presence of a z z component, which is separable from the other components. components. The position vector is r = ρρˆ + z + z zˆ. Using Using Eqns. Eqns. (1.21 (1.21 and 1.26) 1.26) from from the class class notes notes (with r (with r ρ), we have a velocity of
→
v = r˙ = ρ˙ ρˆ + ρ
ρˆ ∂ ˆ ˙ ˆ φ + z + z ˙ zˆ = ρ˙ ρˆ + ρφ ˙ zˆ. ∂t
(6)
However, with no motion in φ, φ , φ˙ = 0, leaving us with velocity components ˙ vρ = ρ,
˙ = 0, = ρ vφ = ρ φ 1
vz = z ˙ .
(7)
Differentiating Eq. (6) once more, we get φ ∂ ρˆ ∂ ˆ ˙ ˆ ¨ ˆ ˙ ˆ + ρ˙ a = v˙ = ρ¨ρ + ρ˙ φφ + ρφφ + ρφ + z¨ ˆz ∂t ∂t ˙ ˆ ˙ 2 ρˆ + z¨ zˆ, = ρ¨ρˆ + 2ρ˙ φ φ + ρφ¨ ˆ φ ρφ
(8)
−
˙ φ = ¨ 0, we have acceleration components and, again with φ = aρ = ρ¨
− ρφ ˙
2
¨ 2ρ˙ φ = ˙ 0, aφ = ρ φ +
= ρ¨,
az = z¨ .
(9)
#3 (10 points) Find the eigenvalues and their corresponding normalized eigenvectors for the following matrix A: A =
−
2 1 0
−1
0 2 0 0 4
Solution To find the eigenvalues, we must solve the characteristic equation det( λI characteristic equation is therefore
λ
−2 1 0
1 λ
−2 0
0 0 λ
−4
= λ3
2
− 8λ
+ 19λ
− A) = 0. The
− 12 = (λ − 1)(λ − 3)(λ − 4) = 0,
(10)
so the eigenvalues are λ1 = 1, λ2 = 3, λ3 = 4. To find the eigenvectors, we substitute the corresponding eigenvalues back into ( A λI)x = 0 and solve for x. For λ 1 = 1,
−
−
1 1 0
−
1 1 0
− 0 0 3
x1 x2 x3
=
0 0 0
.
(11)
The bottom row tells us that 3x3 = 0, and therefore x3 = 0. The first 2 rows agree that x1 = x 2 . We’ll call this p. Then, an eigenvector is [ p, p, 0]T . However, we need to normalize this eigenvector by ensuring that the sum of the squares of all elements is equal to 1. Thus, p2 + p2 = 1, or p = 1/ 2. Finally, the normalized eigenvector corresponding to λ 1 is
−
√
√
1 x1 = 2
1 1 0
For λ2 = 3, the same procedure tells us that x 3 = 0 and x 1 = eigenvector is 1 1 x2 = 1 . 2 0
− √ 2
.
(12)
−x , and thus the normalized 2
(13)
For λ3 = 4, we have
2 1 0 1 2 0 0 0 0
x1 x2 x3
0 0 0
=
.
(14)
This is is a bit trickier - since none of these equations involves x 3 , it is arbitrary; we’ll call it p. Then, by multiplying the second row by 2 and subtracting that from the first row, we get 3x2 = 0, which means that x2 = 0. Finally, substituting our x3 and x2 into the first row tells us that x1 must equal zero. Therefore (setting p = 1 for normalization),
−
x3 =
0 0 1
.
(15)
Note that any of these solutions can be multiplied by any complex value with unit magnitude and they’d still be normalized eigenvectors.
#4 (10 points) Express f (x) = ln(1 x) as a Taylor series in the region around x = 0. Write your answer in summation form, and then write down a 4 th -order polynomial representation of f (x) (that is, the sum of all terms with powers up to x4 .) Using any suitable software, plot these curves on the same graph: f (x), f (1) (x), f (2) (x), f (3) (x), f (4) (x), in the interval 1 x 1 (that is, plot the original function, and the 1 st - through 4th -order Taylor approximations.
−
− ≤ ≤
Solution From eq. (1.47), we have ∞
f (m) (δ ) (x f (x) = m! m=0
− δ )
m
.
Where, in this case, δ = 0. We will start by looking at the zeroth derivative of f (x) at x = 0, which is simply ln(1) = 0. This tells us that the first term of the Taylor series is equal to zero, and that the lower limit of the summation can be changed from 0 to 1. Now, we can calculate the first few derivatives of f (x) and use these results to find a formula for the m th derivative: f (1) (x) = f (2) (x) = f (3) (x) = f (4) (x) = f (5) (x) =
1
−(1 − x) −(1 − x) −2(1 − x) −2(3)(1 − x) −2(3)(4)(1 − x) ··· f (x) = −(m − 1)!(1 − x) −
2
−
3
−
4
−
5
−
(m)
3
m
−
.
Inserting this result in the above summation with δ = 0 gives ∞
f (x) =
(m 1)! m x = m! m=1
−
−
∞
xm . m m=1
−
We can then write down the 4 th -order representation
≈−
1 1 1 ln(1 x) x + x2 + x3 + x4 . 2 3 4 Note that this last step was much easier than that of the example in the class notes; since δ = 0, the mth term in the sum only contributes a term in xm to the polynomial, and not any lower-order terms. f (x) and its approximation are plotted below.
−
1
0
s n o−1 i t a m i x o r −2 p p a d n−3 a f −4
−5 −1
−0.8
−0.6
−0.4
−0.2
0 x
0.2
0.4
0.6
0.8
1
Fig. 1: Problem #4: f (x) (gold) and its approximations f 0 (x) (blue), f 1 (x) (green), f 2 (x) (red), f 3 (x) (cyan), f 4 (x) (magenta)
#5 (15 points) Consider a pair of neutral atoms. They are subject to an attractive force at large separations and a repulsive force at small separations (never mind the details.) A simple model of these effects leads to the Lennard-Jones potential , which describes the electrical potential of the system of the two neutral atoms as a function of their separation r: V (r) = 4A
− β r
12
β r
6
,
where A and β are constants. Calculate the equilibrium position r 0 (that is, the value of r for which V (r) is minimized.) Then, use a Taylor series to express V (r) in a small region 4
around r = r 0 as a second-order polynomial in r. Solution The equilibrium position r 0 is found by setting dV = 4A dr
−
12
6
= 0:
−
6
dV dr
β β β 12 13 + 6 7 = 24A 7 1 r r r
6
β 2 r
√ 6
= 0 when r = r 0 = β 2.
To express V (r) as a second-order polynomial in r in the region of r0 , we can write out explicitly the zeroth, first, and second-order terms of the Taylor series,
≈ V (r ) + V (r )(r − r ) + 12 V (r )(r − r ) . The zeroth-order term is easily calculated as V (r ) = − A. The first-order term is elim′
V (r)
0
′′
0
0
0
0
2
0
inated, since, by definition, V (r) = 0 at the point of equilibrium. We are left with the second-order term, β 12 β 6 V (r) = 4A 156 14 42 8 . r r ′
′′
Evaluated at r = r 0 , this is ′′
V (r0 ) =
4A 156(2) β 2
−
7/3
−
4/3
−
− 42(2)
Combined with the zeroth-order term, we have V (r)
≈−
3
=
36A √ . 2β 3
2
− √
36A A + r 2β 2
√
6
2
β 2 .
We could expand the squared term, but for most applications, the form shown here is the most useful. A plot of V (r) and the second-order approximation is shown in Fig. 2 for A = 1 and β = 1. The curves are nearly identical in the region around r = r 0 .
5
1
0.5
) r ( V
0
−0.5
−1 0.8
1.0
r0
1.2
1.4 r
1.6
1.8
2
Fig. 2: Problem #5: V (r ) (blue) and its second-order approximation (green)
#6 (10 BONUS points) e Consider the function f (x) = cos . Using Taylor series techniques, write down a x th 6 -order polynomial approximation to f (x) around x = 0. x
Solution We could write out the series representations of the numerator and denominator of f (x) and then divide them, but this sort of polynomial division can be quite clumsy. Or, we could hammer through the first 6 derivatives of f (x) = e x sec(x), but this is rather brutal (as many of you figured out). Instead, noting that we only require a 6th-order polynomial and not a complete series solution (in summation notation), we start from our desired point: a function + c6 x6 ; (16) f (x) c0 + c1 x + c2 x2 +
≈
···
we just wish to find the coefficients cn . We see from inside the front cover of the text (or we can easily derive) that ∞
x
e =
n=0
xn x 2 x 3 x 4 x5 x 6 = 1 + x + + + + + + 2! 3! 4! 5! 6! n!
and
∞
cos x =
( 1)n x2n =1 (2n)!
− n=0
−
x 2 x 4 + 2! 4!
−
x 6 + 6!
· · · ,
···
(17)
(18)
(we don’t need to extend these approximations past x6 , since the resulting terms would be neglected in our 6th-order approximation of f (x) anyhow). Therefore, we can write
6
ex = f (x)cos x, or, x 2 x 3 x 4 x 5 x 6 1 + x + + + + + + 2! 3! 4! 5! 6!
···
x 2 x 4 x 6 = (c0 + c1 x + c2 x + c3 x + c4 x + c5 x + c6 x )(1 + + 2! 4! 6! c 0 2 c 1 3 c 2 c0 = c 0 + c1 x + c2 + x + c3 x + c4 x4 2 2 2 24 c 3 c1 c c c 4 2 0 + c5 + + x5 + c6 x6 . 2 24 2 24 720 2
−
3
− −
4
5
6
− −
−
−
−
···)
(19)
Finally, we can solve the cn one by one (starting with c0 ) by equating each coefficient of xn . This can be set up as a simultaneous solution of 7 equations in 7 unknowns, but in our case we can just start from c 0 and solve for the c n individually in numerical order. The result is: 2x3 x 4 3x5 19x6 ex 2 1 + x + x + + + + (20) . cos x 3 2 10 90
≈
−3
2
1
n o1.8 i t a m i x1.6 o r p p a1.4 r e d r o1.2 − h t 6 d 1 n a ) x ( f 0.8
0.5 n o i t a m 0 i x o r p p a n−0.5 i r o r r e −1
−0.5
0 x
0.5
x 10
−1.5 −0.5
0 x
0.5
Fig. 3: Problem #6: (left) f(x) (blue) and its approximation (green); the curves are virtually
indistinguishable. (right) the error in the approximation...as you may have anticipated, it’s a 7th-order polynomial, since the first 6 orders are accounted for in our approximation.
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