Physics 441 Homework III solutions

18.
If you let $ξ = p + i ω x$ , then $\dot{ξ} = ṗ + i ω ẋ = - ω^{2} x + i ω p = i ω (p + i ω x) = i ω ξ$ , which is trivial to integrate; $\frac{d ξ}{ξ} = i ω t$ , $ln \frac{ξ (t)}{ξ (0)} = i ω t$ , so

p + i ω x = ξ (t) = ξ (0) e^{i ω t} = (p (0) + i ω x (0)) e^{i ω t} = (p (0) cos (ω t) - x (0) ω sin (ω t)) + i ω (x (0) cos (ω t) + \frac{p (0)}{ω} sin (ω t))

19.
A.

p_{x} = \frac{\partial L}{ẋ} = m ẋ, ẋ = \frac{p_{x}}{m}, H = p ẋ - L = \frac{1}{2 m} p_{x}^{2} + m g x

B. Look at Eq.4.39, you must Legendre transform all velocity terms out of $L$ in the same way to get $H$ (well, that is pretty obvious, you must treat all variables equally)

p_{x} = \frac{\partial L}{ẋ} = m ẋ, p_{y} \frac{\partial L}{ẏ} = m ẏ, H = p_{x} ẋ - p_{y} ẏ - L = \frac{1}{2 m} (p_{x}^{2} + p_{y}^{2}) - \frac{m M G}{r}

ẋ = \frac{\partial H}{\partial p_{x}} = \frac{p_{x}}{m}, ẏ = \frac{\partial H}{\partial p_{y}} = \frac{p_{y}}{m}, ṗ_{x} = - \frac{\partial H}{\partial x} = - \frac{m M g x}{{(x^{2} + y^{2})}^{\frac{3}{2}}}, ṗ_{y} = - \frac{\partial H}{\partial y} = - \frac{m M g y}{{(x^{2} + y^{2})}^{\frac{3}{2}}}

Polar coordinates; $x = r cos φ, y = r sin φ$

L = \frac{1}{2} m (ṙ^{2} + r^{2} {\dot{φ}}^{2}) + \frac{m M G}{r}, p_{r} = \frac{\partial L}{ṙ} = m ṙ, p_{φ} = \frac{\partial L}{\dot{φ}} = m r^{2} \dot{φ}

H = p_{r} ṙ + p_{φ} \dot{φ} - L = \frac{1}{2 m} (p_{r}^{2} + \frac{1}{r^{2}} p_{φ}^{2}) - \frac{m M G}{r}

ṙ = \frac{\partial H}{\partial p_{r}} = \frac{p_{r}}{m}, \dot{φ} = \frac{\partial H}{\partial p_{φ}} = \frac{1}{m r^{2}} p_{φ}, ṗ_{r} = - \frac{\partial H}{\partial r} = \frac{p_{φ}^{2}}{m r^{3}} - \frac{m M G}{r^{2}}, ṗ_{φ} = - \frac{\partial H}{\partial φ} = 0

so $p_{φ}$ is conserved.
C. I will set this one up in polar only.

L = \frac{1}{2} m (ṙ^{2} + r^{2} {\dot{φ}}^{2}) - \frac{1}{2} m ω^{2} r^{2}, p_{r} = m ṙ, p_{φ} = m r^{2} \dot{φ}

H = p_{r} ṙ + p_{φ} \dot{φ} - L = \frac{1}{2 m} (p_{r}^{2} + \frac{1}{r^{2}} p_{φ}^{2}) + \frac{1}{2} m ω^{2} r^{2}

so once again we see that $p_{φ}$ will be a conserved quantity.
D. We actually did this one in class, see the example of the bead constrained to move on a rotating wire in Ch. 3. This is an important example.

ẋ^{'} = (ẋ cos ω t - ẏ sin ω t) - ω (x sin ω t + y cos ω t), ẏ^{'} = (ẏ cos ω t + ẋ sin ω t) - ω (y sin ω t - x cos ω t)

{ẋ^{'}}^{2} + {ẏ^{'}}^{2} = (ẋ^{2} + ẏ^{2}) + ω^{2} (x^{2} + y^{2}) - 2 ω (y ẋ - x ẏ)

so that in the rotating frame (temporarily leave out the $Ω$ terms, they are just potentials, nothing special)

L = \frac{1}{2} m ((ẋ^{2} + ẏ^{2}) + ω^{2} (x^{2} + y^{2}) - 2 ω (y ẋ - x ẏ))

in which we see ordinary kinetic terms plus centripetal potential plus Coriolis terms.

p_{x} = \frac{\partial L}{\partial ẋ} = m (ẋ - ω y), p_{y} = \frac{\partial L}{\partial ẏ} = m (ẏ + ω x), ẋ = \frac{p_{x}}{m} + ω y, ẏ = \frac{p_{y}}{m} - ω x

H = p_{x} (\frac{p_{x}}{m} + ω y) + p_{y} (\frac{p_{y}}{m} - ω x) - \frac{1}{2} m ({(\frac{p_{x}}{m} + ω y)}^{2} + {(\frac{p_{y}}{m} - ω x)}^{2}) + ω^{2} (x^{2} + y^{2}) - 2 ω (y (\frac{p_{x}}{m} + ω y) - x (\frac{p_{y}}{m} - ω x)))

which collapses to

H = \frac{1}{2 m} (p_{x}^{2} + p_{y}^{2}) + ω y p_{x} - ω x p_{y} + \frac{1}{2} m Ω^{2} (x^{2} + y^{2})

with the $Ω$ terms restored. The Hamiltonian equations of motion are (with the $Ω$ term back in)

\dot{p_{x}} = ω p_{y} - m Ω^{2} x, ṗ_{y} = - ω p_{x} - m Ω^{2} y, ẋ = \frac{1}{m} p_{x} + ω y, ẏ = \frac{1}{m} p_{y} - ω x

(1)

Let $ξ = x + i y$ , $η = p_{x} + i p_{y}$ , then

\dot{ξ} = ẋ + i ẏ = \frac{1}{m} (p_{x} + i p_{y}) + ω (y - i x) = \frac{1}{m} η - i ω ξ

\dot{η} = ṗ_{x} + i ṗ_{y} = ω (p_{y} - i p_{x}) - m Ω^{2} (x + i y) = - i ω η - m Ω^{2} ξ

Use what you learned in elementary calculus, introduce an integrating factor

let ξ = Ξ e^{- i ω t}, η = Υ e^{- i ω t}, then \dot{Ξ} = \frac{1}{m} Υ, \dot{Υ} = - m Ω^{2} Ξ

and let $m = 1$ so it is not ﬂoating around. Now re-use problem 18, let

ζ_{\pm} = Υ \pm i Ω Ξ, {\dot{ζ}}_{\pm} = \dot{Υ} \pm i \dot{Ξ} = - Ω^{2} Ξ \pm i Ω Υ, {\dot{ζ}}_{\pm} = \pm i Ω ζ_{\pm}, ζ_{\pm} = e^{\pm i Ω t} ζ_{\pm} (0)

from which we can recover

ξ = e^{- i ω t} \frac{1}{i Ω} (ζ_{+} - ζ_{-}) = e^{i ω t} \frac{1}{i Ω} (ζ_{+} (0) e^{i Ω t} - ζ_{-} (0) e^{- i Ω t}) = e^{i ω t} (A e^{i Ω t} + B e^{- i Ω t})

Of course you could solve these by more traditional means, but our purpose here is to learn to fully exploit the ﬁrst-ordered-ness of these equations of motion. For example, lets do it with ode

O=1.0
o=0.1
px’=o*py-O*O*x
py’=-o*px-O*O*y
x’=px+o*y
y’=py-o*x
x=1.0
y=0.0
px=0.0
py=0.0
print x,y
step 0, 20

We do this for two sets of initial data just so that you can see how the Coriolis terms affect the motion. They cause precession, the “orbit” of the oscillator does not re-trace its own path. A pendulum swinging at any point on the earth’s surface will exhibit this precession, with a period of

24

hours, which I offer up as proof the it is the earth that rotates, and not the rest of the universe. The rotation of the earth creates the Coriolis force that causes the precession. This is called Foucault’s pendulum.

This problem is very important because of its concrete physical connections, and because (depending on how you work, and how deeply you apply yourself to your work) you can solve it in many ways that appear almost novel. If you study Eqs.1 quite carefully, you can see that they have a block structure; both $ṗ_{x}$ and $ẏ$ depend only on $p_{y}$ and $x$ , and vice versa. Therefore lets invent! (Let $m = 1$ )

(\begin{matrix} ẋ \\ ṗ_{y} \\ ṗ_{x} \\ ẏ \end{matrix}) = (\begin{matrix} 0 & 0 & 1 & ω \\ 0 & 0 & - ω & - Ω^{2} \\ - Ω^{2} & ω & 0 & 0 \\ - ω & 1 & 0 & 0 \end{matrix}) (\begin{matrix} x \\ p_{y} \\ p_{x} \\ y \end{matrix}), Call this \dot{ξ} = M \cdot ξ

Start playing around with this, and you discover that

M \cdot M = - (ω^{2} + Ω^{2}) (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 \end{matrix}) = - (ω^{2} + Ω^{2}) 1

From this we can deduce that

(M)^{2 n} = {(- 1)}^{n} (ω^{2} + Ω^{2}), (M)^{2 n + 1} = {(- 1)}^{n} (ω^{2} + Ω^{2}), M

and insert this into the Taylor series for $ξ (t)$

ξ (t) = \sum_{n = 0}^{\infty} \frac{1}{n!} (\frac{d^{n}}{d t^{n}} ξ (t))_{t = 0} t^{n} = \sum_{n o d d}^{\infty} \frac{1}{n!} (\frac{d^{n}}{d t^{n}} ξ (t))_{t = 0} t^{n} + \sum_{n e v e n}^{\infty} \frac{1}{n!} (\frac{d^{n}}{d t^{n}} ξ (t))_{t = 0} t^{n}

just like you did in our simple series solutions from the ﬁrst assignment. You obtain

ξ (t) = \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)!} (M)^{2 n + 1} ξ (0) t^{2 n + 1} + \sum_{n = 0}^{\infty} \frac{1}{(2 n)!} (M)^{2 n} ξ (0) t^{2 n}

= \sum_{n = 0}^{\infty} \frac{1}{(2 n + 1)!} (- (ω^{2} + Ω^{2}))^{n} M \cdot ξ (0) t^{2 n + 1} + \sum_{n = 0}^{\infty} \frac{1}{(2 n)!} (- (ω^{2} + Ω^{2}))^{n} ξ (0) t^{2 n}

= \frac{sin (\sqrt{ω^{2} + Ω^{2}} t)}{\sqrt{ω^{2} + Ω^{2}}} M \cdot ξ (0) + cos (\sqrt{ω^{2} + Ω^{2}} t) ξ (0)

which is a pretty spiffy route to a solution that is very similar to group-theoretical methods that we will soon apply to quantum systems.

20.
This is an important example. Writing it out explicitly

L = \frac{1}{2} m (ẋ^{2} + ẏ^{2} + ż^{2}) - q \frac{B_{z}}{2} y ẋ + q \frac{B_{z}}{2} x ẏ

this looks suspiciously similar to the previous problem but without the centripetal terms.

p_{x} = m ẋ - q \frac{B_{z}}{2} y, p_{y} = m ẏ + q \frac{B_{z}}{2} x, p_{z} = m ż, ẋ = \frac{p_{x} + \frac{q B_{z}}{2} y}{m}, ẏ = \frac{p_{y} - \frac{q B_{z}}{2} x}{m}, ż = \frac{p_{z}}{m}

The Lagrangian equations of motion are

\frac{d}{d t} \frac{\partial L}{\partial ẋ} = \frac{\partial L}{\partial x}, \frac{d}{d t} (m ẋ - q \frac{B_{z}}{2} y) = q \frac{B_{z}}{2} ẏ, m \ddot{x} = q B_{z} ẏ

\frac{d}{d t} \frac{\partial L}{\partial ẏ} = \frac{\partial L}{\partial y}, \frac{d}{d t} (m ẏ + q \frac{B_{z}}{2} x) = - q \frac{B_{z}}{2} ẋ, m ÿ = - q B_{z} ẋ

\frac{d}{d t} \frac{\partial L}{\partial ż} = \frac{\partial L}{\partial z}, \frac{d}{d t} (m ż) = 0

which are precisely $m \dot{v} = q v \times B$ for $B = B_{z} k$ .

H = p_{x} ẋ + p_{y} ẏ + p_{z} ż - L = \frac{1}{2 m} ({(p_{x} + \frac{q B_{z}}{2} y)}^{2} + {(p_{y} - \frac{q B_{z}}{2} x)}^{2})

In the most general case, with an arbitrary vector potential $A$ we would get

H = \frac{1}{2 m} (p - q A) \cdot (p - q A)

21.Henon-Heiles is important because it behaves chaotically. If you are interested in chaos, see Appendix C, and the books by Drazin and Steeb in the bibliography.

p_{1} = ẋ_{1}, p_{2} = ẋ_{2}, H = \frac{1}{2} (p_{1}^{2} + p_{2}^{2}) + \frac{1}{3} x_{1}^{3} - x_{1} x_{2}^{2}

ẋ_{1} = p_{1}, ẋ_{2} = p_{2}, ṗ_{1} = - (x_{1}^{2} - x_{2}^{2}), ṗ_{2} = 2 x_{1} x_{2}

22.
This is simple, just show that all seven requirements are met;
Closure; $(a + i b) + (c + i d) = (a + c) + i (b + d) \in {x + i y ∣ x, y \in ℝ}$
Unique identity; $(a + i b) + (0 + i 0) = a + i b, 0 + i 0 \in {x + i y ∣ x, y \in ℝ}$
Unique inverses; $(a + i b) + (- a + i (- b)) = 0 + i 0, since a \in ℝ \Rightarrow - a \in ℝ$
Associativity follows from associativity of numerical addition in the Reals.
Monoid/ring properties; $λ (a + i b) = λ a + i (λ b), since λ, a, b \in ℝ \Rightarrow λ a, λ b \in ℝ$ .
$μ (λ (a + i b)) = μ (λ a + i (λ b)) = (μ λ a) + i (μ λ b) \in {x + i y ∣ x, y \in ℝ}$ , and so forth.
23. Using this bilinear form you can show that $T r (e_{i} \cdot f_{j}) = δ_{i j}$ , where $f_{i} = e_{i}^{T}$ (transpose) for example

e_{1} \cdot f_{2} = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}) = (\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}), T r (e_{1} \cdot f_{2}) = 0

e_{1} \cdot f_{1} = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}), T r (e_{1} \cdot f_{1}) = 1

and so one, establishing that $e_{i}^{*} = f_{i} = e_{i}^{T}$ .
24. Simply use the equations of motion.

{\dot{q}}_{1} = p_{1}, {\dot{q}}_{2} = p_{2}, ṗ_{1} = - A q_{1} - q_{1}^{3} - q_{1} q_{2}^{2}, ṗ_{2} = - A q_{2} - q_{2}^{3} - q_{2} q_{1}^{2}

İ_{1} = {\dot{q}}_{1} p_{2} - {\dot{q}}_{2} p_{1} + q_{1} ṗ_{2} - q_{2} ṗ_{1} = p_{1} p_{2} - p_{2} p_{1} + q_{1} (- A q_{2} - q_{2}^{3} - q_{2} q_{1}^{2}) - q_{2} (- A q_{1} - q_{1}^{3} - q_{1} q_{2}^{2}) = 0

and so forth.
25.

\dot{q} = 2 p^{3}, ṗ = - 2 q^{3}

Ḣ = 2 p^{3} ṗ + 2 q^{3} \dot{q} = 2 p^{3} (- 2 q^{3}) + 2 q^{3} (2 p^{3}) = 0, p = \sqrt[4]{2 C - q^{4}}

2 d t = \frac{d q}{{(2 C - q^{4})}^{\frac{3}{4}}}, 2 (t - t_{0}) = \int \frac{d q}{{(2 C - q^{4})}^{\frac{3}{4}}}

This is a nasty one! I’m just kidding, its not as icky as it looks, but the answer is not a function that you are intimately familiar with (yet). Even if you can’t integrate this, you can always get the period of the motion. One period is the time it takes $q$ to go from $- q_{m a x} = - \sqrt[4]{2 C}$ to $q_{m a x} = \sqrt[4]{2 C}$ back to $- q_{m a x}$ , which is $4$ times what it takes to go from $0$ to $q_{m a x}$ ;

2 T = 4 \int_{0}^{\sqrt[4]{2 C}} \frac{d q}{{(2 C - q^{4})}^{\frac{3}{4}}}, Let q = \sqrt[4]{2 C} x; 2 T = 4 \frac{1}{\sqrt{2 C}} \int_{0}^{1} \frac{d x}{{(1 - x^{4})}^{\frac{3}{4}}}

This last integral is not elementary, but it is not too difficult, Maxima gives it as $\frac{1}{4} β (\frac{1}{4}, \frac{1}{4})$ where $β (x, y) = \frac{Γ (x) Γ (y)}{Γ (x + y)}$ . See Lebedev, Special Functions, p.13.

26, 27. These “problems” are simply activities designed to introduce you to ode. Some ode scripts are available for download from the 441 website under computer programs. Ode provides you with a way of solving Hamiltonian equations of motion by computer without the need to learn any programming.

28,29. These “problems” are activities designed to introduce you to Runge Kutta extrapolation, which David will tell you about in 303. Some sample programs in C, Vpython and java are available for download from the 441 website under computer programs. I don’t remember Fortran well enough to provide you with Fortran examples, perhaps you could write some as joint 303/441 activities?

C     Phase space program for quartic oscillator
C     export LD_LIBRARY_PATH=$LD_LIBRARY_PATH:/usr/local/fortran/lib
C                                            :/usr/local/fortran/lib64
C     gfortran osc.f -o osc.exe

      real phase
      real x, v, t, x0, v0,a,b,c,d
      real dt, dev, dev1, q, s
      dt = 0.003
      x0 = 1.0
      v0 = 0.0
      s=0.15

C     Simple RK integration
      do 100 k=1, 5000
         k1=h*p;
         l1=-2.0*h*q*q*q;
         k2=h*(p+0.5*l1);
         l2=-2.0*h*(q+0.5*k1)*(q+0.5*k1)*(q+0.5*k1);
         k3=h*(p+l2);
         l3=-2.0*h*(q+k2)*(q+k2)*(q+k2);
         q=q+(k1+4.0*k2+k3)/6.0;
         p=p+(l1+4.0*l2+l3)/6.0;
         write(6,*) q,p
100     continue
      stop
      end

30. $ℭ (Q_{1, 1})$ has two generators

e_{1} # e_{1} + e_{1} # e_{1} = 2 Q_{1, 1} (e_{1},