Physics 1116 Fall 2011
Due: Wednesday, Sep 7
Reading: K&K §1.9 and Note 1.1; K&K §§2.1-2.4 (The problems in this set focus on K&K §1.9 and Note 1.1. §§2.1-2.4 is mainly an introduction to Newton’s laws that you should read and think about.) 1. Some Mathematical Tools Read Note 1.1 in K&K and (if you need to) review Taylor series. Then do the following exercises. a. Write down the Taylor series for each of the following functions, and give the simplest nontrivial approximation valid for small x. √ √ (i) 3 1 + x − 5 1 + x (ii) ln(1 + x) 2 (iii) e−x (iv) sinh x (v) cosh x Note: sinh x ≡ (ex − e−x )/2; cosh x ≡ (ex + e−x )/2. Solution: (i) (1 + x)1/3 ≈ 1 + 31 x + ... and (1 + x)1/5 ≈ 1 + 15 x + ... so the difference, to lowest order in the expansion, is (2/15)x. (ii) For x ≈ 0, ln(1 + x) ≈ ln(1 + 0) + x(d ln(1 + x)/dx)x=0 = x (iii) You already know the expansion for eu = 1 + u + u2 /2 + ... so just subsitute u → −x2 (iv) Using the expansion for ex just do the algebra. sinh x ≈ x. (v) ditto. cosh x ≈ 1 + x2 /2. b. It’s handy to have a rough idea where common approximations breakdown. Write down simple algebraic expressions for the errors incurred in making the approximations sin x ≈ x and cos x ≈ 1 − 21 x2 . In each case give a numerical value in degrees for the angle at which this error reaches 10% of the correct, unapproximated, value. Solution: Since sin x = x−x3 /3!+x5 /5!−x7 /7!+... what we lose when we truncate the series is everything to the right. When we truncate to sin x = x we are throwing away is −x3 /3! + x5 /5! + .... This is a series which is reasonably approximated by its first term if x is small, so roughly speaking the error in our sin x = x approximation is just −x3 /3!. This is a general way of estimating (when possible) the error one
incurs in truncating an expansion. Here, of course, we can be more exact, and we can evaluate the fractional error f (x) ≡ (x − sin x)/ sin x and observe where it exceeds 0.10. As you can see in Fig. 1 Left, this happens around x = 0.75 radians, which corresponds to about 45◦ . Fig. 1 Right shows the same thing for cos x, which is good to 10% as far as about 60◦ . 0.00 -0.02
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Figure 1: Left: Plot of f (x) ≡ (x − sin x)/ sin x. Red = f (x); Green=x3 /6x. Right: Plot of f (x) ≡ ((1 − 21 x2 ) − cos x)/ cos x. x is in radians in both plots. c. In Special Relativity, as we will see, pthe total energy E of a particle of rest mass m and momentum p is given by E = p2 c2 + m2 c4 . The realm of Newtonian mechanics is where pc mc2 . In this domain, find the simplest nontrivial approximation for E, consisting of two terms, and explain what the two terms mean. Solution: √ Let x ≡ pc/mc2 be the small parameter for expansion. The E = mc2 1 + x2 which after expansion becomes E = mc2 (1 + x2 /2) = mc2 + p2 /2m. The second term you will recognize as the classical kinetic energy p2 /2m = 21 mv 2 , and the first term is a constant that doesn’t correspond to any quantity normally discussed in Newtonian mechanics. You probably recognize it, however, as Einstein’s famous rest mass energy. From a Newtonian point of view, this is just a constant offset, so it never affects one’s use of the energy. In any case, the point is that when pc mc2 , that is, v c, Einstein’s expression reduces to one equivalent to Newton’s. Thus we see that the use of approximation techniques yields not just a simpler expression, but real conceptual insight. 2. Approximate Harmonic Potentials. Harmonic potentials are quadratic functions of the form U (x) = 12 kx2 which play an important role in oscillatory systems. You might recognize this function as the potential energy for a spring, of spring constant k, stretched a distance x beyond its resting
position. (But if you don’t, don’t worry.) In this problem we will find that harmonic potentials turn up in many places, once you know how to look. a. The circular hoop. Picture a bead resting at the bottom of a frictionless circular hoop of radius r. The plane of the hoop is vertical. For definiteness, let the hoop be defined by the equation for the circle x2 + (y − r)2 = r2 , with y measuring vertical height above the minimum point, and x measuring horizontal position. Since gravitational potential energy is given by U (y) = mgy for a bead of mass m, show that for small enough excursions around the bottom of the circle, U (x) ≈ 21 kx2 . Find an expression for k in terms of the available parameters. Solution: Expand and reorganized the equation for the circle to y 2 − 2ry + x2 = 0. A short cut at this point would be to observe that for small oscillations around the bottom of the circle, y r and y x, and hence we can discard the y 2 term, leaving 1 2 y = 2r x . Then U = mgy = 21 mg x2 , so k = mg/r. r A fussier solution would be to start from y 2 −2ry +x2 = 0 and apply the quadratic formula to solve for y. This yields √ y = (2r ± 4r2 − 4x2 )/2 p = r ± r 1 − x2 /r2 ≈ r ± r(1 − x2 /2r2 ). It’s slightly more interesting this way because we get two solutions – one (+) describes motion near the top of the circle, and the other (−) describes motion near the bottom. In this approach, y could be big (y ∼ 2r) or small (y ∼ 0) and the key to approximation is in demanding x/r 1 at the last step. Take the y ∼ 0 solution and you’re done. When we study harmonic oscillators, we’ll learn that the (angular) frequency p of p oscillation is given by ω = k/m. Applied here, that tells us that ω = g/r which is the same as that of a pendulum of length r. ...as we might have expected! b. The Lennard-Jones inter-molecular potential. The interaction between two molecules that attract one another and form a bound state, may be approximated by a potential function U (r) =
a b − 6. 12 r r
(i) If U has units of energy, what are the units of a and b? (ii) Plot U (r) and observe the shape. (For this purpose, set a = 1, b = 2 in their respective units. ) (iii) Compute the location, r0 , of the minimum of U (r) in terms of arbitrary a and b. Page 3
(iv) Calculate the Taylor Series expansion of U (r) about the minimum point, and truncate the series under the assumption that you’re only interested in the region where |r − r0 | is small. This should give something of the form U (r) = U0 + 21 k(r − r0 )2 ; identify the variable k in terms of the given parameters. Solution: (i) If units of energy are E and units of length are L then [a] = EL12 and [b] = EL6 . (ii) See Fig. 2, Left. (iii) Set dU/dr = 0, solve for: r0 = (2a/b)1/6 . (iv) Since we are expanding U(r) about a minimum, we know that dU/dr = 0 and hence the first nontrivial term in the Taylor expansion will be the 2nd order term. Thus we need: U (r) = U (r0 ) + 21 (r − r0 )2 d2 U (r0 )/dr2 . Some algebra leads to: 7 1/3 b b2 U (x) ≈ − + 9 x2 4a 2a4 where x ≡ r − r0 . In this case we would identify k = 29 (b7 /2a4 )1/3 as the “spring constant”... though this is admittedly not very illuminating in the absence further information (or theory) about a and b. -0.98
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Figure 2: Lennard-Jones potential with a=1, b=2. Left: full plot of U (r); Right: zoom in to minimum, with overlay of parabolic approximation (calculated in part iv). c. The Higgs potential. Consider a one-dimensional toy “Higgs potential”, U (x) = − 21 µ2 x2 + 41 λ2 x4 . (i) Plot U (x) for −2 < x < 2 and observe the shape. (For this purpose, set µ2 = λ2 = 1.)
(ii) Compute the location, x0 > 0, of the positive-side minimum of U (x) in terms of arbitrary µ and λ. (iii) Calculate the Taylor Series expansion of U (x) about the minimum point, and truncate the series under the assumption that you’re only interested in the region where |x − x0 | is small. This should give something of the form U (x) = U0 + 12 k(x − x0 )2 ; identify the variable k in terms of the given parameters. Solution: (i) See Fig. 3. Note the minima which are not at x = 0. In the case of a real Higgs potential, this would indicate that the normal state of the Higgs field is to be present everywhere at a constant strength x = xmin . (ii) Set dU/dx = 0 to find xmin = ±µ/λ. (iii) Expand U (x) in a Taylor series about xmin . In terms of η ≡ x − xmin , this is: µ4 U (η) ≈ − 2 + η 2 µ2 4λ In a proper particle-physics treatment of this problem, this √ equation would indicate that the Higgs particle have a mass of mH = 2µ. (This is not especially obvious to us in the Phys 1116 context.) The mass is the manifestation of the oscillation energy of the Higgs field x = µλ + η as it “sloshes” in this parabolic potential well. Unfortunately in practice the parameter µ is itself not predicted so the Higgs mass remains unknown. There is some chance it will be discovered and measured in the next year or so. For more propaganda check out http://cms.web.cern.ch/cms/index.html. 1.0 0.8 0.6 UHxL
0.4 0.2 0.0 -0.2 -0.4 -2
Figure 3: Higgs potential with µ = λ = 1.
3. Complex exponentials and circular motion.
√ a. Write down the Taylor series for eix (i = −1), and then rewrite this as the sum of two other Taylor series, the first one composed of all the even-power terms (x2 , x4 , etc) and the second composed of all the odd-power terms. Observe that this expression has a very simple equivalent in familiar functions. b. A complex number like z = a + ib is conveniently represented in the complex plane in much the same as a vector r = aˆ x + bˆ y is represented in the real xy plane. The 2 2 ∗ analog of r = r · r is |z| = zz = (a + ib)(a − ib). In the complex plane, the horizontal axis measures the real part of a number and the vertical axis measures the imaginary part; in the real plane the horizontal and vertical axes measure the “x” and “y” components of a vector. In both cases we can also use plane polar coordinates, (r, θ). Show by explicit construction that any complex number a + ib can be written as reiθ . In other words, compute r and θ in terms of a and b. c. Since an object in uniform circular motion can be represented in xy Cartesian coordinates by r = r(cos ωtˆ x+sin ωtˆ y), we can see that it might also be conveniently represented by r(t) = r0 eiωt in the complex plane if we let the imaginary axis ˆ axis of the real plane. From this, derive the velocity and play the role of the y acceleration and show that the results agree with standard statements about UCM. Solution: a. The series for the non-complex case is x2 x3 x4 x5 e =1+x+ + + + + ... 2! 3! 4! 5! x
With the factor i in the exponent we get x2 ix3 x4 ix5 − + + + ... 2! 3! 4! 5! x2 x4 ix3 ix5 = (1 − + + ...) + (ix − + + ...) 2! 4! 3! 5! = cos x + i sin x
eix = 1 + ix −
b. z = reiθ = r cos θ + ir sin θ ≡ a + ib. Then, |z|2 = reiθ × re−iθ = r2 ; but also |z|2 = a2 + b2 so we identify r2 = a2 + b2 . Since r cos θ + ir sin θ = a + ib, evidently b/a = tan θ. c. Given r(t) = r0 eiωt , we can easily take time derivatives to find v = dr/dt = iωr0 eiωt and then a = dv/dt = d2 r/dt2 = −ω 2 r0 eiωt . Notice that at t = 0 r = r0 lies along the real (horizontal) axis, while v = iωr0 lies exactly along the imaginary (vertical) axis. Thus we find that v is 90◦ ahead of r, just as we found in analyzing this situation with real 2-dimensional vectors. Similarly, a picks up another factor of i, which rotates it another 90◦ from v, or 180◦ from r. Note also that the magnitudes agree with what we found the traditional way, namely v = rω and a = vω = rω 2 . And finally, notice how each derivative brings
down a factor of i from the exponent (besides factor of ω), which “causes” – ie, is equivalent to – rotating by +90◦ in the complex plane. In general, for a given complex number z = a + ib in the complex plane, zeiθ is just z rotated by θ in the counterclockwise direction. The mathematical representation r(t) = r0 eiωt is thus an excellent match to the physical problem at hand. 4. Spiral motion. K.K. problem 1.20 Solution: From the information provided, we see that the relevant scalar quantities are r = Aθ, ˙ r¨ = Aθ, ¨ θ = 1 αt2 , θ˙ = αt, and θ¨ = α. r˙ = Aθ, 2 a. See Fig. 4. b. Let’s write out the full vector expressions for position, velocity, and acceleration. To do so, we simply start with ~r and systematically take time derivatives – ˆ are time-varying themselves. remembering that the basis vectors ˆr and θ ~r(t) = Aθˆr ˙r + Aθθ˙ θ ˆ ~v (t) = Aθˆ ¨ θ ˆ ~a(t) = A(θ¨ − θθ˙2 )ˆr + A(2θ˙2 + θθ) Plugging in the time dependences given above, these become: ~r(t)
= 12 Aαt2ˆr
ˆ = Aαt ˆr + 12 Aα2 t3 θ
→ 12 t2ˆr ˆ → tˆr + 21 αt3 θ
ˆ ˆ → (1 − 1 α2 t4 )ˆr + 5 αt2 θ ~a(t) = Aα(1 − 21 α2 t4 )ˆr + 52 Aα2 t2 θ 2 2 In the second set of equations above (after the → arrow) I’ve set A = α = 1 for clarity since these just factor out √ anyway. With all this laid out, it is now easy to see that when θ = 21 αt2 = 1/ 2, the ˆr component of ~a vanishes. c. The magnitudes of the radial and tangential accelerations are Aα|1 − 21 α2 t4 | = Aα|1 − 2θ2 | and 25 Aα2 t2 = 5Aαθ. Since 1 − 12 α2 t4 will be positive at early t and negative later, we need to solve two quadratic equations to find all the cases where these two magnitudes are equal: (1 − 2θ2 ) − 5θ = 0
√1 ) 2
corresponding to early t, and (2θ2 − 1) − 5θ = 0
(θ > √12 ) √ corresponding to late t. The solutions are θ = ( 33 ± 5)/4 ≈ 0.19, 2.69 radians. See Fig. 5 for additional info.
x Figure 4: Red: trajectory r(t) for 0 ≤ θ = 12 αt2 ≤ 2π; Blue: velocity vectors are selected points; Green: acceleration vectors at selected points. For this plot, A = 1/π, α = 1, and velocity and acceleration vectors have arbitrary (but fixed) scaling.
Hred, greenL= Hvr,vΘ L
2.0 1.5 1.0 0.5 0.0 0.0
1.0 1.5 2.0 t Hred, greenL= Har,aΘ L
1 0 -1 -2 0.0
Figure 5: Velocity and Acceleration r and θ components, versus time.