Physics 1116 Fall 2011 Homework 2 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 introductionto 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 followingexercises.

a. Write down the Taylor series for each of the following functions, and give thesimplest nontrivial approximation valid for small x.

(i) 3√

1 + x− 5√

1 + x

(ii) ln(1 + x)

(iii) e−x2

(iv) sinhx

(v) cosh x

Note: sinhx ≡ (ex − e−x)/2; cosh x ≡ (ex + e−x)/2.

Solution:

(i) (1 + x)1/3 ≈ 1 + 13x + ... and (1 + x)1/5 ≈ 1 + 1

5x + ... 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 subsituteu→ −x2

(iv) Using the expansion for ex just do the algebra. sinhx ≈ x.

(v) ditto. coshx ≈ 1 + x2/2.

b. It’s handy to have a rough idea where common approximations breakdown. Writedown simple algebraic expressions for the errors incurred in making the approxima-tions sinx ≈ x and cosx ≈ 1− 1

2x2. 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 sinx = x−x3/3!+x5/5!−x7/7!+... what we lose when we truncate the seriesis everything to the right. When we truncate to sinx = x we are throwing away is−x3/3! + x5/5! + .... This is a series which is reasonably approximated by its firstterm if x is small, so roughly speaking the error in our sinx = x approximationis 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, andwe can evaluate the fractional error f(x) ≡ (x − sinx)/ sinx and observe whereit exceeds 0.10. As you can see in Fig. 1 Left, this happens around x = 0.75radians, which corresponds to about 45◦. Fig. 1 Right shows the same thing forcosx, which is good to 10% as far as about 60◦.

0.0 0.2 0.4 0.6 0.8 1.00.00

0.05

0.10

0.15

x

fHxL

0.0 0.2 0.4 0.6 0.8 1.0

-0.12

-0.10

-0.08

-0.06

-0.04

-0.02

0.00

xfH

xL

Figure 1: Left: Plot of f(x) ≡ (x− sinx)/ sinx. Red = f(x); Green=x3/6x. Right: Plot off(x) ≡ ((1− 1

2x2)− cosx)/ cosx. x is in radians in both plots.

c. In Special Relativity, as we will see, the total energy E of a particle of rest mass mand momentum p is given by E =

√p2c2 +m2c4. The realm of Newtonian mechan-

ics is where pc� mc2. In this domain, find the simplest nontrivial approximationfor 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 secondterm you will recognize as the classical kinetic energy p2/2m = 1

2mv2, and the first

term is a constant that doesn’t correspond to any quantity normally discussed inNewtonian mechanics. You probably recognize it, however, as Einstein’s famousrest 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 whenpc � mc2, that is, v � c, Einstein’s expression reduces to one equivalent toNewton’s. Thus we see that the use of approximation techniques yields not just asimpler expression, but real conceptual insight.

2. Approximate Harmonic Potentials.

Harmonic potentials are quadratic functions of the form U(x) = 12kx2 which play an

important role in oscillatory systems. You might recognize this function as the potentialenergy for a spring, of spring constant k, stretched a distance x beyond its resting

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position. (But if you don’t, don’t worry.) In this problem we will find that harmonicpotentials 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 theequation for the circle x2 +(y−r)2 = r2, with y measuring vertical height above theminimum point, and x measuring horizontal position. Since gravitational potentialenergy is given by U(y) = mgy for a bead of mass m, show that for small enoughexcursions around the bottom of the circle, U(x) ≈ 1

2kx2. Find an expression for k

in terms of the available parameters.

Solution:

Expand and reorganized the equation for the circle to y2 − 2ry + x2 = 0. A shortcut at this point would be to observe that for small oscillations around the bottomof the circle, y � r and y � x, and hence we can discard the y2 term, leavingy = 1

2rx2. Then U = mgy = 1

2mgrx2, so k = mg/r.

A fussier solution would be to start from y2−2ry+x2 = 0 and apply the quadraticformula to solve for y. This yields

y = (2r ±√

4r2 − 4x2)/2

= 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 motionnear the bottom. In this approach, y could be big (y ∼ 2r) or small (y ∼ 0) andthe key to approximation is in demanding x/r � 1 at the last step. Take they ∼ 0 solution and you’re done.

When we study harmonic oscillators, we’ll learn that the (angular) frequency ofoscillation 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 boundstate, may be approximated by a potential function

U(r) =a

r12− b

r6.

(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 theirrespective units. )

(iii) Compute the location, r0, of the minimum of U(r) in terms of arbitrary a andb.

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(iv) Calculate the Taylor Series expansion of U(r) about the minimum point, andtruncate the series under the assumption that you’re only interested in theregion where |r − r0| is small. This should give something of the form U(r) =U0 + 1

2k(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 = 0and hence the first nontrivial term in the Taylor expansion will be the 2ndorder term. Thus we need: U(r) = U(r0) + 1

2(r − r0)

2d2U(r0)/dr2. Some

algebra leads to:

U(x) ≈ − b2

4a+ 9

(b7

2a4

)1/3

x2

where x ≡ r − r0. In this case we would identify k = 92(b7/2a4)1/3 as the

“spring constant”... though this is admittedly not very illuminating in theabsence further information (or theory) about a and b.

0.8 1.0 1.2 1.4 1.6 1.8

-1.0

-0.5

0.0

0.5

r

UHr

L

0.98 0.99 1.00 1.01 1.02-1.02

-1.01

-1.00

-0.99

-0.98

r

UHr

L

Figure 2: Lennard-Jones potential with a=1, b=2. Left: full plot of U(r); Right: zoom into minimum, with overlay of parabolic approximation (calculated in part iv).

c. The Higgs potential.

Consider a one-dimensional toy “Higgs potential”, U(x) = −12µ2x2 + 1

4λ2x4.

(i) Plot U(x) for −2 < x < 2 and observe the shape. (For this purpose, setµ2 = λ2 = 1.)

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(ii) Compute the location, x0 > 0, of the positive-side minimum of U(x) in termsof arbitrary µ and λ.

(iii) Calculate the Taylor Series expansion of U(x) about the minimum point, andtruncate the series under the assumption that you’re only interested in theregion where |x− x0| is small. This should give something of the form U(x) =U0 + 1

2k(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 realHiggs potential, this would indicate that the normal state of the Higgs fieldis 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, thisis:

U(η) ≈ − µ4

4λ2+ η2µ2

In a proper particle-physics treatment of this problem, this equation wouldindicate 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 mani-festation of the oscillation energy of the Higgs field x = µ

λ+ η as it “sloshes”

in this parabolic potential well. Unfortunately in practice the parameter µis itself not predicted so the Higgs mass remains unknown. There is somechance it will be discovered and measured in the next year or so. For morepropaganda check out http://cms.web.cern.ch/cms/index.html.

-2 -1 0 1 2

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

x

UHx

L

Figure 3: Higgs potential with µ = λ = 1.

3. Complex exponentials and circular motion.

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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 thisexpression has a very simple equivalent in familiar functions.

b. A complex number like z = a+ ib is conveniently represented in the complex planein much the same as a vector r = ax + by is represented in the real xy plane. Theanalog of r2 = r · r is |z|2 = zz∗ = (a + ib)(a − ib). In the complex plane, thehorizontal axis measures the real part of a number and the vertical axis measuresthe 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 polarcoordinates, (r, θ). Show by explicit construction that any complex number a + ibcan 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 Cartesiancoordinates by r = r(cosωtx+sinωty), we can see that it might also be convenientlyrepresented by r(t) = r0e

iωt in the complex plane if we let the imaginary axisplay the role of the y axis of the real plane. From this, derive the velocity andacceleration and show that the results agree with standard statements about UCM.

Solution:

a. The series for the non-complex case is

ex = 1 + x+x2

2!+x3

3!+x4

4!+x5

5!+ ...

With the factor i in the exponent we get

eix = 1 + ix− x2

2!− ix3

3!+x4

4!+ix5

5!+ ...

= (1− x2

2!+x4

4!+ ...) + (ix− ix3

3!+ix5

5!+ ...)

= cos x+ i sinx

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, evidentlyb/a = tan θ.

c. Given r(t) = r0eiωt, we can easily take time derivatives to find v = dr/dt =

iωr0eiωt and then a = dv/dt = d2r/dt2 = −ω2r0e

iωt. Notice that at t = 0r = r0 lies along the real (horizontal) axis, while v = iωr0 lies exactly alongthe imaginary (vertical) axis. Thus we find that v is 90◦ ahead of r, just as wefound in analyzing this situation with real 2-dimensional vectors. Similarly, apicks up another factor of i, which rotates it another 90◦ from v, or 180◦ fromr. Note also that the magnitudes agree with what we found the traditional way,namely v = rω and a = vω = rω2. And finally, notice how each derivative brings

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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 givencomplex number z = a + ib in the complex plane, zeiθ is just z rotated by θ inthe counterclockwise direction. The mathematical representation r(t) = r0e

iωt isthus 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, r = Aθ, θ = 1

2αt2, θ = αt, and θ = α.

a. See Fig. 4.

b. Let’s write out the full vector expressions for position, velocity, and accelera-tion. To do so, we simply start with ~r and systematically take time derivatives –remembering that the basis vectors r and θ are time-varying themselves.

~r(t) = Aθr

~v(t) = Aθr + Aθθ θ

~a(t) = A(θ − θθ2)r + A(2θ2 + θθ) θ

Plugging in the time dependences given above, these become:

~r(t) = 12Aαt2r → 1

2t2r

~v(t) = Aαt r + 12Aα2t3 θ → tr + 1

2αt3 θ

~a(t) = Aα(1− 12α2t4)r + 5

2Aα2t2 θ → (1− 1

2α2t4)r + 5

2αt2 θ

In the second set of equations above (after the → arrow) I’ve set A = α = 1 forclarity since these just factor out anyway. With all this laid out, it is now easyto see that when θ = 1

2αt2 = 1/

√2, the r component of ~a vanishes.

c. The magnitudes of the radial and tangential accelerations are Aα|1 − 12α2t4| =

Aα|1− 2θ2| and 52Aα2t2 = 5Aαθ. Since 1− 1

2α2t4 will be positive at early t and

negative later, we need to solve two quadratic equations to find all the cases wherethese two magnitudes are equal:

(1− 2θ2)− 5θ = 0 (θ < 1√2)

corresponding to early t, and

(2θ2 − 1)− 5θ = 0 (θ > 1√2)

corresponding to late t. The solutions are θ = (√

33± 5)/4 ≈ 0.19, 2.69 radians.See Fig. 5 for additional info.

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-1.0 -0.5 0.0 0.5 1.0 1.5 2.0

-1.5

-1.0

-0.5

0.0

0.5

x

y

Figure 4: Red: trajectory r(t) for 0 ≤ θ = 12αt2 ≤ 2π; Blue: velocity vectors are selected

points; Green: acceleration vectors at selected points. For this plot, A = 1/π, α = 1, andvelocity and acceleration vectors have arbitrary (but fixed) scaling.

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0.0 0.5 1.0 1.5 2.00.0

0.5

1.0

1.5

2.0

t

Vel

ocity

Hred, greenL= Hvr,vΘL

0.0 0.5 1.0 1.5 2.0-2

-1

0

1

2

t

Acc

eler

atio

n

Hred, greenL= Har,aΘL

Figure 5: Velocity and Acceleration r and θ components, versus time.

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