Accelerated Reference FramesMonday, 25 November 2013

Newton’s equations apply in inertial frames, but some-times it is more convenient to work in an acceleratedframe. In accelerated frames, pseudo forces corre-sponding to the acceleration of the frame must be addedto provide a consistent description of the dynamics.

Ph

ysic

s11

1

This needs to be re-ordered and simplified.

Center of Mass

Before turning to accelerated frames, I will reiterate a point made before aboutthe kinetic energy of a system of mass points. Namely, the kinetic energy maybe expressed in terms of the motion of the center of mass and the motion withrespect to the center of mass. The derivation is important but not difficult, sohere goes.

The kinetic energy of the particles is

T = 1

2

∑α

mαxα · xα = 1

2

∑α

mα(R+ rα) · (R+ rα)

where xα = R+ rα is the position of the αth particle with respect to an inertialframe,

R ≡ 1

M

∑α

mαxα (1)

is the center of mass of the particles, M = ∑αmα, and rα is the position of the

αth particle with respect to the center of mass. Expanding the dot product gives

T = 1

2

∑α

(mαR2 +mαr 2

α+2mαR · rα)= 1

2

(MR2 +∑

αmαr 2

α+2R · d

d t

∑α

mαrα

)but by the definition of the center of mass in Eq. (1), the sum in the final term inparentheses vanishes. Therefore,

T = 1

2MR2 + 1

2

∑α

mαr 2α (2)

which is to say that the kinetic energy may be expressed as that of a single masspoint having the entire mass of the system, located at the center of mass, and

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1. ACCELERATED FRAMES

moving with the center of mass, plus the kinetic energy of particles moving withrespect to the center of mass. We shall see shortly that the summation in Eq. (2)takes a simpler form in the case of a (rotating) rigid body, but the expressiongiven here is perfectly general.

Exercise 1 Show that if the particles interact amongst themselves via forces thatobey Newton’s third law, in addition to any external forces, the acceleration ofthe center of mass is equal to the sum of all external forces divided by the totalmass of the particles.

1. Accelerated Frames

Let’s remind ourselves of Newton’s three laws, first in Motte’s 1729 translation:

I. Every body perseveres in its state of rest, or of uniform motion in a rightline, unless it is compelled to change that state by forces impress’d thereon.

II. The alteration of motion is ever proportional to the motive force impress’d;and is made in the direction of the right line in which that force is im-press’d.

III. To every action there is always opposed an equal reaction: or the mutualactions of two bodies upon each other are always equal, and directed tocontrary parts.

The first law claims that in the absence of external forces, a body either remainsat rest or it moves at constant velocity. In both cases, the claim makes senseonly if we are looking at the body in an inertial frame of reference. Put anotherway, the first law argues that there are inertial frames of reference and motion isparticularly dull in such frames for objects free from external forces.

The second law needs a little translational help. If motion simply means velocity,then the (time rate of) change in velocity is proportional to the impressed force;

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1. ACCELERATED FRAMES

on the other hand, if we understand motion to mean momentum, then we have

dp

d t= F (3)

In either case, the statement presumes that we are looking at the “body” from aninertial frame of reference.

The third law argues that forces come in pairs: the force on A caused by B is equaland opposite to the force on B caused by A. Their vector sum is always zero.

Newton’s laws hold (at least approximately) in inertial reference frames. We havebuilt Lagrangian and Hamiltonian dynamics on the Newtonian bedrock, so youought not to be surprised that they presume that coordinates are measured withrespect to an inertial frame. Sometimes, however, that’s just not particularly con-venient. If we happened to be doing physics inside an elevator—and frankly,who doesn’t from time to time?—then we might find it much more convenient touse the elevator as the reference frame and patch up Newton’s laws by adding inone or more terms to account for the motion of the noninertial (elevator) frame.In general, such a frame may both translate and rotate with respect to an inertialframe. How can we account for this?

Translations are straightforward. Let the coordinates of a point on the elevatorbe R and let us measure positions within the elevator with respect to said point.Then

x′ = R+ r

where x′ is the position of a mass point with respect to an inertial frame, R is theposition of the elevator in the inertial frame, and r is the position of the particlewith respect to the elevator. Since

Fext = mx = mR+mr

the equation of motion in the elevator frame is

mr = Fext −mR

That is, the mass times the acceleration in the elevator frame is equal to the sumof the external forces plus a “pseudo force,” −mR. In particular, if the elevator isin free fall in a uniform gravitational field, then the acceleration (in the elevatorframe) is

mz = mg −mg = 0

and the particle does not accelerate with respect to the elevator.

Physics 111 3 of 11 Peter N. Saeta

2. ROTATION

A

A+δAδA

δθ

Figure 1: The frame in which A is at restrotates through angle δθ, “displacing” Ato A+δA (red). Translating A back to alignthe tail of the vector to the original vectorA, allows us to identify δA, the change inA (green). Note that the vector directionof δθ is perpendicular to the plane of thefigure and determined by the right-handrule.

2. Rotation

Rotation is a more interesting case. Consider a vector A at rest in a frame thatrotates with respect to an inertial frame through an infinitesimal angle δθ. Asshown in Fig. 1, the change in A is perpendicular to both A and δθ and has mag-nitude δθ A. If A had a component parallel to δθ, that component would not bemodified by the rotation; it is only the component in the plane perpendicular tothe rotation axis that counts. The upshot is that the change in A is given by

δA = δθ×A

If the rotation takes place in time δt , then taking the limit as δt → 0 we obtain(dA

d t

)in=ω×A (4)

The total time derivative on the left gives the change in the vector A in the in-ertial frame, the angular velocity vector ω indicates rotation with respect to theinertial frame, and A on the right is the vector in the rotating frame. Of course,at any moment, a given vector in the rotating frame may be expressed in termsof coordinates in the inertial frame. It may be helpful to think of the inertial androtating coordinate systems to coincide at the moment we take the derivative.At that moment, then, the vector ω has the same expression in both coordinatesystems (even though the rotating system is not rotating with respect to itself).

If, in addition, the vector A itself is changing in the rotating frame, then thechange in A in the inertial frame is(

dA

d t

)in=

(dA

d t

)rot

+ω×A (5)

To understand Newtonian dynamics in the rotating frame, we will need to com-pute the acceleration, which requires us to apply Eq. (5) twice: first to get the

Physics 111 4 of 11 Peter N. Saeta

2. ROTATION

velocity v = r and again to get the acceleration a = r:

rin = rrot +ω× r (6)

rin = d

d t(rrot +ω× r)rot +ω× (rrot +ω× r)

so thatrin = rrot + ω× rrot +2ω× rrot︸ ︷︷ ︸

Coriolis

+ω× (ω× rrot)︸ ︷︷ ︸centrifugal

(7)

Once again, it is easiest to think of all the vectors on the right-hand side as ex-pressed in the rotating coordinate system, with the understanding that at themoment we take the derivative we expressω and ω (the angular velocity and an-gular acceleration of the rotating coordinate system with respect to the inertialframe) in the rotating coordinate system.

Expressing Newton’s second law, F = mrin, in terms of the quantities in the rotat-ing frame, we get

mr = F−mω× r−2mω× r︸ ︷︷ ︸Coriolis

−mω× (ω× r)︸ ︷︷ ︸centrifugal

(8)

where the three extra terms on the right are three pseudo forces that arise be-cause we are using coordinates in the (noninertial) rotating frame. The first ofthese is usually negligible in situations of practical interest. The second is calledthe Coriolis pseudo force,1 and the third is the centrifugal pseudo force. Let’sconsider each of them in turn.

To simplify things a bit, we will consider two-dimensional motions on a rotatingturntable in a horizontal plane. The first pseudo force is nonzero during spin-up and points in the azimuthal direction. If you put a penny on a greased-upturntable at rest, start up the turntable in a counterclockwise direction seen fromabove, and watch the penny while standing on the axis of the turntable, you willsee the penny move to the right (as the turntable slides beneath it) under theinfluence of this pseudo force.

Now stop the turntable, strip off the grease, and start the turntable up again.The third term (the centrifugal force) grows like ω2 and points radially outward.Eventually, it exceeds the maximum static friction force the turntable can supplyand the penny begins to slide radially outward.

1Gaspard-Gustave de Coriolis (1792–1843) described this term in a paper of 1835, Sur leprincipe des forces vives dans les mouvements relatifs des machines, although Laplace had obtainedthe same term in connection with tides in 1778, while Giovanni Battista Riccioli and FrancescoMaria Grimaldi had described it in connection with artillery in 1651. Evidently, Coriolis’s namewas first associated with this term early in the 20th century.

Physics 111 5 of 11 Peter N. Saeta

2. ROTATION 2.1 Alternative Derivation

At this point, the Coriolis force comes into play. The radial motion is outward,and the angular velocity vector is vertically up, so the Coriolis force acts to theright as viewed from the center of the turntable. Instead of moving straight out-ward under the centrifugal force, the penny veers to the right.

2.1 Alternative Derivation

I would like to offer an alternative approach to the derivation of the accelerationin the rotating frame, this time using the Lagrangian formalism. To that end, weneed an expression for the kinetic energy, expressed in terms of our chosen gen-eralized coordinates (which are Cartesian coordinates r = (x, y, z) in the rotatingframe, measured with respect to an origin on the axis of rotation). In an infinites-imal time interval δt , this vector changes with respect to an inertial frame for tworeasons:

1. r changes: r → r+δr

2. the frame rotates: r → r+δθ× r

Of course, if both changes happen at once, we have

r → r+δr+δθ× r

so that the rate of change in position (in the inertial frame) is

v = δr+δθ× r

δt= r+ω× r

Since v is measured with respect to an inertial frame, we may use it to computethe kinetic energy of the particle:

T = m

2[r · r+2r ·ω× r+ (ω× r) · (ω× r)]

Note that each value of r and r refers to the generalized coordinates in the rotat-ing frame, and I have colored each term to make it easier to track each term inthe following.

Let us ignore any potential energy for the moment to focus on the kinetic energyin terms of the generalized coordinates (x, y, z) = (r1,r2,r3). We’ll take it step bystep. First, we need to express T in component form:

T = m

2

[ri ri +2εi j k riω j rk + (εi j kω j rk )(εi lmωl rm)

]Now we take the partial with respect to rn :

∂T

∂rn= m

2

[2rn +2εn j kω j rk

]

Physics 111 6 of 11 Peter N. Saeta

2. ROTATION 2.1 Alternative Derivation

The total time derivative of this quantity is

d

d t

(∂T

∂rn

)= m

[rn +εn j k (ω j rk +ω j rk )

]= m [r+ ω× r+ω× r] n

Taking instead the partial with respect to rn gives

∂T

∂rn= m

2

[2εi j n riω j +εi j nω j εi lmωl rm +εi j kω j rkεi lnωl

]I would now like to show that the two terms in red in this last expression are thesame. To manage this, I will use the permutation properties of the Levi-Civitasymbols to bring the n to the front:

∂T

∂rn= m

2

[2εi j n riω j +−εn j iω j εi lmωl rm −εnliωlεi j kω j rk

]= m

[−εn j i riω j − ω× (ω× r)n]

=−m [ω× r+ω× (ω× r)] n

Putting these results together, we have

d

d t

(∂L

∂rn

)− ∂L

∂rn= m [r+ ω× r+2ω× r+ω× (ω× r)] n + ∂U

∂rn= 0

In other words, the “Newtonian” equation of motion for r is

mr =−∇U −mω× r−2mω× r−mω× (ω× r)

Of course, this result is identical to Eq. (8). All terms involving r are expressed inthe rotating frame.

Exercise 2 Show that the angular deviation δ of a plumb line from the true ver-tical at a point on the Earth’s surface at latitude λ (measured from the Equator)is

δ=12 Rω2 sin2λ

g −Rω2 cos2λ

where R is the radius of the Earth. What is the greatest value (in arc-seconds) forthis deviation?

Physics 111 7 of 11 Peter N. Saeta

3. FOUCAULT’S PENDULUM

Exercise 3 Compare the centrifugal acceleration at the Equator due to theEarth’s rotation to the gravitational acceleration. By what factor is the centrifugalacceleration due to the Earth’s rotation greater than the centrifugal accelerationdue to its motion about the Sun. (Light takes about 500 seconds to travel fromthe Sun to the Earth.)

3. Foucault’s pendulum

Jean Bernard Léon Foucault (1819–1868) made the first successful demonstra-tion of the Earth’s rotation using a pendulum at the Paris Observatory in Febru-ary, 1851. Shortly thereafter, he suspended a 28-kg bob from a 67-m long wirefrom the dome of the Panthéon in Paris. The pendulum’s plane of oscillationrotated clockwise 11 per hour. Why?

Let’s assume that the pendulum is at latitude λ (measured with respect to theEquator), as shown in Fig. 2. Then the Earth’s angular velocity is

Ω=Ω(zsinλ− xcosλ) (9)

Physics 111 8 of 11 Peter N. Saeta

3. FOUCAULT’S PENDULUM

λ

Ω

x

z

y

(a) Coordinate system for Foucault’spendulum

x

z

ℓ

y

m

(b) Local coordinates

Figure 2: Foucault’s pendulum.

where z points vertically up (i.e., along the line from the center of the Earth), ypoints east, and x points south. The Coriolis force is thus

FCoriolis =−2mΩ× r =−2mΩ[zsinλ− xcosλ]× (xx+ y y+ zz) (10)

For small-amplitude oscillations, z is negligible. Then

FCoriolis ≈−2mΩ[sinλ(xy− y x)

]+higher order terms (11)

Newton’s equations in the x y plane then become

m(xx+ y y) =−mg( x

`x+ y

`y)−2mΩsinλ

(xy− y x

)(12)

Let ω2 = g /`. Then Eq. (12) becomes the coupled equations

x +ω2x −2Ωsinλ y = 0 (13)

y +ω2 y +2Ωsinλ x = 0 (14)

An elegant approach to solving these coupled equations is to pack x and y intothe complex variable z = x + i y . Multiplying Eq. (14) by i and adding, we have

x + i y +ω2(x + i y)+2Ωz (i x − y) = 0 =⇒ z +ω2z +2iΩz z = 0 (15)

Physics 111 9 of 11 Peter N. Saeta

4. EXERCISES AND PROBLEMS

where Ωz =Ωsinλ is the vertical component of the Earth’s angular velocity. Wenow guess a solution of the form z = z0e i st , substitute into Eq. (15), and convertthe differential equation to a quadratic equation:

(−s2 +ω2 +2iΩz i s)z = 0 =⇒ s2 +2Ωz s −ω2 = 0 (16)

with roots

s =−Ωz ±√Ω2

z +ω2 (17)

Of course, the oscillation frequency of the pendulum is many orders of magni-tude greater than the rotation rate of the Earth, so to first order inΩz

s ≈−Ωz ±ω(1+ Ω2

z

2ω2

)≈−Ωz ±ω (18)

The general solution for z is therefore

z = (Ae iωt +Be−iωt )e−iΩz t (19)

Suppose that at t = 0 the pendulum is aligned with the x direction. Then theterm in parentheses is (and remains) real, since A = B and the term is 2A cosωt .If we wait until Ωz t = π/2, then z will be purely imaginary and the pendulumoscillates in the y direction. Hence, the plane of oscillation rotates at angularfrequencyΩz =Ωsinλ. At the latitude of Claremont, 345′48′′ N, the plane wouldrotate with a period of 42.8h. If we set up a Foucault pendulum with a circle oflittle pins surrounding it, why does this period imply that all the pins would beknocked over in 21.4h?

4. Exercises and problems

Exercise 4 The latitude of the Panthéon in Paris is 4850′46′′ N. Calculate therate of rotation of the plane of Foucault’s pendulum and compare to that re-ported above.

Problem 5 – Dangerous Celebration? If a particle is projected vertically up-ward to a height h above a point on the Earth’s surface at a northern latitude λ,show that it strikes the ground at a point

4/3ωcosλ√

8h3/g

to the west, if you neglect air resistance and assume h is small (compared towhat?).

Physics 111 10 of 11 Peter N. Saeta

4. EXERCISES AND PROBLEMS

Problem 6 – Centrifugal Potential Consider a particle moving in a potentialU (r ). Rewrite the Lagrangian in terms of a coordinate system in uniform rota-tion with respect to an inertial frame. Calculate the Hamiltonian and determinewhether H = E . Is H a constant of the motion? If E is not a constant of motion,why isn’t it? The expression for the Hamiltonian thus obtained is the standardformula 1/2mv2 +U plus an additional term. Show that the extra term is thecentrifugal potential energy.

Problem 7 Chemist Ferdinand Reich conducted early experiments on the Cori-olis acceleration in 1831, shortly before Coriolis made his investigation in 1832–5. Reich dropped pellets down a 158.5-m-deep mine shaft in Freiberg, Germany,and observed a mean eastward deflection of 28.3mm. The latitude of Freiberg isλ= 51.

(a) Calculate the expected eastward deflection, clearly stating any simplifyingassumptions you make.

(b) Reich also found a small southerly deflection, as have subsequent investi-gators. This is a second-order Coriolis effect. The eastward velocity arisesfrom the dominant vertical (z) motion; the eastward velocity (y) producesa Coriolis acceleration in the x direction, proportional to ω2, where ω isthe angular speed of the Earth. You also need to keep track of the variationof g and the centrifugal force with height, since they are also proportionalto ω2. In fact, all terms are proportional to h2ω2

g sinλcosλ. According toMarion and Thornton (10-13), the total southerly deflection is

4h2ω2 sinλcosλ

g

But they’re wrong. The correct leading coefficient is not 4. Find the cor-rect coefficient analytically by keeping track of all terms through quadraticorder in ω. Then check your work with an exact numerical calculation inMathematica.

Physics 111 11 of 11 Peter N. Saeta