Georgi, Howard. Physics 16 - Mechanics and Special Relativity (англ.)

To compute the angular momentum, we need an additional factor of rv = r

2

/` in the integrand

of (9):

Z

`

−`

dr

T r

2

dr

`

q

1 − r

2

/`

2

= T `

2

Z

1

−1

z

2

dz

√

1 − z

2

(12)

The same trigonometric substitution allows us to write this as

L = T `

2

Z

π/2

−π/2

dθ sin

2

θ =

1

2

πT `

2

(13)

Now the point is that we can eliminate ` from (11) and (13) and write this as a relation between the

energy and angular momentum

E

2

= π

2

T

2

`

2

= 2πT L (14)

Sure enough, when we look at quark-antiquark bound states, we see families of particles with

similar properties in which E

2

grows linearly with L within each family.

1

The data shows a slope

of about

2πT ≈

1 GeV

2

¯h

(15)

From this we can extract an approximate value for T , which works out to about 15 tons! These

little particles are pulled by a rather strong force! That’s why quarks and antiquarks don’t get very

far apart!

Good questions

Last time, we saw that a particle of mass m subject to a constant force F

0

in the x direction and

starting from x = 0 at t = 0 satisﬁes

x(t) =

q

F

2

0

t

2

+ m

2

− m

F

0

(16)

I suggested that you put the factors of c back into this and see how the nonrelativistic result

emerges. A couple of people asked about this, so I thought we should just do it.

x(t) =

q

F

2

0

c

2

t

2

+ m

2

c

4

− mc

2

F

0

(17)

There is a very standard trick for dealing with the Taylor expansion in situations like this where

you have a big term and other terms in a square root — factor factor the big term out.

x(t) =

r

m

2

c

4

³

1 +

F

2

0

t

2

m

2

c

2

´

− mc

2

F

0

(18)

x(t) =

mc

2

F

0





Ã

1 +

F

2

0

t

2

m

2

c

2

!

1/2

− 1





(19)

1

See for example http://arxiv.org/abs/hep-ph/0103274.

4

x(t) =

mc

2

F

0

Ã

1

2

F

2

0

t

2

m

2

c

2

+ ···

!

(20)

A number of people asked about the color force, so I thought that if I have time, I should talk

brieﬂy about the “four forces” and discuss why two of them may be unfamiliar to many of you.

You will suffer dire consequences after toying with powerful forces you do not under-

stand, namely gravity.

- The Onion Horoscope

Relativistic trafﬁc

Consider relativistic trafﬁc on a Boston street. We will make the completely unrealistic assumption

that the trafﬁc is ﬂowing smoothly, with evenly spaced cars moving in opposite directions at speed

v as shown below.

v → v → v → v → v → v → v → v → v →

← v ← v ← v ← v ← v ← v ← v ← v ← v

(21)

You will notice that there is something else that is unrealistic about this picture. There are plentiful

parking spaces on the side of the road. This picture is animated in the program TRAFFIC.EXE.

Now let us look at this picture from the inertial frame of a driver in the lower lane of trafﬁc. In

the animation, you can change the inertial frame gradually by pressing the right horizontal arrow.

Each time you press it, you increase the speed of the frame, until you to get to an inertial frame

that is moving along with the lower cars. Of course, as you increase the speed of the inertial frame,

the parking spaces appear to move in the opposite direction, and the cars in the other lane move by

faster. But here, we also see a number of effects of Lorentz contraction. The animation assumes

that the initial speed of the cars is

10

11

c, so that things are very relativistic and Lorentz contraction

is a big effect. I have done this to make the important effects show up better. In the actual physical

demo you will see, the effect is much smaller, but the qualitative message is the same. Notice

that as the lower cars slow down in the moving frame, they also appear longer. This is because

they appeared Lorentz contracted in the initial frame, and going to the moving frame undoes some

of the effect. The parking spaces and the upper cars going in the opposite direction are Lorentz

contracted more and more as the speed of the frame increases. Finally, we get to the frame in

5

which the lower cars are at rest, which looks sort of like this:

v = 0 v = 0 v = 0 v = 0 v = 0 v = 0

←V ←V ←V ←V ←V ←V ←V ←V ←V ←V ←V ←V ←V ←V ←V ←V

(22)

There are several things to notice about this. One is you see all the other cars in your lane farther

apart than they were in the frame of the road, and they don’t seem to ﬁt into the parking spaces

anymore. This is just Lorentz contraction. This is the same thing that is going on the problem of

the relativistic chocolate chip cookies in David Morin’s book. There you remember that the cookie

dough was moving at relativistic speed on a conveyor belt and a circular cookie press was stamping

out cookies - but the because the dough is Lorentz contracted, the cookies come out elongated.

The cars going in the other direction are Lorentz contracted even more, and look shorter and closer

together.

Something may be bothering you about this. How can the density of parking spaces change

relative to the density of the cars? Where do those extra parking spaces come from? If we can

harness this, we will be rich beyond the dreams of avarice. In fact, we can’t make any money on

it, because these extra parking spaces are coming in from the past and the future. This is possible

everywhere if we have an inﬁnite line of them stretching out in both directions, but it is always

possible in some ﬁnite region of the roadway. The density in the rest frame is not the same as the

density in a moving frame because of the relativity of simultaneity (which is the culprit in most

such puzzles). But again, the effect is even larger for the cars going in the opposite direction.

Because they are moving faster, they are even more Lorentz contracted than the parking spaces.

Now for the Demo. To avoid spoiling the fun, I will not tell you exactly what I will do in the

version of these notes that I put up on the web before class. But I will put up a complete version

right after class.

Of course, I don’t really have any relativistic cars available to test these theories. But there

is a very interesting analogy. Instead of two opposing lanes of trafﬁc, consider current in a wire

ﬂowing up and back from a battery.

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.

..

.

− → − → − → − → − → − → − → − → − → − →

← − ← − ← − ← − ← − ← − ← − ← − ← − ← −

+ + + + + + + + + +

.

.....................................................

.

....................................................

−

+

In this analogy, the negatively charged electrons are the cars and the rest of the electrical structure

of the wire form positively charged parking spaces. Of course the actual motion is more chaotic —

more like real Boston trafﬁc. But the net current is as shown. The whole system looks electrically

6

neutral because there is one car for each parking space. Now the question is, what force do the

electrons feel. The force depends on the velocity, and thus it is simplest to think about it in the in-

ertial frame moving along with the electrons. In this frame, any force on the electrons proportional

to velocity goes away, because they are at rest. But in this frame, as the animation or (22) shows,

the electrons in the lower wire see far more negatively charge electrons in the upper wire than they

see positively charged parking spaces for the electrons. Thus they experience a repulsive force.

In the original frame, there is no Coulomb force because everything is neutral. In this frame, the

repulsive force is still there, but it is interpreted as magnetism. When I connect the battery and get

my two lanes of trafﬁc moving, there should be a repulsive force between them.

Thus as you will see in more detail in Physics 15b, magnetism is a relativistic effect. When

you feel the mysterious force between two magnets, that I am sure has fascinated all of you at one

time or another, you are feeling in your bones the effect of special relativity.

7

lecture 15

Topics:

Where are we now?

Review of circular motion

Cross products - introduction

More general motion about an axis

Small rotations

The reference point

Rigid bodies rotating about a ﬁxed axis

The parallel axis theorem

Planar bodies moving in a plane

Appendix - cross products - details

Where are we now?

Now we are going to change gears and talk about angular momentum. We will see that conservation

of angular momentum is associated with the invariance of the laws of physics under rotations. In

that sense, the subject of angular momentum is not so different from relativity. Both are related to

symmetries of the laws of physics.

Review of circular motion

You have all learned in previous physics courses about uniform circular motion. I thought that it

would be useful to review this before going on to discuss rotations in more generality. So let us

consider an object that is moving in a circle of radius R centered at the origin with a constant speed,

v. This is motion in a single plane (because every circle lies in some plane), so it is convenient to

choose our coordinate system so that the motion is in the x-y plane with z = 0. We can also choose

to have the motion in the counterclockwise direction as seen from above (by choosing which is the

positive z direction). Then the motion of our particle can be written as

x(t) = R cos(ωt + φ) y(t) = R sin(ωt + φ) z(t) = 0 (1)

where ω is the angular velocity in radians per unit time. This deﬁnes the position vector of the

particle, ~r (t), in the usual way:

~r (t) = x(t) ˆx + y(t) ˆy + z(t) ˆz = ˆx R cos(ωt + φ) + ˆy R sin(ωt + φ) (2)

This should be familiar. We saw it earlier in the course in another context when we were discussing

the connection between uniform circular motion and complex exponential. This motion in the x-y

plane is precisely the same as the motion of the complex exponential

R e

iωt

(3)

1

in the complex plane.

The velocity of the particle is the derivative

~v (t) =

˙

~r (t) = −ˆx Rω sin(ωt + φ) + ˆy Rω cos(ωt + φ) (4)

Note that because sin

2

θ + cos

2

θ = 1, the length of ~v (t) is Rω, so sure enough, the speed of the

particle is constant,

v = |~v (t)| = Rω , (5)

even though the velocity vector is constantly changing.

The relation between (1) and (4) is illustrated below at time t (for ω > 0):

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.

x

.

..

.

y

.

..

.

..

.

..

.

~r (t)

~v (t)

ωt + φ

(6)

Note that ~v (t) is always perpendicular to ~r (t) and still in the x-y plane (you can see this from the

diagram, but you can also see it by explicitly computing the dot product, ~r (t) · ~v (t) and showing

that it is zero for all t).

1

We can compute the acceleration of the particle by differentiating (4).

~a (t) =

˙

~v (t) = −ˆx Rω

2

cos(ωt + φ) − ˆy Rω

2

sin(ωt + φ) = −ω

2

~r (t) . (7)

The accleration in (7) is always directed to the center of the circle and the magnitude of the accel-

eration is ω

2

R, or, using (5), this gives the (I hope familiar) formula

a = |~a (t)| = ω

2

R =

v

2

R

(8)

1

In fact, ~r (t) · ~v (t) = 0 for any motion on a circle, centered at the origin, whether it is uniform or not. You can

see this by differentiating ~r (t) · ~r (t) = R

2

.

2

Here is (6) again with the acceleration shown this time.

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.

x

.

..

.

y

.

..

.

..

.

..

.

..

.

~r (t)

~v (t)

~a (t)

ωt + φ

(9)

Here is a simple application that is related to something that we know in various other ways.

Suppose that we have a force of the form

~

F (r) = −α r

β

ˆr (10)

directed toward the origin with magnitude r

β

. For what value of β do the circular orbits of a mass

subject to this force have the same angular frequency independent of r? We want

~

F (r) = −α r

β

ˆr = m~a = −m ω

2

r ˆr (11)

so

ω

2

=

α

m

r

β−1

(12)

and if β = 1 the angular frequency is independent of r. This is just a harmonic oscillator in more

than one dimension.

Cross products - introduction

To get a deeper understanding of circular motion in three dimensions, and to ready ourselves to

study the more general problem of rigid body rotations, we will reformulate the discussion of

uniform circular motion in terms of cross products. The advantage of this reformulation is that it

makes the vector structure of the rotations more explicit. This is what we will need to take the next

steps. Cross products are going to be particularly convenient as we talk about rotations and angular

momentum.

Look at (9). As we go from the position, ~r , to the velocity, ~v, to the acceleration, ~a, each time I

just rotate the vector by 90

◦

and change the length by a factor. This is what the cross product does.

The cross product can be deﬁned by

h

~

A ×

~

B

i

x

= A

y

B

z

− A

z

B

y

,

h

~

A ×

~

B

i

y

= A

z

B

x

− A

x

B

z

,

h

~

A ×

~

B

i

z

= A

x

B

y

− A

y

B

x

. (13)

3

This looks complicated (I hope that it will seem simpler after you have read the discussion in the

appendix). The key point is that (4) can be written as

d

dt

~r (t) =

˙

~r (t) = ω ˆz × ~r (t) (14)

because ~r (t) satisﬁes

~r (t) · ˆz = 0 . (15)

The ˆz× produces the 90

◦

twist, and the factor of ω rescales the vector by a factor (necessary in this

case just to get the dimensions right).

You can see from the explicit deﬁnition in the appendix the following property:

If ~α is in the x-y plane, then ˆz × ~α is also in the plane, it has the same magnitude as

~α, and its direction is that of ~α rotated by 90

◦

counterclockwise about the z axis.

(16)

This is just the right-hand rule. If ~α is not in the x-y plane (as you can see in more detail in the

appendix) then the cross product just throws away the component that is not in the plane and acts

in the usual way according to (16) of the component in the place.

The relation (14) is very general. Not only is the time derivative given by this cross product for

the position vector, but any vector undergoing uniform circular motion in the x-y plane behaves

the same way.

For any vector ~u (t) undergoing uniform circular motion in the x-y plane, the time

derivative is given by the cross product

d

dt

~u (t) = ωˆz × ~u (t) .

(17)

In particular, this gives us another way of calculating the acceleration. The velocity vector is also

undergoing uniform circular motion. Thus

~a (t) = ω ˆz × ~v (t) = ω ˆz ×

³

ω ˆz × ~r (t)

´

= −ω

2

~r (t) (18)

Where the last step follows because each cross product with ˆz just rotates ~r by 90

◦

, so the two

cross products give a rotation by 180

◦

, which is just a sign change.

One of the crucial things about describing uniform circular motion in this way with cross prod-

ucts is that (14), (15) and (18) are vector equations. The ˆz appears in (14), (15) and (18) because

rotation in the x-y plane is rotation about the z axis. But because these are vector equations, we

can rotate them to an arbitrary coordinate system. Then we can describe uniform circular motion

in any plane through the origin, about an arbitrary axis through the origin. If the rotation is in some

other plane through the origin, say perpendicular to a unit vector ˆn (it is then rotation about the

axis, ˆn), then the rotated versions of (14), (15) and (18) are

~v (t) =

d

dt

~r (t) = ω ˆn × ~r (t) , (19)

4

~r (t) · ˆn = 0 (20)

and

~a (t) =

d

dt

~v (t) = ω ˆn ×

³

ω ˆn × ~r (t)

´

= −ω

2

~r (t) , (21)

where the last step follows because the cross product with ˆn on a vector perpendicular to it just

produces a rotation of 90

◦

in the plane perpendicular to ˆn. Remember that (19) works only for

motion that is circular! But still as we will see this is very powerful.

More general motion about an axis

Let’s go back to circular motion about the z axis and think brieﬂy about what happens if the vector

r is not in the x-y plane. In this case, uniform circular motion about the z axis looks something

like this:

.

..

.

..

.

ˆz

~r

~v

.

..

.

...

......

.

..

.

...

..

.

..

.

(22)

The velocity is in the x-y plane. What is going on is that the component of ~r perpendicular to ˆz is

undergoing uniform circular motion in the x-y plane while nothing is happening to the component

of ~r in the ˆz direction. This is very elegantly encoded in the cross product formula, (14) if we just

drop the condition (15) that ~r be perpendicular to ˆz and thus in the x-y plane. Thus

~v (t) = ω ˆz × ~r (t) (23)

is more general than we claimed above. Furthermore, there is really no reason to restrict ourselves

to uniform circular motion. If at some time t, the vector ~r is instantaneously rotating about the z

axis with angular velocity ω(t), the velocity is

d~r

dt

= ~v (t) = ω(t) ˆz × ~r (t) (24)

Finally, again because (24) is a vector equation, we can describe instantaneous rotation about

an arbitrary axis ˆn by just replacing ˆz → ˆn:

.

..

.

..

.

......

....

ˆn

~r

~v

.

..

.

...

.

(25)

5

Thus we can make the following very general and very important statement:

If at some time t, the vector ~r is instantaneously rotating about the ˆn axis through

the origin with angular velocity ω(t), the velocity is

d~r

dt

= ~v (t) = ω(t) ˆn × ~r (t )

(26)

This is a very general connection between a rotation and a cross product. Eventually, we will

discuss what happens if the rotation is about an axis that does not go through the origin - but this

is enough for now. We will use (26) very often so make sure that you understand it.

Small rotations

It will also be useful to rewrite (26) as an expression for the instantaneous change in a vector under

an inﬁnitesimal rotation, which we can do by noting that

ω(t) =

dθ

dt

(27)

so that multiplying both sides of (26) by dt gives d~r, the inﬁnitesimal change in the vector ~r under

an inﬁnitesimal rotation dθ,

d~r = dθ ˆn × ~r (28)

One reason for thinking about inﬁnitesimal rotations is that we know that inﬁnitesimal symme-

try transformations lead to conserved quantities by Noether’s theorem. For inﬁnitesimal rotations,

the coresponing conserved quantity is a component of the angular momentum.

For a 3-dimensional system described by some number of 3-vector coordinates ~r

j

, the con-

served quantity associated with invariance under rotations about an axis ˆn is

~

L · ˆn (29)

where

~

L =

X

j

~r

j

× ~p

j

(30)

where ~p

j

is the momentum of the jth particle,

~p

j

=

∂L

∂

˙

~r

j

(31)

If the Lagrangian is invariant under all rotations, all components, and therefore the entire vector,

~

L

is conserved —

~

L is the angular momentum — this is angular momentum conservation.

What I think is worth understanding about this is the way that the appearance of the cross

product in rotations leads to the appearance of the cross product in the expression for the angular

6

Georgi, Howard. Physics 16 - Mechanics and Special Relativity (англ.)

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