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

The free symmetric top ala Euler

For the free symmetric top, ﬂoating out in space and rotating freely about its center of mass with

I

1

= I

2

= I

⊥

, the Euler’s equations take the following form:

I

⊥

δω

1

δt

= −(I

3

− I

⊥

) ω

3

ω

2

I

⊥

δω

2

δt

= (I

3

− I

⊥

) ω

3

ω

1

I

3

δω

3

δt

= 0

(23)

We immediately conclude that ω

3

, the component of angular velocity in the direction of the unequal

principal axis, is constant in time relative to the body frame. With ω

3

then ﬁxed, the other two

equations take the form

δω

1

δt

= −

(I

3

− I

⊥

) ω

3

I

⊥

ω

2

= −Ω ω

2

δω

2

δt

=

(I

3

− I

⊥

) ω

3

I

⊥

ω

1

= Ω ω

1

(24)

where

Ω =

(I

3

− I

⊥

) ω

3

I

⊥

(25)

and Ω is a constant.

We know that we can rewrite (24) as

δ~ω

δt

= (0, 0, Ω) × ~ω (26)

In agreement with the previous analysis in (10), this is what we expect for a vector ~ω undergoing

uniform circular motion with angular velocity Ω about the 3 axis.

You can also see this in a more boring way by solving the coupled differential equations. If we

put the second of the two equations in (24) into the ﬁrst, we get an equation of motion for ω

1

,

δ

2

ω

1

δt

2

= −Ω

2

ω

1

(27)

You can easily check that ω

2

satisﬁes the same equation. (27) is just the equation for simple

harmonic motion, with the general solution

ω

1

(t) = a cos(Ωt + φ) (28)

5

Then (24) implies

ω

2

(t) = a sin(Ωt + φ) (29)

As usual, a and φ must be set by the initial conditions. So, as we saw directly from (26), (28) and

(29) describe the two-dimensional vector,

³

ω

1

(t), ω

2

(t)

´

undergoing uniform circular motion with

angular velocity Ω. Note that the sign of Ω is related to the sign of ω

3

, and also whether I

⊥

> I

3

(which is what we would get for a prolate spheroid) or I

⊥

< I

3

(for an oblate spheroid like the

earth).

Thus the angular velocity in the body frame is ﬁxed in magnitude, but changes direction, mov-

ing with constant angular velocity around a cone.

The tennis racket theorem

Euler’s equations can also be used to understand an interesting fact about rotating rigid bodies with

no two equal moments of inertia. I’m a tennis player, so the tennis racket theorem is a favorite of

mine. The tennis racket, in this case, is just a convenient example of a rigid body with three unequal

moments of inertia. We know that a tennis racket, like any rigid body, is happy to rotate freely about

any one of its principle axes. For such a rotation, the angular momentum and the angular velocity

vector are in the same direction. The theorem concerns the nature of free rotations (no torques) that

begin with an angular velocity almost, but not quite, along one of the principal axes. The statement

of the theorem is that if the axis has the largest or the smallest moment of inertia, the motion is

stable and the angular velocity precesses about the axis, never getting very far away. However,

if the axis has the intermediate moment of inertia, the motion is unstable. The angular velocity

moves away from the axis exponentially and the motion becomes rather complicated. Here is how

it goes. Suppose that we consider motion in which ~ω in the body frame is nearly along principal

axis 3, so that in our nice body frame, ω

1

and ω

2

are much smaller than ω

3

. Euler’s equations look

like

I

1

δω

1

δt

= (I

2

− I

3

) ω

2

ω

3

I

2

δω

2

δt

= (I

3

− I

1

) ω

3

ω

1

I

3

δω

3

δt

= (I

1

− I

2

) ω

1

ω

2

(30)

For the assumed initial conditions, we can generally ignore the time dependence of ω

3

, because

its time derivative is proportional to the product of two small numbers, ω

1

and ω

2

. This argument

could go wrong if I

3

is very close to either I

1

or I

2

, but for a tennis racket in which the three

principle moments are very different, it is a good approximation.

If we then treat ω

3

as a constant, the other two Euler equations are linear in the small quantities,

6

ω

1

and ω

2

, and we can write them (approximately) as.

δω

1

δt

=

1

I

1

(I

2

− I

3

) ω

3

ω

2

δω

2

δt

=

1

I

2

(I

3

− I

1

) ω

3

ω

1

(31)

Differentiate the ﬁrst of these and use the second to get

δ

2

ω

1

δt

2

=

1

I

1

(I

2

− I

3

) ω

3

δω

2

δt

=

1

I

1

I

2

(I

3

− I

1

) (I

2

− I

3

) ω

2

3

ω

1

(32)

Now here is the point. If I

3

is the largest or smallest moment, then

1

I

1

I

2

(I

3

− I

1

) (I

2

− I

3

) ω

2

3

< 0 (33)

But then (32) describes an oscillation. ω

1

oscillates about zero and remains small. The same thing

happens to ω

2

. In this case, the motion of the system remains basically about axis 3 with small

oscillating wobbles. However if I

3

is the intermediate moment, then

1

I

1

I

2

(I

3

− I

1

) (I

2

− I

3

) ω

2

3

> 0 (34)

Then (32) describes an unstable equilibrium and ω

1

and ω

2

grow exponentially. Then our approx-

imation quickly breaks down. We can’t tell what happens next without a more detailed analysis.

But what is clear is that the motion around the axis with the intermediate moment of inertia is

much more complicated. You can easily see this yourself by playing with a tennis racket, or any

similarly shaped object.

7

lecture 22

Topics:

Where are we now?

Derivation of ﬁctitious forces

1/r

2

forces

The force between spheres

Tidal forces

Extra Dimensions

Where are we now?

This week, the general subject is gravity (and more generally “central” forces) and “ﬁctitious

forces.” Fictitious forces are convenient constructions that allow us to continue to use

~

F = m~a

in non-inertial frames of reference. This is actually familiar to all of us. Our brain makes the

same kind of construction whenever we turn a corner in a moving car and we feel a centrifugal

force. Gravity is also familiar, but seems rather different. Einstein however, would tell us that the

difference between the two is not so obvious as it seems. It is not an accident, he would say, that

both forces are proportional to the mass of the particle on which they act. We will begin to see

why this is important when we talk about the tides. And it will come back again later when we talk

about cosmology.

We will probably not get to everthing in the notes in the lecture today, but some of it is better to

read about, and some of it I will do next time. I will try to take a little time at the end to talk a bit

about the subject of “extra dimensions” and the connection with gravity, because this crazy idea is

sort of fun and something that many people are interested in today.

1/r

2

forces

Let us start our discussion of gravity by talking about what is so special about 1/r

2

forces? We

have talked about this a little before discussing relativistic strings. Then, I mentioned ﬁeld lines

and contrasted the 1/r

2

Coulomb force between electrically charged particles with the color force

between quarks. Gravity is really quite different from the Coulomb force. But the mathematics is

pretty similar. The potential energy associated with the gravitational force between two massive

particles in Newton’s theory of gravity is

V (~r

1

, ~r

2

) = −

G m

1

m

2

|~r

1

−~r

2

|

(1)

where ~r

1

and ~r

2

are the positions of the particles and m

1

and m

2

are their masses. The force on

particle 1 from particle 2 is

~

F

from m

2

on m

1

= −

∂

∂~r

1

V (~r

1

, ~r

2

) = −G m

1

m

2

~r

1

−~r

2

|~r

1

−~r

2

|

3

(2)

1

Because the potential is invariant under space translation because (1) just depends on the vector

~r

12

= ~r

1

−~r

2

(3)

from ~r

2

to ~r

1

, the satisﬁes Newton’s third law:

~

F

from m

1

on m

2

= −

∂

∂ ~r

2

V (~r

1

, ~r

2

) = −G m

1

m

2

~r

2

−~r

1

|~r

1

−~r

2

|

3

= −

~

F

from m

2

on m

1

(4)

It is also no surprize that the force vector points from particle 1 to particle 2 because the potential

just depends on the distance between the two particles and is therefore invariant under rotations in

addition to space translations. So the force has to be proportional to ~r

12

because there is no other

vector in the problem.

The potential energy associated with the Coulomb force between two charged particles is (in

cgs units)

q

1

q

2

|~r

1

−~r

2

|

(5)

where q

1

and q

2

are the charges. Clearly the dependence on the positions is exactly the same.

Both potentials lead to a 1/r

2

force. So we can understand some of the special properties of the

gravitational force by thinking about the Coulomb force instead. Before we start, perhaps I should

also emphasize the differences. The most obvious difference is that while the electric charge can

have either sign, the mass is always positive, so the gravitational force is always attractive. Less

obvious, but even more important, is the effect of the fact that the strength of the gravitational force

between two masses is proportional to the masses themselves.

1

This has far-reaching effects, a few

of which we will see later.

Now for the similarities and special properties. Newtonian Gravity and the Coulomb force

share the crucial property of linearity.

2

The force on a particle due to an arbitrary number of

masses or charges is just the sum over the forces from each of the other masses or charges. For

gravity

~

F

on m

1

=

X

j6=1

~

F

from m

j

on m

1

= −G m

1

X

j6=1

m

j

~r

1

−~r

j

|~r

1

−~r

j

|

3

(6)

We can extend this in an obvious way to continuous distributions of masses or charges. The

sums simply become integrals. In fact, as far as we can tell today, all the masses and charges in

the universe are essentially point masses and charges. So our integrals really are approximations

to sums. But the individual masses and charges are so small that the so-called continuum ap-

proximation, in which we replace a collection of lots of tiny masses or charges with a continuous

distribution is often an essentially perfect approximation.

One thing I want to do today is to go over Newton’s theorem. It is sufﬁciently important that

it doesn’t hurt to look at it several times. In fact, you will see it again in a different and more

1

Note that we are talking here about Newtonian gravity rather than Einstein’s General Relativity. Both of these

statements must be made more carefully in General Relativity, as we will glimpse after break.

2

Again this is true of Newtonian gravity but not General Relativity. In General Relativity, nonlinearities are present

because the gravitational ﬁeld carries energy, which in turn produces gravity. We will come back to this later.

2

sophisticated form if you take Physics 15b next semester. The theorem is that the force on a point

mass or charge due to a uniform shell of mass or charge is zero inside the shell and outside the

shell is just what you would get if all the mass or charge were concentrated at the center. I am not

so much interested in a rigorous proof as in making it obvious why the theorem is true.

First let’s show directly that there is no force inside a shell. There is a beautiful geometrical

argument that shows that this is a consequence of the 1/r

2

nature of the force. I hope that it will

help you remember the theorem. Consider the force on some point P inside the sphere of charge

or mass from some inﬁnitesimal area A anywhere on the surface of the sphere. The point is that

we can construct an inﬁnitesimal area A

0

on the other side of the sphere by drawing a straight line

from every point on the boundary of A through the point P . The locus of points where these lines

intersect the sphere on the other side is the boundary of A

0

. Now we can show (and will do so

momentarily) that the ratio of the area of A

0

to the area of A is equal to the square of the ratio of

the distance from A

0

to P to the distance from A to P . Thus because the force is proportional to

1/r

2

, the force from A

0

just cancels the force from A. If A is closer to P , it is also smaller in just

the right ratio to cancel the force. This why the theorem works.

Now let us see why the areas are in the right ratio. The situation is illustrated below in the

plane deﬁned by the point P , the center of the sphere, C, and the center of the region A:

.

..

.......

..

.

..

...

.....

..

.

A

0

•

P

•

C

(7)

The boundary of A seen from P subtends some inﬁnitesimal solid angle dΩ. By deﬁnition, A

0

subtends the same solid angle. The area on a sphere of radius ρ around P subtended by dΩ is

ρ

2

dΩ. This is the deﬁnition of solid angle. But because dΩ is inﬁnitesimal, this area is nearly ﬂat,

so it is also the area of the region subtended by dΩ on a plane perpendicular to the line from P a

distance ρ away. Thus the areas of the regions subtended on the perpendicular planes shown below

are in the ratio of the squares of the distances.

.

A

0

•

P

(8)

Now the areas of A and A

0

are larger than this because they are slanted. But both are slanted

at exactly the same angle because the triangle formed by the center of sphere and A and A

0

is

3

isosceles, and the tangents to the sphere are perpendicular to the radii to A and A

0

.

..

.

..

.........

..

.

..

....

..

.

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

.

θθ

A

0

•

P

(9)

Thus both areas are larger than the area of the corresponding perpendicular regions by the same

factor of 1/ cos θ. Thus the areas of A and A

0

are in the ratio of the squares of the distances from

P , as promised.

This shows that for every inﬁnitesimal area on the spherical shell, there is a region on the other

side that produces an equal and opposite force at P . This is not quite enough, because we need to

be sure that when we integrate over the whole sphere, each of these little areas appears only once.

Fortunately, this is easy. We can consider a plane through P and the center of the circle. Then we

can ﬁnd the total force at P by integrating only over the areas on one side of the plane. The areas

that cancel the force are all on the other side of the plane. Thus the whole sphere is accounted for,

and we have proven the ﬁrst part of the theorem.

The key idea that we will use to understand the second part of the theorem is something we

discussed when we talked about the electromagnetic force. The idea there was “ﬁeld lines.” The

electromagnetic force can be thought of in the following way. One charge produces an electro-

magnetic ﬁeld that affects other charges, producing the forces (of course, all charges produce these

ﬁelds, and we don’t talk about the effect of the part of the ﬁeld due to a given charge on the charge

that produces it — we will come back to this later in the course). The electric ﬁeld, in turn, can

be thought of as associated with ﬁeld lines, which begin on positive charges and end on negative

charges — they never begin or end in empty space — only on charges. The ﬁeld lines point in the

direction of the electric ﬁeld, and the density of the ﬁeld lines is proportional to the strength of the

electric ﬁeld. A single charge sends out ﬁeld lines symmetrically in all directions. Since the lines

can never end if there are no other charged particles around, they continue to spread. On a surface

of radius r, the lines are spread out over an area 4π r

2

. Thus the density of the ﬁeld lines falls off

as 1/r

2

. Thus the electric ﬁeld and therefore the force falls off as 1/r

2

. This is the basic reason for

the Coulomb force.

4

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

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

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

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

...

..

.

..

.

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

.

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

.

For gravitation, ﬁeld lines are very similar. Since there is only one kind of charge (which is just

mass) and like charges attract, we can think of ﬁeld lines as coming in from inﬁnity and ending on

masses. Like electric ﬁeld lines, gravitational ﬁeld lines cannot end where there is no mass. So all

the rest of the argument goes through in the same way.

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

.

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

.

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

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

...

..

.

..

.

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

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

...

..

.

..

.

Now what does this mean for Newton’s theorem? Consider now a uniform shell of mass. This

mass distribution has spherical symmetry. Let us ﬁrst show that this implies that the force must

be radial and equal in all directions. First consider the force at some point ~r. This force must be

the same if we rotate the mass distribution around the ~r axis, which implies that it must be in the

~r direction because any transverse component would rotate when we rotate the mass distribution.

Once we know that the force is radial, it is clear, again because of rotation invariance, that it is the

same in all directions, with the magnitude being only a function of |~r|.

5

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

.

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

.

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

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

...

.

..

.

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

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

.

..

...

.

....

.

Since the force vanishes inside the shell, there is no gravitational ﬁeld, and thus no ﬁeld lines

inside. All the ﬁeld lines go from inﬁnity to the shell. But the total mass produces the same number

of ﬁeld lines, whatever the radius of the shell. This must be true because very far away from the

shell, it looks just like a point mass, so the number of gravitational ﬁeld lines coming in from

inﬁnity must be the same independent of the radius of the shell. But the ﬁeld line cannot appear or

disappear outside the shell, because there is no mass there. Thus outside the shell, the gravitational

ﬁeld is completely independent of the radius of the shell, and therefore the same as that produced

by a point mass at the center. This ﬁnishes the theorem.

There are many important consequences of Newton’s theorem. One that we will come back

to soon is the gravitational force on a point mass m inside a spherically symmetric distribution of

masses centered at the origin. If the point mass is at ~r from the center of the distribution, Newton’s

theorem implies that the force is

− ˆr

G m M

r

2

(10)

where M is the mass inside ~r. This will be very important for one of the more mysterious facts

about the universe — the existence of dark matter.

You can think about these arguments involving ﬁeld lines in two different ways. What is more

fundamental - the ﬁeld lines or the 1/r

2

force? The straightforward way to interpret this is to say

that we know that the force goes like 1/r

2

and that allows us to show that the ﬁeld line picture

makes sense. Alternatively, we might say that the ﬁeld lines are the fundamental things, and the

ﬁeld line picture implies (because the ﬁeld lines spread out over the surface of a sphere which

grows like r

2

) that the force falls off like 1/r

2

. In fact, what we believe from a more fundamental

description of these forces is that the ﬁeld lines give the more fundamental description. This

doesn’t matter at all if we stick to three dimensions. But it will be important when we talk later

about theories of extra dimensions. In n dimensions, because the area of a sphere of radius r

increases like r

n−1

, the electric and gravitational forces fall like 1/r

n−1

!

6

The force between spheres

Newton’s theorem implies the gravitational force between two spherically symmetric bodies is the

same as if all the mass were concentrated at their centers. This follows from linearity and Newton’s

third law. Consider two spherical bodies, A and B. The force on each of the point masses in B

due to A is the same as the force on due to a point at the center of A with the same total mass. But

by Newton’s third law, this means that the force on A from each of the point masses in B is the

same as the force on a point mass of the center of A. But we know that this is the same as a force

between a point mass at the center of B and a point mass at the center of A.

This means, among other important things, that spherical bodies exert no torques on one an-

other.

Central forces

One of the most important examples of gravity in action is the solar system, where the sun is much

more massive than all the other bodies, so the center of mass is essentially at the position on the

sun. Then to a good approximation, all the other planets orbit the sun. The gravitational forces of

the planets on each other are very small compared to the force of the sun and can be ignored to

ﬁrst approximation. This mean that we can put the sun at the origin and look at the orbit of one

planet at a time. This in turn is an example of a more general system with a central potential, with

Lagrangian

L(

˙

~r, ~r) =

1

2

m

˙

~r

2

− V (r) (11)

where

r =

¯

¯~r

¯

¯ (12)

This is invariant under rotations about any axis through the origin so there is a conserved angular

momentum,

~

L = ~r × ~p = m ~r ×

˙

~r = m ~r ×~v (13)

While we know formally that the angular momentum is conserved because of Noether’s theorem,

it is instructive to see how the conservation arises in this particular case. Using the product rule,

we can write

d

dt

~

L = m

d

dt

~r ×

˙

~r = m

˙

~r ×

˙

~r + m ~r ×

¨

~r (14)

The ﬁrst term vanishes trivially because

~

A ×

~

A = 0 for any vector

~

A because of the antisymmetry

of the cross product. The second term vanishes because

m

¨

~r =

~

F = −

∂

∂ ~r

V (r) = −

∂r

∂ ~r

V

0

(r) = −

1

r

~r V

0

(r) = −ˆr V

0

(r) (15)

so the second term is proportional to ~r ×~r = 0. The crucial step here is

∂r

∂ ~r

=

1

r

~r (16)

7

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

Подождите немного. Документ загружается.