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

to the Euler-Lagrange equation. Often, we can identify this quantity with the energy. We will do

this ﬁrst for a single degree of freedom, and then extend the result to systems with more degrees

of freedom.

Consider a system with a single degree of freedom described by the Lagrangian

L(q, ˙q, t) (54)

for some single coordinate q, and construct the quantity

F = ˙q

∂L

∂ ˙q

− L (55)

In general, F may depend on q, ˙q and t. Let us now ask how F changes with time, by taking the

total derivative

d

dt

F =

d

dt

Ã

˙q

∂L

∂ ˙q

!

−

d

dt

L (56)

In the ﬁrst term on the right hand side of (56), we use the product rule to write

d

dt

Ã

˙q

∂L

∂ ˙q

!

= ¨q

∂L

∂ ˙q

+ ˙q

d

dt

∂L

∂ ˙q

(57)

In the second term on the right hand side of (56), we use the fact that the t dependence of L comes

from the explicit t dependence, and also from the implicit dependence on t through q and ˙q:

d

dt

L = ¨q

∂L

∂ ˙q

+ ˙q

∂L

∂q

+

∂L

∂t

(58)

The relation (58) is an example of one of those multivariable calculus things that will make your

eyes glaze over if you just stare at the symbols. But if you translate it into words, it makes perfect

sense. It says that the total rate of change of F with t is the rate of change of ˙q times the rate at

which F changes with ˙q plus the rate of change of q times the rate at which F changes with q

plus the rate of change from the explicit time dependence. It is simply a matter of adding up all

the possible sources of time variation of the function F . Subtracting (58) from (57) and using the

Lagrange equation

d

dt

∂L

∂ ˙q

=

∂L

∂q

(59)

we get

d

dt

F = −

∂L

∂t

(60)

Thus if L does not depend on time EXPLICITLY, but only implicitly through the time dependence

of q and ˙q, the function F is constant for the trajectory.

It is important to understand what is meant here by the words “explicit” and “implicit”. Explicit

time dependence occurs only if there is some physics in the problem that changes with time. For

example, the Lagrangian we discussed above in (27) for the particle sliding on a vertically rotating

frictionless rod depends explicitly on time — the Lagrangian has a sin ωt in it! On the other hand,

13

ANY function of q and ˙q depends implicitly on time, because q and ˙q for the trajectory depend on

time.

Usually, this function F is the Energy! For example, suppose we look at the Lagrangian for a

particle moving in a potential

L(x, ˙x) =

m

2

˙x

2

− V (x) (61)

For this Lagrangian, the function F is

˙x

∂

∂ ˙x

L(x, ˙x) − L(x, ˙x) = ˙x m ˙x − L(x, ˙x) =

m

2

˙x

2

+ V (x) (62)

which is the energy, as promised.

For example, in the example of the bead on the horizontally rotating rod, where the Lagrangian

is

L(`,

˙

`) =

1

2

m

³

˙

`

2

+ `

2

ω

2

´

(63)

the construction of (55) gives

F =

˙

` m

˙

` −

1

2

m

³

˙

`

2

+ `

2

ω

2

´

=

1

2

m

³

˙

`

2

− `

2

ω

2

´

(64)

You can check explicitly that this quantity is conserved for the general solution to (21).

These two examples are rather different. In the second, the fact that the Lagrangian has no

explicit time dependence is an accident, arising from the cancellation of the ωt dependence that we

talked about earlier. This is related to the fact that F in this case is not the kinetic plus the potential

energy, but rather a curious combinations of the terms in the kinetic energy. It is conserved, but the

physical interpretation is obscure.

In the ﬁrst example, however, the particle moving in a potential, the Lagrangian does not de-

pend explicitly on t because there is a symmetry of the system. The symmetry in this case is time

translation invariance. In this case, F really is the physical energy. This is the ﬁrst of several impor-

tant examples we will see in this course of the connection between a symmetry and conservation

law. We will explore this further on Thursday.

For the vertically rotating rod, because L in (27) depends on time explicitly because of the

factor of sin ωt, F is not conserved.

The construction of the energy function makes it clear how to deal with more degrees of free-

dom. If q has an index, q

j

where j goes from 1 to n, the analogous construction for the function F

is

F =

X

j

˙q

j

∂L

∂ ˙q

j

− L (65)

It must have this form for the analog of (56)-(60) to be valid, because the analog of (58) for more

degrees of freedom is

d

dt

L =

X

j

¨q

j

∂L

∂ ˙q

j

+

X

j

˙q

j

∂L

∂q

j

+

∂L

∂t

(66)

This requires that the same sum over degrees of freedom appears in the ﬁrst term in (65). With F

deﬁned as in (65), (60) is still valid.

14

Thus the construction (65) is very general. In fact, it reproduces the usual expression for the

energy as the kinetic plus the potential energy whenever the potential depends only on q

j

and

the kinetic energy is proportional to two powers of the velocity. However, it is really even more

general than that. For any Lagrangian that does not depend explicitly on t, (65) deﬁnes a conserved

quantity. And if the explicit t dependence vanishes because of time translation invariance, the

conserved quantity is the energy.

For the system we discussed earlier with two masses connected by a string, one sliding on and

the other hanging below a frictionless table, with Lagrangian (37),

L(`, θ,

˙

`,

˙

θ) =

1

2

(m

1

+ m

2

)

˙

`

2

+

1

2

m

1

`

2

˙

θ

2

− m

2

g` (67)

again there is no explicit time dependence, so we expect a conserved energy. Because there are

two degrees of freedom, we have to use the construction (65), which gives for the conventional

energy, T + U,

F =

˙

` (m

1

+ m

2

)

˙

` +

˙

θ m

1

`

2

˙

θ −

1

2

(m

1

+ m

2

)

˙

`

2

−

1

2

m

1

`

2

˙

θ

2

+ m

2

g`

=

1

2

(m

1

+ m

2

)

˙

`

2

+

1

2

m

1

`

2

˙

θ

2

+ m

2

g`

(68)

Example - Atwood’s machine

Atwood’s machines are collections of massless ropes, massless pulleys, and masses, like that

shown below:

.

m

1

m

2

•

.

...

.

...

.

...

.

...

.

. .

.

. .

.

(69)

15

The motion of this Atwood’s machine can be found the old fashioned way, by assigning unknowns

to the various physical quantities in the system, the tension in the ropes at various points.

T

T T

2T

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

.

m

1

m

2

•

.

....

.

....

.

....

.

....

.

. .

.

. .

.

(70)

Note that because the pulleys are massless, the tension T is the same throughout the long rope, and

the rope connected to mass 1 has tension 2T , as shown.

We can now apply F = ma to each of the two masses:

m

1

a

1

= 2T − m

1

g m

2

a

2

= T − m

2

g (71)

Finally, we have to realize that the two accelerations are related, because if mass 1 moves up a

distance x, mass 2 must move down a distance 2x, because the length of the rope does not change.

Thus if we allow only vertical motions, this is a system with only one degree of freedom, because

x

2

is actually determined once we ﬁx x

1

.

x

2

= −2x

1

+ constant (72)

and the constant can be ignored, or absorbed into the deﬁnitions of x

1

and x

2

. It doesn’t affect the

accelerations or the forces. This is illustrated in the diagram below:

2x

1

x

1

.

m

1

m

2

•

.

....

.

..

.

....

.

..

.

. .

.

. .

.

(73)

Thus we can write

a

2

= −2a

1

(74)

16

so that (71) becomes

m

1

a

1

= 2T − m

1

g − 2m

2

a

1

= T − m

2

g (75)

We can now solve these two equations for T and a

1

, and conclude that

a

1

=

2m

2

− m

1

m

1

+ 4m

2

g (76)

and

T =

3m

1

m

2

g

m

1

+ 4m

2

(77)

There is a politically incorrect but amusing story about Atwood’s machine. A physics professor

in an introductory course was describing this derivation to his introductory mechanics class (very

much like Harvard’s Physics 1a), when one of the students asked why anyone should care about

such a random collection of ropes and pulleys and masses. Without missing a beat, the professor

replied that Atwood’s machine had saved more lives than Penicillin — by keeping dumb people

out of medical school!

When you think about it though, this derivation is a bit indirect, because you have to introduce

the quantity T, which you don’t care about. It is no wonder that premedical students ﬁnd this chal-

lenging. Note how much simpler this would be for them if they were armed with the Lagrangian!

Here they just have to identify the kinetic and potential energies and turn the crank. The potential

energy in this system is

V (x

1

) = m

1

gx

1

+ m

2

gx

2

= (m

1

− 2m

2

)gx

1

(78)

The kinetic energy is

T ( ˙x

1

) =

m

1

2

˙x

2

1

+

m

2

˙x

2

=

m

1

+ 4m

2

˙x

2

1

(79)

Thus

L(x

1

, ˙x

1

) =

m

1

+ 4m

2

˙x

2

1

− (m

1

− 2m

2

)gx

1

(80)

Then

∂L

∂x

1

= −(m

1

− 2m

2

)g

∂L

∂ ˙x

1

= (m

1

+ 4m

2

) ˙x

1

(81)

So the Lagrange equation is

0 = −(m

1

− 2m

2

)g − (m

1

+ 4m

2

)¨x

1

(82)

which immediately gives (76). We are lucky that this is not taught in premed physics around the

world — we might end up with a lot of dumb doctors!

17

lecture 8

Topics:

Where are we now?

When is F the energy?

Symmetries and transformations

Example: space translations for one particle

Space translations for two particles

Space translations for many particles

Finding Symmetries

Rotations

Noether’s theorem

Momentum conservation from Noether’s theorem

More on rotations

What functions are invariant?

Where are we now?

We have seen how conserved quantities can arise as the generalized momenta associated with vari-

ables that do not appear in the Lagrangian. Here we will discuss a generalization of this fact, that

is an even more important principle. In Lagrangian mechanics, continuous symmetries lead to

conserved quantities. We have already seen one example of this in our discussion of energy, and

we will begin by making the connection with time translation invariance. We will go on to dis-

cuss symmetries more generally, and also in more detail the speciﬁc example of space translation

symmetry, which leads to the conservation of total momentum.

When is F the energy?

Last time we talked about the quantity F , deﬁned by

F =

X

j

˙q

j

∂L

∂ ˙q

j

− L (1)

which when evaluated for solutions to Euler-Lagrange equations satisﬁes

d

dt

F = −

∂L

∂t

(2)

We noted that F is often the energy. I said this in lecture 7, but I don’t think it got into the notes, so

I thought it worth saying again. Of course, to even get started on this, we have to restrict ourselves

to situations where there are no frictional forces, and therefore nothing is converting kinetic and

potential energy into heat. Since heat is just kinetic energy of random particle motion, this really

just means that we need to look at a system at a sufﬁciently fundamental level to see all the kinetic

energy explicitly. For now we will take care of this by restricting ourselves to systems described

1

by Lagrangians of the form T − V . Then the ﬁrst answer is that F is the energy whenever the

kinetic energy T is quadratic in the velocities, ˙q and the potential energy V does not depend on the

velocities. The reason is that the differential operator

z

∂

∂z

(3)

counts the degree in z, because

z

∂

∂z

z

n

= z n z

n−1

= n z

n

(4)

That is, this operator acting on a term that is some power of z just gives the term back multiplied

by the power. When we sum over all the velocities, the differential operator

X

j

˙q

j

∂

∂ ˙q

j

(5)

adds up the powers of all the velocities. When this acts on a function quadratic in the velocities,

every term in the function just gets multiplied by 2. For example

X

j

˙q

j

∂

∂ ˙q

j

˙q

2

1

= ˙q

1

∂

∂ ˙q

1

˙q

2

1

= 2 ˙q

2

1

(6)

and

X

j

˙q

j

∂

∂ ˙q

j

˙q

1

˙q

2

= ˙q

1

∂

∂ ˙q

1

˙q

1

˙q

2

+ ˙q

2

∂

∂ ˙q

2

˙q

1

˙q

2

= ˙q

1

˙q

2

+ ˙q

1

˙q

2

= 2 ˙q

1

˙q

2

(7)

Thus if T is quadratic in the velocities and V is independent of the velocities,

X

j

˙q

j

∂L

∂ ˙q

j

=

X

j

˙q

j

∂

∂ ˙q

j

(T − V ) = 2T (8)

and then

F =

X

j

˙q

j

∂L

∂ ˙q

j

− L = 2T − (T − V ) = T + V = E (9)

and as promised, F is the energy.

At the fundamental level, the kinetic energy is always quadratic in the velocities, because it is

just a sum over all the parts of the system of

1

2

mv

2

. But what can go wrong with this can be seen

in examples like the bead on a rod rotating with ﬁxed angular velocity ω. Sometimes, we have to

separate the variables that describe the system into those that are dynamical - like the position of

the bead along the rod - and those that are imposed by some “external” constraint - like the angle

of the rod, which is ﬁxed whatever is causing the rod to rotate. Then we include include in our list

of coordinates only the dynamical coordinates and put the effect of the others into the Lagrangian

by hand. Then if our external constraint is causing the motion, some of the velocities that appear

in the kinetic energy do not correspond to dynamical coordinates in our Lagrangian, and they are

not included in the sum,

X

j

˙q

j

∂

∂ ˙q

j

(10)

2

In such a case, (9) is not correct, and F is not the energy.

In a system with time translation invariance, however, there can be no time dependent external

constraint. Such a thing would look different at different times and break time translation invari-

ance. Thus time time translation invariance does two things. Not only does it ensure that the

Lagrangian does not depend explicitly on time - which implies that F is conserved. But is also

ensures that the kinetic energy is quadratic in the dynamical velocities, which ensures that F is the

energy. In fact, time translation invariance is the more general answer to the question. Even if L is

not in the form T − V , time translation invariance implies that F is the energy.

Symmetries and transformations

What is a symmetry? We have talked about several examples, so perhaps we should deﬁne pre-

cisely what we mean by it. Symmetry is a mathematical statement of some very speciﬁc regularity

in a system. A system has a symmetry if there is some transformation you can make that leaves the

system looking exactly as it did before. We tend to regard things with many symmetries as pretty,

like the kalaidoscope that we saw at the beginning of lecture, which has many planes of symmetry.

In the case of mechanics, we have an even more speciﬁc meaning in mind. Let us now consider

a class of symmetries in which we make some transformation of the coordinates describing a

system at a ﬁxed time. What this means mathematically is that we deﬁne a new set of coordinates

as functions of the original coordinates. The transformation is then a symmetry if the physics looks

exactly the same in terms of the new coordinates as it did in the old coordinates.

We talked brieﬂy about such a transformation when we discussed the double pendulum with

two equal masses in lecture 2. The Lagrangian for the double pendulum for small oscillations

looks approximately like

m

2

³

˙x

2

1

+ ˙x

2

´

−

g

`

(x

2

1

+ x

2

) −

K

2

(x

1

− x

2

)

2

(11)

This has the property that it is unchanged if we interchange x

1

and x

2

. This is the mathematical

statement of the obvious physical symmetry of the system.

The symmetry of the double pendulum is an example of a discrete symmetry, so-called because

the symmetry is an all or nothing sort of thing. The transformation cannot be made bigger or

smaller - it is ﬁxed by the structure of the symmetry.

It is more even interesting to consider symmetries in which the symmetry transformation can be

made arbitrarily small. Such a thing is called a continuous symmetry, because the transformation

can change the system continuously. By putting arbitrarily small transformations together, we can

get a whole set of transformations which, unlike the symmetry of the double pendulum, depend on

a parameter that can be continuously varied.

An example is translations. We think that space probably looks the same everywhere, and we

could describe this by saying that there is a symmetry in which we move everything by the same

arbitrary vector and we would end up with a completely equivalent physical system.

Here is the general theoretical setup (we’ll discuss examples in more detail shortly). Consider

a system of n degrees of freedom described by coordinates q

j

for j = 1 to n. Let’s assume

3

that there is a symmetry in which each of the coordinates changes only a tiny bit, proportional

to an inﬁnitesimal parameter, ² and that the changes involve only the current conﬁguration of the

system. What we mean precisely by inﬁnitesimal is that ² is sufﬁciently small that we can always

ignore terms of order ²

2

. Translating what we have just assumed into mathematics, we consider a

symmetry in which the coordinates q

j

are transformed as follows:

q

j

→ ˜q

j

= q

j

+ ² κ

q

j

(q) . (12)

That is each of the coordinates changes by ² times a function κ

q

j

(q) of the qs. The κ

q

j

(q) tells you

how the variable q

j

changes under the transformation. The transformation (12) is a symmetry of

the Lagrangian if

L(˜q,

˙

˜q) = L(q, ˙q) (13)

Example: space translations for one particle

Here is a simple (perhaps even boring) example. Consider a particle with mass m moving along

the x-axis in a potential. When does this system have a symmetry under the inﬁnitesimal transfor-

mation

x → ˜x = x + ² ? (14)

This transformation has the form of (12) with κ(x) = 1. The Lagrangian looks like this:

L(x, ˙x) =

1

2

m ˙x

2

− V (x) (15)

The condition that (14) is a symmetry is then

L(˜x,

˙

˜x) = L(x, ˙x) (16)

Because κ(x) is just a constant, we have

˙

˜x =

d

dt

˜x = ˙x (17)

and so (16) becomes

L(x + ², ˙x) = L(x, ˙x) (18)

for inﬁnitesimal ² . Because the ˙x doesn’t change and kinetic energy is the same on both sides, so

this condition only effects V -

V (x + ²) = V (x ) (19)

But because this is supposed to be true for any inﬁnitesimal ², we can use the Taylor expansion

(surprise, surprise) to rewrite (19) as

V (x + ²) = V (x ) + ² V

0

(x) + O(²

2

) = V (x) (20)

If this is to be satisﬁed for inﬁnitesimal ², we must have

V

0

(x) = 0 (21)

4

so that V (x) is just a constant and the particle has no force on it at all. In this case, mv is a con-

served momentum. We will see how this connection between symmetry and conserved momentum

generalizes to more complicated (and more interesting) situations.

Another reason that I wanted to look at this simple system in detail is to emphasize the dif-

ference between continuous and discrete symmetries. Suppose that instead of being constant, the

potential in (15) is

V (x) = − E

0

cos(x/`) (22)

where E

0

and ` are constants. This system also has a symmetry under space translations of the

form

x → ˜x = x + 2πn` (23)

for any integer n. But here we clearly cannot conclude that V (x) is constant because we started

with an example with the symmetry that is not constant. Except at special points where the particle

is in equilibrium, there is a force on it. There is no conserved momentum (though energy is still

conserved because the Lagrangian does not depend explicitly on t). The difference between this

and the previous example is that this is a discrete symmetry. The changes in x that leave the

Lagrangian invariant are a discrete set. They cannot be varied continuous, and they cannot be

made inﬁnitesimally small. Thus we cannot use the Taylor expansion argument to conclude that

V (x) is constant.

Space translations for two particles

Space translation symmetry becomes interesting and important when there is more than one par-

ticle. Let us now consider a one-dimensional system of 2 particles, with positions x

1

and x

2

, so

that q

j

= x

j

for j = 1 to 2. A space translation in the x direction just adds the same inﬁnitesimal

constant, ², to both x

1

and x

2

— so the transformation has the form

x

1

→ ˜x

1

= x

1

+ ² , x

2

→ ˜x

2

= x

2

+ ² . (24)

Notice that this satisﬁes (12), with q

1

= x

1

, q

2

= x

2

and κ

x

1

(x) = κ

x

2

(x) = 1. This is a symmetry

of any Lagrangian that depends only on ˙x

1

and ˙x

2

and the difference between x

1

−x

2

, for example

L(x, ˙x) =

m

1

2

˙x

2

1

+

m

2

˙x

2

− V (x

1

− x

2

) (25)

The transformation (24) is a symmetry of the kinetic energy because ² is a constant, so that

˙

˜x

1

= ˙x

1

,

˙

˜x

2

= ˙x

2

. (26)

It is a symmetry of the potential energy because the ²s cancel when we subtract one coordinate

from another, so that

˜x

1

− ˜x

2

= (x

1

+ ²) −(x

2

+ ²) = x

1

− x

2

. (27)

Putting (26) and (27) together implies

L(˜x,

˙

˜x) = L(x, ˙x) . (28)

5

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

Подождите немного. Документ загружается.