Georgi, Howard. Physics 16 - Mechanics and Special Relativity (англ.)
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This system has a conserved momentum,
p = m
1
˙x
1
+ m
2
˙x
2
(29)
because the forces that come from the potential energy obey Newton’s third law. This in turn is
related to the fact that the potential energy depends only on x
1
−x
2
, which in turn is related to the
symmetry. We will see that this connection between a continuous symmetry and the existence of a
conserved momentum is a general thing.
Space translations for many particles
Let us now consider a one-dimensional system of n particles, with positions x
j
, so that q
j
= x
j
for
j = 1 to n. A space translation in the x direction adds the same infinitesimal constant, ², to each
x
j
— so the transformation has the form
x
j
→ ˜x
j
= x
j
+ ² ∀n . (30)
Notice that this satisfies (12), with q
j
= x
j
and all of the κ
x
j
(x) = 1. This is a symmetry of any
Lagrangian that depends only on ˙x
j
and differences between two x
j
s, for example
L(x, ˙x) =
X
j
m
j
2
˙x
2
j
− V (x
1
−x
2
, x
2
−x
3
, ··· , x
j
−x
j+1
, ··· , x
n−1
−x
n
) . (31)
Again, (30) is a symmetry of the kinetic energy because ² is a constant, so that
˙
˜x
j
= ˙x
j
. (32)
It is a symmetry of the potential energy because the ²s cancel when we subtract one coordinate
from another, so that
˜x
j
− ˜x
j+1
= (x
j
+ ²) −(x
j+1
+ ²) = x
j
− x
j+1
. (33)
Here the total momentum,
X
j
m
j
v
j
(34)
is conserved because the forces that come from the potential energy obey Newton’s third law.
Finding Symmetries
So far, we have looked at systems in which it is pretty obvious what the symmetry transformation
is. But when come upon some Lagrangian, you may want to find what transformation the La-
grangian is invariant under. The way to do this is to write down how your Lagrangian trasforms
under a general infintesimal transformation of the form
q
j
→ ˜q
j
= q
j
+ ² κ
q
j
(q) . (35)
6
and then require that it be invariant - that is that the coefficient of the ² term in
L(˜q,
˙
˜q) − L(q, ˙q) (36)
vanishes. Here is a simple example, where you could probably guess the answer, but it will illus-
trate the technique.
L(x, ˙x) =
m
1
2
˙x
2
1
+
m
2
2
˙x
2
2
− V (x
1
+ 2x
2
) (37)
Does this Lagrangian has a symmetry? To see, we want to find L(˜x,
˙
˜x) where
x
j
→ ˜x
j
= x
j
+ ² κ
x
j
(x) . (38)
L(˜x,
˙
˜x) =
m
1
2
µ
˙x
1
+ ² ˙κ
x
1
¶
2
+
m
2
2
µ
˙x
2
+ ² ˙κ
x
2
¶
2
− V (x
1
+ ²κ
x
1
+ 2x
2
+ 2²κ
x
2
) (39)
=
m
1
2
µ
˙x
2
1
+ 2²x
1
˙κ
x
1
+ ···
¶
+
m
2
2
µ
˙x
2
2
+ 2²x
2
˙κ
x
2
+ ···
¶
−V
µ
x
1
+ 2 x
2
+ ²(κ
x
1
+ 2 κ
x
2
)
¶
(40)
The ² term will vanish if
˙κ
x
1
= ˙κ
x
2
= κ
x
1
+ 2κ
x
2
= 0 (41)
One way to satisfy the first two conditions is to take the κs to be constant independent of x (if they
depend on x, the time derivatives would be nonzero). Then the last condition implies that we can
take
κ
x
1
= −2 κ
x
2
= 1 (42)
which gives a transformation of the form
x
1
→ x
1
− 2² x
2
→ x
2
+ ² (43)
You see that it is sometimes easy to find symmetry transformations. It is harder to show that you
have found them all. This depends on details, in this case the precise form of the function V . We
won’t talk about that now.
Rotations
Another simple example can be found in the example we discussed in lecture 7 of the mass m
1
sliding on a horizontal frictionless table, connected to a string that goes through a hole at the origin
to a mass m
2
hanging below the table. In terms of the length, `, of string on the table and the angle,
θ, of the string on the table from the x axis, the Lagrangian for this system looks like
L(`, θ,
˙
`,
˙
θ) =
1
2
(m
1
+ m
2
)
˙
`
2
+
1
2
m
1
`
2
˙
θ
2
− m
2
g` (44)
This physical system has a symmetry under rotations about the z axis, which add a constant to the
angle θ without changing `,
` →
˜
` = ` , θ →
˜
θ = θ + ² . (45)
7
This symmetry is responsible for the fact that the Lagrangian does not depend on θ at all, but only
on
˙
θ, because the condition for symmetry, that
L(
˜
`,
˜
θ,
˙
˜
`,
˙
˜
θ) = L(`, θ,
˙
`,
˙
θ) (46)
in this case becomes
L(`, θ + ²,
˙
`,
˙
θ) = L(`, θ,
˙
`,
˙
θ) (47)
This just says that if we make a little change in θ in the function, nothing happens, so the function
must not depend on θ. If we wanted to say the same thing in fancier mathematics, we could say
that because ² is infinitesimal, we can reliably Taylor expand the left-hand-side of (47) and keep
only the first two terms,
L(`, θ + ²,
˙
`,
˙
θ) = L(`, θ,
˙
`,
˙
θ) + ²
∂
∂θ
L(`, θ,
˙
`,
˙
θ) (48)
Putting (48) into (47) implies
∂
∂θ
L(`, θ,
˙
`,
˙
θ) = 0 (49)
which as promised is the statement that L does not depend on θ.
As we saw in lecture 7, this system also has a conserved quantity, the angular momentum
L = m
1
`
2
˙
θ . (50)
So again, we see a connection between a symmetry and a conserved quantity.
Noether’s theorem
Having seen a correlation between symmetry and conservation laws in a couple of examples, let
us now consider how this works in more generality and see if we can pin down the connection
precisely. Putting (12) into (13) and Taylor expanding gives
L(q, ˙q) = L(q + ²κ, ˙q + ² ˙κ) = L(q, ˙q) + ²
X
q
j
˙κ
q
j
∂L
∂ ˙q
j
+ ²
X
q
j
κ
q
j
∂L
∂q
j
. (51)
Thus the condition that (12) is a symmetry of the Lagrangian is equivalent to
X
q
j
˙κ
q
j
∂L
∂ ˙q
j
+
X
q
j
κ
q
j
∂L
∂q
j
= 0 (52)
For example, for space translations, (30),
κ
x
j
= 1 (53)
for all j, because all the x
j
s are translated in the same way. Thus ˙κ = 0 and the condition of
symmetry becomes
X
x
j
κ
x
j
∂L
∂x
j
=
X
x
j
∂L
∂x
j
= 0 (54)
8
This is equivalent to the statement that L depends only on differences x
j
− x
k
.
Now return to the general case and consider the quantity
X
q
j
κ
q
j
∂L
∂ ˙q
j
(55)
Consider the time derivative of this quantity
d
dt
X
q
j
Ã
κ
q
j
∂L
∂ ˙q
j
!
=
X
q
j
˙κ
q
j
∂L
∂ ˙q
j
+
X
q
j
κ
q
j
d
dt
∂L
∂ ˙q
j
(56)
If we apply the Lagrange equations of motion to the second term, it becomes
d
dt
X
q
j
Ã
κ
q
j
∂L
∂ ˙q
j
!
=
X
q
j
˙κ
q
j
∂L
∂ ˙q
j
+
X
q
j
κ
q
j
∂L
∂q
j
= 0 (57)
which vanishes because of the condition (52) that the Lagrangian is symmetric. This is a very
important general theorem. It is the precise connection between a continuous symmetry and a
conservation law. For every continuous symmetry of the form (12), there is a conservation law —
the quantity of the form (55) is constant.
Now we can find the conserved quantity for any symmetry. For example for the Lagrangian of
(37), where the κs are given by (42), the conserved quantity is
κ
x
1
∂L
∂ ˙x
1
+ κ
x
2
∂L
∂ ˙x
2
= −2m
1
˙x
1
+ m
2
˙x
2
(58)
This theorem is a special favorite of mine because a lot of my own work in particle physics
is based on it. It was worked out early in this century by the great woman mathematician and
theoretical physicist Emmy Noether.
1
The quantity
∂L
∂ ˙q
j
(59)
that appears in Noether’s theorem is a very important one in classical mechanics. It is the “gen-
eralized momentum” we talked about earlier, corresponding to the coordinate q
j
, we discussed in
the last lecture. Of course, if q
j
is a normal space coordinate (like x
j
), and the kinetic energy has
the standard form, it is the momentum (or a component of it). In fact, one way of thinking about
(57) is to recognize that in a sense (which I will not explain in detail) (55) is the generalized mo-
mentum associated with, ², the infinitesimal variable that describes how all the q
j
s change under
the symmetry.
1
For more information about Noether, see The Life and Times of Emmy Noether by Nina Byers —
http://xxx.lanl.gov/abs/hep-th/9411110 and on the 16 website in handouts/noether.pdf.
9
Momentum conservation from Noether’s theorem
For space translation symmetry, because all the κ
x
j
are equal to 1, the quantity (55) becomes
X
x
j
κ
x
j
∂L
∂ ˙x
j
=
X
x
j
∂L
∂ ˙x
j
(60)
which is just the sum over the momenta of all the individual particles. Space translation invari-
ance implies that the total momentum is conserved.
This analysis can obviously be extended to three dimensions, where the coordinates and the
corresponding momenta become vectors.
More on rotations
Here is a more involved example that we can discuss if we have time. When we thought about
rotations for the bead sliding on the table, we were already using polar coordinates. This makes
rotations easy, because an infinitesimal rotation is just a translation of θ,
θ → θ
0
= θ + ² . (61)
This is an example with two particles in Cartesian coordinates. Consider two particles, with masses
m
1
and m
2
, which move in a plane with coordinates ~r
1
= (x
1
, y
1
) and ~r
2
= (x
2
, y
2
). Consider the
following Lagrangian:
1
2
m
1
³
˙x
2
1
+ ˙y
2
1
´
+
1
2
m
2
³
˙x
2
2
+ ˙y
2
2
´
− V
³
x
2
1
+ y
2
1
+ x
2
2
+ y
2
2
− 2(x
1
x
2
+ y
1
y
2
)
´
(62)
This Lagrangian is invariant under the following infinitesimal transformation:
x
1
→ ˜x
1
= x
1
− ²y
1
y
1
→ ˜y
1
= y
1
+ ²x
1
x
2
→ ˜x
2
= x
2
− ²y
2
y
2
→ ˜y
2
= y
2
+ ²x
2
(63)
If you have never seen anything like this before, it is not obvious at all, but you can see it by explicit
calculation. For example, look at the m
1
terms in the kinetic energy. First note that
x
1
→ ˜x
1
= x
1
− ²y
1
y
1
→ ˜y
1
= y
1
+ ²x
1
(64)
then substitute
˙
˜x
2
1
+
˙
˜y
2
1
=
³
˙x
1
− ² ˙y
1
´
2
+
³
˙y
1
+ ² ˙x
1
´
2
= ˙x
2
1
+ ˙y
2
1
+ O(²
2
) (65)
The ²
2
terms do not cancel, but we don’t care about them because ² is infinitesimal - we only
consider linear terms in Noether’s theorem. So this is good enough. All the other terms work
10
similarly. You can see what the rotation looks like pictorially in the figure below.
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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•
(˜x, ˜y)
²
θ
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(x, y)
The heavy lines represent the vector ~r = (x, y) from the origin to the original point, and the smaller
vector from ~r to
~
˜r = (˜x, ˜y). Because ² is infinitesimal, these two are nearly perpendicular, and the
ratio of their lengths is about ². It is helpful to complete these two lines into a similar triangle.
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
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11
Then blowing up the box, you can see that the change in x under the rotation is −²y and the change
in y is ²x.
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.............................................................................................................
²y
²x
θ
Now let’s see what Noether’s theorem looks like. The κs are
κ
x
1
= −y
1
, κ
y
1
= x
1
, κ
x
2
= −y
2
, κ
y
2
= x
2
. (66)
and the momenta are
∂L
∂ ˙x
j
= m
j
˙x
j
,
∂L
∂ ˙y
j
= m
j
˙y
j
. (67)
Thus the conserved quantity is
X
j
m
j
(x
j
˙y
j
− y
j
˙x
j
) (68)
As we will see in more detail later in the course, this is the angular momentum.
Notice also that the complicated function of the variables that appears in V can be written as
(~r
1
−~r
2
) · (~r
1
−~r
2
) = |~r
1
−~r
2
|
2
(69)
just the square of the length of the difference ~r
1
−~r
2
.
What functions are invariant?
We have seen some examples of functions that are invariant under transformations, such as the
potential in (31),
V (x
1
−x
2
, x
2
−x
3
, ··· , x
j
−x
j+1
, ··· , x
n−1
−x
n
) . (70)
It always straightforward to check that a given function is invariant under a specific transforma-
tion. And we have seen how to find transformations that are symmetries, if they exist. It is also
important to be able to go the other way and to construct the most general function invariant under
a transformation or set of transformations.
12
To illustrate what I am talking about, let’s show that a function of F (x
1
, x
2
) that is invariant
under the symmetry transformation
x
j
→ x
j
+ ² (71)
for j = 1 and 2 actually only depends on the difference, x
1
− x
2
. One way to do this is to change
variables to include the variable x
1
− x
2
, and eliminate x
1
—
y ≡ x
1
− x
2
x
1
= x
2
+ y (72)
Then we can define a new function
G(y, x
2
) ≡ F (x
2
+ y, x
2
) (73)
in terms of the original function. But now, in terms of the new variables the transformation (71) is
y → y , x
2
→ x
2
+ ² . (74)
Now since F is invariant, G must be also, because we have only relabeled things. Thus
G(y, x
2
+ ²) = G(y, x
2
) (75)
for infinitesimal ² . In words, this says that making an infinitesimal change in x
2
doesn’t affect the
fucntion, so it is probably obvious that this means that G(y, x
2
) is independent of x
2
. But if we
want to be more formal about it, we can use the Taylor expansion to get
G(y, x
2
+ ²) = G(y, x
2
) + ²
∂
∂x
2
G(y, x
2
) + ··· = G(y, x
2
) (76)
and thus
∂
∂x
2
G(y, x
2
) = 0 (77)
which means that G doesn’t depend on x
2
, so we can take x
2
to be anything in G. Thus using (72)
F (x
1
, x
2
) = G(x
1
− x
2
, x
2
) = G(x
1
− x
2
, 0) = f(x
1
− x
2
) . (78)
It is easy to extend this proof to show that the most general function invariant under the transfor-
mation (71) for n variables, x
j
for j = 1 to n is given by (70).
Notice that what we are doing here makes Noether’s theorem seem a little trivial. If all we
do with invariance is to show that there is some variable that the function doesn’t depend on,
then we could have changed variables first and then found the conserved quantity by just using
the statement that the generalized momentum associated with a variable that doesn’t appear in
the Lagrangian is conserved. And in fact, in this course, you can always do that. But there is
actually more to Noether’s theorem, because transformations that leave a system invariant have an
additional interesting property. They form what mathematicians form a group. The group property
is quite powerful and often allows you to extend the infinitesimal transformations that we start with
to a much larger set. For example, in the case of translations, group theory can be used to show
13
that if things are invariant under (71) for infinitesimal ², they are also invariant for finite ². Then
we don’t need the Taylor expansion any more. For example, in (75), we could first set x
2
to zero
to get
G(y, ²) = G(y, 0) (79)
and then relabel ² → x
2
to get
G(y, x
2
) = G(y, 0) (80)
which directly gives (78). We won’t pursue this approach much in the course but I may not be able
to restrain myself from talking more about it, because it forms the basis of much of the research
that I have done in my scientific career.
14
lecture 9
Topics:
The structure of science and common sense
The speed of light
Time dilation
The twin paradox
The Doppler effect
The twin paradox and inertial frames
The structure of science and common sense
from www.raremaps.com
The beautiful edifice of Newtonian mechanics, which we have seen a bit of in the last few
weeks, provides a wonderful precise mathematical description of most of the things we see in our
everyday world. It is obviously right. But it is also wrong. We have discussed qualitatively the
underlying quantum mechanical reality from which Newton’s mechanics emerges as an approx-
imation. In the next few weeks, we will discuss in quantitative detail the bizarre things that go
wrong with Newton’s picture at large velocities. What is going on here? Newtonian mechanics
beautifully captures the mathematical essence of what we know about the world we have grown
1