Wolfram Alpha:

```Symmetry and Conservation Laws - Noether's Theorem
--------------------------------------------------

Calculus of Variations
----------------------

Consider a critical point (minima).  The definition
is that ∂f(x)/∂x = 0.  A small displacement from x0
does not change the value of the function to a first
order in the displacement from that point.  We can
see this from Taylor's theorem:

f(x) = f(x0 + δx) = f(x0) + f'(x0)δx + 1/2f''(xo)δx2

+ ...

where δx is referred to as the variation of x.  In
general:

δf(x) = f'(x)δx    (i.e. the change in f(x) = the
rate of change of f(x) wrt x * the change in x).

There is a multi-variable version of the Taylor
series:

f(x,y) = f(x0,y0) + (∂f/∂x)δx + (∂f/∂y)δy

For the Lagrangian we can write:
.                  .   .
L(qi,q) = L + Σ{(∂L/∂qi)δqi + (∂L/∂qi)δqi}
i

So the change in L is just:

.   .
δL =  Σ{(∂L/∂qi)δqi + (∂L/∂qi)δqi}
i

and the change in the Action is given by:
.   .
δS = ∫Σ{(∂L/∂qi)δqi + (∂L/∂qi)δqi}dt
i
This is referred to as the first variation in the
action.

FROM THIS POINT ON WE WILL DROP THE Σ FROM THE
EQUATIONS TO MINIMIZE COMPLEXITY ON THE
UNDERSTANDING THAT SUMMATION OVER i IS IMPLIED.

Integration by parts of the second term in the
integral yields the following result:
.   .                               .
(∂L/∂qi)δqi = (∂L/∂qi)δqi - d/dt(∂L/∂qi)δqi

Substituting back into the equation and rearranging
gives:

.               .     t2
δS = ∫dt{(∂L/∂qi) - d/dt(∂L/∂qi)}δqi + [(∂L/∂qi)δqi]
t1

In the case where the endpoint of the trajectories
are the same, the last term vanishes and we are left
with the E-L equations.  Now, if the original
trajectory is a solution to the equations of motion,
the variation in the E-L equation is equal to zero
and there is no variation in the action.

Linear Translational Symmetry
-----------------------------

But what happens if we shift the entire trajectory
so the endpoints change as in the following diagram?

Now we need to figure out what we need to do with
the last term.  If we assume that any symmetry
operation doesn't change the action, then the last
term must also vanish when the endpoints change.
This implies that the quantity at t2 is equal to
the quantity at t1 or in other words, the quantity
is conserved.  Thus,
.
(∂L/∂qi)δqi = 0

and so,
.
δS = ∫dt{(∂L/∂qi) - d/dt(∂L/∂qi)}δqi

Now the term inside the {} is the Euler-Lagrange
equation:
.
d(∂L/∂qi)/dt - ∂L/∂qi = 0

Consequently δS = 0 and there is a conserved quantity.

To find the conserved quantity we need to look at
δL.
.   .
δL = (∂L/∂qi)δqi + (∂L/∂qi)δqi

.                     .
Now, (∂L/∂qi) = pi and ∂L/∂qi = pi.  pi (ofter
written as Πi) is called the "canonical momentum
conjugate to the coordinate qi".  For example,
.          .    .
L = mx2/2 ∴ ∂L/∂x = mx = p

Using the Euler-Lagrange equation:
.
d(∂L/∂qi)/dt - ∂L/∂qi = 0

It follows that:

dpi/dt = ∂L/∂qi

Therefore, δL becomes:
.     .
δL =  piδqi + piδqi

Now d(FG)/dt = FG + FG.  Therefore, we can write
δL as:

δL = d(piδqi)/dt

For symmetry we require that δL = 0

Therefore,

d(piδqi)/dt = 0

In general, we can replace δqi with εfi(q) where
fi(q) is a function that defines the particular
symmetry operation (in this case a linear
translation).  Thus, we can write:

d(piεfi(q))/dt = 0

But ε is just a small number (i.e. a constant)
so,

d(pifi(q))/dt = dQ/dt = 0

Where Q = pifi(q) is the NOETHER CHARGE equal
to the conserved quantity.

Example:

Consider a linear translation along the x axis.
.
L = (1/2)mx2  f(x) = 1
.    .
Q = Πxf(x) => ∂L/∂x = mx = the conserved quantity

Rotational Symmetry
-------------------

Rotate axes by small angle ε.  The 2D
rotation matrix is:

- -     -          -  - -
| x' | = | cosε -sinε || x |
| y' |   | sinε  cosε || y |
- -     -          -  - -

Thus,

x' = xcosε - ysinε

y' = xsinε + ycosε

If ε = 0 => (x,0), if ε = 90° => (0,-x)

Basically, the x direction becomes the y direction
and the y direction becomes the = -x direction. Now
if ε is very small and is in radians then sinε ~ ε
and cosε ~ 1.  Therefore we can write:

fx = δx = -εy

fy = δy = εx

Q = pxfx + pyfy
.     .
= -ypx + xpy => xmy - ymx = z component of the
angular momentum, L = r ^ p.

The conserved quantity is the angular momentum, L.

Time Translation Symmetry
-------------------------

Time translation t -> t + δt where δt = εT (T is
the generator of time evolution, ε is infinitesimal)

A positive time shift in the trajectory cause q
to move backwards.  The change in q can be viewed
as being in the vertical, t, or the horizontal,
q. Let's look at the horizontal shift.

The translation is q(t) -> q(t - ε)

The trajectories can be viewed as being the same
as a coordinate transform with overhanging pieces
A and B.

Note: A shift in q represents a negative coordinate
shift.
.             .     tB
δS = ∫dt{(∂L/∂q) - d/dt(∂L/∂q)δq} + [(∂L/∂q)δq]|
tA

. tA
= 0 + δq∂L/∂q|
tB

Now we need to include the contributions from
A and B,

. tA
δS = δq∂L/∂q| + A - B
tB

Using Taylor series we can write,
.                  .
q(t - ε) = q(t) - εq therefore δq = -εq

If ε is sufficiently small, the action for A and
B is ∫Lε.  We can write,

SA = εL(tA) and SB = εL(tB).  Therefore, we end up
with:

. .  tA
δS = -ε(∂L/∂q)q}| + εL(tA) - εL(tB)
tB
. .               . .
= ε[L - (∂L/∂q)q] - ε[L - (∂L/∂q)q]
tA                tB

As in the case of linear translation, this term
must equal 0 on the assumption that the symmetry
operation does not change the action.  Its value
at tA must equal its value at tB.

Again, ε is just a small constant that is irrelevant.
Therefore, we can conclude from the above that the
conserved quantity is the energy:
.
Q = L - qipi = -H

Or,
.
The HAMILTONIAN, H = -L + qipi

Example:
.                        .
L = 1/2mx2 - U(x) therefore Π = mx
.           . .
H = -1/2mx2 + U(x) + xmx
.
= 1/2mx2 + U(x)    QED

Inclusion of Explicit Time Dependence
-------------------------------------
.
The Lagrangian has an implicit time dependence (qi).
We now want to consider an explicit time dependence
i.e.,
.
L(q,q,t).

Example:

L = (1/2){mv2 - k(t)x2} ... k changes with time.

To see if L changes with time we look at dL/dt.

.         . ..
dL/dt = (∂L/∂qi)qi + (∂L/∂qi)qi + ∂L/∂t
\
Now,                          explicit time term
.     . .
(∂L/∂qi)(qi) = piqi

and,
.   ..     ..
(∂L/∂qi)(qi) = piqi

Therefore,
. .     ..
dL/dt = piqi + piqi + ∂L/∂t

Which can be written as:
.                                 .   .
dL/dt = d(piqi)/dt + ∂L/∂t  using d(FG)/dt = FG + FG
.
From before  piqi = H + L

dL/dt = dH/dt + dL/dt + ∂L/∂t

0 = dH/dt + ∂L/∂t

H varies with time only if the Lagrangian has an
explicit time dependence. If a system is time
translation independent then ∂L/∂t = 0 and H is
conserved.

Summary
-------

In general Noether's theorem can be written as:

t -> t + δt  where δt = ΣεT

Where T is the generator of time evolution.

q -> q + δq  where δq = ΣεQ

Where Q is the generator of coordinates.
. .              .
{(∂L/∂q)q - L}T - (∂L/∂q)Q

For translational and rotational invariance,
T = 0, Q = 1:
.
p = ∂L/∂q
.     .
L = -ypx + xpy => xmy - ymx = z

For time invariance, T = 1 and Q = 0:
. .
H = (∂L/∂q)q - L

In summary, Noether's theorem states that each
symmetry of a system leads to a physically conserved
quantity.  Symmetry under translation corresponds
to the conservation of momentum, symmetry under
rotation corresponds to the conservation of angular
momentum, symmetry in time corresponds to the
conservation of energy.```