Redshift Academy


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''(x_o)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(q_i,q) = L + Σ{(∂L/∂q_i)δq_i + (∂L/∂q_i)δq_i}   
              ⁱ

So the change in L is just:


               .   .
  δL =  Σ{(∂L/∂q_i)δq_i + (∂L/∂q_i)δq_i}  
        ⁱ

and the change in the Action is given by:
                             .   .
  δS = ∫Σ{(∂L/∂q_i)δq_i + (∂L/∂q_i)δq_i}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/∂q_i)δq_i = (∂L/∂q_i)δq_i - d/dt(∂L/∂q_i)δq_i   

Substituting back into the equation and rearranging 
gives:


                              .               .     t₂
 δS = ∫dt{(∂L/∂q_i) - d/dt(∂L/∂q_i)}δq_i + [(∂L/∂q_i)δq_i]
                                                    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/∂q_i)δq_i = 0

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

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

Consequently δS = 0 and there is a conserved quantity.

To find the conserved quantity we need to look at 
δL.
          .   .
δL = (∂L/∂q_i)δq_i + (∂L/∂q_i)δq_i  
      
          .                     .
Now, (∂L/∂q_i) = p_i and ∂L/∂q_i = p_i.  p_i (ofter 
written as Π_i) is called the "canonical momentum 
conjugate to the coordinate q_i".  For example, 
     .          .    .
L = mx²/2 ∴ ∂L/∂x = mx = p

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

It follows that:

dp_i/dt = ∂L/∂q_i

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

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

δL = d(p_iδq_i)/dt

For symmetry we require that δL = 0

Therefore, 

d(p_iδq_i)/dt = 0

In general, we can replace δq_i with εf_i(q) where 
f_i(q) is a function that defines the particular 
symmetry operation (in this case a linear 
translation).  Thus, we can write:
       
d(p_iεf_i(q))/dt = 0
      
But ε is just a small number (i.e. a constant)
so,

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

Where Q = p_if_i(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 = Π_xf(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:

f_x = δx = -εy 
                               
f_y = δy = εx 

Q = p_xf_x + p_yf_y  
                   .     .                
  = -yp_x + xp_y => 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.
                            .             .     t_B          
δS = ∫dt{(∂L/∂q) - d/dt(∂L/∂q)δq} + [(∂L/∂q)δq]| 
                                                t_A 
                                                          
               . t_A
   = 0 + δq∂L/∂q| 
                 t_B
 
Now we need to include the contributions from 
A and B,
                                                   
           . t_A
δS = δq∂L/∂q| + A - B
             t_B

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,

S_A = εL(t_A) and S_B = εL(t_B).  Therefore, we end up 
with:

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

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 t_A must equal its value at t_B.
                                                                           
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 - q_ip_i = -H

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

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 (q_i).  
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/∂q_i)q_i + (∂L/∂q_i)q_i + ∂L/∂t 
                                    \
Now,                          explicit time term
         .     . .
(∂L/∂q_i)(q_i) = p_iq_i

and,
     .   ..     ..
(∂L/∂q_i)(q_i) = p_iq_i

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

Which can be written as:
            .                                 .   .
dL/dt = d(p_iq_i)/dt + ∂L/∂t  using d(FG)/dt = FG + FG
          .
From before  p_iq_i = 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 = -yp_x + xp_y => 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.