Noether Math 1:
ε−Transformation
In 1918 Emmy Noether rigorously proved that the conservation laws are consequences of certain continuous symmetries of time, space and other symmetries appearing in classical mechanics, field theory, quantum mechanics, general relativity and other parts of theoretical physics.
In a very simplified form she proved that, if the laws of physics are the same today and tomorrow, or here and there, then some physical characteristics of a mechanical system must remain invariant, that is unchanged.
Symmetry
Imagine a physicist performs an experiment in a box.
We then rotate the box. If the physicist repeats this experiment with exactly the same initial configurations and finds no difference in the results, we say the laws governing the experiment are rotationally symmetric.
Similarly, if there are no differences in experiments under spatial or temporal translation, then the laws describing the experiment are spatially or temporally invariant.
Spatial translation is when the physicist repeats the experiment at a different location.
Temporal translation is when the physicist repeats the same experiment after a certain waiting period.
The mechanical laws are represented by the equations of motion.
So, "the same results of an experiment" means that the laws (that is, the equations of motion) are the same (invariant) relative to a corresponding transformation.
In Lagrangian Mechanics, this idea of symmetry takes a precise formulation.
A symmetry is a continuous transformation of time t, space coordinates q={qi}, or other dynamical variables under which the Lagrangian L describing the mechanical system remains unchanged or changes only by a total time derivative dF(t,q(t),q'(t))/dt of some smooth function F(t,q(t),q'(t)), where q'(t) signifies a time-derivative of a coordinate q, that is velocity.
A nuance related to a total time derivative of some smooth function F(t,q(t),q'(t)) might need an explanation.
If the Lagrangian will change by dF(t,q(t),q'(t))/dt, the action functional
Φ[L(t)] = ∫ab[L(t,q,q')+dF/dt]dt
defined on a trajectory with endpoints q(a)=A and q(b)=B fixed will change by the boundary term F(b)−F(a).
Considering the end points of a trajectory are fixed as initial conditions, the extremals will be preserved, leading to the same Euler–Lagrange equations and the same physical trajectories.
Let's rigorously define the symmetry as a continuous transformation we are talking about.
Consider a configuration space of n degrees of freedom with generalized coordinates
Consider a family of transformation functions parameterized by variable ε
t(ε)=T(ε,t,q1,...,qn)=T(ε,t,q)
and (for each i∈[1,n])
qi(ε)=Qi(ε,t,q1,...,qn)=Qi(ε,t,q)
define a transformations from original space-time coordinates {t,q} to the new ones {t(ε),q(ε)} if t(0)=t and qi(0)=qi for each i∈[1,n].
Sometimes for brevity we will use
q(ε)=Q(ε,t,q)
implying qi(ε) on the left and Qi(ε,t,q) on the right side for all i from 1 to n.
Functions T(ε,t,q) and Qi(ε,t,q) are assumed to be continuous with respect to the first argument ε for any t and q.
Usually, we will require even a stronger requirements for transformational functions T(ε,t,q) and Qi(ε,t,q) to uniformly converge to T(0,t,q) and Qi(0,t,q) as ε→0.
In addition, to define it more rigorously, we require that all transformations make up a group (more precisely, the Lie group) in a sense that transformations can be combined:
{t,q}→{t(α),q(α)}→{t(β),q(β)}
should be equivalent to
{t,q}→{t(α+β),q(α+β)}
In particular, any
(ε)+(−ε)=(0).
and, as we stated above,
t = t(0) = T(0,t,q)
q = q(0) = Q(0,t,q)
Functions T(ε,t,q) and Q(ε,t,q) define the parameterized by ε transformation of time and position.
Since, ultimately, we need to check the Lagrangian for invariance to transformations, we also need to know how velocities are changing with transformation because Lagrangian depends on them.
The derivation of a formula that defines this change is rather complex and the rest of this lecture is dedicated precisely to dealing with transformation of velocity.
If transformation does not involve time then the transformed velocity is just a time-derivative of the transformed position
qi'(ε) = dqi(ε)/dt = qi(ε)'
Notice the order of ε-transformation and apostrophe that indicates a time derivative.
Thus, in this particular case
qi'(ε) = dQ(ε,t,q)/dt
The complication in our case is, not only coordinates are changing, but time as well.
So, the correct definition of velocity transformation is
qi'(ε) = dqi(ε)/dt(ε)
Let's start with some simple manipulation.
dqi(ε)/dt(ε) =
= [dqi(ε)/dt] / [dt(ε)/dt] =
= [dQi(ε,t,q)/dt] / [dT(ε,t,q)/dt]
The numerator, by chain rule, equals to
∂Qi/∂t+Σk(∂Qi/∂qk)·qk'
The denominator, by chain rule, equals to
∂T/∂t+Σk(∂T/∂qk)·qk'
Therefore,
| qi'(ε) = |
|
Qi = Qi(ε,t,q) = qi(ε)
T = T(ε,t,q) = t(ε)
As ε→0, we can represent
t(ε)= t + ε·[∂t(ε)/∂ε|ε=0] +o(ε)
qi(ε)= qi + ε·[∂qi(ε)/∂ε|ε=0] +o(ε)
An expression
τ(t,q) = ∂t(ε)/∂ε|ε=0 =
= ∂T(ε,t,q)/∂ε|ε=0
is a derivative of the time transformation function by parameter of transformation ε at the initial point of transformation ε=0 and is called the infinitesimal generator of time transformation.
So, in terms of this generator,
t(ε)= t + ε·τ(t,q) +o(ε)
An expression
ξi(t,q) = ∂qi(ε)/∂ε|ε=0 =
= ∂Qi(ε,t,q)/∂ε|ε=0
is a derivative of the space transformation function by parameter of transformation ε at the initial point of transformation ε=0 and is called the infinitesimal generator of space transformation.
So, in terms of this generator,
qi(ε)= qi + ε·ξi(t,q) +o(ε)
Using the above representations, we can evaluate the terms of the formula for qi'(ε).
∂T(ε,t,q)/∂t =
= ∂t(ε)/∂t =
= ∂[t+ε·τ(t,q)+o(ε)]/∂t
derivative of t by t is 1;
ε is an independent of t parameter
= 1 + ε·∂τ(t,q)/∂t
∂T(ε,t,q)/∂qk =
= ∂t(ε)/∂qk =
= ∂[t+ε·τ(t,q)+o(ε)]/∂qk
derivative of t by qk is 0;
ε is an independent of qk parameter
= ε·∂τ(t,q)/∂qk
∂Qi (ε,t,q)/∂t =
= ∂qi(ε)/∂t =
= ∂[qi+ε·ξi(t,q)+o(ε)]/∂t
qi is the initial position in space (before transformation is applied) and, when used in partial derivatives of function that depends on it and other parameters, like time t, is treated as independent of other parameters, thus derivative
ε is an independent of t parameter
= ε·∂ξi(t,q)/∂t
∂Qi (ε,t,q)/∂qk =
= ∂qi(ε)/∂qk =
= ∂[qi+ε·ξi(t,q)+o(ε)]/∂qk
derivative of qi by qk is 1 for i=k and 0 otherwise ⇒ (using math symbol δ) derivative is equal to δik;
ε is an independent of t parameter
= δik + ε·∂ξi(t,q)/∂qk
We are ready to evaluate numerator and denominator of the above expression for qi'(ε).
Numerator =
= ε·∂ξi(t,q)/∂t +
+ Σk[δik+ε·∂ξi(t,q)/∂qk]·qk' =
= ε·∂ξi(t,q)/∂t +
+ qi' + ε·Σk[∂ξi(t,q)/∂qk]·qk' =
= qi' +
+ε·{∂ξi(t,q)/∂t+ Σk[∂ξi(t,q)/∂qk]·qk'}
Denominator =
= 1 + ε·∂τ(t,q)/∂t +
+ Σk[ε·∂τ(t,q)/∂qk]·qk' =
= 1 +
+ε·{∂τ(t,q)/∂t+ Σk[∂τ(t,q)/∂qk]·qk'}
Now the formula for a transformation of velocity looks like
| qi'(ε) = |
|
A = ∂ξi(t,q)/∂t +
+ Σk[∂ξi(t,q)/∂qk]·qk'
and
B = ∂τ(t,q)/∂t +
+ Σk[∂τ(t,q)/∂qk]·qk'
Let's not forget our goal - to establish how the velocity is transformed by
If the above expression would be of the form
qi'(ε) − qi' = ε·X + o(ε)
then we could say that
limε→0 (qi'(ε) − qi')/ε =
= d(qi'(ε))/dε|ε=0 = X
To achieve this goal, we will convert our formula
| qi'(ε) = |
|
(1+ε·B)·[qi' + ε·(A−B·qi')] =
= qi' + ε·A + o(ε)
Dividing both parts by (1+ε·B) will result in
qi' + ε·(A−B·qi') =
= [qi' + ε·A + o(ε)] / (1+ε·B) =
= [qi' + ε·A] / (1+ε·B) + o(ε) =
= qi'(ε) + o(ε)
Therefore,
qi'(ε) = qi' + ε·(A−B·qi') + o(ε)
Hence, the instantaneous rate of change of velocity at point {t,q} is
d(qi'(ε))/dε|ε=0 = A−B·qi' =
= {∂ξi(t,q)/∂t +
+ Σk[∂ξi(t,q)/∂qk]·qk'} −
− qi'·{∂τ(t,q)/∂t +
+ Σk[∂τ(t,q)/∂qk]·qk'}
In the above expression we can recognize two full time derivatives
dξi(t,q)/dt = ∂ξi(t,q)/∂t +
+ Σk[∂ξi(t,q)/∂qk]·qk'
and
dτ(t,q)/dt = ∂τ(t,q)/∂t +
+ Σk[∂τ(t,q)/∂qk]·qk'
which makes our formula for a rate of change of velocity much simpler
d(qi'(ε))/dε|ε=0 =
= dξi(t,q)/dt − qi'·dτ(t,q)/dt
where
represents the rate of change of position by infinitesimal ε-transformation, whose time derivative (the first term in the above formula) is the rate of change of velocity without taking into consideration the transformation of time
and
whose time derivative multiplied by qi' (the second term in the above formula) represents an adjustment caused by ε-transformation of time.
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