How the definition of work is derived from Noether theorem? I cite the following phrases from an answer to the Phys.SE question Why does a force not do any work if it's perpendicular to the motion?

...an alternative would be to treat the work-energy theorem
  Wnet=ΔT,
  (with T the kinetic energy) as an expected behavior (because it is integral in building the conservation law from Newtonian principles and the conservation principle is so useful) and use that to deduce the form that work must take and thus show the reason for the scalar product....At a higher level of sophistication one would employ Noether's theorem as a postulate and work from there.

How the definition of work is related to the theorem of Noether?
Any link, comment or answer, no matter the level of mathematical abstraction, would be welcome.
 A: Partial answer, specifically in relation only to the part, "Why does a force not do any work ... ?", I believe this is as easy as noting that the only way the Lagrangian does not depend on a parameter (alpha) is if the potential also does not depend on that parameter (alpha).  Let's use position in the x direction as our parameter (alpha=x) , as a simple example.  Only the Potential Energy (V) term in the Lagrangian (T-V) can depend on x.  The Kinetic Term (T) in the Lagrangian (T-V) depends on x_dot, or velocity.  If the potential term does not depend on that parameter, the gradient of the potential term will be zero, which means zero force (F = -dV/dx = dp/dt = 0), meaning a conservation property (px in this simple example) related to x, .  Further, if W = F*x, then W=0 here, which is why a force would do no work ... all starting the reasoning from Neother's Theorem, i.e one can start from Neother's Theorem by noting (dL/dalpha = 0, where alpha = x in our example) within her famous equation and go from there.
A: I'll give it an informal try. Just try to go easy on the downvotes. Anyway, Noether's theorem as stated here
essentially says that if we find the lagrangian of a system which remains invariant with respect to some transformation(time evolution or other coordinate changes), then we will find a conserved quantity with respect to this transformation. We represent the transformation as small perturbations. Mathematically, if $\delta t$ denotes the change in time and $\delta q$ the change in genralised coordinates, then:
$\delta t = \Sigma_r.\epsilon_r.T_r$ [where $T_r$ denote the time generator which you can take to be some sequence which give the time of the evolved system]
similarly $\delta q = \Sigma_r.\epsilon_r.Q_r$
The Lagrangian is given by:
$(\frac{\partial L}{\partial q'}.q' - L).T_r - \frac{\partial L}{\partial q'}.Q_r$
Now when the force is perpendicular to the an infinitesimal trajectory of the particle/system, it cannot affect the change in coordinates along the trajectory, i.e no change is $q_r$ is expected from the force, if $q_r$ is used as a parameter to denote the trajectory.(just assume this intuitively obvious fact. Please let me skip the rigor here.)
So we now search for a quantity in the lagrangian, which is invariant with respect to changes in $q_r$. Clearly, this means for the transformation $q_r \to q_r + \delta q_r$ , all other factors are invariant. Namely, the generator $T_r = 0, Q_r = 1$. Thus from the equation, above, the conserved quantity is :
$p = \frac{\partial L}{\partial q'_r}$ One will immediately recognise this as the momentum component along $\delta q_r$. So, the momentum is constant. But this means the quantity $K = \frac{p^2}{2m}$ is also conserved. We define K to be the kinetic energy, and note that this is also conserved. We say work is done by a force $\vec F$ acting on a particle $m$ if there is some change in velocity/momentum along $\vec F$. Since we know that the kinetic energy is conserved there is no work done. 
This explanation may be wrong. Please point out while reading, and i will edit/remove the answer accordingly. However, i am more or less confident about the approach, though i may have messed up near the end.
A: Let's take a look at how the Lagrangian changes with time:
$$
\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{\partial L}{\partial q}\frac{dq}{dt} + \frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dt}
$$
Then by substituting in $\frac{\partial L}{\partial q}$ for the Euler-Lagrange equation:
$$
\frac{\partial L}{\partial q} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)
$$
we get:
$$
\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)\dot{q} + \frac{\partial L}{\partial \dot{q}}\frac{d\dot{q}}{dt}
$$
Noting that the last two terms are simply the product rule applied to $\frac{\partial L}{\partial \dot{q}}\dot{q}$, we get:
$$
\frac{dL}{dt} = \frac{\partial L}{\partial t} + \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\dot{q}\right)
$$
Now, recall that the total energy of a system, $H$, can be given in terms of the Lagrangian by the Legendre transformation:
$$
H = \frac{\partial L}{\partial \dot{q}}\dot{q} - L
$$
Hence, the rate of change in the energy of the system over time is given by:
$$
\frac{dH}{dt} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\dot{q}\right) - \frac{dL}{dt}
$$
from which it is clear that:
$$
\frac{dH}{dt} = - \frac{\partial L}{\partial t}
$$
The quantity $W = \int_{0}^{t} \frac{dH}{dt} dt$, the total change in energy, is the work done on the system (if it is positive) or by the system (if it is negative). If the system does no work, then $W=0$ for any $t$, and so $\frac{dH}{dt} = 0$. This in turn means the Lagrangian can have no explicit time dependence.
If the Lagrangian has no explicit time dependence, then that means the potential, $V$, has no explicit time dependence (since $L = T - V$ and $T$ is the kinetic energy which only depends on $\dot{q}$). This in turn implies that the system must be isolated.
