I Taylor expansion about lagrangian in noether

AI Thread Summary
The discussion centers on the mathematical derivation of Noether's theorem, particularly how Lagrangians transform under infinitesimal transformations. It highlights that when Lagrangians differ by a total time derivative of a function, the first-order Taylor expansion suffices to establish equivalence between transformed and original Lagrangians. The key point is that the first-order term leads to a conservation law derived from the Euler-Lagrange equations. The conversation also touches on the possibility of extending these concepts to more general transformations, maintaining the symmetry condition and resulting conservation laws. Understanding these transformations is crucial for grasping the implications of Noether's theorem in physics.
gionole
Messages
281
Reaction score
24
I was studying a derivation of noether's theorem mathematically and something struck my eyes.

Suppose you have ##L(q, \dot q, t)## and you transform it and get ##L' = L(\sigma(q, a), \frac{d}{dt}\sigma(q,a), t)##. ##\sigma## is a transformation function for ##q##

Let's represent ##L'## by taylor around point 0, which gives us:

##L(q, \dot q) + a\frac{\partial L}{\partial a}\Bigr|_{a=0} + a^2\frac{\partial^2 L}{\partial a^2}\Bigr|_{a=0} + ...##

Now, here is the tricky part: If lagrangians(before and after transformation) are differed by total time derivative of some function, we can say that:
##\frac{\partial L}{\partial a}\Bigr|_{a=0} = \frac{d}{dt}\Lambda##

Note that first order from taylor turned out to be enough, even though taylor only with first order is not the exact(100%) approximation of any ##L##. So it turns out that we use: ##a\frac{\partial L}{\partial a}\Bigr|_{a=0} = a\frac{d}{dt}\Lambda## instead of ##a\frac{\partial L}{\partial a}\Bigr|_{a=0} + a^2\frac{\partial^2 L}{\partial a^2}\Bigr|_{a=0} + ... = \frac{d}{dt}\Lambda##

I am told that we can do this because ##a## is infinetisemal transformation, but still don't get why this works. I know I need to know lie theory to understand this, but isn't there really any other way to somehow grasp it without bunch of math ?
 
Last edited:
Physics news on Phys.org
You just need that ##L'## and ##L## should be equivalent Lagrangians, i.e., that there exists a function ##\Omega(q,t)## such that
$$L'[\sigma(q,a),\dot{\sigma}(q,a),t]=L(q,\dot{q},t]+\dot{\Omega}(q,t),$$
where the dots mean total derivatives (symmetry condition).

For an infinitesimal transformation, i.e., ##q'=q+\epsilon \sigma(q,t)## you can make the ansatz ##\Omega(q,t)=-\epsilon \tilde{\Omega}(q,t)## and expand the LHS of the symmetry condition to first order in ##\epsilon##, which then leads to a conservation law for the solutions of the Euler-Lagrange equations.

BTW: That's only the most simple special form of the transformations considered in Noether's theorem. You can extend it to more general transformations,
$$t'=t+\epsilon \Theta(q,\dot{q},t), \quad q'=q+\epsilon Q(q,\dot{q},t).$$
Then
$$\frac{\mathrm{d}}{\mathrm{d} t'} L[q+\epsilon Q,\dot{q}+\epsilon \dot{Q},t+\epsilon \Theta(q,\dot{q},t)]=L(q,\dot{q},t) -\epsilon \dot{\tilde{\Omega}}(q,t).$$
Now again you can expand everything to 1st order in ##\epsilon## to get the symmetry condition, and then there follows again a conservation law for the solutions of the Euler-Lagrange equations.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Hello! I am generating electrons from a 3D gaussian source. The electrons all have the same energy, but the direction is isotropic. The electron source is in between 2 plates that act as a capacitor, and one of them acts as a time of flight (tof) detector. I know the voltage on the plates very well, and I want to extract the center of the gaussian distribution (in one direction only), by measuring the tof of many electrons. So the uncertainty on the position is given by the tof uncertainty...
Back
Top