Noether’s theorem -- Question about symmetry coordinate transformation

[gionole]

TL;DR

Question about symmetry coordinate transformation

Let's say we got one particle in x direction only and we got some motion x(t) which we figured it out through Lagrangian.

In Noether's theorem for coordinate transformation symmetry, we start with the following:

x(t)' = x(t) +εf(t) (ε - some number) - I denoted new path with x(t)'

I'm focusing now on εf(t). In Noether's theorem, εf(t) can't really be any arbitrary function(note that for 2 particles, it must be the same, but I'm not asking that). The reason it can't be arbitrary is if it is kind of wiggling function, then adding it to x(t) will result in a x(t)' where every coordinate of the old path didn't move by the same distance to result in the new path.

Q1: Is the εf(t) really arbitrary and can be anything ?(I know it's small, but that doesn't mean it can't be such a function where middle of it is much bigger than left one, which could cause the new path trajectory definitely not symmetrically moved from old one). Namely, I thought what we mean by coordinate transformation is moving every bit of point on the current trajectory by the same amount. I'm asking as I haven't seen such restrictions in textbooks and if it can be anything, then trajectory has not moved symmetrically - while in Lagrangian, it can be any wiggling function.

Could you say/explain where I'm making wrong statements ? what would you feel like I'm missing ? The best way would be to follow the single particle example first.

Q2: If every point on the trajectory didn't move by the same distance because of εf(t), then we wouldn't have δS = 0 right ? How do I prove that ONLY the symmetric movement of the whole trajectory would cause δS = 0 ? If you assume that δS = 0, then you can get the momentum conservation, but I'm more interested in the proof of δS = 0 in terms of εf(t) and Noether's theorem.

 

[wrobel]

Well, let's divide Noether into parts.

1) $L(x,\dot x)=L\Big(g^s(x),\frac{\partial g^s}{\partial x}\dot x\Big),\quad \forall s,$
here $g^s$ is a group of symmetry generated by a vector field $v(x)$. In this case, Noether says that there exists a first integral of the Lagrangian system:
$$I(x,\dot x)=\frac{\partial L}{\partial \dot x}v.$$
Direct calculation checks it.
2) The general case when the Lagrangian depends on t and the symmetry group is defined on the space (t,x) is reduced to 1) by the following trick.

Assume that $L=L(t,x,\dot x)$; and perform a change of time $t=t(\tau)$. From the Least Action Principle, it follows that;
in this new time, the system is described by the following autonomous Lagrangian
$$\tilde L(t,x,x',t')=L(t,x,x'/t')t',\quad '=\frac{d}{d\tau}$$

 

[gionole]

Well, let's divide Noether into parts.

1) $L(x,\dot x)=L\Big(g^s(x),\frac{\partial g^s}{\partial x}\dot x\Big),\quad \forall s,$
here $g^s$ is a group of symmetry generated by a vector field $v(x)$. In this case, Noether says that there exists a first integral of the Lagrangian system:
$$I(x,\dot x)=\frac{\partial L}{\partial \dot x}v.$$
Direct calculation checks it.
2) The general case when both the Lagrangian and the symmetry group depend on the time is reduced to 1) by the following trick.

Assume that $L=L(t,x,\dot x)$; and perform a change of time $t=t(\tau)$. From the Least Action Principle, it follows that;
in this new time, the system is described by the following autonomous Lagrangian
$$\tilde L(t,x,x',t')=L(t(\tau),x,x'/t')t',\quad '=\frac{d}{d\tau}$$

I am sorry but this does not help me as I really cant imagine what g and s are even though you said it. Could you read my question and explain it in my words ? Thank you

 

[wrobel]

explain it in my words ?

Your words are harder to understand than mine, especially the hardness increases due to relaxed language common for physics textbooks.
Good luck!

 

[vanhees71]

TL;DR Summary: Question about symmetry coordinate transformation

Let's say we got one particle in x direction only and we got some motion x(t) which we figured it out through Lagrangian.

In Noether's theorem for coordinate transformation symmetry, we start with the following:

x(t)' = x(t) +εf(t) (ε - some number) - I denoted new path with x(t)'

I'm focusing now on εf(t). In Noether's theorem, εf(t) can't really be any arbitrary function(note that for 2 particles, it must be the same, but I'm not asking that). The reason it can't be arbitrary is if it is kind of wiggling function, then adding it to x(t) will result in a x(t)' where every coordinate of the old path didn't move by the same distance to result in the new path.

Q1: Is the εf(t) really arbitrary and can be anything ?(I know it's small, but that doesn't mean it can't be such a function where middle of it is much bigger than left one, which could cause the new path trajectory definitely not symmetrically moved from old one). Namely, I thought what we mean by coordinate transformation is moving every bit of point on the current trajectory by the same amount. I'm asking as I haven't seen such restrictions in textbooks and if it can be anything, then trajectory has not moved symmetrically - while in Lagrangian, it can be any wiggling function.

Could you say/explain where I'm making wrong statements ? what would you feel like I'm missing ? The best way would be to follow the single particle example first.

Q2: If every point on the trajectory didn't move by the same distance because of εf(t), then we wouldn't have δS = 0 right ? How do I prove that ONLY the symmetric movement of the whole trajectory would cause δS = 0 ? If you assume that δS = 0, then you can get the momentum conservation, but I'm more interested in the proof of δS = 0 in terms of εf(t) and Noether's theorem.

In Noether's theorem it's sufficient to consider "infinitesimal transformations" of time and configuration-space variables. In the Lagrange formalism it's of the form
$$q^{\prime j}=q^j + \delta \epsilon Q^j(q,\dot{q},t), \quad t'=t+\delta \epsilon \Theta(q,\dot{q},t).$$
Then you demand that the new Lagrangian is equivalent to the old, i.e., that there exists a function $\Omega(q^j,t)$ (it must NOT be dependent on the $\dot{q}^j$!)
$$L'(q',\dot{q}',t) = \frac{\mathrm{d} t'}{\mathrm{d} t} L[q^{\prime j}(q,\dot{q},t),\dot{q}^{\prime j}(q,\dot{q},t)) + \frac{\mathrm{d}}{\mathrm{d} t} \Omega(q^j,t).$$
Then the above infinitesimal transformation is a symmetry transformation.

Then, using the equations of motion (Euler-Lagrange equations) you can show that this implies the existence of a conserved quantity, which is given in terms of $Q^j$, $T$, and $\Omega$.

I think there's a good treatment in Scheck's textbook on mechanics:

F. Scheck, Mechanics, Springer (2010)