Conserved quantity when potential is invariant under transformation

Question

I am very new to Noether's theorem and in our (first!) mechanics class it was proved using generators $X_i$,$X$ of a Lie group. Because I didn't really understand this proof I have trouble solving the following problem:

A particle of mass $m=1$ moves through a potential $V(\vec{r})$ which is invariant under the transformation

$$\left(\begin{array}{c}x \\ y \\ z \end{array}\right) \mapsto \left(\begin{array}{c} x\cos\varphi-y\sin\varphi\\ x\sin\varphi + y\cos\varphi\\ z+c\varphi\end{array}\right)$$

Now I have to calculate the conserved Noether current

$$J=\frac{\partial L}{\partial\dot{q}_i}X_i+ \left(L-\frac{\partial L}{\partial\dot{q}_i}\dot{q}_i\right)X.$$

My problem is finding the generators $X$ and $X_i$.

Every advice, hint, tip, etc. is very much appreciated!

$\textbf{Edit:}$ The action is thought to be invariant under the 1-parameter group $t\rightarrow t'=g(t,s)$, $q_i\rightarrow q_i'=g(q_i,s)$ where the depency on $s$ is continuous.

Infinitessimal transformations are defined as $t'=t+\delta t$ with $\delta t = X\delta s$ and $q_i'=q_i+\delta q_i$ with $\delta q_i=X_i\delta s$.

$\textbf{Edit 2:}$ Building on the answer of lux the transformation in matrix form looks like $$\left(\begin{array}{ccc}\cos\varphi&&-\sin\varphi&&0\\ \sin\varphi&&\cos\varphi&&0\\ 0&&0&&1+\frac{c\varphi}{z}\end{array}\right).$$ Differentiating this with respect to $\varphi$ gives $$A(\varphi)=\left(\begin{array}{ccc} -\sin\varphi&&-\cos\varphi&&0\\ \cos\varphi&&-\sin\varphi&&0\\0&&0&&\frac{c}{z} \end{array}\right).$$ Then

$$A(0)=\left(\begin{array}{ccc}0&&-1&&0\\ 1&&0&&0\\0&&0&&\frac{c}{z}\end{array}\right).$$

(The following is just speculation which is inspired from the example of the rotation around the $z$-axis which I looked at to know how to continue in this case.)

The generator would the be $X_i =A(0)\cdot\vec{x}$, where $\vec{x}=(x,y,z)^T)$.

Then using the above expression for the Noether current $J$ I get that

$$\frac{\partial L}{\partial \dot{q}_i}X_i=L_z+p_z\cdot c,$$ where $L_z$ is the $z$-component of the angular momentum and $p_z$ is the $z$-component of the momentum. This is based on the matrix $A$ for which I am not very sure about the entry $a_{33}$...

Answer 1

The generators of the Lie algebra of a Lie group are defined (slight conventional dependence) as follows:

If an element of the group $g( \boldsymbol \alpha) = \exp(i \boldsymbol \alpha \cdot \mathbf T)$ is parameterised by a set of parameters $\alpha_i$ then generator $T^i$ can be found by differentiation:

$$T^i = - i \frac{\partial}{\partial \alpha_i} g( \boldsymbol \alpha) \bigg|_{ \boldsymbol \alpha =0} $$

In your case, there is one parameter, $\varphi$ and one generator, that you can find by differentiation of the transformation if you write it in matrix$\times$vector form.

On the other hand, as you say in your edit, an equally viable way to proceed is to Taylor expand your transformation to first order in the parameter which will define $\delta x$ as the first order change under the transformation. Of course the above formula shows you that the generators "generate" precisely this first order change, and that the "large" transformation corresponds to simply applying an infinite number of infinitesimal transformations via the exponential map.

Edit: Your answer is essentially correct but there are some things that need fixing.

In fact the case of translations is a little non-trivial because it is desirable (indeed necessary) to represent the translation as a linear operator acting on the vector. Unfortunately your suggestion of $1 + \frac{c \varphi}{z}$ is not linear. Moreover, it is sufficiently strange that the translation is by an amount equal to the angle of rotation, so in fact I would prefer to set for the moment $z \rightarrow z + c \omega$ and promote your transformation to a two-parameter one (you can always fix the value of $\omega$ to be equal to $\varphi$ at the end.

As for representing the translation there are two choices. Firstly, in my opinion the best, is to recall that the "momentum operator" $-i \frac{d}{dx}$ in the sense that $$\begin{align}f(x + a) = \exp(i(-i a \frac{d}{dx}))f(x) &= (1 + a \frac{d}{dx} + \frac{1}{2} a^{2} \frac{d^{2}}{dx^{2}} + \cdots )f(x) \\&= f(x) + af'(x) + \frac{1}{2}a^{2}f''(x) + \cdots\end{align}$$ simply reproduces the Taylor expansion of the function about $x$.

In this way I could replace your (3,3) component with a linear operator $\exp(i(-i c \omega \partial_{z}))$ and the generator is simply $T_{p} = -i \partial_{z}$ (I take $\alpha = c \omega$ for this transformation). Note this is just the momentum operator, $p_{z}$! Operating with $T_{p}$ now gives you the conserved current $p_{z}. \omega c$.

The other way is to express the translation using the results of this page on Affine Transformations https://en.wikipedia.org/wiki/Affine_transformation#Representation but it requires you extending your 3-vector to an "augmented" 4-vector; that's the cost of linearising the transformation using matrices and vectors....the result is the same.