# Infinity times zero, rotational symmetry

1. Feb 27, 2016

### Happiness

To show that the Lagrangian $L$ is invariant under a rotation of $\theta$, it is common practice to show that it is invariant under a rotation of $\delta\theta$, an infinitesimal angle, and then use the fact that a rotation of $\theta$ is a composite of many rotations of $\delta\theta$. But a rotation of $\theta$ is a composite of an infinite number of rotations of $\delta\theta$. If $L$ is invariant under a transformation $R$, is it still invariant under an infinite composite of $R$?

Is 0 + 0 + ..., added infinitely, or $\infty\times0$ still 0?

2. Feb 28, 2016

### Simon Bridge

... the second line is not equivalent to the first.
I guess you are thinking that an infinitesimal has zero size, but that is not correct.

The logic goes like this:
If a process leaves an object unchanged, then repeating the process will still leave the object unchanged - it does not matter how many times you repeat the process.

3. Feb 28, 2016

### Happiness

The logic is clear when the process is repeated a finite number of times. But to get $\theta$, the process must be repeated an infinite number of times.

The Lagrangian $L$ after a transformation $R = L + \delta L = L + 0 = L$.
The Lagrangian $L$ after a composite transformation $RR = L + 2(\delta L) = L + 2(0) = L$.
The Lagrangian $L$ after a composite transformation $RR...R\,($with an infinite number of $R) = L + \infty(\delta L) = L + \infty(0) = L$?

4. Feb 28, 2016

### Simon Bridge

Bad notation... infinity is better understood as a limit, not as a number. ie. Evaluate:
$$\lim_{N\to\infty} N(\delta L) : \delta L = 0$$ ... this works because it's defined, while $\infty(0)$ is undefined.

Perhaps if we switch notation a bit:
A process R acting on L would be run in operator notation like $L' = RL$ ...
If L is invarient under R, then $L'=RL=L$

If we do it again: $L' = R^2L = RRL = R(RL) = R(L) = L$ so we see it is also invarient when the operation is repeated once.

For N (positive integer) operations we write: $L' = R^N L = R^{N-1}(RL) = \cdots$
If we do it infinite times then we are evaluating: $$L' = \lim_{N\to\infty} R^NL$$

5. Feb 28, 2016

### haushofer

How do you calculate derivatives explicitly? Aren't you encountering then the same issue?

6. Feb 28, 2016

### Happiness

I believe explicitly it is as follows:

The infinity $\infty$ here is the number of infinitesimal rotation in the composite. It is of order $\frac{1}{\delta\theta}$. The infinitesimal $\delta$ here is the $\delta L$ under an infinitesimal rotation. It is at most of order $(\delta\theta)^2$. So in this case, the $\infty\times\delta$ is at most $\lim_{\delta\theta\rightarrow0}\frac{1}{\delta\theta}(\delta\theta)^2=\lim_{\delta\theta\rightarrow0}\delta\theta=0$.