1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinity times zero, rotational symmetry

  1. Feb 27, 2016 #1
    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. jcsd
  3. Feb 28, 2016 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ... 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.
     
  4. Feb 28, 2016 #3
    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##?
     
  5. Feb 28, 2016 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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: $$L' = \lim_{N\to\infty} R^NL$$
     
  6. Feb 28, 2016 #5

    haushofer

    User Avatar
    Science Advisor

    How do you calculate derivatives explicitly? Aren't you encountering then the same issue?
     
  7. Feb 28, 2016 #6
    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##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Infinity times zero, rotational symmetry
  1. Rotational symmetry (Replies: 2)

Loading...