Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to show that 2 seccessive Lorentz Trasformations ,in general,are non-commutative?

  1. Jan 10, 2007 #1
    thank you for giving me the answer :)
  2. jcsd
  3. Jan 10, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    I doubt anyone will just give you the answer without you putting some effort in. For example, try writing what you know about lorentz transformations, how you might go about answering the question, etc..

    Why should someone spend time answering a question that you have not even asked properly?
    Last edited: Jan 10, 2007
  4. Jan 11, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    HINT: Lorentz boosts along the same space axis do form an uniparameter group.

  5. Jan 13, 2007 #4


    User Avatar
    Homework Helper

    long way of doing things: Lorentz transformations form a group, if this group is non-abelian then you original problem is answered. may need to work out multiplication table of this group to show that... or perhaps there are smarter way to do things? I'll leave that to you.... but one final remark: if all you need to show is that they are in general such as such....just show by examples
  6. Jan 13, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    From the way you posed the question, you just need to find one counterexample to commutativity.

    Note that a spatial rotation is a particular Lorentz transformation. Hopefully you know that spatial rotations in three-dimensional [Euclidean] space are also generally noncommutative. There's a famous example with rotating a book by 90 degrees along two different axes successively, then comparing the results when the order is reversed.

    You're probably asking about Lorentz boosts. You might try the analogue of the preceding example... implicit in the hint by dextercioby.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook