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Why does the Lorentz transformations form a group?

  1. Oct 23, 2006 #1
    Is the reason why the Lorentz transformations form a group because of the reason on this website
    http://en.wikipedia.org/wiki/Lorentz_transformation_under_symmetric_configuration [Broken]

    So the group consits of 3 matrices, {identity, forward transformations, inverse transformations}?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 23, 2006 #2
    There are four conditions which must be satisfied for a set of objects to form a groupt:

    1) If A and B are elements of the group, then C = AB must also be an element of the group.

    2) If A, B, and C are elements of the group, then A(BC) = (AB)C.

    3) There exists an identity element, I, such that AI = IA = A.

    4) Every element has an inverse which is also an element of the group.

    You can show that the Lorentz transformations in 1-dimension obey all of these. And, in fact, if you include the general 3-d rotations, you'll find that the set of all Lorentz transformations together with all rotations form a group.

    So, there are quite a few more than 3 matrices. In fact, there are an infinite number, since [tex]\beta[/tex] can take on any value from 0 to 1. And, this is just for the 1-d transformations. In general, you can have up to 6 free parameters (3 representings boosts in the three spacial directions and 3 representing rotations).
  4. Oct 23, 2006 #3
    So the group contains an infinite number of matrices like the ones on the site with different [tex]\beta[/tex] and [tex]\gamma[/tex]
    values. Correct?
  5. Oct 23, 2006 #4
    Quite right. And, I think I need to make a little correction. [tex]\beta[/tex] can also be negative, indicating a boost in the opposite direction. So, it can really run from -1 to 1. As I wrote it above, I managed to leave out all the inverse boosts. Guess I wasn't quite awake enough yet.
  6. Nov 19, 2006 #5

    Chris Hillman

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    Lorentz group

    Oh my, oh my. I only skimmed the Wikipedia article you linked to, because:

    1. I quit WP some time ago,

    2. as of the time of this post, it has been edited by only one registered but apparently inexperienced Wikipedia user (the individual who created this article) and one anon editor who appears likely to represent the same individual. This is a clue that the article has probably not been read/corrected by WikiProject Physics members, and from experience I know that horribly incoherent or wrong articles are not uncommon, particularly now that so many physicist wikipedians have quit WP.

    Don't forget that the slogan of the Wikipedia includes that little phrase "that anyone can edit".

    Did you try searching WP for "Lorentz group"? You should have found http://en.wikipedia.org/wiki/Lorentz_group, which HAS been read closely by several WikiProject Physics members and as of the version listed at http://en.wikipedia.org/wiki/User:Hillman/Archive was, as far as I know, correct. I can be pretty confident about this because I wrote almost all of that version myself :-/ So my opinion is hardly unbiased, but compare http://www2.corepower.com:8080/~relfaq/penrose.html [Broken]. Even better, compare my version of "Lorentz group" with a good book such as Needham, Visual Complex Analysis or Jones and Silverman, Complex Functions.

    To answer your question, the Lorentz group has infinitely many elements. It is a Lie group, i.e. not only a group but a (finite dimensional) smooth manifold, in fact it is a six dimensional Lie group. See the above cited sources for more details.

    Chris Hillman
    Last edited by a moderator: May 2, 2017
  7. Nov 21, 2006 #6
    Although its not important in this context, this definition is not quite correct. A group is a set of objects and a binary operator. So C=AB should be C=A.B where . is some binary operator.

    So we have, for example, that the real numbers form a group under addition, but not under division.
  8. Nov 22, 2006 #7


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    I agree that the Wiki article on the Lorentz group is well written.

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