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I Proof that Galilean & Lorentz Ts form a group

  1. Dec 8, 2017 #1
    The Galilean transformations are simple.

    x'=x-vt
    y'=y
    z'=z
    t'=t.

    Then why is there so much jargon and complication involved in proving that Galilean transformations satisfy the four group properties (Closure, Associative, Identity, Inverse)? Why talk of 10 generators? Why talk of rotation as part of the proof?

    Can anyone point to a simple formal proof that Galilean transformations form a group, without unnecessary jargon.

    Once we have a simple proof that Galilean transformations form a group why can THAT not be extended to prove, in a simple way, that Lorentz transformations also form a group. Thanks.
     
  2. jcsd
  3. Dec 8, 2017 #2

    haushofer

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    Because the isometries of Galilean spacetime don't merely exist of 3 boosts, but also of 3 spatial rotations and 4 spacetime translations.Those give 3+3+4=10 generators.

    You should start from the action of a point particle and from there derive all the symmetries.
     
  4. Dec 8, 2017 #3

    pervect

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    Formal proofs need very precise language, also known as jargon. So I don't think you'll find a jargon-free formal proof, because in order to state the problem and results precisely enough, you need the precise language that jargon offers.

    You might be able to find an informal jargon-free discussion. I find those illuminating to read when they work, but sometimes the omission of all the formal details causes communications and understanding issues :(. Informal discussions are also arguably easier to write, though perhaps not easier to write well.

    As far as rotation goes, it's part of the group. Without rotations, you have a subgroup of the Lorentz group (or the Galilean group. I'm not sure if this subgroup has a name or not. If you think, informally, of the Galilean group as preserving distances, the Lorentz group preserves the Lorentz interval. But if you want a more formal statement , you really need the formal language (aka jargon), and the informal statements may get the broad picture across, but not necessarily all the fine details.
     
  5. Dec 8, 2017 #4
    Why do proofs of the Galilean transformations seem to approach it as a special case within Lorentz group, rather than give a proof independent on any jargon that arose from special relativity. A 1903 the proof that Galilean transformations form a group would not involve first understanding the proof that Lorentz transformations form a group. What would a 1903 simple proof look like? Any link is appreciated.
     
  6. Dec 8, 2017 #5

    PeterDonis

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    If by "rotations" you mean just spatial rotations, that's actually not true unless you restrict to boosts in a single spatial direction. The set of boosts in all three spatial directions does not form a group, since it is not closed under composition (the composition of two boosts in different directions is not a pure boost but a boost plus a spatial rotation).

    The translations, i.e., the subgroup not including spacetime "rotations" (spatial rotations and boosts) is a subgroup, which I believe is called the translation group.
     
  7. Dec 9, 2017 #6

    haushofer

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    Don't know if it's a 1903 proof, but a simple proof would be to parametrize infinitesimal transformations on the coordinates and check all the commutators, showing that the algebra closes. That's how I used to check the closure of non-rel. algebras.
     
  8. Dec 9, 2017 #7

    haushofer

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  9. Dec 9, 2017 #8

    haushofer

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    Indeed, this is an important difference between Poincare and Galilei-groups; galilean boosts do form a subgroup, while Lorentz boosts do not. This also follows from the particular rescaling of generators in the contraction; see the paper I mentioned.
     
  10. Dec 9, 2017 #9

    pervect

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    Ah yes, Thomas precession. Well, I was at the time of writing thinking of restricting the boosts to a single direction, but being sloppy I didn't say that, and it turns out to make a difference. An excellent point. That's a good illustration of the perils of being informal, I thi9nk.
     
  11. Dec 9, 2017 #10

    pervect

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    I think the point Peter is making is worth expanding. Though I hope I'm not making another error in doing so, sloppily (and from memory). Let X represent the group operation of a Lorentz boost in the x direction. And let ##X^{-1}## represent the inverse operation. Similaryly, let Y and ##Y^{-1}## be the group members for a boost in the Y direction. We can represent X and Y with matrices. (If we were being formal, I'd have to prove this, but I'm just going from memory. Oh, and in the limit of "small" boosts, we can consider X and Y to be the "generaotrs" of the group that the original poster was talking about.)

    Then we need to ask if the commutator ##X \circ Y \circ X^{-1} \circ Y^{-1}## vanishes. And if we compute it using the matrix representation (for example), we find it doesn't. This is not very intuitive, and results in the physical effect called Thomas Precession, There's a wiki article on the topic, https://en.wikipedia.org/wiki/Thomas_precession.

    As far as the four axioms go, I have totally skipped over proving that our proposed group operators, X and Y, are associative, which is something that one might want to prove if there were some question about representing the group operators via matrices, which I simply assumed (without proof) was possible from my recollections.

    Groups must be associative, but they don't have to be commutative, and in general are not. Matrix multiplications have the same properties, which is why they work for representing groups.

    The non-vanishing of the commutator demonstrates the lack of commutation.
     
  12. Dec 9, 2017 #11

    PeterDonis

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    With this definition of "commutator", I think the question is whether it is the identity operator or not. The other common definition of "commutator" is ##XY - YX##, which will vanish if ##X## and ##Y## commute; but if you apply ##X^{-1} Y^{-1}## from the right to both terms, so that the ##XY## term becomes the "commutator" you defined, then the other term becomes the identity if and only if ##XY - YX## vanishes.
     
  13. Dec 11, 2017 #12

    Orodruin

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    Indeed.

    This definition of a commutator requires the difference of the two group elements to hold a meaning, which is usually fine if you are considering the action of the group on some vector space, but does not necessarily hold general meaning. Assuming a Lie group, you could look at the commutator for the elements of the Lie algebra. Of course, the commutator in the Lie algebra is really nothing but the infinitesimal version of ##X^{-1} Y^{-1} X Y = 1##.

    I am not sure it is any less intuitive than rotations not commuting. It is essentially the same effect.
     
  14. Dec 11, 2017 #13

    pervect

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    I wouldn't say rotations are very intuitive either :). But the math isn't terribly difficult, either - matrix notation suffices in many cases. I hope we're not drifting too far afield from the OP's original question. I suspect the OP was finding the formal treatment of Lie groups and generators and whatnot a bit overwhelming in it's detail. (I could be wrong, that was just the impression I had). The meta-point is that all this seemingly abstract math about groups is actually physically quite useful, and at least some degree of familiarity with it is going to be very helpful if one is interested in physics.
     
  15. Dec 11, 2017 #14

    Orodruin

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    Perhaps not very intuitive (depending on who's intuition probably), but probably easier to visualise for most people. It is rather easy to convince oneself that the commutator of rotations in different directions is a rotation in the third direction.

    To return to the OPs questions:

    To be honest, I have not seen this much. Perhaps in literature on special relativity it may be more common to look at the Galilei transformations as a limiting case of the Lorentz transformations, but showing that Galilei transformations form a group would be a standard exercise i classical mechanics without the need to introduce the Lorentz group at all.
     
  16. Dec 11, 2017 #15

    pervect

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    I agree, for a non-mathematical demonstration I personally like to rotate a book around various axes in 90 degree increments. The cover can face in any of six directions, and for each cover facing there are 4 rotational states, for a total of 24 members in the discreete group this generates. Of course, one can carry out the group operations by twisting a physical book with zero math, and demonstrate that the order in which one performs the twists matters. An example of a group operation would be "twist the book 90 degrees clockwise around the z-axis".
     
  17. Dec 20, 2017 #16
    Can you post 1-2 such places where one can find the standard exercise of showing that Galilei transformations form a group - without introducing Lorentz group. Thanks.
     
  18. Dec 20, 2017 #17

    Orodruin

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    This should be available in any basic treatment of symmetries in connection to classical mechanics. For example, I have some discussion on this in my book. The proof itself is straight-forward. Just derive the composition rule and show that it satisfies the group axioms.
     
  19. Dec 20, 2017 #18
    Can you suggest a good book or link for a straight-forward proof? Also which is your book?
     
  20. Dec 20, 2017 #19

    Orodruin

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    There is a discussion on the wikipedia page on Galilean transformations, with a representation of the Galilean group given. To show that it satisfies the group axioms is trivial and is something that you should be easily able to do given that representation.

    See my avatar.
     
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