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Invariant functions of the four-momenta

  1. Sep 3, 2009 #1
    Why is the following statement true?

    The only functions of p^mu that are left invariant under proper proper, orthochronous Lorentz transformations are p^2 = p_mu p^mu and for p^2<=0 also the sign of p^0.

    I can see that they are invariant, but why are these the only invariants?
     
  2. jcsd
  3. Sep 3, 2009 #2
    Any scalar product of four-vectors is an invariant but p^2 is the only combination not involving other vectors.

    Any invariant can be calculated in the rest frame where solely the temporal component is different from zero. It determines the sign of this invariant.
     
  4. Sep 3, 2009 #3
    Well, i should have written "functions of p only". But this still doesn't answer my question.
     
  5. Sep 3, 2009 #4
    In fact, any function of p^2 is an invariant.
     
  6. Sep 3, 2009 #5
    Well then once again: The invariants are the functions of p^2 and the sign of p0 only.
     
  7. Sep 4, 2009 #6

    PeterDonis

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    Well, because a proper orthochronous Lorentz transformation can change everything else. :)

    Seriously, I'm not sure what kind of answer you're looking for. It's kind of like asking why the length of a 3-vector is the only invariant under 3-D rotations; well, of course it is, that's how the group of 3-D rotations is defined. Similarly, proper orthochronous Lorentz transformations (which are the 4-D Minkowski spacetime analogue of rotations in 3-D Euclidean space) are *defined* as the set of all transformations that leave the squared length of 4-vectors and the sign of the 0-component of 4-vectors invariant, and nothing else. (If we imposed other invariants, we'd be looking at a subset of all the transformations, not the full set.)
     
  8. Oct 15, 2009 #7
    You have asked a good question evilcman. In general given a group (in this case the Lorentz group) and a representation (in this case the Lorentz group acting on a four-vector in the usual way,) it is often difficult to say exactly what the invariant functions are. Typically one tries to describe the "ring of invariant polynomials".

    As PeterDonis pointed out, sometimes this is straightforward. When you take a 3-vector (x,y,z) and consider 3d rotations (via actions of SO(3)), you find that the ring of invariant polynomials is generated by x^2+y^2+z^2 which is of course r^2. The ring then has elements like r^2, r^4, r^2 +10r^12, for example. The fact that r^2 is the only generator is essentially due to the fact that the 3d rotation group can take you from any point on a given sphere (centred at the origin) to any other point on the same sphere.

    If you were to restrict to a subgroup though, like rotations around the z-axis, then you would find the ring of invariant polynomials to be generated by x^2+y^2 and z. Lets define R^2:=x^2+y^2. then the ring would have elements like z, R^2, zR^2, 7z+R^4 +z^3R^8, etc.

    There are many instances of groups with a specific representation where we know exactly what the ring of invariant polynomials is and there are plenty where we don't. Broadly speaking this is the branch of mathematics known as classical invariant theory.

    All kinds of techniques from algebra, topology, group theory, representation theory, differential geometry,... are used to try and work out how many invariants there are, how to describe them and then to work out what the relations between them are.

    It is not my experience (recollection?) that the "and nothing else" part of PeterDonis's definition is included in the definition of Lorentz transformations. Do you have a reference PD?

    Now you didn't ask about the invariant polynomials, you asked about the invariant functions. This is a slightly more complicated question in general, but often there are theorems like "any smooth function that is invariant is a smooth function of the invariant *polynomial* functions". There are some examples where this is not true though.

    You posted this a few months ago, but I thought it was worth drawing attention to the fact that it is often not known what the full set of invariants are for a given group and representation and that people actively work on "Invariant theory". I would be happy to share more of my knowledge about this if people were curious. I have seen quite a few times now a publication will present a set of functions as a a complete set of invariants without convincing me they can justify the statement and I'd like to see that change.
     
  9. Oct 15, 2009 #8

    PeterDonis

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    The definitions of the groups we're talking about are http://en.wikipedia.org/wiki/Lorentz_group" [Broken].

    The fact that Lorentz transformations in general (without the "proper orthochronous" part) preserve only p^2 (or only the squared length of 4-vectors in general) follows from their definition by the same form of argument that you gave for 3-D rotations; just substitute the group SO(3,1) for SO(3).

    The OP said "proper orthochronous" Lorentz transformations, which are those that lie in the identity component of the general group of Lorentz transformations. That's why they also preserve the sign of p_0, but again, by the same argument as above, they can't preserve anything else; just use the identity component of SO(3,1) instead of the full SO(3,1) in the argument.
     
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  10. Oct 15, 2009 #9

    PeterDonis

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    Oops--this should be O(3,1), not SO(3,1), in both places (but the identity component *will* be a subgroup of SO(3,1)).
     
  11. Oct 15, 2009 #10

    dx

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    Just think about Euclidean space. It is easy to see that the only invariant associated with a vector is its length. This doesn't prove anything about Minkowski space, but they are similar enough that this analogy can be used to work out many things with confidence if used properly.
     
  12. Oct 15, 2009 #11
    Yes but you will notice that the definition says it preserves the quadratic form, not that it *only* preserves the quadratic form.

    This is an important distinction to be made because sometimes you will formulate a system to have a particular kind of invariance only to discover there are additional invariant quantities.


    Anyway, to answer the original question, it is not quite "just substitute the group SO(3,1) for SO(3)" but it is not much harder.

    If (orthochronous) SO(3,1) leaves a function invariant than so does the SO(3) subgroup of (orthochronous) SO(3,1) acting on the spatial part of the 4-vector. Now because SO(3) is completely transitive on the sphere, only functions of r are invariant (like mentioned above.)

    Hence one can just consider (orthochronous) SO(1,1) acting on (t,r). (Orthochronous) SO(1,1) can be represented by a 2-by-2 Lorentz matrix transformation that you would no doubt be familiar with. (orthochronous) SO(1,1) will leave any hyperbola of the form t^2-r^2 = constant invariant so t^2-r^2 is an invariant. However SO(1,1) is not completely transitive on each of these hyperbola. It will not map a point on the one branch of the hyperbola to the other but on each branch is easily shown to be completely transitive. Thus every orbit of orthochronous SO(3,1) on a four vector is completely specified by the sign of the zeroth component and the Lorentzian norm of the four-vector.

    Anything else that is invariant must at least be able to be written in a relation with these quantities (even if it is complicated) since it is not specifying any new information (refining the orbit space.)
     
  13. Oct 15, 2009 #12
    Thanks for the input. Sometimes I do miss explanations like this from physics texts...
     
  14. Oct 16, 2009 #13

    Fredrik

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    Very good posts firearrow. I hope you stick around. We always need more people like you here.

    For the rest of you: Note that there actually is another invariant. If we consider 1+1-dimensional proper and orthochronous Lorentz transformations, for [itex]p^2>0[/itex], the sign of [itex]p^1[/itex] is invariant. And even if there had been no other invariants (or if you were just ignoring this invariant because we're not interested in tachyons) evilcman's question was still a good question to ask.

    The key to understanding these things is to understand that a group action on a set partitions the set into "orbits". In this case the set is [itex]\mathbb R^4[/itex], and the orbits are defined as [itex]\mathcal O_p=\{\Lambda p|\Lambda\in SO(3,1)\}[/itex] for each [itex]p\in\mathbb R^4[/itex]. Note that if [itex]p'\in\mathcal O_p[/itex], then [itex]\mathcal O_{p'}=\mathcal O_p[/itex]. From this it's easy to see that each p belongs to exactly one orbit.

    Let's focus on SO(1,1) again. When we discover that [itex]p^2=-(p^0)^2+(p^1)^2[/itex] is invariant, we know that each orbit must be a subset of a hyperbola, but we don't know which subset until we have noticed (as firearrow mentioned) that the group action restricted to a branch of one of these hyperbolas is transitive. "Transitive" means that for each p and p' on the branch, there's a [itex]\Lambda[/itex] such that [itex]p'=\Lambda p[/itex]. If we consider a specific p with [itex]p^0>0[/itex], then transitivity on the branch [tex]p^0=\sqrt{(p^1)^2+m^2}[/tex] is proved by the fact that we can choose [itex]\Lambda[/itex] to give [itex](\Lambda p)^1[/itex] any value.

    Now what that means is that we have completely determined the orbits of SO(1,1), which is what we're really interested in. This is much more interesting (to a physicist) than finding all the invariant functions, because it's a huge step towards finding all the irreducible representations of the group.
     
  15. Oct 16, 2009 #14

    PeterDonis

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    I wondered about this when I was reminded, by reading up on Lorentz transformations, that proper orthochronous Lorentz transformations preserve the handedness (parity) of the spatial vectors as well as the direction of time. But I wasn't able to find an expression for the corresponding invariant for 3+1 dimensions. Is there one?
     
  16. Oct 16, 2009 #15

    PeterDonis

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    True, and your post captured what I was trying to get at much better than mine did.
     
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