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Only Galilei and Poincaré

  1. May 27, 2010 #1

    Fredrik

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    Suppose that we write a function that changes coordinates from one inertial frame to another in the form

    [tex]x\mapsto\Lambda x+a[/tex]

    where [itex]\Lambda[/itex] is linear, with components

    [tex]\Lambda=\gamma\begin{pmatrix}1 & \alpha\\ -v & \beta\end{pmatrix}[/tex]

    in the standard basis. (This is the most general form of a 2×2 matrix with the upper left component non-zero. I'm writing the matrix this way because this v can be interpreted as the velocity difference between two inertial frames). Suppose that we now impose the requirement that the substitution [itex]v\rightarrow -v[/itex] must give us [itex]\Lambda^{-1}[/itex]. (This can be thought of as a mathematical statement that expresses one aspect of Galileo's principle of relativity). We can show that

    [tex]\Lambda^{-1}=\frac{1}{\gamma(\beta+\alpha v)}\begin{pmatrix}\beta & -\alpha\\ v & 1\end{pmatrix}[/tex]

    so the requirement I just mentioned gives us the following information:

    [tex]\beta=1[/itex]

    [tex]\gamma=\pm\frac{1}{\sqrt{1+\alpha v}}[/tex]

    [itex]\alpha[/itex] is an odd function of v, and not equal to -1/v.

    I would like to impose one more mathematical requirement that also expresses an aspect of the principle of relativity and is sufficient to imply that [itex]\alpha(v)=-Kv[/itex], where K is a constant (and the sign is just a convention). I would appreciate if someone could help me find an appropriate axiom, or to justify one that I already know would work. For example, it's sufficient to require that "velocity addition" is commutative, but is there a way to think of that as a consequence of the principle of relativity?

    Once we have the condition [itex]\alpha(v)=-Kv[/itex], it's not too hard to show that K=0 gives us the Galilei group, that K=1 gives us the Poincaré group, and that all other choices of K gives us a group that's isomorphic to the Poincaré group.
     
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  3. May 27, 2010 #2

    Ben Niehoff

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    Given the symmetry between v and -v, I think it is sufficient to demand that any given frame is unique, once its velocity relative to some reference frame A is specified.

    Given frames A and D, it is possible to construct intermediate frames B and C, and use the v/-v symmetry to guarantee that the transitions A -> B -> D and A -> C -> D correspond to velocity additions U + V and V + U. Since frame D must be unique, this establishes the commutativity of velocity addition.
     
  4. May 27, 2010 #3

    Fredrik

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    I don't quite get it. Let's write the velocity of a transformation from A to B (i.e. the speed of the spatial origin of B in the coordinates of A) as vAB. Then we have

    [tex]v_{AD}=v_{AB}\oplus v_{BD}=v_{AC}\oplus v_{CD}[/tex]

    where I'm using "\oplus" for the relativistic velocity addition (which hasn't been completely determined yet, since we don't know [itex]\alpha[/itex]). You seem to be saying that if we choose vCD=vAB, we automagically get vAC=vBD, and therefore commutativity.

    This looks like an approach that might work, but I still don't see how. I also don't see how you think we should use that v and -v thing.


    For completeness, here's my argument for why commutativity of velocity addition implies [itex]\alpha(v)=-Kv[/itex]. We can easily show (just by multiplying two Lambdas together) that velocites "add" according to

    [tex]u\oplus v=\frac{u+v}{1-\alpha(u)v}[/tex]

    If this "addition" is commutative, this is equal to

    [tex]v\oplus u=\frac{v+u}{1-\alpha(v)u}[/tex]

    That implies [itex]\alpha(v)u=\alpha(u)v[/itex] and

    [tex]\alpha'(v)=\frac{d}{dv}\left(\frac{\alpha(u)v}{u}\right)=\frac{\alpha(u)}{u}[/tex]

    which is independent of v.
     
  5. May 27, 2010 #4

    Fredrik

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    Wait, I just realized something:

    [tex]\Lambda(u)\Lambda(v)=\gamma(u)\gamma(v)\begin{pmatrix}1 & \alpha(u)\\ -u & 1\end{pmatrix}\begin{pmatrix}1 & \alpha(v)\\ -v & 1\end{pmatrix}=\gamma(u)\gamma(v)(1-\alpha(u)v)\begin{pmatrix}1 & *\\ * & \frac{1-u\alpha(v)}{1-\alpha(u)v}\end{pmatrix}[/tex]

    (An asterisk means "irrelevant"). The entry on the lower right in that last matrix is the "beta" of the product of two "Lambdas", and we have already seen that beta must be =1. So this gives us the condition [itex]u\alpha(v)=\alpha(u)v[/itex], and that solves the problem completely.

    In other words, we don't need a new axiom. We just use the same one as before, but this time on [itex]\Lambda(u)\Lambda(v)[/itex] instead of [itex]\Lambda(v)[/itex].
     
  6. May 27, 2010 #5

    DrGreg

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    On a minor technicality, I think you meant to define

    [tex]\Lambda=\gamma\begin{pmatrix}\beta & \alpha\\ -v & 1\end{pmatrix}[/tex]​

    for v to be velocity, assuming you are following the standard convention that [itex]\Lambda[/itex] acts on the vector

    [tex]\begin{pmatrix}t\\ x\end{pmatrix}[/tex]​

    That doesn't affect the details of your argument, the same conclusion holds.

    You've probably realised this already, but if you can establish that K ≥ 0, then the next step would be to observe that [itex]\Lambda[/itex]'s eigenvectors are

    [tex]\begin{pmatrix}\pm \sqrt{K}\\ 1\end{pmatrix}[/tex]​

    which are independent of v. That's another way of saying that [itex]\pm K^{-1/2}[/itex] are invariant speeds (abusing language a little by allowing for an "infinite speed" when K = 0).
     
  7. May 28, 2010 #6

    Fredrik

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    This is a copy-and-paste from my personal notes about this. The very first comment is of course only true when the 00 (top left) component is non-zero. The possibility that the 00 component is zero is ruled out [strike]e.g. by the requirement that we get the identity matrix when we set v=0[/strike]. Edit: It's probably better to just say that we interpret the principle of relativity as a statement about frames with finite speeds relative to each other. If we set the 00 component of a 2×2 Lambda to zero, it takes the 0 axis to the 1 axis, meaning that my speed in the other guy's frame is infinite.


    __________________________________________________


    Any 4×4 matrix [tex]\Lambda[/tex] can be written as

    [tex]\Lambda=\gamma\begin{pmatrix}1 & \alpha^T \\ -v & \beta\end{pmatrix}[/tex]

    where [tex]v[/tex] and [tex]\alpha[/tex] are 3×1 matrices, [tex]\beta[/tex] is a 3×3 matrix and [tex]\gamma[/tex] is a number. The reason why we have taken the 00 component outside the matrix is that the v defined this way, i.e. the v defined by

    [tex]v=-\frac{1}{\Lambda_{00}}\begin{pmatrix}\Lambda_{10}\\ \Lambda_{20}\\ \Lambda_{30}\end{pmatrix}[/tex]

    has a natural interpretation as a velocity. To understand why, note that all Lorentz transformations take straight lines through the origin to straight lines through the origin. In particular, the 0 axis is mapped to some straight line through the origin. We can find out which one by letting [tex]\Lambda[/tex] act on (1,0,0,0).

    [tex]\Lambda\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix} =\begin{pmatrix}\Lambda_{00}\\ \Lambda_{10}\\ \Lambda_{20}\\ \Lambda_{30}\end{pmatrix} = \Lambda_{00}\begin{pmatrix}1\\ \Lambda_{10}/\Lambda_{00}\\ \Lambda_{20}/\Lambda_{00}\\ \Lambda_{30}/\Lambda_{00}\end{pmatrix} =\Lambda_{00}\begin{pmatrix}1\\ -v_1\\ -v_2\\ -v_3\end{pmatrix}[/tex]

    We see that the 0 axis is mapped to the line

    [tex]\tau\mapsto\tau\begin{pmatrix}1\\ -v_1\\ -v_2\\ -v_3\end{pmatrix}[/tex]

    When [tex]\Lambda[/tex] represents a coordinate change from an inertial frame S to another inertial frame S', this line can be thought of as the world line of S in S', or if you prefer, as the world line of the physical observer who's using the coordinates S. We choose to call the velocity represented by this line -v, to make v the velocity of S' in S. We will refer to v as the velocity of [tex]\Lambda[/tex].
     
    Last edited: May 28, 2010
  8. May 28, 2010 #7

    Fredrik

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    I thought I had, but now that I'm looking at it again I don't find my argument convincing. I'm going to have to think about that.

    Hm, I just noticed that the choice K=-1 is just a Euclidean rotation. Other negative values of K give us something similar to a rotation, but different lines through the origin are rotated by different angles. (The 0 axis is mapped to a line with slope -1/v and the 1 axis is mapped to a line with slope -Kv).

    This is odd...I tried to find the proof that K≥0 online, but so far I have only found an article that used the same method that I used in my notes. This is how that argument goes: The "gamma" of [itex]\Lambda(u)\Lambda(v)[/itex] is [itex]\gamma(u)\gamma(v)(1+Kuv)[/itex], and if we only admit positive gammas (because we want the group to be path connected), then a negative K is a problem because large enough velocities will make that new gamma negative. But wouldn't that only mean that speeds have to stay below [tex]1/\sqrt{-K}[/tex]?

    So I'm still confused. Are rotations ruled out by the principle of relativity or not?
     
    Last edited: May 28, 2010
  9. May 28, 2010 #8

    Fredrik

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    I had another look, and I think I got it. The proof goes roughly like this: Suppose that K<0. If there's no upper bound on the set of velocities, [itex]\Lambda(u)\Lambda(v)[/itex] may have a negative gamma, contradicting the axiom that the group is path connected. If there is an upper bound on the set of velocities, we can show no matter what that upper bound is, the velocity addition rule can give us a velocity that's even higher, contradicting the assumption that there's an upper bound.
     
  10. May 29, 2010 #9
    Are you working with a specified set of axioms? Could you list them?
     
  11. May 29, 2010 #10

    Fredrik

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    Sure. There's more than one way to do this. I think the axioms can be much weaker than I've made them, but I'm not looking for the weakest possible axioms. I'd rather find the simplest possible proof from some set of axioms that can be described as a mathematical version of Galileo's "principle of relativity".

    The goal is to find every subset G of the set of functions from [tex]\mathbb R^2[/tex] into itself that has the following properties:

    1. The members of G are all smooth bijections.
    2. The members of G all take straight lines to straight lines.
    3. G is closed under the composition of functions.
    4. (G,[itex]\circ[/itex]) is a topological group.
    5. G is connected.
    6. If you take any member [tex]\Lambda[/tex] of G and change the sign of the "velocity" [tex]v=-\Lambda_{10}/\Lambda_{00}[/tex], you get the inverse [tex]\Lambda^{-1}[/tex].

    Note that the point of a "principle" is to eliminate as many possibilities as possible when we need to find an appropriate mathematical structure to use in a new theory. In this case, we're assuming that the mathematical representation of space and time is going to be a mathematical structure that has [tex]\mathbb R^2[/tex] as the underlying set, and that the remaining detail that must be specified to completely identify the structure that we're going to call "spacetime" is that group G.
     
    Last edited: May 29, 2010
  12. May 29, 2010 #11

    DrGreg

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    By the way, Fredrik, have you noticed the following post in the General Physics forum?
    The arXiv link gives an argument very similar to yours.
     
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  13. May 29, 2010 #12

    Fredrik

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    Thanks. I didn't see that thread, but I found that article too. I didn't use it much because I like to do these things my own way, but I did look inside it when I wanted to find a better argument than the one I already had for why my K must be non-negative. And now I also see that I could have saved some time by reading his argument for why my alpha must be a first-degree polynomial. That one took me some time.

    Now I would like to do this whole thing in 3+1 dimensions, and obtain the proper orthochronous Poincaré group. That looks much harder. This time I won't mind "cheating" by peeking into a good article if I can find one.

    Edit: I'm actually making some progress. I started with

    [tex]\Lambda=\gamma\begin{pmatrix}1 & \alpha^T\\ -v & \beta\end{pmatrix}[/tex]

    where [itex]\alpha[/itex], [itex]\beta[/itex] and v are matrices of the appropriate sizes, and was able to show that

    [tex]\Lambda^{-1}=\frac{1}{\gamma(1+\alpha^T\beta^{-1}v)}\begin{pmatrix}1 & -\alpha^T\beta^{-1}\\ \beta^{-1}v & \beta^{-1}(1+\alpha^T\beta^{-1}v)(1+v\alpha^T\beta^{-1})^{-1}\end{pmatrix}[/tex]

    Don't bet any money on this being correct just yet. I still have some double-checking to do. Then I need to figure out how to change the axioms in #10 so that they'll work in the 3+1-dimensional case. Axiom 6 will definitely be more complicated, and I'll have to add at least one more. Something about including rotations as a subgroup.
     
    Last edited: May 29, 2010
  14. May 31, 2010 #13

    Fredrik

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    I have kept working on this, and it's starting to look pretty good. I'm still not sure what's the best way to include a condition that corresponds to the [itex]\Lambda^{-1}(v)=\Lambda(-v)[/itex] that I used in the 1+1-dimensional case, but an acceptable one is to use the definition of a "pure boost" that DrGreg showed me in another thread, and to require that the above identity holds when [itex]\Lambda[/itex] is a pure boost.

    A pure boost is a [itex]\Lambda[/itex] with its [itex]\beta[/itex] equal to

    [tex]B=\frac{I}{\gamma}+\left(1+\frac{1}{\gamma}\right)\frac{vv^T}{v^Tv}[/tex]

    I found that

    [tex]B^{-1}=\gamma I+(1-\gamma)\frac{vv^T}{v^Tv}[/tex]

    Both of these are obviously even functions of v. The above implies Bv=v and a few similar identities that are useful. For a pure boost, we have

    [tex]\Lambda(-v)=\gamma\begin{pmatrix}1 & \alpha^T(-v)\\ v & B\end{pmatrix}[/tex]

    [tex]\Lambda^{-1}(v)=\frac{1}{\gamma(1+\alpha^Tv)}\begin{pmatrix}1 & -\alpha^T B^{-1}\\ v & (1+\alpha^Tv)(1+v\alpha^T)^{-1}B^{-1}\end{pmatrix}[/tex]

    Setting these two equal (and using the forgotten axiom mentioned at the end of this post) gives me

    [tex]\gamma=\frac{1}{\sqrt{1+\alpha^Tv}}[/tex]

    and

    [tex]\alpha(-v)=-B^{-1}\alpha(v)[/tex]

    To find the exact form of [itex]\alpha[/itex], we use that the "beta" of [tex]\bar\Lambda\Lambda[/tex] must have the pure boost form given above, with [itex]\gamma'=\bar\gamma\gamma(1-\bar\alpha^Tv)[/itex] instead of [itex]\gamma[/itex]. That condition gave me

    [tex]\bar\alpha^Tv\frac{vv^T}{v^Tv}=\bar v\alpha^T[/tex]

    which implies that [itex]\alpha_{i,j}(v)[/itex] is independent of v. (A comma denotes partial differentiation). That gives us [itex]\alpha(v)=(k_1v_1,k_2v_2,k_3v_3)[/itex] (plus a constant, which must be zero because [itex]\alpha(-v)=-B^{-1}\alpha(v))[/itex], and isotropy (which was also used to derive the pure boost form of [itex]\beta[/itex]) now implies that all the k's are equal. So [itex]\alpha(v)=-Kv[/itex].

    Oh yeah, I forgot an axiom in my list above:

    7. [tex]\left[\Lambda\begin{pmatrix}1\\ 0\end{pmatrix}\right]_0=\left[\Lambda^{-1}\begin{pmatrix}1\\ 0\end{pmatrix}\right]_0[/tex]

    This one is essentially saying that my predictions about the ticking rates of your clocks will be the same as your predictions about mine. We need this condition to see that [itex]\gamma[/itex] is an even function.
     
    Last edited: May 31, 2010
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