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Composition of Lorentz Transformations

  1. May 18, 2006 #1
    It is not intuitive, for me at least, why when relating the velocity of 3 inertial frames (Say F1, F2 moving at v1 with respect to F1, and F3 moving at v2 with respect to F2), one mulitplies the transforms of F2 and F3 to get the transform for F1 with respect to F3 to get v3. I understand why v3 does not equal v1+v2 and have seen some very understandable derivations of the velocity composition formula. But I have not been able to find the derivation of how mupltiplying two transforms leads to the third (and thus leads to v3).

    Thanks, Howard
    Last edited: May 18, 2006
  2. jcsd
  3. May 18, 2006 #2
    Remember that you can make rotations in 3D multiplying diferent matrix, each one corresponding to a rotation along any selected axis, this is totally analogous, since lorentz transformations corresponds to rotations in four dimensions, with the lorentz metric.
    Last edited: May 18, 2006
  4. May 18, 2006 #3

    Tom Mattson

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    Remember that a Lorentz transformation is just a linear transformation, like the ones you encounter in linear algebra. And the Lorentz transformation, like all linear transformations, has a matrix. You perform successive Lorentz transformations the same way you perform any successive linear transformations: By multiplying their matrices together.
  5. May 18, 2006 #4
    Applying the transform once transforms all your vectors to the new co-ordinate system. So:

    [tex]\vec{v'} = T_1\vec{v}.[/tex]

    The next transform will now transform our new vectors, so:

    [tex]\vec{v''} = T_2\vec{v'} = T_2 T_1 \vec{v}.[/tex]

    I hope that explains it.
  6. May 18, 2006 #5


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    The strategy is essentially laid out here [using rapidities in 1+1 spacetime]
    ...but you have to finish it off recognizing that the relative spatial velocity is c*tanh(theta_2-theta_1)... then use the hyperbolic trig identity. If you wish to avoid rapidities, you can use betas and gammas and various identities among them... but they are really hyperbolic trig identities.

    The quickest way to motivate the velocity-composition [without explicitly using the transformations] uses the k-calculus.

    For non-parallel velocity composition, you'll have to work with a larger more general version of the LT matrices (or else work with infinitesimals, or else do trig on a hyperbolic space).
  7. May 19, 2006 #6


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    Do you understand how, in general, to compose arbitrary transforms?

    Say transform T1 maps x->x', and transform T2 maps x'->x'', where x is a vector.

    We can say x' = T1(x), and x''=T2(x').

    The "composed" transform is a map from x->x''

    x'' = T2(T1(x))

    All you need to do for the specific case of the Lorentz transforms is work out the details.
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