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Orthochronous subspace of Lorentz group.

  1. Apr 14, 2014 #1
    In a Lorentz group we say there is a proper orthochronous subspace. How can I prove that the product of two orthchronous Lorentz matrices is orthochronous? Thanks. Would appreciate clear proofs.
     
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  3. Apr 14, 2014 #2

    robphy

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    Sounds like homework...

    As a start, what is your definition of orthochronous?
     
  4. Apr 14, 2014 #3
    [itex] \Lambda^0_0 \geq 1 [/itex]
     
  5. Apr 14, 2014 #4

    samalkhaiat

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    Consider two such matrices, [itex]A[/itex] and [itex]B[/itex], call their product [itex]C[/itex]. Now take the [itex](0,0)[/itex] entries of the following matrix equations
    [tex]A B = C , \ \ \ A \ \eta \ A^{ T } = B^{ T } \eta \ B = \eta .[/tex]
    Now, if you use the Schwarz inequality
    [tex]
    - \sqrt{ ( A^{ 0 }{}_{ i } )^{ 2 } ( B^{ i }{}_{ 0 } )^{ 2 } } \leq A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 } ,
    [/tex]
    and
    [tex]A^{ 0 }{}_{ 0 } \geq 1 , \ \ B^{ 0 }{}_{ 0 } \geq 1 ,[/tex]
    you should be able to show that [itex]C^{ 0 }{}_{ 0 } \geq 1[/itex]
     
  6. Apr 14, 2014 #5
    Thanks for reply. Why is there a minus sign in Schwartz inequality? Isn't it [itex] \sqrt{ ( A^{ 0 }{}_{ i } )^{ 2 } ( B^{ i }{}_{ 0 } )^{ 2 } } \geq |A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 }| [/itex]?

    [tex]

    C^0 {}_0 - A^0 {}_0 B^0 {}_0 = A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 }

    [/tex]

    How to proceed?
     
  7. Apr 15, 2014 #6

    Fredrik

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    I don't (yet) see what Sam is doing there. But what you wrote isn't the standard Cauchy-Schwarz inequality either. I think what you're looking for is
    $$\left|\sum_i A^0{}_i B^i{}_0\right|\leq \sqrt{\sum_i(A^0{}_i)^2}\sqrt{\sum_j(B^j{}_0)^2}.$$ Do you know how to prove that if ##\Lambda## is orthochronous, its 00 component is either ##\geq 1## or ##\leq -1##? If you have proved that, it will be sufficient to prove that the 00 component of the product transformation is ##\geq 0##. I think that makes the problem much easier. One approach is to try to show that (in a notation with all indices of matrix components written downstairs) ##(\bar\Lambda\Lambda)_{00}=\bar\Lambda_{00} \Lambda_{00} X,## where X is a non-negative real number.
     
    Last edited: Apr 15, 2014
  8. Apr 15, 2014 #7
    So how does it work out?
     
  9. Apr 15, 2014 #8

    Fredrik

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    Have you tried this approach? (I have, so I know that it works). What do you get?
     
  10. Apr 15, 2014 #9

    micromass

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    Moved to homework since it is a homework type question.

    Please provide an attempt at the solution.
     
  11. Apr 15, 2014 #10
    [tex] |C^0 {}_0 - A^0 {}_0 B^0 {}_0 | \leq \sqrt{\sum_i(1-B^0{}_0{}^2)}\sqrt{\sum_j(A^0{}_i)^2} [/tex]. I don't know how to express $$ A^0{}_i^2$$ in terms of A00
     
    Last edited: Apr 15, 2014
  12. Apr 15, 2014 #11

    Fredrik

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    I don't see what you're doing there.

    I worked through the solution by my method again. I had to use Cauchy-Schwartz, the familiar version for vectors on ##\mathbb R^3##, but not until the very end.

    You didn't answer if you have already proved that the 00 component of a Lorentz transformation can't be in the interval (-1,1). You should start with that, because it makes your task easier. You will find that result if you look at the 00 component of the matrix equation ##\Lambda^T\eta\Lambda=\eta##. (Use the definition of matrix multiplication). The next step is to use the definition of matrix multiplication on the 00 component of ##\bar\Lambda\Lambda##. Feel free to use the notation AB instead of ##\bar\Lambda\Lambda## if you prefer.
     
  13. Apr 15, 2014 #12
    Yes, that's clear.
     
  14. Apr 15, 2014 #13

    Fredrik

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    OK, I should also mention that I find it useful to know that if ##\Lambda## is the transformation from the coordinate system S to the coordinate system S', then (in my notation, where all matrix indices are downstairs), ##\Lambda_{i0}/\Lambda_{00}## is the velocity of S in S'. It's not absolutely necessary to know that, but it makes some steps easier to see.
     
  15. Apr 15, 2014 #14

    samalkhaiat

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    Ok, doesn’t this imply that
    [tex]- \sqrt{ ( A^{ 0 }{}_{ i } )^{ 2 } ( B^{ j }{}_{ 0 } )^{ 2 } } \leq A^{ 0 }{}_{ i } \ B^{ i }{}_{ 0 } .[/tex]

    Now, from the other two equations, you have (sums over indices are implies)
    [tex]( A^{ 0 }{}_{ i } )^{ 2 } = ( A^{ 0 }{}_{ 0 } )^{ 2 } - 1 ,[/tex]
    [tex]( B^{ j }{}_{ 0 } )^{ 2 } = ( B^{ 0 }{}_{ 0 } )^{ 2 } - 1 .[/tex]
    Using these, the Schwarz inequality becomes equivalent to
    [tex]0 \leq | A^{ 0 }{}_{ 0 } | \ | B^{ 0 }{}_{ 0 } | \left( 1 - \sqrt{ ( 1 - | A^{ 0 }{}_{ 0 } |^{ - 2 } ) ( 1 - | B^{ 0 }{}_{ 0 } |^{ - 2 } ) } \right) \leq C^{ 0 }{}_{ 0 } .[/tex]
    Now you can check that [itex]C^{ 0 }{}_{ 0 } \geq 1[/itex].
     
  16. Apr 15, 2014 #15
    How did you get this: [itex] ( A^{ 0 }{}_{ i } )^{ 2 } = ( A^{ 0 }{}_{ 0 } )^{ 2 } - 1 [/itex]? I understand it for B. how are [itex] A^0{}_j [/itex] and [itex] A^j{}_0 [/itex] related?
     
  17. Apr 15, 2014 #16

    samalkhaiat

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    Lower indices number the rows in the matrix. Choose any convention you like and stick with it.
     
  18. Apr 15, 2014 #17
    I still don't understand. They are different parts of lorentz matrix. One can only use the definition of lorentz group, I.e. orthogonality.
     
  19. Apr 15, 2014 #18

    samalkhaiat

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    I don't realy do homework. Any way, orthogonality is what I used in writing the matrix equations
    [tex]\eta = A \eta A^{ T }, \ \ \eta = B^{ T } \eta B[/tex]
    Now, forget about upper and lower indices, Can you find the first matrix element, i.e. the [itex](0,0)[/itex] entry in each of the above equations?
     
  20. Apr 16, 2014 #19

    Fredrik

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    OK, I see now that I got confused by the notation. The inequality that samalkhaiat wrote down is an almost immediate consequence of Cauchy-Schwartz for vectors in ##\mathbb R^3##. If we use your inequality from the quote above, and then use that ##|x|\geq -x## for all real numbers x,...

    Regarding notational conventions: Some people would write ##x_i x_i## rather than ##(x_i)^2## to ensure that the index to be summed over explicitly appears twice, but I suppose that can be weird too when we use a convention that typically has one of the indices upstairs and the other downstairs. I would denote the component on row ##\mu##, column ##\nu## of an arbitrary matrix X by ##X_{\mu\nu}##, ##X^\mu_\nu## or ##X^\mu{}_\nu##. Samalkhaiat apparently uses ##X^\nu{}_\mu##. That's fine too. As he said, "choose any convention you like and stick with it".
     
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