1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Special relativity

  1. May 10, 2009 #1

    martyf

    User Avatar

    1. The problem statement, all variables and given/known data

    L1 and L2 are two lorentz trasformation.
    show that L3=L1 L2 is a lorentz trasformation too.


    2. Relevant equations



    3. The attempt at a solution
     
    martyf, May 10, 2009
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  3. May 10, 2009 #2

    phsopher

    User Avatar

    I could be wrong but I think it would be enough to show the invariance of the dot product under L3.
     
    phsopher, May 10, 2009
  4. May 10, 2009 #3

    martyf

    User Avatar

    what is the dot product?
     
    martyf, May 10, 2009
  5. May 10, 2009 #4

    phsopher

    User Avatar

    The scalar product, or inner product or whatever it's called of 4-vectors. In Minkowski space it's [tex]r_1 \cdot r_2 = c^2 t_1 t_2 - x_1 x_2 - y_1 y_2 - z_1 z_1[/tex] or with the signs reversed depending on the convention. But if you haven't yet done the Minkowski notation then you could write L1 and L2 as Lorentz transformations with some velocities and show by applying them consecutively that L3 has the same form.
     
    phsopher, May 10, 2009
  6. May 11, 2009 #5

    martyf

    User Avatar

    I write better :

    I have: [tex]\Lambda [/tex] [tex]^{\mu}_{\nu}[/tex] and [tex]\Lambda\widetilde{} [/tex] [tex]^{\mu}_{\nu}[/tex] : lorentz trasformations.

    Show that [tex]\Lambda\overline{}[/tex] [tex]^{\sigma}_{\rho}[/tex] = [tex]\Lambda\widetilde{} [/tex] [tex]^{\sigma}_{\mu}[/tex] [tex]\Lambda [/tex] [tex]^{\mu}_{\rho}[/tex] is a lorentz trasformation
     
    martyf, May 11, 2009
  7. May 11, 2009 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What is the definition of Lorentz transformation given in your notes and/or text?
     
    George Jones, May 11, 2009
  8. May 11, 2009 #7

    martyf

    User Avatar

    a trasfonmation that not change the space-time distance of a point to the origin.
     
    martyf, May 11, 2009
  9. May 11, 2009 #8

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Can you write a definition in terms of mathematics, i.e., [tex]\Lambda^\mu{}_\nu[/itex] is a Lorentz transformation iff ... ?
     
    George Jones, May 11, 2009
  10. May 11, 2009 #9

    martyf

    User Avatar

    ...if :

    x[tex]^{2}_{0}[/tex] - r [tex]^{2}[/tex]= ([tex]\Lambda[/tex] [tex]^{0}_{\nu}[/tex] x[tex]_{0}[/tex])[tex]^{2}[/tex] -( [tex]\Lambda[/tex] [tex]^{\mu}_{\nu}[/tex] r[tex]_{\mu}[/tex])[tex]^{2}[/tex]
     
    martyf, May 11, 2009
  11. May 11, 2009 #10

    martyf

    User Avatar

    is it right?
     
    martyf, May 11, 2009
  12. May 11, 2009 #11

    phsopher

    User Avatar

    You don't have the indices right, your right hand side depends on on [tex]\nu[/tex] whereas the left hand side doesn't. It should be:

    [tex](x^0)^2 - r^2 = (\Lambda^0{}_\nu x^\nu)^2 - (\Lambda^j{}_\nu x^\nu)^2[/tex]

    Where j goes from 1 to 3. More compactly:

    [tex]g_\alpha_\beta x^\alpha x^\beta = g_\mu_\nu \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta x^\alpha x^\beta[/tex]

    Where g is the metric. Use the fact that L1 and L2 satisfy this to show that L3 satisfies it as well.
     
    phsopher, May 11, 2009
  13. May 11, 2009 #12

    martyf

    User Avatar

    L1 and L2 are two generical lorents tranformation. I must demostrate that their product is also a lorentz transormation.
    Can I demostrate :

    [tex]g_\alpha_\beta x^\alpha x^\beta = g_\mu_\nu \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta x^\alpha x^\beta[/tex]

    with \Lambda =L3= L1 L2 ?
     
    martyf, May 11, 2009
  14. May 11, 2009 #13

    phsopher

    User Avatar

    Yes you can.
     
    phsopher, May 11, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Special relativity

  1. Special Relativity (Replies: 9)

  2. Special Relativity ! (Replies: 12)

  3. Special Relativity (Replies: 3)

  4. Special Relativity (Replies: 1)

  5. Special relativity and speed of body (Replies: 15)

Loading...