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Vectors in Special Relativity

  1. Apr 26, 2010 #1
    I've been reading Barut's "Electrodynamics and Classical Theory of Fields and Particles" and he derives de inverse Schwarz inequality for two time-like vectors [itex]<z_1,z_2>^2\ge z_1^2z_2^2[/itex] in the folowing way : " To show this we can assume without loss of generatlity z_2 to be (1,0,0,0). Then [itex](z_1^0)^2\ge(z_1^0)^2-(z_1^1)^2-(z_1^2)^2-(z_1^3)^2[/itex] which proves the statement".
    My question is because I don't see how this particular case can prove the general statement
  2. jcsd
  3. Apr 26, 2010 #2


    Staff: Mentor

    If you are given that z_2 is a timelike vector then, by definition, there exists some inertial reference frame where z_2 = (ct,0,0,0), then by appropriate choice of units you can have z_2 = (1,0,0,0). Then since the left side and the right side are each quantities which are invariant under the Lorentz transform we immediately know that the result applies for all inertial frames.
  4. Apr 27, 2010 #3
    This can be confusing. To add to what Dale said,

    Both sides of the inequality are Lorentz invariant scalars. If the left side is strictly greater than the right side,

    [itex]\left<z_1,z_2\right>^{2} > \left<z_1,z_1 \right>\left<z_2,z_2 \right>[/itex]

    then under a Lorentz transformation the inequality still holds. That was supposed to be a <z1,z2>2, on the right, but didn't come out well.

    If the left and right are equal, they remain equal under a Lorentz transform. This is the case of parallel vectors.
  5. Apr 27, 2010 #4
    Thank you DaleSpan, now I see the book's argument.
    I don't unterstand Phrak when you say it didn't come out well, in this case we are talking about the "inverse" Shchwarz inequality it's supposed to be on the "wrong" side
    Last edited: Apr 27, 2010
  6. Apr 27, 2010 #5


    Staff: Mentor

    I think Phrak just meant that the LaTeX didn't display the way he wanted it to.
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