Vectors in Special Relativity

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  • #1
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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
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
 
  • #4
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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
 
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  • #5
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I don't unterstand Phrak when you say it didn't come out well
I think Phrak just meant that the LaTeX didn't display the way he wanted it to.
 

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