Lorentz invariance

  • Thread starter parton
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  • #1
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Homework Statement



I have two four vectors v and w with [tex] v^{2} = m^{2} > 0, v_{0} > 0 [/tex] and [tex] w^{2} > m^{2}, w_{0} > 0 [/tex]. Now we consider a system with
[tex] w' = (w_{0}', \vec{0}) [/tex] and [tex] v' = (v_{0}', \vec{v} \, ') [/tex] and in addition we consider the quantity [tex] \lambda = \vert \vec{v}' \vert \, \sqrt{ w_{0}'^{2} - m^{2}} [/tex]. Now I should find a Lorentz invariant expression of [tex] \lambda [/tex] only using the invariants [tex]v^{2}, w^{2}, vw[/tex].

Homework Equations





The Attempt at a Solution



I think I've found a solution: [tex] t = \sqrt{\dfrac{(vw)^{2} - v^{2} w^{2}}{v^{2}} (v^{2} - w^{2})} [/tex].
But I'm not really sure if this "solution" is really Lorentz invariant (my problem is the square root). Could anyone confirm this solution or is there any mistake?
 

Answers and Replies

  • #2
Avodyne
Science Advisor
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The square root is OK as long as what you're taking the root of is positive. I didn't check your math, but the answer should look something like this.
 

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