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Lorentz product on hyperboloid

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    (a) Let p(t) be a smooth curve in the hyperboloid S = {(x, y, z) ∈ R3 |z^2 − x^2 − y^2 = 1, z > 0}. Prove that p(t) .L p'(t) = 0, where .L is the Lorentz product.
    (b) Prove that any non-zero tangent vector to S has positive (Lorentz)length. (Hint: Use the Cauchy-Schwarz inequality for the dot product.)

    2. Relevant equations
    x .L y = x1y1+x2y2-x3y3 (the lorentz product)



    3. The attempt at a solution
    (a)
    Let p(t) = (a(t),b(t),c(t))
    c(t) =(a^2+b^2+1)^.5
    p'(t) = (a',b',.5*(2aa'+2bb')*(a^2+b^2+1)^-.5)

    p(t) .L p'(t) = aa'+bb'-((a^2+b^2+1)^.5)*(aa'+bb')*((a^2+b^2+1)^-.5)
    =aa'+bb'-aa'-bb'=0

    (b)
    |<p,p'>| <= ||p||*||p'||
    from (a)
    0 <= ||p||*||p'||
    and we know p' is non zero so p must also
    so 0 < ||p||*||p'||
    now ||p||^2 = p .L p = aa+bb -(aa+bb+1) = -1

    By doing the lorentz product of p with itself I was hoping to show that ||p|| > 0
    and hence that ||p'|| must also be greater than 0 to satisfy the inequality, but it seems I've instead shown that ||p|| = -i.

    Any help would be appreciated.
     
  2. jcsd
  3. Nov 5, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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