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How the surface area of a cone changes; special relativity

  1. Jul 3, 2017 #1
    1. The problem statement, all variables and given/known data
    A cone has half angle θ0 and lateral surface area S0 in the frame in which the cone is at rest. If someone moves at relative speed β=v/c along the cones symmetry axis, what surface area will they see for the cone?

    2. Relevant equations
    I believe the lateral surface area of a cone is S=πh2tanθ, where θ is the half angle and h is the height of the cone.
    Length contraction is also relevant; h1=h0√(1-β2)

    3. The attempt at a solution
    S0 = πh02tanθ0
    S1 = πh12tanθ1

    If the cone had maximum radius R in the rest frame, it should remain R in the frame moving along the symmetry axis (because R is measured normal to this axis).
    So we have tanθ0=R/h0, and also tanθ1=R/h1, so we see tanθ1=(h0/h1)tanθ0

    Putting all this together (with the length contraction relation) it seems to be that S1=S0√(1-β2)

    However my book claims S1=S0√(1-(βcosθ)2); so where is my mistake?
     
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  3. Jul 3, 2017 #2

    Orodruin

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  4. Jul 3, 2017 #3
    Ahh. For some reason, when I thought over that formula, I had in my mind that "R=h*sinθ" instead of "R=h*tanθ"
    (I even remember thinking this formula over a couple times, and each time I kept incorrectly saying this to myself! I am tired.)

    So then the correct formula is actually S=πh2tanθ/cosθ.

    So then my answer is off by (needs to be multiplied by) a factor of cosθ0/cosθ1

    If you take the ratio of these two equations
    cosθ0=1/√(1+(R/h0)2)
    cosθ1=1/√(1+(R/h1)2)=1/√(1+(R/h0)2/(1-β2))
    and simplify a fair amount (then use R/h0=tanθ as well as 1/(1+tan2θ)=cos2θ) then multiply it by my incorrect answer then it indeed becomes the correct answer, so that was the only mistake.

    Thank you Orodruin
     
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