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B Derivation of special relativity formulae

  1. Jun 8, 2016 #1
    I was bored, so I decided to derive the special relativity formulae.
    I drew the following diagram of a light clock:
    In order to find t, I did sinθ=d/ct
    Which gives tsinθ=d/c
    Which gives t=d/csinθ

    If v = 0, vt = 0, and θ = 90
    sin90 = 1
    t = d/csinθ = d/c
    We call this t0

    If v is greater than 0, vt is greater than zero, and θ is less than 90
    sin90 is less than1
    t = d/csinθ is greater than d/c

    We use Pythagoras to get t0=t(1- (v2/c2))-1/2

    What are the steps involved in getting to this answer?

    Attached Files:

  2. jcsd
  3. Jun 8, 2016 #2


    Staff: Mentor

    Usually both t and θ are considered unknowns. So writing one equation in two unknowns doesn't help. You should use the Pythagorean theorem instead to get one equation in one unknown.
  4. Jun 8, 2016 #3
    Valid point :wink:

    so (ct)2 = d2 + (vt)2
    Last edited: Jun 8, 2016
  5. Jun 8, 2016 #4


    Staff: Mentor

    Yes. Then rearranging and using ##d/c=t_0## gives you the desired formula
  6. Jun 9, 2016 #5
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