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Special Relativity Derivation

  1. Dec 14, 2014 #1
    Taken from Steane's "Relativity made relatively easy" equation 4.8

    2h3bigj.png

    I have been trying to show (4.8) using these relations earlier on in the book:

    2u5ymw4.png

    Tried most means (rearranging, taking dot products) but can't seem to make it work. Is there an easy method I'm missing out?
     
    Last edited: Dec 14, 2014
  2. jcsd
  3. Dec 14, 2014 #2

    Dale

    Staff: Mentor

    Try expanding it out in the basis where x is the "u parallel" direction.
     
  4. Dec 15, 2014 #3
    Sorry, I don't understand what you mean by "it". Which term should I expand?
     
  5. Dec 15, 2014 #4

    Dale

    Staff: Mentor

    Expand both sides of 4.8.
     
  6. Dec 15, 2014 #5
    I'm trying to derive eq 4.8 starting with the ones below, 2.27 and 2.28.
     
  7. Dec 15, 2014 #6

    Dale

    Staff: Mentor

    Then expand all 3. The point is to get them all into the same form.
     
  8. Dec 15, 2014 #7
    I wouldn't know what RHS of eq 4.8 would look like initially. I am trying to derive its form starting from 2.27 and 2.28.
     
  9. Dec 15, 2014 #8

    jtbell

    User Avatar

    Staff: Mentor

    Often, when you want to find a derivation from A to B, it helps to start by working from both A and B to find some equation in the "middle", so to speak. Then you can reverse the steps that took you from B to the "middle." Or, the working might give you enough insight that you can come up with a better way to write the complete derivation.
     
  10. Dec 15, 2014 #9
    Point is, I wouldn't even know what RHS of 4.8 would look like.

    OK here is the challenge, rephrased:

    Without manipulating RHS of 4.8, use eq 2.27 and 2.28 to show that form.
     
  11. Dec 15, 2014 #10

    Dale

    Staff: Mentor

    Why would you add that restriction?

    Just expand it. I don't know why you keep on saying that you don't know what the RHS of 4.8 is. u' and v are vectors, just write them as components and expand in the basis where x is the parallel direction.
     
  12. Dec 15, 2014 #11
    I believe unscientific doesn't want to manipulate the RHS of 4.8 because he considers that he isn't supposed to know what it is before deriving it from 2.27 and 2.28 in the first place.

    @unscientific: even though you want to derive 4.8 directly from the other two equations, expanding 4.8 will help you understand how you can do that, because some steps may not be obvious. Then you can do the whole derivation starting from 2.27 and 2.28
     
  13. Dec 15, 2014 #12

    Ibix

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    Science Advisor

    It may help to note that the parallellity (is that a word?) and perpendicularity are with reference to the relative velocity of the moving frame. In that case, u.v=u||×v + u×0.
     
  14. Dec 15, 2014 #13

    ChrisVer

    User Avatar
    Gold Member

    Start from:

    [itex] \frac{1}{1 - (u \cdot v)/c^2 } = ...[/itex]
    and you will reach the result. It's straightforward.
     
    Last edited: Dec 15, 2014
  15. Jan 8, 2015 #14
    Thanks, got it!
     
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