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Solving for a variable help

  1. Jul 31, 2014 #1
    Hey guys, this is a little silly question but it bothers me. Im not a math genius (yet i hope) and im still in elementary school so theres alot to learn. But i just read about the lorenz factor in this example he basically used pythagoras of this light clock in a train, so it started of as

    [tex](ct)^2 = (cx)^2 + (vt)^2[/tex]

    and he derived it into:

    [tex]t = \frac{x}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    I would have posted an attemp to solve it but i really just dont know how to crack it and get started

    Pleeeaaase help it would be really nice :D
    Last edited: Jul 31, 2014
  2. jcsd
  3. Jul 31, 2014 #2


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    That is nothing you can derive (at least not in the way you ask for here), that is a definition of γ.
  4. Jul 31, 2014 #3
    right its me god im stupid! he solved for t not gamma dont really know what went through my head while i wrote it. i corrected it in the post now
  5. Jul 31, 2014 #4
    OK, I'll help you get started: bring all terms which have a ##t## to one side of the equation and the other terms on the other side. Isolate ##t^2## so you have ##t^2 = \text{something}##. Then take roots.
  6. Jul 31, 2014 #5
    ok ill try:

    [itex]c^2t^2 = c^2x^2 + v^2t^2[/itex]
    [itex]t^2 = \frac{c^2x^2 + v^2t^2}{c^2}[/itex]
    Dividing both sides by t^2
    [itex]1 = \frac{c^2x^2 + v^2t^2}{c^2t^2}[/itex]

    im stuck... lol
    normally i dont really have trouble when solving for variables but this one irritates me.. can i have another hint ? :)
  7. Jul 31, 2014 #6
    I can see a ##t^2## on the LHS and on the RHS. The idea is to have all occurences of ##t^2## on the LHS.
  8. Jul 31, 2014 #7
    Exacly thats the though part, because normally i would just divide out the t^2 but that wont help in this example as it would leave me with a 1 on the LHS.. and that wouldnt help much,, is there some mechanism or method that i am missing that could solve this? i feel like theres something i havent learned that could allow this to be solved.. or is it just me thats blind?
  9. Jul 31, 2014 #8
    If you have ##2x^2 = 5 - 3x^2##, can you solve that? You just need to do the same thing here. Isolate ##x##.
  10. Jul 31, 2014 #9
    Yea no problem but there i can just add 3x^2 to both sides, in my example its [itex]v^2t^2[/itex] so i cant do the same thing here?
  11. Jul 31, 2014 #10


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    You could try subtracting ##v^2t^2## from both sides.
  12. Aug 1, 2014 #11


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    Why do you think v2 is different from 3?
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