How to Derive the Lorentz Factor from Pythagoras?

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Hey guys, this is a little silly question but it bothers me. I am not a math genius (yet i hope) and I am still in elementary school so there's a lot 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

(ct)^2 = (cx)^2 + (vt)^2

and he derived it into:

t = \frac{x}{\sqrt{1-\frac{v^2}{c^2}}}

I would have posted an attemp to solve it but i really just don't know how to crack it and get started

Pleeeaaase help it would be really nice :D
 
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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.
 
right its me god I am stupid! he solved for t not gamma don't really know what went through my head while i wrote it. i corrected it in the post now
 
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.
 
ok ill try:

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

im stuck... lol
normally i don't really have trouble when solving for variables but this one irritates me.. can i have another hint ? :)
 
MathiasArendru said:
ok ill try:

c^2t^2 = c^2x^2 + v^2t^2
t^2 = \frac{c^2x^2 + v^2t^2}{c^2}

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.
 
Exacly that's the though part, because normally i would just divide out the t^2 but that won't help in this example as it would leave me with a 1 on the LHS.. and that wouldn't help much,, is there some mechanism or method that i am missing that could solve this? i feel like there's something i haven't learned that could allow this to be solved.. or is it just me that's blind?
 
MathiasArendru said:
Exacly that's the though part, because normally i would just divide out the t^2 but that won't help in this example as it would leave me with a 1 on the LHS.. and that wouldn't help much,, is there some mechanism or method that i am missing that could solve this? i feel like there's something i haven't learned that could allow this to be solved.. or is it just me that's blind?

If you have ##2x^2 = 5 - 3x^2##, can you solve that? You just need to do the same thing here. Isolate ##x##.
 
Yea no problem but there i can just add 3x^2 to both sides, in my example its v^2t^2 so i can't do the same thing here?
 
  • #10
MathiasArendru said:
Yea no problem but there i can just add 3x^2 to both sides, in my example its v^2t^2 so i can't do the same thing here?

You could try subtracting ##v^2t^2## from both sides.
 
  • #11
MathiasArendru said:
Yea no problem but there i can just add 3x^2 to both sides, in my example its v^2t^2 so i can't do the same thing here?

Why do you think v2 is different from 3?
 

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