How to Derive the Lorentz Factor from Pythagoras?

[MathiasArendru]

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

 

[mfb]

That is nothing you can derive (at least not in the way you ask for here), that is a definition of γ.

 

[MathiasArendru]

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

 

[micromass]

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.

 

[MathiasArendru]

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 ? :)

 

[micromass]

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.

 

[MathiasArendru]

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?

 

[micromass]

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 $x$.

 

[MathiasArendru]

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?

 

[Nugatory]

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.

 

[Borek]

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?