# Explanation and derivation of Lorentz equations

I've read about special relativity recently and I've learned about the Lorentz transformation. I get what time dilation and length contraction are and I think I understand what the transformation does (at least to the distances and time intervals) but I want to be able to show it mathematically in a simple and intuitive form. I've read a lot of papers yet it is quite hard and unintuitive to me to derive the formula for time coordinate. I don't know if such a complex topic can be explained simply but what I'm looking for is a way to derive these equations not by squaring, substracting the sides etc. which can't be showed visually and doesn't represent any "ideas" in physics but by explaining which part of the formula means what.

I appreciate all the help.

Well, there are many ways to show it. I find none of them really intuitive, but this is my best attempt.

First, we have to assume that Lorent's transformations are linear. HEre you have a proof: Link to the proof

So, if Lorentz transformation is linear, then

$$\exists \gamma \qquad | \qquad x'=\gamma(x-vt)$$

where $x$ is the direction of movement. $y, z$ should remain invariant, that is, $y'=y; z'=z$.

And now it's just about finding what $\gamma$ is.

For the same argument, we can say

$$x=\gamma' (x'+vt')$$ and you can substitute:

$$x'=\gamma'[\ \gamma(x-vt)+vt']$$

Now, postulate I implies that $\gamma=\gamma'$. Then

$x=\gamma^2(x-vt)+\gamma vt'$, from which you get

$$t'=\frac{x-\gamma^2x+\gamma^2vt}{\gamma v}=\gamma t+ \frac{x}{\gamma v} (1-\gamma^2)$$

Now, 2nd postulate implies that, for both observers, the particular case of light is $x=ct, \ x'=ct'$, with the same speed $c$ !

So, substitute $x=ct$ and $x'=ct'$ and you get

$$ct=\gamma(ct-vt)$$

$$c\cdot\left[ \gamma t + \frac{x}{\gamma v} (1-\gamma^2\right] = \gamma (c-v)t$$

Cancelling $t$, this expression depends only on $v$ and $c$,

$$c\gamma + \frac{c^2}{\gamma v} (1-\gamma^2)=\gamma(c-v)$$

and, if you keep on rearranging it, you'll get

$$\gamma^2=\frac{c^2}{c^2-v^2}= \frac{1}{1-v^2/c^2}$$

So here are Lorentz Transformations:

$$x'=\frac{x-vt}{\sqrt{1-v^2/c^2}}; \qquad y'=y;\ \ z'=z$$

If you now substitute this $\gamma$ in $'$, you'll get

$$t'=\gamma \left( t-\frac{xv}{c^2}\right)$$

• Note: there are easier forms of getting them, but they imply defining Minkowski's metric, and then deriving the form of the boosts. It's easier but less intuitive. – FGSUZ Jul 17 '18 at 22:49