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How to solve for x from$$ \frac{1}{\sqrt{1-x^2/m^2}}=\frac{1}{\sqrt{1-y^2/m^2}} \cdot\frac{1}{\sqrt{1-z^2/m^2}}\cdot(1+\dfrac{yz}{m^2}) $$ to $$ x = \frac{y+z}{1+yz/m^2}$$

There is also a tip, which says $$\dfrac{x^2}{m^2}=\frac{\left(\frac{1}{\sqrt{1-x^2/m^2}}\right)^2-1}{\left(\frac{1}{\sqrt{1-x^2/m^2}}\right)^2}$$

epzylVan
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  • What is your aim? – 1.414212 Feb 13 '20 at 05:32
  • It looks like an object is moving to the right in frame $S_1$ with velocity $z$ and frame $S_1$ is moving to the right relative to frame $S_2$ with velocity $y$ and you are trying to find an expression for the Lorentz factor $\gamma=\frac1{\sqrt{1-x^2/c^2}}$ for the object as seen by frame $S_2$, is that right? If $m=c$ your first expression seems to be equivalent to the second. – user5713492 Feb 13 '20 at 05:55
  • @1.414212 I want to know how the author solved for $x$ from the formula above, so basically the calculation path for that. – epzylVan Feb 13 '20 at 06:11
  • @user5713492 Exactly, that's it, I am looking for the Velocity-addition formula (often called $w$ (or $u$ on Wikipedia)), I wrote with different variables just to minimalize plagiarism when my work gets checked. – epzylVan Feb 13 '20 at 06:18
  • I would try to substitute $x=m\sin\alpha, y=c\sin\beta, z=c\sin\gamma$ and then simplify. – Sam Feb 13 '20 at 06:37

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I took both sides to the $-2$ power: $$1-\frac{x^2}{c^2}=\frac{\left(1-\frac{y^2}{c^2}\right)\left(1-\frac{z^2}{c^2}\right)}{\left(1+\frac{yz}{c^2}\right)^2}$$ Then took $1$ minus both sides: $$\begin{align}\frac{x^2}{c^2}&=\frac{\left(1+\frac{yz}{c^2}\right)^2-\left(1-\frac{y^2}{c^2}\right)\left(1-\frac{z^2}{c^2}\right)}{\left(1+\frac{yz}{c^2}\right)^2}\\ &=\frac{\color{red}{1}+\frac{2yz}{c^2}\color{blue}{+\frac{y^2z^2}{c^4}}\color{red}{-1}+\frac{y^2}{c^2}+\frac{z^2}{c^2}\color{blue}{-\frac{y^2z^2}{c^4}}}{\left(1+\frac{yz}{c^2}\right)^2}\\ &=\frac{\frac{(y+z)^2}{c^2}}{\left(1+\frac{yz}{c^2}\right)^2}\end{align}$$ The terms with corresponding $\color{red}{\text{c}}\color{orange}{\text{o}}\color{yellow}{\text{l}}\color{green}{\text{o}}\color{blue}{\text{r}}\color{violet}{\text{s}}$ canceling. Then multiply by $c^2$ and take square root to get $$x=\frac{y+z}{1+\frac{yz}{c^2}}$$ Is that what you wanted, or did you rather want to start with the last expression for $x$ and substitute into $\gamma=\left(1-\frac{x^2}{c^2}\right)^{-1/2}$?

user5713492
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  • That's exactly what I needed, thank you a lot good Sir! And sorry for the mistake with $m=c$, my bad. – epzylVan Feb 13 '20 at 11:06