Hey there I would just like some help in rearranging the formula: [tex] z=\frac{1+\frac{v\cos \theta}{c}}{\sqrt{1-\frac{v^2}{c^2}}}-1 [\tex] for v. Thanks, PP
That is [tex]z= \frac{1+ \frac{v cos(\theta)}{c}}{\sqrt{1- \frac{v^2}{c^2}}}- 1[/tex] You solve an equation by "undoing" what was done to the unknown, step by step. The last thing done there is "-1" so we undo that by adding 1 to both sides: [tex]z+ 1= \frac{1+ \frac{v cos(\theta)}{c}}{\sqrt{1- \frac{v^2}{c^2}}}[/tex] Now, multiply on both sides by the denominator: [tex](z+1)\left(\sqrt{1- \frac{v^2}{c^2}}\right)= 1+ \frac{v cos(\theta)}{c}[/tex] Get rid of that square root by squaring both sides (which might introduce false solution- you will need to check for that after finding v): [tex](z+1)^2\left(1- \frac{v^2}{c^2}\right)= 1+ 2\frac{v cos(\theta)}{c}+ \frac{v^2 cos^2(\theta)}{c^2}[/tex] Now that is a quadratic equation in v. Combining like terms, [tex]\left(\frac{1}{cos^2(\theta)}+ \frac{(z+1)^2}{c^2}\right)v^2+ \frac{2vcos(\theta)}{c}v+ 1- (z+1)^2= 0[/tex] which can be solved using the quadratic formula.