
#1
Jun2311, 12:31 AM

P: 8

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 



#2
Jun2311, 06:40 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

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. 



#3
Jun2411, 02:52 PM

P: 8

Thanks for your help!
:) 


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