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Rearranging redshift formula for v

by Pretty Pony
Tags: formula, rearranging, redshift
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Pretty Pony
#1
Jun23-11, 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
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HallsofIvy
#2
Jun23-11, 06:40 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,497
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.
Pretty Pony
#3
Jun24-11, 02:52 PM
P: 8
Thanks for your help!
:)


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