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Rearranging redshift formula for v

  1. Jun 23, 2011 #1
    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. jcsd
  3. Jun 23, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
  4. Jun 24, 2011 #3
    Thanks for your help!
    :)
     
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