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Gamma plus or minus gamma-beta, two different outcomes?

  1. Mar 7, 2014 #1
    In Wikipedia it reads that γ(1+β) = [itex]\sqrt{\frac{1+β}{1-β}}[/itex], however, if I did my homework correctly I get γ(1+β) = [itex]\sqrt{\frac{1+β^{2}}{1-β^{2}}}[/itex]. Digging more deeply into why Wikipedia is listing it as such I found that it is based on the hyperbolic angles:γ=coshΘ. But it leads to definitions, not an explanation. More interestingly, WikiPedia is equating γ(1-β) with [itex]\sqrt{\frac{1-β}{1+β}}[/itex]. However, my calculations lead to


    γ(1-β) =γ-γβ=[itex]\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\sqrt{\frac{1-β^{2}}{1-β^{2}}}[/itex],

    which is equal to unity. I guess I am doing something utterly wrong and it feels like as if I have to start over learning calculus. I truly hope someone is setting me straight,

    thanks,
    Alfred
     
  2. jcsd
  3. Mar 7, 2014 #2

    PeterDonis

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    Staff: Mentor

    $$
    \gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{\left( 1 - \beta \right) \left( 1 + \beta \right)}}
    $$

    $$
    \gamma \left( 1 + \beta \right) = \gamma \sqrt{\left( 1 + \beta \right) \left( 1 + \beta \right)} = \sqrt{\frac{\left( 1 + \beta \right) \left( 1 + \beta \right)}{\left( 1 - \beta \right) \left( 1 + \beta \right)}} = \sqrt{\frac{1 + \beta}{1 - \beta}}
    $$
     
  4. Mar 7, 2014 #3

    George Jones

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    This is where a mistake is; look at what Peter wrote. I am not sure, but did you make the mistake

    $$\left( 1 + \beta \right)^2 = 1^2 + \beta^2?$$

    In general, ##\left( x + y\right)^2 = x^2 + 2xy + y^2 \ne x^2 + y^2##.

    Also, below, you seem to be saying ##\sqrt{x} - \sqrt{y} = \sqrt{x -y}##, which is not correct.

     
  5. Mar 7, 2014 #4

    robphy

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    Transcribing into trigonometry...

    [tex]\cosh\theta
    =\frac{1}{\sqrt{1-\tanh^2\theta}}
    =\frac{1}{\sqrt{(1-\tanh\theta)(1+\tanh\theta)}}
    [/tex]

    [tex]
    \cosh\theta(1+\tanh\theta)
    =\cosh\theta\sqrt{(1+\tanh\theta)(1+\tanh\theta)}
    =\sqrt{\frac{ {(1+\tanh\theta)(1+\tanh\theta)}}{{(1-\tanh\theta)(1+\tanh\theta)}}}
    =\sqrt{\frac{ {1+\tanh\theta}}{{1-\tanh\theta}}}
    [/tex]


    However, thinking trigonometrically...

    Strating from the LHS:
    [tex]
    \cosh\theta(1+\tanh\theta)
    =\cosh\theta(1+\frac{\sinh\theta}{\cosh\theta})
    =\cosh\theta+\sinh\theta
    =e^{\theta}
    [/tex]


    Starting from the RHS:
    [tex]
    \sqrt{\frac{ {1+\tanh\theta}}{{1-\tanh\theta}}}
    =\sqrt{\frac{ 1+\frac{\sinh\theta}{\cosh\theta}}{{1-\frac{\sinh\theta}{\cosh\theta}}}}
    =\sqrt{\frac{ \cosh\theta+\sinh\theta}{\cosh\theta-\sinh\theta}}
    =\left(\frac{e^\theta}{e^{-\theta}}\right)^{1/2}=e^{\theta}
    [/tex]

    Note
    [tex]
    \cosh\theta(1-\tanh\theta)
    =\cosh\theta-\sinh\theta
    =\cosh(-\theta)+\sinh(-\theta)
    =e^{-\theta}
    =1/e^{\theta}=\sqrt{\frac{1-\tanh\theta}{1+\tanh\theta}}
    [/tex]
     
  6. Mar 7, 2014 #5
    Try this:


    γ(1-β) =γ-γβ=[itex]\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\left(\frac{1}{\sqrt{1-β^2}}-\frac{β}{\sqrt{1-β^2}}\right)=\frac{(1-β)}{\sqrt{1-β^2}}=\frac{\sqrt{(1-β)(1-β)}}{\sqrt{(1-β)(1+β)}}=\frac{\sqrt{1-β}}{\sqrt{1+β}}[/itex]

    and


    γ(1+β) =γ+γβ=[itex]\sqrt{\frac{1}{1-β^{2}}}+\sqrt{\frac{β^{2}}{1-β^{2}}}=\left(\frac{1}{\sqrt{1-β^2}}+\frac{β}{\sqrt{1-β^2}}\right)=\frac{(1+β)}{\sqrt{1-β^2}}=\frac{\sqrt{(1+β)(1+β)}}{\sqrt{(1-β)(1+β)}}=\frac{\sqrt{1+β}}{\sqrt{1-β}}[/itex]
     
  7. Mar 7, 2014 #6

    pervect

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  8. Mar 8, 2014 #7
    I am grateful to all of you. Yes, I made a big mistake, thanks to using Microsoft Equation Editor, where I fooled myself. Always use pen and paper. This is not meant as an excuse, I stand corrected.
     
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