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

1. Mar 7, 2014

ANvH

In Wikipedia it reads that γ(1+β) = $\sqrt{\frac{1+β}{1-β}}$, however, if I did my homework correctly I get γ(1+β) = $\sqrt{\frac{1+β^{2}}{1-β^{2}}}$. 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 $\sqrt{\frac{1-β}{1+β}}$. However, my calculations lead to

γ(1-β) =γ-γβ=$\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\sqrt{\frac{1-β^{2}}{1-β^{2}}}$,

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. Mar 7, 2014

Staff: Mentor

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

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

3. Mar 7, 2014

George Jones

Staff Emeritus
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.

4. Mar 7, 2014

robphy

Transcribing into trigonometry...

$$\cosh\theta =\frac{1}{\sqrt{1-\tanh^2\theta}} =\frac{1}{\sqrt{(1-\tanh\theta)(1+\tanh\theta)}}$$

$$\cosh\theta(1+\tanh\theta) =\cosh\theta\sqrt{(1+\tanh\theta)(1+\tanh\theta)} =\sqrt{\frac{ {(1+\tanh\theta)(1+\tanh\theta)}}{{(1-\tanh\theta)(1+\tanh\theta)}}} =\sqrt{\frac{ {1+\tanh\theta}}{{1-\tanh\theta}}}$$

However, thinking trigonometrically...

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

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

Note
$$\cosh\theta(1-\tanh\theta) =\cosh\theta-\sinh\theta =\cosh(-\theta)+\sinh(-\theta) =e^{-\theta} =1/e^{\theta}=\sqrt{\frac{1-\tanh\theta}{1+\tanh\theta}}$$

5. Mar 7, 2014

yuiop

Try this:

γ(1-β) =γ-γβ=$\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\left(\frac{1}{\sqrt{1-β^2}}-\frac{β}{\sqrt{1-β^2}}\right)=\frac{(1-β)}{\sqrt{1-β^2}}=\frac{\sqrt{(1-β)(1-β)}}{\sqrt{(1-β)(1+β)}}=\frac{\sqrt{1-β}}{\sqrt{1+β}}$

and

γ(1+β) =γ+γβ=$\sqrt{\frac{1}{1-β^{2}}}+\sqrt{\frac{β^{2}}{1-β^{2}}}=\left(\frac{1}{\sqrt{1-β^2}}+\frac{β}{\sqrt{1-β^2}}\right)=\frac{(1+β)}{\sqrt{1-β^2}}=\frac{\sqrt{(1+β)(1+β)}}{\sqrt{(1-β)(1+β)}}=\frac{\sqrt{1+β}}{\sqrt{1-β}}$

6. Mar 7, 2014

pervect

Staff Emeritus

7. Mar 8, 2014

ANvH

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.