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

[ANvH]

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

 

[PeterDonis]

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

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

 

[George Jones]

I get γ(1+β) = [itex]\sqrt{\frac{1+β^{2}}{1-β^{2}}}[/itex].

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.

[itex]\sqrt{\frac{1}{1-β^{2}}}-\sqrt{\frac{β^{2}}{1-β^{2}}}=\sqrt{\frac{1-β^{2}}{1-β^{2}}}[/itex]

 

[robphy]

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

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

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]

 

[yuiop]

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,

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]

 

[pervect]

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].

Sorry, you didn't do you homework correctly :-(

##\gamma(1+\beta) = \sqrt{\frac{1}{1-\beta^2}}(1+\beta) = \sqrt{\frac{(1+\beta)^2}{1-\beta^2}} = \sqrt{\frac{(1+\beta)(1+\beta)}{(1+\beta)(1-\beta)}} = \sqrt{\frac{1+\beta}{1-\beta}}##

[add]
I see a bunch of posts snuck in before mine

 

[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.