Special relativity — Observer measuring the velocity of a passing rocket

[LCSphysicist]

Homework Statement

There is a print just below

Relevant Equations

\n

I am a little confused with the text above. Actually, all is ok until 2.51, i think i am missing something.
What is this? Another way to define gamma? (The doubt is not about the raising indices, neither about "n00 = -1", is just where does the equation came from. That is, the equation is easy to understand, but i don't know from where this come from)

 

[PeroK]

Homework Statement:: There is a print just below
Relevant Equations:: \n

View attachment 272605
I am a little confused with the text above. Actually, all is ok until 2.51, i think i am missing something.
What is this? Another way to define gamma? (The doubt is not about the raising indices, neither about "n00 = -1", is just where does the equation came from. That is, the equation is easy to understand, but i don't know from where this come from)

Equation 2.51 is simply highlighting that $-\gamma$ equals the inner product of the two four vectors. It's easy to work out.

By noticing this you have $\gamma$ (and hence $v$) expressed in a manifestly covariant form. As the inner product is independent of coordinates, this is useful for generalising toall coordinate systems and to curved spacetime.

 

[robphy]

It might be helpful to consider the Euclidean version of (2.51).
$\gamma_{euc}=\hat U \cdot \hat V =\cos\theta_{\scriptsize{\rm between\ U\ and\ V}}<br /> =\displaystyle\frac{1}{\sqrt{1+\tan^2\theta_{rel}}}<br /> =\displaystyle\frac{1}{\sqrt{1+m_{rel}^2}}$

(2.51) interpreted with rapidities (and the $(-,+,+,+)$ signature)
$\gamma=-\hat U \cdot \hat V =\cosh\phi_{rel}<br /> =\displaystyle\frac{1}{\sqrt{1-\tanh^2\phi_{rel}}}<br /> =\displaystyle\frac{1}{\sqrt{1-v_{rel}^2}}$