Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm working in a book on relativity. The author states that if $u$ and $u'$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by

$$ (l, m, n) = \frac{1}{|u|}(u_x, u_y, u_z) $$

and

$$ (l', m', n') = \frac{1}{|u'|}(u'_x, u'_y, u'_z) $$

are related by

$$ (l', m', n') = \frac{1}{D}(l - \frac{v}{u}, m\gamma ^{-1}, n\gamma ^{-1}) $$

and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.

$$ D = \frac{u'}{u}\left( 1 - \frac{u_xv}{c^2}\right) = \left[1 - 2l\frac{v}{u} + \frac{v^2}{u^2} - (1 - l^2)\frac{v^2}{c^2} \right]^{\frac{1}{2}} $$

My question is why the author finessess the expression into the third/final form. I was able to get it but it seems like a pain. Why is that better than the second expression? It seems more difficult to calculate and not at all clear or meaningful.


Also, in case it's not clear, $\gamma =1/ \sqrt{1 - \frac{v^2}{c^2}}$ and $|u| = |(u_x, u_y, u_z)| = \sqrt{u_x^2 + u_y^2 + u_z^2}$

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

The last form avoids any usage of $u'$ as well as $u_x$ which may be considered derived or non-covariant quantities, respectively, so the last form may be superior in many contexts. Moreover, it makes it much easier to imagine how far the result is from one - because it resembles a Taylor expansion of a sort.

But even if one didn't write the comments above, why would it be a problem to write a result in yet another mathematically equivalent way?

share|improve this answer
    
I don't think he felt there was a problem; he just wanted to understand the motivation for writing it in that way. That sort of understanding often helps follow the author's reasoning, particularly when learning an unfamiliar subject. –  Colin K May 6 '11 at 13:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.