Physical interpretation of equation for relativistic aberration

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 $|u| = |(u_x, u_y, u_z)| = \sqrt{u_x^2 + u_y^2 + u_z^2}$

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1 Answer

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?

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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