I'm following a Electrodynamics course and I'm currently stack in the following calculation:

$\int{ \frac{ \vec{n} \times (\vec{n}-\vec{\beta}(t)) \times \vec{\beta'}(t) }{(1-\vec{n} \cdot \vec{\beta}(t) )^2 } dt}$

that the book calculates as

$\int{ \frac{ \vec{n} \times (\vec{n}-\vec{\beta}(t)) \times \vec{\beta'}(t) }{(1-\vec{n} \cdot \vec{\beta}(t) )^2 } dt}= \frac{ \vec{n} \times \vec{n} \times \vec{\beta}(t) }{(1-\vec{n} \cdot \vec{\beta}(t)) }$

I have been unable to reproduce that calculus without proving the reverse ( that the derivative of the right side is the integrand of the left side). I would like to calculate the integral without using a coordinate basis if possible. Some details of the notation:

$\frac{d \vec{\beta}(t)}{d t} = \vec{\beta'}(t)$

$\vec{n}$ is a constant vector that does not depend of the integration variable t.

This is from The Physics of Synchrotron Radiation by Albert Hofmann, section 2.8. The Fourier transform of the radiation field, page 36.

  • $\begingroup$ Can you give the title of the book and the page ? To see the context can be useful. $\endgroup$ – Tony Piccolo Sep 5 '15 at 5:34
  • $\begingroup$ @TonyPiccolo The book is ''The physics of syncrotron radiation'' page 36 in ''Fourier Transform of the fields". $\endgroup$ – Dargor Sep 5 '15 at 11:20
  • $\begingroup$ Till now, after reading several sources, I found they always give the result by inspection, just the thing you don't want. Sorry. $\endgroup$ – Tony Piccolo Sep 5 '15 at 17:06
  • $\begingroup$ @TonyPiccolo Thank you anyway :) $\endgroup$ – Dargor Sep 5 '15 at 17:14
  • $\begingroup$ @PabloGalindoSalgado I have added reference and link to the book based on your comment. (It is more visible in post than in comments, so this improves context of the post.) Please check whether this is the place you meant. $\endgroup$ – Martin Sleziak Sep 8 '15 at 5:24

If you expand out the integrand you see one term is zero, since the cross product of two parallel vectors is zero:

\begin{align} I &= \int \frac{\vec{n}\times(\vec{n}-\vec{\beta(t)})\times\vec{\beta'(t)}}{(1-\vec{n}\cdot \vec{\beta(t)})^2}dt\\ &= \int \frac{\vec{n}\times\vec{n}\times\vec{\beta'(t)}-\vec{n}\times\vec{\beta(t)}\times\vec{\beta'(t)}}{(1-\vec{n}\cdot \vec{\beta(t)})^2}dt\\ &= \int \frac{\vec{n}\times\vec{n}\times\vec{\beta'(t)}}{(1-\vec{n}\cdot \vec{\beta(t)})^2}dt \end{align}

This gets you halfway there. I'll try to come back to this to finish it soon!


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