I'm working through a book on relativity so this may end up being a physics question but I'm pretty sure that my problem is mathematical so I'm asking here. In deriving the "special Lorentz transformations", (where the axes of the second coordinate system, $S'$ are parallel to the those of the first, $S$, and movement is along the $x$, $x'$ axes with velocity $v$) the author notes that because $y = y'$ and $z = z'$, the requirement

$$ x_1^2 + x_2^2 + x_3^2 - (ct)^2 = x_1'^2 + x_2'^2 + x_3'^2 - (ct')^2 $$ reduces to $$ x_1^2 - (ct)^2 = x_1'^2 - (ct')^2 $$ and introduces the variables $$ x_4 = ict, x_4' = ict' $$ to get $$ x_1^2 + x_4^2 = x_1'^2 + x_4'^2 $$ giving $$ \begin{eqnarray} x_1' &=& x_1\cos\phi + x_4\sin\phi\\ x_4' &=& -x_1\sin\phi + x_4\cos\phi\\ \end{eqnarray}. $$

A point at rest in $S$, i.e. for which $\frac{dx_1'}{dx_4'} = 0$ must then have $\frac{dx_1}{dx_4} = \frac{-iv}{c}$. The author is using the notation for single variable derivatives here. It seems to be that they should be partials. From here, he says that the transformation rules given above imply $$ \frac{dx_1}{dx_4'} = \frac{\frac{dx_1}{dx_4}\cos\phi + \sin\phi}{-\frac{dx_1}{dx_4}\sin\phi + \cos\phi} $$

I can't get this. I've tried inverting the transformation and taking the total derivative $$ \frac{dx_1}{dx_4'} = \frac{\partial x_1}{\partial x_4'} + \frac{\partial x_1}{\partial x_1'}\frac{dx_1}{dx_4'} = -\sin\phi $$ because $\frac{dx_1'}{dx_4'} = 0$.

Treating it like a partial gets the same result. Implicitly differentiating and solving for the derivative gets the same result as well. I can't get any result other than $-\sin\phi$

How an I supposed to take this derivative? Thanks for any pointers!

  • $\begingroup$ If this is better suited to physics.SE, I'd love to have it moved. Like I said, I think that my problem here is mostly mathematical. $\endgroup$ – knucklebumpler Apr 21 '11 at 19:40
  • $\begingroup$ I think there's a mistake in the transformation relations from $(x_1,x_4)$ to $(x_1',x_4')$. There is a minus sign too much. $\endgroup$ – Raskolnikov Apr 21 '11 at 20:38
  • $\begingroup$ Also, you probably just dropped an accent, it should be $dx_1'/dx_4'$ which you want to relate to $dx_1/dx_4$, for obvious reasons: the goal is to connect the angle $\phi$ to $iv/c$. $\endgroup$ – Raskolnikov Apr 21 '11 at 20:44
  • $\begingroup$ @Raskolnikov. Obviously correct about the rotation being off. The book writes it as the reverse of a normal rotation matrix and I'm in the other habit. I accidentally did both and didn't notice! That problem wasn't present in my work on paper though. The accent got dropped in the book (if that is what happened) and it's explicitly stated that $dx_1'/dx_4' = 0$ in the book. I can prove (didn't mention in the question) that the side on the left is equal to zero so that would make sense. Thanks for the input. $\endgroup$ – knucklebumpler Apr 21 '11 at 21:01

First, about the total/partial derivatives: He's talking about the trajectory of a point at rest, not about a coordinate transformation, so it's appropriate to take the total derivative with respect to $x_4'$ -- that's like taking a total time derivative along a trajectory in classical mechanics. $\partial x_1'/\partial x_4'$ would only make sense if you specified which other coordinates are meant to be held fixed; if you apply the usual interpretation that $\partial x_4'$ means that the other primed coordinates are held fixed, then $\partial x_1'/\partial x_4'=0$ by definition, independent of any points being at rest or not.

I agree with Raskolnikov that there's probably a prime missing in the derivative, since this is the correct result for $\mathrm dx_1'/\mathrm dx_4'$. (You can get it by differentiating both transformation laws with respect to $x_4$.) There's no contradiction with this being $0$ for a point at rest -- the one is a general transformation law for the coordinates, the other is a statement about a certain trajectory. I always find it surprising how little care is taken in books and the like to distinguish these two concepts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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