0
$\begingroup$

One of my textbooks says that a contravariant vector $a^\lambda$ transforms according to $$a'^\mu=\frac{\partial x'^\mu}{\partial x^\lambda} a^\lambda,$$ when changing the inertial frame of reference ($x^\mu\rightarrow x'^\mu$). At the same time $$a'^\mu=\Lambda_{\mu\lambda}a^\lambda,$$ where $\Lambda$ is the Lorentz transformation matrix.

This kind of confuses me as it implies the Jacobi-matrix w.r.t. to the basis vectors of the two frames of reference equals the Lorentz transformation. Shouldn't the first equation be $$da'^\mu=\frac{\partial x'^\mu}{\partial x^\lambda} da^\lambda\mathrm{~?}$$

$\endgroup$
  • 1
    $\begingroup$ I don't know what the "$da$" are supposed to be in your second equation - what is $d$? Also, since a transformation between inertial frames is just a Lorentz transformation, $x' = \Lambda x$ and the Jacobian is the Lorentz transformation, I'm not sure what the problem is. $\endgroup$ – ACuriousMind Nov 20 '16 at 13:37
  • $\begingroup$ By $da$ i mean an infinitesimal change of the vector components. Your last sentence confuses me. As far as I know the term Jacobian refers to the determinant of the Jacobi matrix. How can this be the lorentz transformation? $\endgroup$ – OD IUM Nov 20 '16 at 14:39
2
$\begingroup$

Actually Lorentz transform = Jacobi matrix. This is because Lorentz transformations are linear. Probably your textbook is going to discuss nonlinear transformations, like the ones encountered in general relativity. In that case, the equation you wrote is the transformation law of the coordinates of a tangent vector.

$\endgroup$
  • $\begingroup$ Thx for making this clear $\endgroup$ – OD IUM Nov 20 '16 at 14:40

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.