Tag Info

New answers tagged

3

It's almost the defition. A tensor $T_{ab}$ of rank $2$ is symmetric if, and only if, $T_{ab}=T_{ba}$, and antisymmetric if, and only if, $T_{ab}=-T_{ba}$. So from this definition you can easily check that this decomposition indeed yields a symmetric and antisymmetric part. Edit: Let $S_{bc}=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)$. Then ...


5

The statement from the Wikipedia articles is, as written, wrong. The EM field tensor - as a tensor - does change under change of reference frames. It is covariant, but not invariant under the Lorentz group, while the electric and magnetic field are neither, but they are covariant under the rotation group. The electric and magnetic fields are ordinary, ...


2

In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent. Question: why is this true? Actually this isn't quite true. General relativity doesn't have frames of reference (except locally, which is trivially true because GR is the same as SR locally). A better way of ...


0

$\sigma_{\vec{\imath}}$ is the rotation matrix about the vector $\vec{\imath}$: $$\sigma_{\vec{\imath}} = \imath_x\sigma_{x}+\imath_y\sigma_{y}+\imath_z\sigma_{z}$$ After expanding it out completely and after some shuffling around, the solution is as follows: $$e^{-i\theta/2 \sigma_{\vec{\imath}}^A} \otimes e^{-i\theta/2 \sigma_{\vec{\imath}}^B} ...



Top 50 recent answers are included