Co-variant Derivative of a Complex Vector

thehangedman
Messages
68
Reaction score
2
What is the form of the co-variant derivative for a vector with complex elements (such as the Electromagnetic field vector A)?
 
Physics news on Phys.org
Usually, those complex terms come from treating time as imaginary. THis is frequently done in SR, but is rarely done in GR, where one usually sees the covariant derivative.

There are occasions when one does use complex coordinates in GR - I've never done it. (It's common enough that GRtensor provides allowanes for it, though).

As far as I know it doesn't affect taking the covariant derivative at all there are several ways of looking at it, the way I prefer to describe the covariant derivative is that it's just like the ordinary derivative, except that you have to parallel transport vectors (and/or tensors) to the same location before you subtract them and divide by the delta.
 
Last edited:
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top