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As I recall, covariant refers to how an object transforms when you boost to another inertial frame. An example would be the relativistic 4-momentum $P^{\mu}$. Invariant refers to quantities which are unchanged under boosts to different frames. For example the product $P^{\mu}P_{\mu}=m$ has the same numerical value in any frame. Sometimes a relativistic ...


3

First, they do not transform in an actual "representation" in the sense of a linear representation of the group of coordinate transformation since their behaviour under a coordinate transformations $x\mapsto y(x)$ is given as $$ {\Gamma^\alpha}_{\beta\gamma} \overset{y(x)}\mapsto \frac{\partial x^\mu}{\partial y^\beta}\frac{\partial x^\nu}{\partial ...


4

The Christoffel symbols do not transform under any representation. The reason for this is that they do not transform linearly, which puts them out of the game altogether. The transformation law is $$ \tilde \Gamma^{\mu}_{\nu\kappa} = {\partial \tilde x^\mu \over \partial x^\alpha} \left [ \Gamma^\alpha_{\beta \gamma}{\partial x^\beta \over \partial \tilde ...


2

Nothing wrong with the dimension. Once you put into the index and summation form, then it's basically adding up a bunch of numbers. Note that $$g_{\alpha \beta}\Delta x^{\beta}=\Delta x_{\alpha}$$ then your equation will be trivial.


4

When people study continuum mechanics they usually do so at first in $\mathbb{R}^3$ where we have usually implied the usual metric tensor $(g_{ij}) = \operatorname{diag}(1,1,1)$ and the Levi-Civita connection associated with it. In that case vectors and covectors are equivalent: the metric tensor induces the musical isomorphism and allows one to convert ...


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Disclaimer: this is a brief answer, dealing with classical fields. To have a thorough understanding of the subject you should refer to books on GR and QFT in curved spacetime. A good read is the book General Relativity by Robert Wald. You want to write down general covariant equations for some fields that reduce to the familiar ones (KG, Maxwell etc) when ...


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You should first prove that $\omega_{\mu\nu}$ is antisymmetric (from the properties of Lorentz transforms). Then your equality is obviously correct if $\mu=\nu$ (both parts vanish). Then you can assume that $\mu\ne\nu$. Then you can use, e.g., your third formula (by the way, please use either $g$ or $\eta$, not both), and use $(\gamma^\mu)^2=g^{\mu\mu}$, ...



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