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First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces. If we have a spacetime manifold $\mathcal{M}$, then the fields are functions $$ f : \mathcal{M} \to V$$ where $V$ is some vector space upon which a representation $\rho : \mathrm{SO}(1,3)\to ...


2

The easiest trick is to write (say, in three dimensions. For higher dimensions, add indices to the Levi-Civita symbol, and factors of g): $$g = \frac{1}{3!}\epsilon^{abc}\epsilon^{xyz}g_{ax}g_{by}g_{cz}$$ Then, the variation is easy. I'll leave it as an excersise to work out the variation, and how to translate the result into factors of $g_{ab}$ and $g$


2

Use the identity that if $M$ is invertible and $\delta M$ is "small" compared to $M$, then we have $$ \det (M + \delta M) = \det(M) \det( 1 + M^{-1} \delta M) \approx \det(M) \left[ 1 + \text{tr} (M^{-1} \delta M) \right]. $$ In the case of the metric, this implies that $$ -\det(g + \delta g) \approx -\det(g) \left[ 1 + g^{ab} \delta g_{ab} \right] $$ and ...


1

Timaeus's answer could be correct. The $\Lambda$ matrix from your book may have been intended as a passive transformation (one that acts on the coordinate system) and you mistakenly used it as an active transformation (one that acts on the object). Alternatively, the $\Lambda$ matrix from your book may have been intended as the active transformation of a ...


0

The first thing that bothers me is that for positive ${v}$ I get negative component of the velocity vector If your object was at rest then to someone moving to the right the object at rest is moving to the left in their frame. As for the second question it is same issue. Someone at rest will be moving to the left relative to someone moving faster than ...


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I believe this is the "missing link", stated in a less abstract fashion than in the above comment: https://en.wikipedia.org/wiki/Quadratic_form . Some programs in physics cover that in undergraduate algebra courses, some leave it for later. Notably, this method doesn't apply just to tensors, it's a general connection between symmetric matrices (of spaces ...


0

For intuition, consider the Inertia Tensor, which is symmetric and hence has six components. The scalar |0,0> part (~trace) is just the mass of the object: clearly rotationally invariant, and hence, physically significant. That leaves us with the five |2, m> parts. If you rotate to the principle axes, the |2, +/-1> are zero, and you are left with m=0 and ...



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