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5

Recall that the Faraday tensor in this form is a linear mapping that maps a charged particle's contravariant four-velocity to the latter's rate of change, wrt proper time (modulo scaling by invariant rest mass $m$ and invariant charge $q$): $$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu\tag{1}$$ Now let's think of a particle's ...

2

In my experience, reading the indices left to right and top to bottom, the first index is the row and the second is the column. Your screenshot from Carroll doesn't have to be contradictory (although it's definitely confusing/doesn't make rigorous sense). You can just imagine he omits a little "$_{\mu \nu}$" on the matrix: $$F_{\mu \nu}=\Bigg( \cdots ... 3 Your example is an outlier, in my experience (personally, I would have written (F_{\mu\nu})^T instead of F_{\nu\mu}). Almost always, it's the order of the indices that determines row vs. column. If someone writes T^i_j, then while technically there's no way to tell, I would say that it would be far less confusing to make the upper index label the rows ... 3 First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces. If we have a spacetime manifold \mathcal{M}, then the fields are functions$$ 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 ...

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