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Notes
- 1.
Note that we have lowered the index μ in the term \(\frac {1}{\mathbf {A} ^{2}}h_{\mu }^{\;\;\alpha }A^{\beta }T_{\alpha \beta }\) therefore we take the transpose matrix of \(\frac {1}{{\mathbf {A}}^{2}}h^{\mu \alpha }A^{\beta }T_{\alpha \beta }\) i.e. the \(\frac {1}{25}\left ( \begin {array}{lll} -6 & 3 & 0 \end {array} \ \ \right ).\)
- 2.
We consider c = 1. Otherwise we have u a u a = −c 2.
- 3.
- 4.
Recall: First index row second index column! Also \(u_{a}=(-\sqrt {3},1,1,1)!\)
- 5.
The proof has as follows:
$$\displaystyle \begin{aligned} T_{ab}& =\delta _{a}^{c}\delta _{b}^{d}T_{cd}=\left( h_{a}^{c}+\frac{ \varepsilon (A)}{A^{2}}A^{c}A_{d}\right) \left( h_{b}^{d}+\frac{\varepsilon (A)}{A^{2}}A^{d}A_{b}\right) T_{cd} \\ & =\frac{1}{A^{4}}\left( T_{cd}A^{c}A^{d}\right) A_{a}A_{b}+\frac{ \varepsilon (A)}{A^{2}}\left( h_{a}^{\;\;c}A^{d}A_{b}T_{cd}+h_{b}^{\;\;d}A^{c}A_{a}T_{cd}\right) +h_{a}^{c}h_{b}^{d}T_{cd}. \end{aligned} $$ - 6.
Because Minkowski space is flat, it is possible to transport a four-vector from one point to another along any path. This implies that the four-vectors need not have a common point of application. Simply they must define a 2-plane. We shall use this observation in the derivation of the covariant Lorentz transformation.
- 7.
The proof is easy: \(p_{a}^{c}(A,B)h_{c}^{b}=p_{a}^{c}(A,B)\left ( \delta _{c}^{b}+A_{c}A^{b}\right ) =p_{a}^{b}(A,B)\).
- 8.
You can find the result by writing C a = η ab C b and using (12.20) to replace η ab.
- 9.
The last requirement means that all three four-vectors have the same sign of their zero component, that is, they point in the same part of the light cone.
- 10.
The first two conditions mean that the energy is positive. The third is the restriction that the measure of the 3-momentum is non-negative.
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Tsamparlis, M. (2019). Irreducible Decompositions. In: Special Relativity. Undergraduate Lecture Notes in Physics. Springer, Cham. https://doi.org/10.1007/978-3-030-27347-7_12
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DOI: https://doi.org/10.1007/978-3-030-27347-7_12
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