[OTO] Solved a bonus exercise and improved certain cancelling out in ex 3

2 files changed, 19 insertions, 2 deletions

diff --git a/Year 3/General theory of relativity/Project.pdf b/Year 3/General theory of relativity/Project.pdf
index 53f4b8d..3b44c1e 100644
--- a/Year 3/General theory of relativity/Project.pdf
+++ b/Year 3/General theory of relativity/Project.pdf
diff --git a/Year 3/General theory of relativity/Project.tex b/Year 3/General theory of relativity/Project.tex
index 2ad12e6..3498177 100644
--- a/Year 3/General theory of relativity/Project.tex
+++ b/Year 3/General theory of relativity/Project.tex
@@ -167,8 +167,8 @@ $00800028 \; mod \; 3 = 0$, тоест ще докажем, че ковариа�
 	& \qquad - \left( \tilde{\Gamma}^a {}_{nc} \frac{\partial x^i}{\partial y^a} \frac{\partial y^n}{\partial x^m} \frac{\partial y^c}{\partial x^k} + \frac{\partial x^i}{\partial y^p} \frac{\partial^2 y^p}{\partial x^m \partial x^k} \right)
 	\left( \tilde{\Gamma}^n {}_{bd} \frac{\partial x^m}{\partial y^n} \frac{\partial y^b}{\partial x^j} \frac{\partial y^d}{\partial x^l} + \frac{\partial x^m}{\partial y^p} \frac{\partial^2 y^p}{\partial x^j \partial x^l} \right) \\
 	& \qquad \bigg) \frac{\partial}{\partial y^a} \frac{\partial y^a}{\partial x^i} \frac{\partial x^j}{\partial y^b} \der y^b \frac{\partial x^k}{\partial y^c} \der y^c \frac{\partial x^l}{\partial y^d} \der y^d \\
-	& = \left( \frac{\partial}{\partial y^d} \tilde{\Gamma}^a {}_{bc} \left( \frac{\partial y^d}{\partial x^l} \frac{\partial x^i}{\partial y^a} \frac{\partial y^b}{\partial x^j} \frac{\partial y^c}{\partial x^k} \right) + \cancel{\frac{\cancel{\partial y^d}}{\partial x^l} \frac{\partial}{\cancel{\partial y^d}} \frac{\partial x^i}{\partial y^p} \frac{\partial^2 y^p}{\partial x^j \partial x^k}} \right. \\
-	& \qquad - \frac{\partial}{\partial y^c} \tilde{\Gamma}^a {}_{bd} \left( \frac{\partial y^c}{\partial x^k} \frac{\partial x^i}{\partial y^a} \frac{\partial y^b}{\partial x^j} \frac{\partial y^d}{\partial x^l} \right) - \cancel{\frac{\cancel{\partial y^c}}{\partial x^k} \frac{\partial}{\cancel{\partial y^c}} \frac{\partial x^i}{\partial y^p} \frac{\partial^2 y^p}{\partial x^j \partial x^l}} \\
+	& = \left( \frac{\partial}{\partial y^d} \tilde{\Gamma}^a {}_{bc} \left( \frac{\partial y^d}{\partial x^l} \frac{\partial x^i}{\partial y^a} \frac{\partial y^b}{\partial x^j} \frac{\partial y^c}{\partial x^k} \right) + \cancel{\frac{\bcancel{\partial y^d}}{\partial x^l} \frac{\partial}{\bcancel{\partial y^d}} \frac{\partial x^i}{\partial y^p} \frac{\partial^2 y^p}{\partial x^j \partial x^k}} \right. \\
+	& \qquad - \frac{\partial}{\partial y^c} \tilde{\Gamma}^a {}_{bd} \left( \frac{\partial y^c}{\partial x^k} \frac{\partial x^i}{\partial y^a} \frac{\partial y^b}{\partial x^j} \frac{\partial y^d}{\partial x^l} \right) - \cancel{\frac{\bcancel{\partial y^c}}{\partial x^k} \frac{\partial}{\bcancel{\partial y^c}} \frac{\partial x^i}{\partial y^p} \frac{\partial^2 y^p}{\partial x^j \partial x^l}} \\
 	& \qquad + \tilde{\Gamma}^a {}_{nd} \frac{\partial x^i}{\partial y^a} \cancel{\frac{\partial y^n}{\partial x^m}} \frac{\partial y^d}{\partial x^l} \tilde{\Gamma}^n {}_{bc} \cancel{\frac{\partial x^m}{\partial y^n}} \frac{\partial y^b}{\partial x^j} \frac{\partial y^c}{\partial x^k}
 	+ \tilde{\Gamma}^a {}_{nd} \frac{\partial x^i}{\partial y^a} \frac{\partial y^n}{\partial x^m} \frac{\partial y^d}{\partial x^l} \frac{\partial x^m}{\partial y^p} \frac{\partial^2 y^p}{\partial x^j \partial x^k} \\
 	& \qquad + \tilde{\Gamma}^n {}_{bc} \frac{\partial x^m}{\partial y^n} \frac{\partial y^b}{\partial x^j} \frac{\partial y^c}{\partial x^k} \frac{\partial x^i}{\partial y^p} \frac{\partial^2 y^p}{\partial x^m \partial x^l}
@@ -203,4 +203,21 @@ $00800028 \; mod \; 3 = 0$, тоест ще докажем, че ковариа�
 	& = \tilde{R}^a {}_{bcd} \frac{\partial}{\partial y^a} \der y^b \der y^c \der y^d
 \end{align*}
 
+\paragraph{Бонус задача} Докажете Тъждеството на Бианки
+\paragraph{Решение} Използвайки формулата за смяна на променливи при кривини, виждаме че:
+
+\begin{align*}
+	{\Gamma^i}_{ab} = {\Gamma^i}_{ba} \frac{\partial y^i}{\partial y^i} \frac{\partial y^b}{\partial y^a} \frac{\partial y^a}{\partial y^b} + \frac{\partial y^i}{\partial y^p} \frac{\partial^2 y^p}{\partial y^a \partial y^b} = {\Gamma^i}_{ba}
+\end{align*}
+
+Следователно:
+
+\begin{align*}
+	& {R^i}_{abc} + {R^i}_{bca} + {R^i}_{cab} \\
+	& = 	\cancel{\frac{\partial}{\partial x^c} {\Gamma^i}_{ab}} - \bcancel{\frac{\partial}{\partial x^b} {\Gamma^i}_{ac}} \quad + \cancel{{\Gamma^i}_{mc} {\Gamma^m}_{ab}} - \bcancel{{\Gamma^i}_{mb} {\Gamma^m}_{ac}} \\
+	& \quad + \xcancel{\frac{\partial}{\partial x^a} {\Gamma^i}_{bc}} - \cancel{\frac{\partial}{\partial x^c} {\Gamma^i}_{ba}} \quad + \xcancel{{\Gamma^i}_{ma} {\Gamma^m}_{bc}} - \cancel{{\Gamma^i}_{mc} {\Gamma^m}_{ba}} \\
+	& \quad + \bcancel{\frac{\partial}{\partial x^b} {\Gamma^i}_{ca}} - \xcancel{\frac{\partial}{\partial x^a} {\Gamma^i}_{cb}} \quad + \bcancel{{\Gamma^i}_{mb} {\Gamma^m}_{ca}} - \xcancel{{\Gamma^i}_{ma} {\Gamma^m}_{cb}} \\
+	& = 0
+\end{align*}
+
 \end{document}
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