[OTO] Fixed up ex 1 and bonus

2 files changed, 20 insertions, 8 deletions

diff --git a/Year 3/General theory of relativity/Project.pdf b/Year 3/General theory of relativity/Project.pdf
index 7e981eb..ac49b05 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 94da9a2..31f9d5b 100644
--- a/Year 3/General theory of relativity/Project.tex
+++ b/Year 3/General theory of relativity/Project.tex
@@ -41,7 +41,7 @@
 \paragraph{1. Решение}
 
 Собствения часовник в собствената координатна система винаги върви най-бързо, защото отнема време на светлината от единия брат да достигне другия брат.
-Тоест един брат винаги вижда остарял облик на другия брат (при крайна скорост на светлината).
+Всеки брат изминава еднакво разстояние с еднаква скорост, значи за всеки брат е изминало еднакво количество време.
 
 \paragraph{2. Решение}
 
@@ -204,20 +204,32 @@ $00800028 \; mod \; 3 = 0$, тоест ще докажем, че ковариа�
 \end{align*}
 
 \paragraph{Бонус задача} Докажете Тъждеството на Бианки
-\paragraph{Решение} Използвайки формулата за смяна на променливи при кривини, виждаме че:
+\paragraph{Решение}
+
+\begin{align*}
+	{R^i}_{abc} + {R^i}_{bca} + {R^i}_{cab} & = \frac{\partial}{\partial x^c} {\Gamma^i}_{ab} - \frac{\partial}{\partial x^b} {\Gamma^i}_{ac} \quad + {\Gamma^i}_{mc} {\Gamma^m}_{ab} - {\Gamma^i}_{mb} {\Gamma^m}_{ac} \\
+	& \quad + \frac{\partial}{\partial x^a} {\Gamma^i}_{bc} - \frac{\partial}{\partial x^c} {\Gamma^i}_{ba} \quad + {\Gamma^i}_{ma} {\Gamma^m}_{bc} - {\Gamma^i}_{mc} {\Gamma^m}_{ba} \\
+	& \quad + \frac{\partial}{\partial x^b} {\Gamma^i}_{ca} - \frac{\partial}{\partial x^a} {\Gamma^i}_{cb} \quad + {\Gamma^i}_{mb} {\Gamma^m}_{ca} - {\Gamma^i}_{ma} {\Gamma^m}_{cb} \\
+\end{align*}
+
+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:
+Тоест:
 
 \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
+	 & \frac{\partial}{\partial x^c} {\Gamma^i}_{ab} - \frac{\partial}{\partial x^c} {\Gamma^i}_{ba} = \frac{\partial}{\partial x^c} {\Gamma^i}_{ab} - \frac{\partial}{\partial x^c} {\Gamma^i}_{ab} = 0 \\
+	 & \text{Аналогично } \frac{\partial}{\partial x^b} {\Gamma^i}_{ca} - \frac{\partial}{\partial x^b} {\Gamma^i}_{ac} = 0 \quad \frac{\partial}{\partial x^a} {\Gamma^i}_{bc} - \frac{\partial}{\partial x^a} {\Gamma^i}_{cb} = 0
 \end{align*}
 
+\begin{align*}
+	& {\Gamma^i}_{mc} {\Gamma^m}_{ab} - {\Gamma^i}_{mc} {\Gamma^m}_{ba} = {\Gamma^i}_{mc} {\Gamma^m}_{ab} - {\Gamma^i}_{mc} {\Gamma^m}_{ab} = 0 \\
+	& \text{Аналогично } {\Gamma^i}_{mb} {\Gamma^m}_{ca} - {\Gamma^i}_{mb} {\Gamma^m}_{ac} = 0 \quad {\Gamma^i}_{ma} {\Gamma^m}_{bc}  - {\Gamma^i}_{ma} {\Gamma^m}_{cb} = 0
+\end{align*}
+
+Следователно ${R^i}_{abc} + {R^i}_{bca} + {R^i}_{cab} = 0$
+
 \end{document}
\ No newline at end of file