User talk:AnakinLee0708

Covariant Derivative of a contravariant vector : $$\nabla_a{T^i} = \frac{\partial{T^i}}{\partial{x^a}} + \Gamma^i_{ab}{T^b}$$

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Here is a derivation of Covariant Derivative. Consider a $${\mathbf{T}}=T^i\mathbf{e}_i$$, a type(1,0) tensor. By taking Covariant Derivative: $$\nabla_a\left({T^i\mathbf{e}_i}\right)=\frac{\partial{T^i}}{\partial{x^a}} \mathbf{e}_i + T^i \frac{\partial\mathbf{e}_i}{\partial{x^a}}$$, here we used product rule.

Now consider the term $$\frac{\partial\mathbf{e}_i}{\partial{x^a}}$$, it can be written as $$\partial_a \mathbf{e}_i$$.

When this term is expressed in terms of origin basis, it is linked by a connection "Christoffel Symbol", such that $$\partial_a \mathbf{e}_i = \Gamma^j_{ai}\mathbf{e}_j$$, substitute back into the equation and change the indices.

$$\nabla_a\left({T^i\mathbf{e}_i}\right)=\frac{\partial{T^i}}{\partial{x^a}} \mathbf{e}_i + T^b \Gamma^i_{ab}\mathbf{e}_i$$

So, in component form we have $$\nabla_a{T^i} = \frac{\partial{T^i}}{\partial{x^a}} + \Gamma^i_{ab}{T^b}$$

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Also, Absolute Derivative of a contravariant vector w.r.t arc length parameter : $$\frac{\delta}{\delta{t}} (T^i) = \frac{d{T^i}}{dt} + \Gamma^i_{ab}{T^a}{\frac{d{x^b}}{dt}}$$, where $$x^b$$ is the arbitrary coordinate. $$\forall x^b=x^b(t)$$

In $$\mathbf{R^3}$$, the wedge product is equal to cross product, as : $$\mathbf{A} \wedge \mathbf{B} = \mathbf{A} \times \mathbf{B} = \begin{vmatrix}\mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3} \end{vmatrix} = ({a_2}{b_3}-{a_3}{b_2}) {\mathbf{e}_2} {\wedge} {\mathbf{e}_3} + ({a_3}{b_1}-{a_1}{b_3}) {\mathbf{e}_3} {\wedge} {\mathbf{e}_1} + ({a_1}{b_2}-{a_2}{b_1}) {\mathbf{e}_1} {\wedge} {\mathbf{e}_2} $$