My question is related to this link (18:10) https://www.youtube.com/watch?v=oQZTYt_Pxcc&list=PLJHszsWbB6hpk5h8lSfBkVrpjsqvUGTCx&index=28
Recently i have watched a video about volume derivation. In this video,the presenter tried to relate the ordinary derivative of the Volume to the Ricci tensor but i get stuck when i try to understand it.
My problem is i dont understand why $\ddot{S^{\mu_j} _j} $ is equal to $ -R^{\mu_j}{}_{xyz} s^y_j v^z v^x$ at 18:10
The reason why $\ddot{S^{\mu_j} _j} $ should not be equal to $ -R^{\mu_j}{}_{xyz} s^y_j v^z y^x$ where $$[\nabla_v\nabla_vS]^{\mu^j}= -R^{\mu_j}{}_{xyz} s^y_j v^z v^x $$ is because they are different things. The former is just second ordinary derivative and the latter is the full-form second covariant derivative.
i cannot see why it is not second ordinary derivative since volume itself a scalar. if you take derivative of the volume with ordinary derivative,then the logic should follow in a similar manner and the second derivative of the vector component (part of the decomposition because of product rule) should be the ordinary type derivative
I would now give an example why they are not equal.
Given a 2D-Volume(area) which is spanned by vector $\vec{A} $ and $\vec{B}$, we can take the derivative:
The volume is $ V=\epsilon_{kj} A^kB^j$. If you take the ordinary derivative of the volume $$\dfrac {dV} {d\lambda}=\epsilon_{jk} [\dfrac{dB^j}{d\lambda}]A^k\epsilon_{jk} [\dfrac{dA^k}{d\lambda}]B^j$$
and this implies you will take the ordinary derivative of [$A^k$](as a result of product rule) which is $$\dfrac{dA^k}{d\lambda} $$ however this is just part of the $\nabla_vA$:
$[∇vA]^k=\dfrac{dA^k}{d\lambda}[1.]+v^j\gamma^{k}_{ij}$[2]=$v^j\dfrac{dA^k}{dx^j} + v^j\gamma^{k}_{ij}$
where$\gamma^{k}_{ij} $is the connection factor that is missed out and $ v^j \dfrac{d}{dx^j}$ is equal to $\frac{d}{d\lambda}$.
It is shown that the ordinary derivative of $A^i$ is just[1.]and[2] is neglected. [1.] is just the ordinary derivative part of the volume derivation [2] is the whole derivative which includes connection factor it is clear that [1.] is not equal to [2]. Therefore$[∇vA]^k$ is not equal to $\dfrac{dA^k}{d\lambda}$ . Follow the similar line of reasoning,i think $\ddot{S^{\mu_j} _j} $=$\dfrac{d^2(s^{\mu_j})}{d\lambda^2}$ should not be equal to$[\nabla_v\nabla_vS]^{\mu^j}$ ,i am not sure why they are equal
$\ddot{S^{\mu_j} _j}$= $\dfrac{d^2(s^{\mu_j})}{d\lambda^2}$=?? $[∇_v∇_vS]^{\mu_j}= R^{\mu_j}{}_{xyz} s^y_j v^z v^x$
could anyone help me please
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