- #1

TimeRip496

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Source:

Basically the video talk about how moving from A to A'(which is basically A) in an anticlockwise manner will give a vector that is different from when the vector is originally in A in curved space.

$$[(v_C-v_D)-(v_B-v_A)]$$ will equal zero in flat space.

$$[(v_C-v_D)-(v_B-v_A)]-[(v_C-v_D)-(v_B-v'_A)]=v_A-v'_A$$

$$v_A-v'_A=dv$$

That is the difference, that is going to be the change in the vector as it is parallel transport around that parallelogram. In flat space, dv will be zero as the vector will be parallel transported and come back in the same magnitude and direction. but in curved space, there will be a difference, not in length but in angle of the vector.

$$v_C-v_D=\frac{∂v}{∂x^μ}dx^μv⇒∇_μdx^μv$$

change in vector between c and d is the gradient multiplied by the distance times the value of the vector. We need to use the covariant derivatives so instead of using the partial derivative, we use this ∇ instead.

$$(v_C-v_D)-(v_B-v'_A)=∇_μdx^μ ∇_νdx^ν v$$

**I don't really understand this part. Like so just by derivative can we actually change the vector twice?**

$$v_A-v'_A=dV=dx^μdx^ν v [∇_v.∇_μ]$$

And the commutator between the two covariant derivatives will give the Riemannian Tensor, which can be regarded as the Ricci tensor.

So how does parallel transportation relates to the Riemannian manifold? Does the Riemannian manifold exhibit the above properties? Do I need to learn about differential geometry in order to understand Riemannian manifold?