Let, we are in 1d cartesian space with metric $g_{xx} = x^2$. Let we have a vector $v = 1/x e_x$. Since the vector is designed to shrink its components as the basis grows - its total length will remain constant - meaning that it will be parallel transported along $x$ axis, and its covariant derivative will be zero. But what about its partial derivative (here just derivative because of 1d)?
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$\begingroup$ What do you mean by partial derivative of a vector field? $\endgroup$– MBNCommented May 30, 2023 at 8:43
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