I have a one-form field on Euclidean space. Suppose we integrate it over a loop around the specific point $x$. $$I(x)=\int_xU.$$ I want to calculate the partial derivatives of this integral respect to the $x$. $$\frac{\partial}{\partial x^\alpha}I(x) ??$$

Consider that the closed curves at each point ( which we integrated over them) are identical.

$$I(x+\delta x)-I(x)=\int_{x+\delta x}U_{\beta}(t'(s))\frac{\partial t'^{\beta}(s)}{\partial s}ds - \int_x U_{\beta}(t(s))\frac{\partial t^{\beta}(s)}{\partial s}ds$$ In the above formula, $t'^{\beta}(s)$ is parametrization of the curve at point $x^{\alpha}+\delta x^{\alpha}$ and $t^{\beta}(s)$ is at point $x^{\alpha}$. Because of the similarity of the curves at each point, their parametrization differ just by a translation, so l have the following relation

$$t'^{\beta} (s) = t^{\beta} (s) +{\delta}x^{\beta}$$ By inputing the above relation at third equation and using Taylor expansion, we can find explicit expression for derivative

$$\frac{\partial}{\partial x^\alpha}I(x) = \int_x \frac{\partial U_{\beta} (t(s))}{\partial t^{\alpha}(s)} \frac{\partial t^{\beta}(s)}{\partial s} ds $$

But the left hand side of the last equation has a well defined behaviour under coordinate transformation, and the right hand side has not. One of my friends tell me that, under general coordinate transformation the fourth relation is not meaningful and only works for constant transformations.

I think this is true. How can l change my calculations to achieve a covariant version of the last equation?


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