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Let $U\subset \mathbb{R}^3$ be an open subset endowed with a triple orthogonal coordinate system $\{x^1,x^2,x^3\}$ and let $X$ be a smooth vector field on $U$. The vector field rot$X$ (or curl$X$) which is called a rotor (or curl) of a vector field $X$ can be defined in coordinates $x^1,x^2,x^3$ by means of the forms $\omega_X,\theta_X$, which are defined as $$(\omega_X)_p(v_p)=\langle X_p,v_p\rangle,$$ $$(\theta_X)_p(v_p,w_p)=\langle X_p,v_p\times w_p\rangle,$$ in a following way $$d\omega_X=\theta_{\text{rot}X}$$ where $d$ is the exterior derivative. Find the rotor of the vector field $X$ in coordinates $x^1,x^2,x^3$ and then in particular cases of Cartesian cylindrical and spherical coordinates.

Should I first prove that $\omega$ is a 1-form and $\theta$ is a 2-form? Any ideas on how to approach this problem?

user332350
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1 Answers1

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Hint: There is no need to formally prove that $\omega_X$ is a one-form and $d\omega_X$ is two-form by definition. To solve the problem, you on the one hand have to compute functions $a_i$ for $i=1,2,3$ such that $\omega_X=a_1dx^1+a_2dx^2+a_3dx^3$ from the components of $X$ (which is rather easy for orthogonal coordinates). This also gives you an explicit formula for $d\omega_X$.

It is also easy to see that the formula you give defines a two-form $\theta_Y$ for any vector field $Y$, and again you can compute its coefficients $b_{ij}$ in an expansion as $b_{12}dx^1\wedge dx^2+b_{13}dx^1\wedge dx^3+b_{23}dx^2\wedge dx^3$ explicitly in terms of the coefficients of $Y$. Then you can compute the coefficients of $rot X$ from the equation $d\omega_X=\theta_{rot X}$.

Andreas Cap
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