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We did in lectures gradient of a scalar field and I am wondering how is the grad of a vector field. I tried the following Let

$$V = f(x,y,z)\hat{\imath} + g(x,y,z)\hat{\jmath} + h(x,y,z)\hat{k} \, .$$

Then by definition

$$\mathrm{grad}(V)= \hat{\imath} V_x + \hat{\jmath}V_y + \hat{k}V_z \, ,$$

where $V_x$, $V_y$, $V_z$ denote the partial derivatives of $V$ with respect to to $(x,y,z)$. But $i V_x = f(x,y,z)x$ since $\hat{\imath}\cdot\hat{\imath}=1$, $\hat{\imath}\cdot\hat{\jmath}=0$, and $\hat{\imath}\cdot\hat{k}=0$.

Similarly, $\hat{\jmath}V_y = g(x,y,z)y$ and $\hat{k}V_z=h(x,y,z)$. So we end up with

$$\mathrm{grad}(V) = f(x,y,z)x + g(x,y,z)y + h(x,y,z)z = \mathrm{div}(V) \, .$$

I guess that is wrong but what is wrong with the method.

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1 Answer 1

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The gradient of a vector field in ordinary $R^3$ is a tensor field. You will need to use the tensor product to represent it $\otimes$.

$$ \nabla V = \left( \hat{e}_i \frac{\partial}{\partial x_i} \right ) (V_m \hat{e}_m) \\ = \frac{\partial V_m}{\partial x_i} \hat{e}_i \otimes \hat{e}_m $$

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  • $\begingroup$ Basically, the matrix of partial derivatives of the components. $\endgroup$ Oct 30, 2014 at 16:13

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