TITLE: Difference between gradient and Jacobian
QUESTION [93 upvotes]: Could anyone explain in simple words (and maybe with an example) what the difference between the gradient and the Jacobian is? 
The gradient is a vector with the partial derivatives, right?

REPLY [2 votes]: The gradient in a general coordinate system depends on the metric tensor but the Jacobian matrix consists of only the partial derivatives.
The gradient of a vector field is given by:
$\nabla\mathbf{f}=g^{jk}\frac{\partial f^{i}}{\partial x^{j}}\mathbf{e}_{i}\otimes\mathbf{e}_{j}$,
where the Einstein summation notation is implied, $g^{jk}$ are the metric tensor elements evaluated from the Jacobian matrix consisting of the partial derivatives of the coordinate transformation from the Cartesian coordinate system. In Cartesian coordinate system, this equals to exactly the transpose of the Jacobian matrix, which is given by, regardless of the metric tensor,
$\left\{ J\mathbf{f}\right\} _{i,j}=\frac{\partial f^{i}}{\partial x^{j}}$.
For example, for the spherical coordinate system with the coordinate $\mathbf z$, we have
$x_0=z_0\sin z_1\sin z_2,\ x_1=z_0 \cos z_1 \sin z_2,\ x_2=z_0\cos z_2$, where $\mathbf x$ is the coordinate in the Cartesian coordinate system.
Given a vector function $f$, Each column of gradient is given by
$\left\{ \nabla\mathbf{f}\right\} _{:,i}=\frac{\partial f^{i}}{\partial\mathbf{z}_{0}}\hat{\mathbf{z}}_{0}+\frac{1}{\mathbf{z}_{0}}\frac{\partial f^{i}}{\partial\mathbf{z}_{1}}\hat{\mathbf{z}}_{1}+\frac{1}{\mathbf{z}_{0}\sin\mathbf{z}_{2}}\frac{\partial f^{i}}{\partial\mathbf{z}_{2}}\hat{\mathbf{z}}_{2}$
or
$\begin{bmatrix}\frac{\partial f^{i}}{\partial\mathbf{z}_{0}} & \frac{1}{\mathbf{z}_{0}}\frac{\partial f^{i}}{\partial\mathbf{z}_{1}} & \frac{1}{\mathbf{z}_{0}\sin\mathbf{z}_{2}}\frac{\partial f^{i}}{\partial\mathbf{z}_{2}}\end{bmatrix}^\mathrm{T}$
but the each row of Jacobian is still given by
${\left\{ J\mathbf{f}\right\} _{i,:}}=\begin{bmatrix}\frac{\partial f^{i}}{\partial\mathbf{z}_{0}} & \frac{\partial f^{i}}{\partial\mathbf{z}_{1}} & \frac{\partial f^{i}}{\partial\mathbf{z}_{2}}\end{bmatrix}$
The reason for this is because the Jacobian matrix is applied to solve integrals by substitution where the determinant of the Jacobian matrix is needed. It is also used to transform partial derivatives into partial derivatives of another coordinate system. Another application is to evaluate the metric tensor as mentioned before.
PS: I hope this answer doesn't get deleted because I elaborated the one liner I initially had but probably was not very convincing.