how to prove gradients vectors are the same in polar and cartesian coordinates.

Question

Suppose $T=T(r,\theta)=G(x,y)$

How do you prove $\nabla T(r,\theta)=\nabla G(x,y)$?

I can think of some arguments in favor of this equality, but I want an actual proof or a very good intuitive argument. My arguments in favor go something like this:

-Gradient vectors should be the same because if my directional derivative is taken parallel to the gradient vector then I get its maximum/minimum value and if these two gradient vectors are the same then everything will be consistent .

Thanks.

Answer 1

The intuitive explanation is:

the gradient is function on the set of scalar fields, and it has an existence independent of coordinate systems.

It's the same as defining a linear transformation using a matrix, but then later realizing that the transformation exists independently of the basis you originally used to make the matrix.

Geometrically, the direction of greatest increase of a scalar field is not influenced by coordinates. Each different coordinate system is just a different frame of reference for the same object.

---

Here's an example: take $f(x,y)=x^2+y^2=r^2$ in the plane. The gradient in $(x,y)$ coordinates is $(2x,2y)$, and the gradient in $(\theta,r)$ is $(0,2r)$.

Let's test points on the unit circle. The coordinate system tangent to the coordinate lines through each point is rectilinear just like the old coordinate system, and the gradient arrows point radially away from the origin. This makes sense because the square of the distance is increasing most quickly as you move radially away from the origin.

The coordinate system tangent to the coordinate lines in $(\theta,r)$ are no longer horizontal and vertical since we're using a curvilinear coordinate system. The tangent coordinate systems along the unit circle have an "x-axis" tangent to the circle, and the "y-axis" perpendicular to the other axis. In these coordinates, the gradient $(0,2r)$ is pointing "straight up" in these tangential coordinate systems, which is the same as radially outward.

Answer 2

The gradient of a function $f:\mathbb{R}^n\to\mathbb{R}$ at a point $x=(x_1,\ldots,x_n)$ is defined to be the unique vector $v\in\mathbb{R}^n$that satisfies: $<v,u>=Df_x(u)$ for all $u\in\mathbb{R}^n$, where $Df_x(u)$ denotes the directional derivative of $f$ at $x$ in the direction $u$.

Note that the gradient is indeed well defined, since $Df_x$ is a linear functional on $\mathbb{R}^n$, and the inner product induces an isomorphism between $\mathbb{R}^n$ and ${\mathbb{R}^n}^*$.

Also note that this definition is intrinsic, meaning that it does not depend on coordinates in any way, as the inner product doesn't. Hence the gradient's direction as well as magnitude are well defined.