# Why does the gradient vector change with coordinate transformation?

I know that a vector is invariant under coordinate transformation, that is:

$$\vec{v} = v^i e_i = \tilde{v}^i\tilde{e_i}$$

where here, I'm implying einstein's summation convention.

When I transform the gradient vector however, it does not follow that rule. For example:

$$\vec\nabla f = e_i g^{ij} \frac{\partial f}{\partial x^j} = e_r\frac{\partial f}{\partial r} + e_\theta\frac{1}{r}\frac{\partial f}{\partial \theta} \neq g^{rr}e_r\frac{\partial f}{\partial r} + g^{\theta \theta}e_\theta\frac{\partial f}{\partial \theta}$$

Why is this? Is this because I'm not using the covariant derivative? I'm not well versed in using it but I suspect the problem is because the partial derivatives do not transform like a vector. In other words, is this correct for all my choice of basis?:

$$\vec \nabla f = e_i g^{ij}D_jf$$

• people should understand that $\vec\nabla$ isn't a vector neither a covector: it is a operator Commented Dec 21, 2022 at 16:33
• $\vec\nabla f$ is a vector Commented Dec 21, 2022 at 16:37
• No, it is a covector field Commented Dec 21, 2022 at 16:38
• You're treating $\vec \nabla f$ as a covector quantity even though it has no indices. What you're referring to as covector field is $\partial_i f$, at least for cartesian coordinates. Commented Dec 21, 2022 at 16:40
• What I am suspecting is that the covector field is $D_i f$, which transforms as a covector. Commented Dec 21, 2022 at 16:41

## 1 Answer

$$\newcommand\R{\mathbb R} \newcommand\PD[2]{\frac{\partial#1}{\partial#2}}$$You're using the wrong basis vectors; you have to use those that are directly associated with the coordinates.

We need to distinguish between points, vectors, and coordinates; let $$M$$ be our space of points, $$V$$ our space of vectors, and $$C$$ our space of coordinates. All of these happen to be modeled as $$\R^n$$, but that does not mean they are literally the same thing. If our manifold $$M$$ were not flat, then instead of $$V$$ we would need a tangent space $$T_pM$$ for each $$p \in M$$. A choice of coordinate space $$C$$ consists of two things: a point function $$P : C \to M$$ taking coordinates $$(x^1,\dotsc,x^n)$$ to the corresponding point $$P(x^1,\dotsc, x^n)$$, and coordinate functions $$X^i : M \to C$$ taking a point $$p$$ to its $$i^{\text{th}}$$ coordinate $$X^i(p) = x^i$$. These are inverses of each other: $$P(X^1(p), \dotsc, X^n(p)) = p,\quad X^i(P(x^1,\dotsc, x^n)) = x^i.$$

The coordinate basis of $$V$$ is then defined as $$e_i = \PD{P(x^1,\dotsc,x^n)}{x^i}.$$ Derivatives of $$M$$-valued functions are always valued in $$V$$. Note that $$e_i$$ is actually a function $$C \to V$$, but we will usually not write this coordinate dependence explicitly. Using the coordinate functions $$X^i$$, these can also be viewed as function $$M \to V$$.

If we also have a metric on $$V$$, in our case the usual dot product, then there is a reciprocal basis $$e^i$$ uniquely defined such that $$e^i\cdot e_j = \delta^i_j$$. Since $$v\cdot\nabla$$ ought to be the derivative in the $$v$$ direction, we see that $$e_i\cdot\nabla X^j = \PD{X^j}{x^i} = \delta_i^j$$ so we must have $$\nabla X^j = e^j$$. This gives us the coordinate expression for $$\nabla$$: $$\nabla = \sum_{i=1}^ne^i\PD{}{x^i}.$$

Now let's apply this to the specific case where $$n = 2$$, $$M$$ is modeled with Cartesian coordinates $$C_{xy}$$ so that $$C_{xy} = M$$, and we are interested in polar coordinates $$C_{r\theta}$$. The point function $$P : C_{r\theta} \to M$$ is $$P(r, \theta) = (r\cos\theta, r\sin\theta),$$ and the coordinate functions $$R : M \to C_{r\theta}$$ and $$\Theta : M \to C_{r\theta}$$ are $$R(x, y) = \sqrt{x^2 + y^2},\quad \Theta(x, y) = \arctan\frac yx.$$ (Or something like that, I am not being particularly careful.) The coordinate basis is $$e_r = (\cos\theta)e_x + (\sin\theta)e_y,\quad e_\theta = -(r\sin\theta)e_x + (r\cos\theta)e_y.$$ Notice that $$e_\theta$$ is not a unit vector. If $$\hat e_\theta = e_\theta/|e_\theta|$$ then $$\hat e_\theta = e_\theta/r = -(\sin\theta)e_x + (\cos\theta)e_y.$$ We can then compute the reciprocal basis to be $$e^r = e_r,\quad e^\theta = e_\theta/r^2 = \hat e_\theta/r.$$ Hence the gradient expressed in polar coordinates is $$\nabla = e^r\PD{}r + e^\theta\PD{}\theta = e_r\PD{}r + e_\theta\frac1{r^2}\PD{}\theta = e_r\PD{}r + \hat e_\theta\frac1r\PD{}\theta.$$

The short of it is: $$\{e_r, \hat e_\theta\}$$ is the wrong basis to transform; it needs to be $$\{e_r, e_\theta\} = \{e_r, r\hat e_\theta\}$$.