# Why is this step in proving that a partial derivative of a vector is not a tensor illegal?

If I want to prove that the derivative of a vector $$V$$, $$\partial_\mu V^\nu$$, is not a tensor, I can apply the tensor transformation rule to $$V^\nu$$:

$$\partial_\mu V^\nu \rightarrow \partial_{\mu\prime} V'^{\nu\prime} = \partial_{\mu\prime}\left( \frac{\partial x'^{\nu'}}{\partial x^\nu}V^\nu\right)$$

Now the next correct step would be to apply the chain rule to $$\partial_{\mu'}=\frac{\partial x^\mu}{\partial x'^{\mu'}}\frac{\partial}{\partial x^\mu}$$ and then expand. Everything from this point is clear to me.

But my course notes mention that it would not be correct to differentiate the expression in the brackets with $$\partial_{\mu\prime}$$ right away, but I do not understand why.

Why would that step be illegal? Why did we have to apply the chain rule first?

• and what if it is a tensor, indeed? Nov 8, 2022 at 14:31
• apologies, i should have said that V is a vector, I will edit it in Nov 8, 2022 at 14:32
• this is not the way you transform tensors. An answer will follow soon Nov 8, 2022 at 14:35

The shorthand notation is tripping you up. Let's say that in a particular coordinate chart $$x$$, your vector components take the form $$V^\mu (x)$$. You now decide to differentiate these components, to obtain an expression $$\frac{\partial}{\partial x^\nu}\big[ V^\mu(x)\big]$$

Now you wish to change to a different coordinate chart $$y$$. The components of $$V$$ transform as $$V^\mu(x) \mapsto J^\mu_{\ \ \alpha}(y) \hat V^\alpha(y)$$ where $$J^\mu_{\ \ \alpha}(y)\equiv \frac{\partial x^\mu (y)}{\partial y^\alpha}$$ is the Jacobian of the transformation. Plugging that in yields

$$\frac{\partial}{\partial x^\nu}\left[J^\mu_{\alpha}(y)\hat V^\alpha(y)\right]$$

The problem now is that the quantity inside the brackets is a function of the coordinates $$y$$, so you cannot differentiate them with respect to the coordinates $$x$$. Instead, you need to transform your derivative operator: $$\frac{\partial}{\partial x^\nu} \mapsto \big(J^{-1}(y)\big)^\beta_{\ \ \nu} \frac{\partial}{\partial y^\beta}$$

Now you can expand everything out using the product rule. To recover the shorthand notation, you can drop the arguments $$(x)$$ and $$(y)$$ from all of the functions, write the Jacobian as $$\partial x/\partial y$$, and replace $$y$$ with $$x'$$. From there, replace $$\frac{\partial}{\partial x^\mu}$$ with $$\partial_\mu$$ and $$\frac{\partial}{\partial x'^\mu}$$ with $$\partial_{\mu'}$$. (Can you see how this gets confusing for beginners?)

As an example, we can transform the vector field $$\mathbf V = \frac{x}{\sqrt{x^2+y^2}}\frac{\partial}{\partial x}+\frac{y}{\sqrt{x^2+y^2}}\frac{\partial}{\partial y}$$ from Cartesian coordinates $$(x,y)$$ to polar coordinates $$(r,\theta)$$. Clearly this is simply equal to $$\frac{\partial}{\partial r}$$, so this is what we should expect to find at the end.

Because $$x(r,\theta)=r\cos(\theta)$$ and $$y(r,\theta) = r\sin(\theta)$$, the Jacobian and its inverse take the form $$J(r,\theta) = \pmatrix{\cos(\theta) & -r\sin(\theta)\\ \sin(\theta) & r\cos(\theta)} \qquad J^{-1}(r,\theta) = \pmatrix{\cos(\theta) & \sin(\theta)\\ -\frac{1}{r}\sin(\theta) & \frac{1}{r}\cos(\theta)}$$

The initial Cartesian components are $$V^x(x,y) = \frac{x}{\sqrt{x^2+y^2}}$$ and $$V^y(x,y) = \frac{y}{\sqrt{x^2+y^2}}$$. Expressed in terms of $$r$$ and $$\theta$$, we have that $$V^x = \cos(\theta)$$ and $$V^y = \sin(\theta)$$. According to the transformation rule,

$$\pmatrix{V^x\\V^y} = J(r,\theta) \pmatrix{\hat V^r\\ \hat V^\theta} \implies \pmatrix{\hat V^r\\ \hat V^\theta} = J^{-1}(r,\theta) \pmatrix{\cos(\theta)\\\sin(\theta)} = \pmatrix{1\\0}$$ And so $$\hat V^r(r,\theta) = 1$$ and $$\hat V^\theta(r,\theta) = 0$$, as anticipated.

In this language, the question in the OP relates to expressions of the form

$$\frac{\partial}{\partial x} \pmatrix{V^x(x,y)\\ V^y(x,y)} = \frac{\partial}{\partial x} \left[ J(r,\theta) \pmatrix{\hat V^r(r,\theta)\\\hat V^\theta(r,\theta)}\right]$$

and the fact that it is inappropriate to differentiate the RHS before transforming the derivative operator.

• "The problem now is that the quantity inside the brackets is a function of the coordinates y, so you cannot differentiate them with respect to the coordinates x." Ah I think I see. I missed that the Jacobian is actually a function of x, not x'. Nov 8, 2022 at 15:43

What you write $$A_{\mu}^{\nu} = \partial_{\mu} V^{\nu}$$ is the component representation of the tensor

$$\mathbb{A} = \mathbf{b}^{\mu}\otimes\mathbf{b}_{\nu} A_{\mu}^{\nu} = \mathbf{b}^{\mu}\otimes\mathbf{b}_{\nu} \partial_{\mu} V^{\nu}$$,

assuming $$\mathbf{b}_{\nu}$$ as the base vectors you're referring to, and $$\mathbf{b}^{\mu}$$ the vectors of the reciprocal basis, s.t. $$\mathbf{b}_{\nu} \cdot \mathbf{b}^{\mu} = \delta^{\mu}_{\nu}$$.

We can write the components of your tensor in a different basis, whose vectors $$\mathbf{b}'_{\alpha}$$ can be as a linear combination of the vectors of the original basis, using the

$$\mathbf{b}'_{\alpha} = T_{a}^{\nu} \mathbf{b}_{\nu}\qquad , \qquad\mathbf{b}'^{\beta} = \tilde{T}^{\beta}_{\mu} \mathbf{b}^{\mu}$$
$$\mathbf{b}_{\nu} = \tilde{T}^{a}_{\nu} \mathbf{b}'_{\alpha}\qquad , \qquad\mathbf{b}^{\mu} = {T}_{\beta}^{\mu} \mathbf{b}'^{\beta}$$ ,

being $$\tilde{T}^{\beta}_{\mu}$$ the inverse matrix $$T_{a}^{\nu}$$. Replacing the expression of the vectors of the basis in the expression of the tensor $$\mathbb{A}$$ we get

$$\mathbb{A} = \mathbf{b}^{\mu}\otimes\mathbf{b}_{\nu} A_{\mu}^{\nu} = \mathbf{b}'^{\beta}\otimes\mathbf{b}'_{\alpha} \underbrace{ {T}_{\beta}^{\mu} \tilde{T}^{a}_{\nu} A_{\mu}^{\nu}}_{{A'_{\beta}}^{\alpha}} = \mathbf{b}'^{\beta}\otimes\mathbf{b}'_{\alpha} {A'_{\beta}}^{\alpha}$$

• I am sorry I do not understand. My course notes and everywhere else I looked at say that the partial derivative of a vector is not a tensor. I am only familiar with the component representation of tensors, and I've never seen the 'cross in a circle' operation that you used here, what is it? Nov 8, 2022 at 15:02
• @NX37B The cross in a circle is a tensor product. Nov 8, 2022 at 18:12
• @NX37B. Technically any linear combination of the basis vectors, or tensor products of the basis vectors, is a tensor. However the differential of a vector field is not a “good” tensor in that $\partial_j v^i \pmb{e}_i \otimes \pmb{e}^j \neq \partial’_j v’^i \pmb{e}’_i \otimes \pmb{e}’^j$. Nov 17, 2022 at 4:26