# Equality of mixed partial derivatives from different coordinates

Suppose I have some coordinates $$𝑞_𝑖 (𝑖=1,2,...,𝑛)$$ and consider an invertible transformation thereof to some other set of coordinates, $$𝑥_𝑗=𝑥_𝑗(𝑞_1,...,𝑞_𝑛), (𝑗=1,2,...𝑛)$$ (and assume the requisite smoothness of the transformation).

Suppose I want to take the derivative $$\left( \frac{\partial}{\partial q_i}\left(\frac{\partial q_k}{\partial x_j} \right)_x \right)_q$$

where the subscripts are meant to show what is being held constant (every other $$q_m$$ or $$x_l$$). May I use the "equality of mixed partials" to obtain $$\left( \frac{\partial}{\partial q_i}\left(\frac{\partial q_k}{\partial x_j} \right)_x \right)_q=\left( \frac{\partial}{\partial x_j}\left(\frac{\partial q_k}{\partial q_i} \right)_q \right)_x=0$$.

if $$i\neq j$$?

I study physics so my math is typically of that hand-waving sort, but I am uncomfortable with this particular process since typically we apply the theorem to partial derivatives taken amongst the same set of independent coordinates? Note again that you may assume the transformation is as well-behaved as required.

I am using this in solving Exercise 2.7.8 of Shankar's quantum mechanics textbook, if that helps anyone who has the text.

No, you can't swap because they're different sets of coordinates. If you have a function $$f$$, then $$\frac{\partial f}{\partial q^i}=\sum\limits_{l=1}^n\frac{\partial f}{\partial x^l}\frac{\partial x^l}{\partial q^i}$$. Applying this to the function $$f=\frac{\partial q^k}{\partial x^j}$$, you get $$\frac{\partial}{\partial q^i}\left(\frac{\partial q^k}{\partial x^j}\right)=\sum\limits_{l=1}^n\frac{\partial^2q^k}{\partial x^l\partial x^j}\frac{\partial x^l}{\partial q^i}$$.

You should think of this as having a diffeomorphism $$\phi:U\to V$$ where $$U,V\subset\Bbb{R}^n$$ are open sets (diffeomorphism means $$\phi$$ is smooth and $$\phi^{-1}$$ exists and is smooth). Then, we denote a general point in $$U$$ as $$x$$, and a general point in $$V$$ as $$q$$, so $$\phi$$ maps '$$x$$-coordinates to $$q$$-coordinates'. The symbol $$\frac{\partial q^k}{\partial x^j}$$ then denotes the $$(k,j)$$ entry of the Jacobian matrix $$\phi'$$, i.e at a point $$a\in U$$, the number $$\frac{\partial q^k}{\partial x^j}\bigg|_a$$ denotes the number which resides in the $$(k,j)$$-entry of the matrix $$\phi'(a)$$. Likewise, when we do $$\frac{\partial x^l}{\partial q^i}$$, it means the $$(l,i)$$ entry of the Jacobian matrix of the inverse map $$(\phi^{-1})'$$. So, the above computation can also be written as \begin{align} D_i\left((D_j\phi^k)\circ \phi^{-1}\right)=\sum\limits_{l=1}^n[(D_lD_j\phi^k)\circ \phi^{-1}]\cdot D_i(\phi^{-1})^l. \end{align} (The composition by $$\phi^{-1}$$ is because we have to express $$D_j\phi^k\equiv\frac{\partial q^k}{\partial x^j}$$ first 'as a function of $$q$$' so that we can apply $$\frac{\partial}{\partial q^i}$$ to it).

Now, regarding the exercise, where one has to show the 'cancellation of dots' property, I think it's much clearer if the problem is phrased as follows. Let $$\phi:U\to V$$ be a diffeomorphism as above. Now, consider the following function, $$\Phi:U\times\Bbb{R}^n\to V\times\Bbb{R}^n$$, defined as $$\Phi(x,v)=(\phi(x), \phi'(x)\cdot v)$$; here $$\phi'(x)\cdot v$$ means the product of the matrix $$\phi'(x)$$ with the column vector $$v$$. In classical notation, we say that we have a coordinate transformation $$x\mapsto q(x)$$, which induces a mapping on the space of positions and velocities (more technically called the tangent bundle), $$(x,\dot{x})\mapsto (q,\dot{q})$$; this is what the map $$\Phi$$ does.

Then, you can easily convince yourself that the Jacobian matrix of $$\Phi$$ has the following block structure: \begin{align} \Phi'(x,v)&= \begin{pmatrix} \phi'(x)&0_{n\times n}\\ \star & \phi'(x) \end{pmatrix}, \end{align} where $$\star$$ denotes some irrelevant (for our purposes) $$n\times n$$ matrix. So, the 'cancellation of dots' property corresponds to the fact that the top left block and the bottom right block are both $$\phi'(x)$$ (the top left is the matrix $$\frac{\partial q}{\partial x}$$, and the bottom right block is the matrix $$\frac{\partial\dot{q}}{\partial\dot{x}}$$).

• This answer is at a bit too high of a level mathematically for me. I might need to read Loomis and Sternberg before understanding it :) At any rate, are you saying that it's indeed possible to show that $\frac {\partial \overline{p_j}}{\partial q_i}=0$ in the problem which seems to be required? That is, some correct way other than what I've done?
– EE18
Jun 16, 2022 at 21:24
• @EE18 which part of the problem are you referring to? Jun 16, 2022 at 22:33
• We need to show that the PB $\{\overline{p_i},\overline{p_j}\}=0$ as part of the proof that a point transformation is canonical, which necessitates showing the derivative above is zero.
– EE18
Jun 16, 2022 at 22:49