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If we have a (non-singular) coordinate transformation from the coordinates $\{x^i\}$ to the coordinates $\{x^{\prime i}\}$, what is the relation between the partial derivatives $\frac{\partial x^{\prime j}}{\partial x^i}$ and $\frac{\partial x^i}{\partial x^{\prime j}}$?

Or where can I get more information about this?

Thanks for reading!

Larara
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  • Is this what you are looking for: {∑i}{∂(x′^j)/∂(x^i)}{∂(x^i)/∂(x′^h)}=δ(jh)? – Urgje May 28 '14 at 21:18
  • Not exactly... For example, if ${x^i}={x,y}$ are cartesian coordinates and ${x^{\prime i}}={r,\theta }$ are polar coordinates, we have $\frac{\partial r}{\partial x}=\frac{\partial x}{\partial r}=\cos(\theta)$... – Larara May 28 '14 at 21:23

1 Answers1

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I assume that as a coordinate transformation you understand a $ C^1$-Diffeomorphism

$$ \phi: U \rightarrow V, \{x^i\} \mapsto \{x^{'i}\}, $$

where $U,V$ are open subsets of $\mathbb{R}^n$. Then $\phi$ has per definition a differentiable inverse and we get

\begin{align*} \phi \circ \phi^{-1} & = id_V\\ \phi^{-1} \circ \phi & = id_U \end{align*}

Differentiating both sides and using the chain rule we yield \begin{align*} J\phi \circ J(\phi^{-1}) & = id_{\mathbb{R}^n} \\ J(\phi^{-1}) \circ J\phi & = id_{\mathbb{R}^n}, \end{align*} where $J\phi$ denotes the Jacobi-Matrix, i.e. $J \phi = \left ( \frac{\partial x^{'i}}{\partial x^{j}} \right )_{i,j} $ and $J (\phi^{-1}) = \left ( \frac{\partial{x^i}}{\partial x^{'j}} \right )_{i,j}$.

If you want to omit the Matrices you get the relation mentioned in the comments above, which is

$$ \sum_{k = 1}^n \frac{\partial x^{'i}}{\partial x^{k}} \cdot \frac{\partial x^{k}}{\partial x^{'j}} = \delta^i_j $$

EDIT:

$$ \frac{\partial r}{\partial x} = \frac{\partial \sqrt{x^2 + y^2}}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r} = \frac{r \cos{\phi}}{r} = \cos(\phi), $$

which shows, that your computation was correct. Sorry, for that.

Nevertheless in general you do not have the equation $\frac{\partial x}{\partial x'} = \frac{\partial x'}{\partial x}$. One counterexample is the following:

\begin{align*} x' & = 2\cdot x \\ y' & = 3\cdot y \end{align*}

This is a $C^1$-Diffeomorphism with partial derivatives:

\begin{align*} \frac{\partial x'}{\partial x} = 2 \neq \frac{1}{2} = \frac{\partial x}{\partial x'} \\ \frac{\partial y'}{\partial y} = 3 \neq \frac{1}{3} = \frac{\partial y}{\partial y'} \end{align*}

I hope I could help you.

Emrys-Merlin
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  • Thank you @Emrys-Merlin! In the last equation, if we substitute $x$ by $r\cos(\theta)$, we obtain $\frac{\partial r}{\partial x}=\cos(\theta)$ ... – Larara May 28 '14 at 22:30
  • Of course, you are totally right... Thank you very much^^ – Emrys-Merlin May 28 '14 at 22:31
  • So we have $\frac{\partial r}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial r}{\partial y}\frac{\partial y}{\partial r}=1$; but in more general cases we can't say that $\frac{\partial x^i}{\partial x^{\prime j}}=\frac{\partial x^{\prime j}}{\partial x^i}$ or $\frac{\partial x^i}{\partial x^{\prime j}}=(\frac{\partial x^{\prime j}}{\partial x^i})^{-1}$? (I mean, it's a coincidence that this works in polar coordinates?) – Larara May 28 '14 at 22:47
  • I added a counter example in my post. – Emrys-Merlin May 28 '14 at 23:00
  • Thanks for the counter example. I noted that, $\frac{\partial x^{\prime}}{\partial x}=(\frac{\partial x}{\partial x^{\prime}})^{-1}$ and the same for $y$ and $y^{\prime}$, but that's because $x^{\prime}$ depends on $x$ only (same for $y$ and $y^{\prime}$). Your answer was very enlightening, thanks! – Larara May 28 '14 at 23:09