I am having problems finding the Jacobian matrix of a transformation in a exercise assigned by my professor. I was wondering if anyone could help me solve this problem:
Suppose a contravariant tensor $A^i$ in space $\mathbb{R^2}$ where $A^i=(q_{2},q_{1})$ in $q_{i}$ coordinates. Calculate $\overline{A^i}$ in $\overline{q_{i}}$ under the transformation $\overline{q_{1}}=(q_{2})^2$ and $\overline{q_{2}}=q_{1}q_{2}$. Obtain the Jacobian matrix of the transformation.
My main problem is that I am not sure how to apply the Jacobian to this problem. I am confused about how to write the equation of $\overline{A^i}$ so that I may take the partial derivatives.
Any help is appreciated. Thanks.
Hello MathLearner and welcome to math here at stackexchange!
A contravariant vector is a type (1,0)-tensor. Using any of the standard setups one can show that the components, $A^j$, of this tensor transform according to the law
$$\bar{A}^i=\frac{\partial \bar{q}^i}{\partial q^j } A^j\tag{1}$$
Sometimes this is even used as the defining property of the quantities $A^j$, especially in older literature.
Clearly the values of $q^j$ and $\bar{q}^j$ must somehow be related. Usually a compressed notation is used:
$$\bar{q}^j=\bar{q}^j(q^h)\tag{2}$$ with inverse $$q^h=q^h(\bar{q}^j)\tag{3}$$
In your specific case you have $$\begin{align}\bar{q}^1&=(q^2)^2\\ \bar{q}^2&=q^1q^2\end{align}$$
$$\frac{\partial \bar{q}^i}{\partial q^j }=\begin{pmatrix}\frac{\partial \bar{q}^1}{\partial q^1} & \frac{\partial \bar{q}^1}{\partial q^2}\\ \frac{\partial \bar{q}^2}{\partial q^1} & \frac{\partial \bar{q}^2}{\partial q^2} \end{pmatrix}=\begin{pmatrix}0&2q^2\\q^2 & q^1\end{pmatrix}$$
The Jacobian of the transformation $(2)$ is
$$J=\frac{\partial(\bar{q}^1,\dots,\bar{q}^n)}{\partial(q^1,\dots,q^n)}=\det\left(\frac{\partial \bar{q}^i}{\partial q^j}\right)=-2(q^2)^2$$
With $A^j=(q^2,q^1$)
$$\bar{A}^i=\frac{\partial \bar{q}^i}{\partial q^j } A^j=\left(2q^1 q^2, (q^1)^2 + (q^2)^2\right)$$
Note that the question wants you to express this in $\bar{q}^i$ rather than in $q^j$ which is left as an exercise.