Prueba $dy_1 \, dy_2=|J|\,dx_1 \, dx_2$ , donde $|J|$ es el determinante del jacobiano.

Question

Supongamos, $y_1=y_1(x_1,x_2)$ y $y_2=y_2(x_1,x_2)$ , tal que,

$$dy_1=\frac{\partial y_1}{\partial x_1}dx_1+\frac{\partial y_1}{\partial x_2} \, dx_2$$

$$dy_2=\frac{\partial y_2}{\partial x_1}dx_1+\frac{\partial y_2}{\partial x_1} \, dx_2$$

Entonces, he tomado el producto de los dos anteriores, pero no puedo llegar al resultado.

Answer 1

Dejemos que $y^1 = y^1(x^1,x^2)$ y $y^2 = y^2(x^1,x^2)$ . Supongamos también que;

$$\\$$

$$dy^1=\frac{\partial y^1}{\partial x^1}dx^1+\frac{\partial y_1}{\partial x^2}dx^2$$

$$dy^2=\frac{\partial y^2}{\partial x^1}dx^1+\frac{\partial y^2}{\partial x^1}dx^2$$

Entonces tenemos;

$$dy^1dy^2 = dy^1 \wedge dy^2$ = \left ( \frac { \partial y^1}{ \partial x^1}dx^1+ \frac { \partial y^1}{ \partial x^2}dx^2 \right ) \wedge \left ( \frac { \partial y^2}{ \partial x^1}dx^1+ \frac { \partial y^2}{ \partial x^1}dx^2 \right )$$

Lo anterior da lo siguiente;

$$\\$$

$$\left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}\right) dx^1\wedge dx^1 + \left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}\right) dx^1\wedge dx^2 + \left(\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1}\right) dx^2\wedge dx^1 + \left(\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1}\right) dx^2\wedge dx^2$$

$$\\$$

Recordemos que $dx^i \wedge dx^i = 0, dx^j \wedge dx^i = -dx^i \wedge dx^j$ y así lo has hecho;

$$\\$$

$$\left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}\right) dx^1\wedge dx^2 + \left(\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1}\right) dx^2\wedge dx^1 = \underbrace{ \left(\frac{\partial y^1}{\partial x^1} \frac{\partial y^2}{\partial x^1}-\frac{\partial y^1}{\partial x^2} \frac{\partial y^2}{\partial x^1} \right)}_{\textbf{Jacobian}} \ dx^1\wedge dx^2$$

Answer 2

No necesita el producto habitual, sino el exterior producto de los dos (véase Arnol'd Métodos matemáticos de la mecánica clásica ): $$dy_{1} \wedge dy_{2} = \left({\partial y_{1} \over \partial x_{1}} {\partial y_{2} \over \partial x_{2}} - {\partial y_{1} \over \partial x_{2}} {\partial y_{2} \over \partial x_{1}} \right) dx_{1} \wedge dx_{2}.$$ La expresión en () es su $J$ .