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Suppose,$y_1=y_1(x_1,x_2)$ and $y_2=y_2(x_1,x_2)$, such that,

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

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

Then,I've taken the product of the above two,but unable to reach to the result.

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2 Answers 2

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Let $y^1 = y^1(x^1,x^2)$ and $y^2 = y^2(x^1,x^2)$. Suppose also that;

$$\\$$

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

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

Then we have;

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

The above gives the following;

$$\\$$

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

$$\\$$

Now recall that $dx^i \wedge dx^i = 0, dx^j \wedge dx^i = -dx^i \wedge dx^j$ and so you have;

$$\\$$

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

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  • $\begingroup$ This is the reason for when you are integrating and do a coordinate change, the Jacobian pops up after the parametrization in which you transition to the new (hopefully more simple) integral. $\endgroup$ Aug 25, 2016 at 18:17
  • $\begingroup$ You have some indices wrong in this. $\endgroup$ Aug 26, 2016 at 2:36
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You need not the usual product, but the exterior product of the two (see Arnol'd's Mathematical Methods of Classical Mechanics): $$ dy_{1} \wedge dy_{2} = \left({\partial y_{1} \over \partial x_{1}} {\partial y_{2} \over \partial x_{2}} - {\partial y_{1} \over \partial x_{2}} {\partial y_{2} \over \partial x_{1}} \right) dx_{1} \wedge dx_{2}. $$ The expression in ()'s is your $J$.

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  • $\begingroup$ The expression in ()'s is $J$, not $|J|.$ $\endgroup$
    – zhw.
    Aug 25, 2016 at 18:08
  • $\begingroup$ @zhw., thanks, corrected. $\endgroup$
    – avs
    Aug 25, 2016 at 20:57

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