# Derivation of Jacobian Determinant

1. Sep 5, 2009

### alt20

Hi,

I'm having some problems with the derivation of the Jacobian determinant when used to describe co-ordinate transformations. As I understand it, the Jacobian determinant should relate the areas defined by two vectors in both co-ordinate systems. As the vectors are not necessarily perpendicular, the area is calculated using the cross-product, giving:

| dx x dy | = J dudv

(Examples of the derivation can be found http://books.google.co.uk/books?id=...obian determinant area cross product&f=false".)

My problem is that dudv is not a cross product and so doesn't describe the area of a parallelogram in the u-v co-ordinate system. So, as far as I can see it, one of three things is happening:

1) du and dv are assumed to be perpendicular, and so the area is just the product of the sides of the rectange, dudv.

2) du and dv are assumed to be very small, so that the area approximates a rectangle

3) there's something funny about dudv that I haven't spotted - it *is* the product of two vectors, which doesn't actually mean anything I guess. Someone mentioned something about it being a wedge or exterior product...

So yeah, any ideas? What am I missing?

Thanks

James

Last edited by a moderator: Apr 24, 2017
2. Sep 5, 2009

### arildno

You have some severe misunderstandings, here!

Let $$\vec{i},\vec{j}$$ be unit vectors in the x,y directions, and let $\vec{u},\vec{v}[/tex] be unit vectors in u,v-directions. Thus, we have that: $$d\vec{x}=dx\vec{i}=\frac{\partial{x}}{\partial{u}}du\vec{u}+\frac{\partial{x}}{\partial{v}}dv\vec{v}$$ $$d\vec{y}=dy\vec{j}=\frac{\partial{y}}{\partial{u}}du\vec{u}+\frac{\partial{y}}{\partial{v}}dv\vec{v}$$ Compute the length of the vector $$d\vec{x}\times{d}\vec{y}$ and see what you get! 3. Sep 5, 2009 ### alt20 Okey dokey. Adding a third dimension to allow the cross product to be calculated: [tex]d\vec{x} \times d\vec{y} = \left[\left(\frac{\partial x}{\partial v}dv\right)(0)-(0)\left(\frac{\partial y}{\partial v}dv\right)\right] \vec{i} + \left[(0)\left(\frac{\partial y}{\partial u}du\right)-(0)\left(\frac{\partial x}{\partial u}du\right)\right] \vec{j} + \left[\left(\frac{\partial x}{\partial u}du\right)\left(\frac{\partial y}{\partial v}dv\right) - \left(\frac{\partial x}{\partial v}dv\right)\left(\frac{\partial y}{\partial u}du\right)\right] \vec{k}$$

$$= \left[\left(\frac{\partial x}{\partial u}du\right)\left(\frac{\partial y}{\partial v}dv\right) - \left(\frac{\partial x}{\partial v}dv\right)\left(\frac{\partial y}{\partial u}du\right)\right] \vec{k}$$

Taking the absolute value of the vector:

$$\left(\frac{\partial x}{\partial u}du\right)\left(\frac{\partial y}{\partial v}dv\right) - \left(\frac{\partial x}{\partial v}dv\right)\left(\frac{\partial y}{\partial u}du\right)$$

$$= \left[\left(\frac{\partial x}{\partial u}\right)\left(\frac{\partial y}{\partial v}\right) - \left(\frac{\partial x}{\partial v}\right)\left(\frac{\partial y}{\partial u}\right)\right]dudv$$

$$= J dudv$$

Ah I see, $$du$$ and $$dv$$ aren't vectors, they're just magnitudes, right? Is that what you mean?

4. Sep 5, 2009

### arildno

QUITE SO!

The AREA dx*dy equals the AREA J*du*dv.

5. Sep 5, 2009

### alt20

Ok, but surely that still assumes that $$\vec{u}$$ and $$\vec{v}$$ are perpendicular?