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In basic courses on analysis, one learns that the Jacobian determinant represents the change of differential volume under a coordinate transformation. For example in 2 dimensions:

If

$$u = f_1(x,y) \\ v = f_2(x,y)$$

our Jacobian matrix would be

$$J = \left( \begin{array}{cc} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{array}\right)$$

and it's determinant

$$|J| = \frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} - \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}$$ should now be the ratio of the infinitesimal areas $dudv = J dxdy$.

We get the same result if we form the differentials $$du = \frac{\partial f_1}{\partial x} dx + \frac{\partial f_1}{\partial y} dy \\ dv = \frac{\partial f_2}{\partial x} dx + \frac{\partial f_2}{\partial y} dy$$ interpret the $d$-terms as forms and multiply these equations with the wedge product.

$$ du \wedge dv = \frac{\partial f_1}{\partial x}\frac{\partial f_2}{\partial x} dx \wedge dx + \frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} dx \wedge dy + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x} dy \wedge dx + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial y} dy \wedge dy $$ Using the properties of the wedge product:

$$ du \wedge dv = \left(\frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} - \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}\right) dx \wedge dy $$

My question now is: Why does the "naive" approach lead to a wrong result:

$$ dudv = \frac{\partial f_1}{\partial x}\frac{\partial f_2}{\partial x} dx^2 + \left(\frac{\partial f_1}{\partial x} \frac{\partial f_2}{\partial y} + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial x}\right) dxdy + \frac{\partial f_1}{\partial y}\frac{\partial f_2}{\partial y} dy^2 $$

Why do I need forms (with the concept of orientation) to get the correct result? My math classes have always treated differentials as a quick and "dirty" way to cut short calculations, but are there any texts that provide a consistent and rigorous treatment of differentials, and their interpretation as forms?

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    1. I'm not sure how this is a physics question rather than one for [math.SE]. 2. You are effectively asking "why do I need to be rigorous to get the correct result", since without notion of differential forms and wedge products, $\mathrm{d}u\mathrm{d}v$ is just notation without any formal meaning. Why do you expect the "quick and dirty" manipulations to yield the "correct" result? – 2017-02-02
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    @ACuriousMind 1. I agree. 2. I guess OP means $\mathrm du\mathrm dv\equiv\mathrm du\otimes\mathrm dv$, which is well defined as a tensor (but not a differential form). – 2017-02-02

1 Answers 1

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In essence, the problem is that the unit volume spanned by $\mathrm du$ and $\mathrm dv$ does not need to be a rectangular parallelogram, in which case its volume is obviously no longer given by the product of the lengths of the sides.

Thus this is an issue with the concept of volume itself, and it will remain even if your transformation $(x,y)\mapsto(u,v)$ is a general linear transformation. The outer product of covectors was built to capture the concept of volume in this context, and the wedge product of one-forms is its transparent generalization to the case of curvilinear coordinates.