# Why isn't there an extra term in the jacobian to account for how much du and dv are perpendicular?

I wanted to derive the formula for the multivariable change of basis in an integral on my own (for the 2 by 2 case). What I did was:

$$x=f(u,v)$$

$$y=g(u,v)$$

so $$dx = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv$$ $$dy = \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv$$

Then, $$dx \wedge dy = (\frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv) \wedge (\frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv) = (\frac{\partial f}{\partial u}\frac{\partial g}{\partial v} - \frac{\partial g}{\partial u}\frac{\partial f}{\partial v}) du \wedge dv$$

I recognize that term as the Jacobian. Then: $$dx\wedge dy = J(u,v) du \wedge dv$$

but I don't want to be working with bivectors, I want to work with scalars. I take the absolute value on both sides and since dx is perpendicular to dy:

$$dx dy = J(u,v) \sin(\theta) du dv$$

where $$\theta$$ is the angle between the two vectors. This is not the formula I learned in my undergraduate studies. How did the original formula work even if dx and dy were scalars?

• If you imagine/consider/treat $\text{d}x$ and $\text{d}y$ as perpendicular vectors with $x,y$ coordinates, so are $\text{d}u$ and $\text{d}v$ with $u,v$ coordinates. Commented Feb 7, 2023 at 14:18

Long story short; you need to take the determinant in $$u-v$$ space, and the $$-$$ in the usual formula saves you.

Long story long; lets think about the extreme cases. Firstly, if $$du$$ and $$dv$$ are perpendicular in $$x-y$$ space, then no problem. In the other extreme case, let us consider $$v(x,y)=u(x,y)+\epsilon \delta(x,y)$$ where we will eventually take $$\epsilon$$ to be very small to investigate what happens when $$du$$ and $$dv$$ are almost colinear. We can now write $$dx = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv=\frac{\partial f}{\partial u}du + \left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial f}{\partial \delta}\frac{\partial \delta}{\partial v}\right)dv$$ $$dy = \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv= \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial u}\left(\frac{\partial g}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial g}{\partial \delta}\frac{\partial \delta}{\partial v}\right)dv$$

such that $$dx \wedge dy = \left(\frac{\partial f}{\partial u}\left(\frac{\partial g}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial g}{\partial \delta}\frac{\partial \delta}{\partial v}\right) - \frac{\partial g}{\partial u}\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial f}{\partial \delta}\frac{\partial \delta}{\partial v}\right)\right) du \wedge dv$$ Then we can do some calculations, and hand wave that $$\epsilon^2$$ will be way too small to consider, to get $$dx \wedge dy = \epsilon\left(\frac{\partial f}{\partial u}\frac{\partial g}{\partial \delta}-\frac{\partial g}{\partial u}\frac{\partial f}{\partial \delta}\right)\frac{\partial \delta}{\partial v}du \wedge dv+\mathcal{O}(\epsilon^2).$$

By again arguing that for very small $$\epsilon$$ we have $$\sin(\epsilon)=\epsilon+\mathcal{O}(\epsilon^2)$$ we can see that in this case actually

$$dx \wedge dy \approx J(u,\delta)\sin(\epsilon)du \wedge dv$$

So for both extreme cases we can see that your intuition for how these measures should act is actually spot on in a sense, the only thing is that the usual formula already 'protects' itself by having the difference of partial derivatives (so if the $$u$$ and $$v$$ functions are very similar, then they largely cancel), so by tacking on an additional $$\sin$$ you are overcorrecting.

Hope this helps, and apologies of using physicist level hand-waving of infinitesimals.

• It looks this answer is trying to confuse /fool me. Commented Feb 12, 2023 at 7:21