Why does (dx₁)² equal zero in coordinate transformation with Jacobian 1?

[xuphys]

Hi,

When I do the following transformation:

$$
X_1=x_1+x_2 \\
X_2=x_2
$$

It turns out that the Jacobian $\partial (X_1,X_2)/\partial (x_1,x_2)$ is 1. But we have:

$$
dx_1dx_1+dx_1dx_2=d(x_1+x_2)dx_2=dX_1dX_2=|\partial (X_1,X_2)/\partial (x_1,x_2)|dx_1dx_2=dx_1dx_2
$$

So we have $dx_1dx_1=0$. Is this kind of weird? Why does $(dx_1)^2$ have to be 0?

Thank you!

 

[jambaugh]

It represents the differential area for the parallelogram formed by varying by dx1 and then dx1. Since both sides are the same direction, the area is zero.

At a higher level the differentials are treated as Grassmann variables (like cross product but yielding tensor instead of vector). Then the Jacobian is built into the algebra.

 

[HallsofIvy]

Strictly speaking "$dx_1dx_1$" has no meaning! How did it get in that problem?