I have been trying to understand Jacobian Determinant. I hope someone is able to give me a pointer.
Most material that I found on Internet didn't provide derivation of Jacobian Determinant.
One such web site is: http://tutorial.math.lamar.edu (Which I find quite good, otherwise.)
I spent a lot of time trying to deepen my understanding of Jacobian Determinant.
I played with Transformations that define uv-axes and how integration of a function over a Region/area would work with the Transformations.
For example, when I started with simple Transformations of:
u = ( x - y )/√2
v = ( x + y )/2√2
which is uv-axes rotated -45° from Cartesian xy-axes, and with v-axis at 2 times the scale, that is, v = 1 maps to 2 units length in xy-coords.
So, I say that uscale = 1, vscale = 2, for the above transformations.
With this uv-axes, I can simplify a 10x20 rectangle Region which is rotated at 45° from x-axis, such that the longer dimension points at 45° from x-axis.
With such examples, I begin to develop intuition how Jacobian Determinant works.
I understand Jacobian Determinant to be a Scaling Factor to convert area measurement in uv-axes to xy-dimensions.
Area measurement in uv-axes is given simply by formula Δu x Δv, where Δu = 10, Δv = 10, because vscale = 2).
Jacobian Determinant Scaling Factor = uscale x vscale (quite intuitively).
Area in xy-dimensions = Δu x Δv x (uscale x vscale) = 10 x 10 x 1 x 2 = 200.
Integration of volume over such a simpler uv Square, could be easier than over the same xy Region, appearing at an angle.
With the above initial understanding, I am trying to work out how Jacobian Determinant is derived.
Deriving from the above Transformations formula:
dx/du = √2 / 2
dx/dv = √2
dy/du = -√2 / 2
dy/dv = √2
I can also derive from Geometry that:
dx/du = uscale cos Θ
dy/du = uscale sin Θ
dx/dv = vscale cos (90° - Θ)
dy/dv = vscale sin (90° - Θ)
I could get:
areaInXY / areaInUV = uscale x vscale
which matches my understanding.
However, Jacobian Determinant formula is:
∂(x, y) / ∂(u, v) = ∂x/∂u ∂y/∂v - ∂x/∂v ∂y/∂u
= uscale * vscale * cos 2Θ
This leaves me quite puzzled why I have the extra cos 2Θ factor which isn't making intuitive sense -- why would the area Scaling Factor depends on how the rectangle is rotated and thus how uv-axes are rotated?!
Anybody can see where my reasoning went wrong above?
Let me try to explain what basically the Jacobian determinant does. This is true in general for smooth functions mapping from R^n to R^n, but for the sake of simplicity, assume we are working on R^2. Let F(x,y) a smooth R^2 to R^2 function. Then we can say that F(x,y) sends the x coordinate to f1(x,y) and the y coordiate to f2(x,y) at point (x,y). Then think about an infinitesimal rectangular area, defined by the points (x,y),(x+dx,y),(x,y+dy) and (x+dx,y+dy). Now, the area of this infinitesimal rectangle is dxdy. What happens to this rectangle when it goes through the F(x,y) transformation? We apply F(x,y) to each of the four coordinates and obtain the following points:
A:(x,y)->(f1(x,y),f2(x,y))
B:(x+dx,y) -> (f1(x+dx,y),f2(x+dx,y)) (approx.)= (f1(x,y) + (∂f1/∂x)dx,f2(x,y) + (∂f2/∂x)dx)
C:(x,y+dy) -> (f1(x,y+dy),f2(x,y+dy)) (approx.)= (f1(x,y) + (∂f1/∂y)dy,f2(x,y) + (∂f2/∂y)dy)
D:(x+dx,y+dy) -> (f1(x+dx,y+dy),f2(x+dx,y+dy)) (approx.)=(f1(x,y) + (∂f1/∂x)dx + (∂f1/∂y)dy,f2(x,y) + (∂f2/∂x)dx + (∂f2/∂y)dy)
The equalities are approximately equal and exactly hold in the limit where dx and dy goes to 0, they are the best linear approximation to the function F at new points. (We obtain these from the first order parts of the Taylor approximation of the functions f1 and f2).
If we look to the new (approximated) area under the transformation F(x,y), we see the new distance vectors between the transformed points a:
B-A:((∂f1/∂x)dx,(∂f2/∂x)dx)
C-A:((∂f1/∂y)dy,(∂f2/∂y)dy)
D-C:((∂f1/∂x)dx,(∂f2/∂x)dx)
D-B:((∂f1/∂y)dy,(∂f2/∂y)dy)
As you can see, the newly transformed infinitesimal area is a parallelogram. Let:
u=((∂f1/∂x)dx,(∂f2/∂x)dx)
v=((∂f1/∂y)dy,(∂f2/∂y)dy)
These vectors constitute the edges of our parallelogram. It can be shown with the help of the cross product between u and v, that the area of the parallelogram is:
area^2 = (u1v2 - u2v1)^2 = ((∂f1/∂x)(∂f2/∂y)dxdy - (∂f2/∂x)(∂f1/∂y)dxdy)^2
area^2 = ((∂f1/∂x)(∂f2/∂y) - (∂f2/∂x)(∂f1/∂y))^2 (dxdy)^2
area = |(∂f1/∂x)(∂f2/∂y) - (∂f2/∂x)(∂f1/∂y)|dxdy (dx and dy are positive)
area = |det([∂f1/∂x, ∂f1/∂y],[∂f2/∂x, ∂f2/∂y])|dxdy
So, the matrix we are going to take the determinant of is simply the Jacobian matrix. Like I said in the beginning, this derivation can be extended to arbitrary dimensions of n,given the coordinate transformation function F is smooth and the Jacobian matrix is hence invertible, with non-zero determinant.
A good visual explanation of this is given at: http://mathinsight.org/double_integral_change_variables_introduction