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Jacobian Matrices in Transformations


Date: 08/05/99 at 00:58:31
From: Mike Smith
Subject: Determinants

I have a question involving determinants. In calculus, to make an 
integration simple we do transforms on x, y, z. We use this formula we 
were given, without proof of its correctness: 

   partial x/partial U  partial x/partial v. 

This is called the Jacobian matrix. Then to find the Jacobian, you 
take the determinant of the matrix.

1. Why do we need to set up this matrix when doing transforms?

2. Why does the determinant of this matrix give us an expression that 
   we multiply into the integral? For example, if 

     x = rcos(theta), y = rsin(theta) and z = z 

then the transform gives

     dx.dy.dz = r.dr.d(theta).dz

What does r mean geometrically?

Is there something more general about transforms and determinants I 
should understand? I hope you can help me!

Thanks, Mike


Date: 08/06/99 at 16:54:24
From: Doctor Anthony
Subject: Re: Determinants

The Jacobian
-------------

When we make a change of variables in multiple integrals the Jacobian 
gives us a general technique for making the change. A very simple 
example is when we change from Cartesian (x,y) coordinates to polar 
(r,theta) coordinates. The element of area in Cartesian coordinates is 
the little rectangle dx.dy, while in polar coordinates the small 
elementary portion of area is r.d(theta).dr. So in carrying out an 
integration over an area, and changing from Cartesian to polar 
coordinates, we do not replace dx.dy by d(theta).dr, but by  
r.d(theta).dr, and that extra factor r is the Jacobian in this case.

In above example, simple geometry is enough to show what is required 
in making the change of coordinate system, but we could for example be 
changing to a system where x = u^2-v^2, y = 2uv and having to carry 
out the integration over a given area A. In such a situation we have:

       INT.INT[F(x,y)dx.dy]
        area A

     = INT.INT[F(u^2-v^2,2uv).J.du.dv]      
        area A

where J is the Jacobian of the transformation.

We must now find a general method for determining J.

We start by reminding you of the formula for the area of triangle ABC 
with coordinates of the vertices (x1,y1), (x2,y2) and (x3,y3)

                 |1     1     1|
     Area = (1/2)|x1   x2    x3|
                 |y1   y2    y3|

The determinant is positive if the points A, B, C are taken in an 
counterclockwise direction.

Curvilinear Coordinates.

Consider the double integral  INT.INT[phi(x,y)dx.dx]
                               Region R

We introduce two new independent variables, u and v, with

     x = f(u,v), y = g(u,v)

where f(u,v) and g(u,v) are single-valued functions such that every 
point P(x,y) within region R maps to a unique point P1(u,v) in the 
corresponding region R1 in the uv-plane.

Consider now an element of area in the xy-plane and the corresponding 
element in the uv-plane.

Draw a rectangle in the xy-plane with corners at P, Q, R, and S, and 
the corresponding element in the uv-plane with corners at P1, Q1, R1, 
and S1. The sides of this second figure will now be curved, but keep 
its shape like that of a distorted rectangle. Let the point P(x0,y0) 
have the value P1(u0,v0) in the new variables. If the transformation 
equations are x = f(u,v) and y = g(u,v) these can be treated as a pair 
of simultaneous equations and solved for u and v, giving:

     u = F(x,y)   and   v = G(x,y)

Then u = F(x,y) will be a family of curves in the xy-plane depending 
on the particular constant value given to u in each case. So u = 1, 
u = 2, u = 3 will produce a family of curves in the xy-plane. 
Similarly, v = G(x,y) will be a family of curves depending on the 
particular constant value given to v in each case.

These two sets of curves will therefore cover the region R and form a 
network and to the point P(x0,y0) there will be a pair of curves 
u = u0 (constant) and v = v0 (constant) that intersect at that point.

The u and v values relating to any particular point in the xy-plane 
are known as its curvilinear coordinates and x = f(u,v) and y = g(u,v) 
are the transformation equations between the two systems.

In the (x,y) system, the element of area is dA = dx.dy and is the area 
bounded by the lines x = x0, x = x0+dx, y = y0 and y = y0+dy.

In the new system of curvilinear coordinates (u,v) the element of area 
dA1 can be taken as that of the figure P1,Q1,R1,S1, i.e. the area 
bounded by the curves u = u0, u = u0+du, v = v0 and v = v0+dv.

Since dA1 is small, P1,Q1,R1,S1 may be regarded as a rectangle,

     i.e. dA1 = 2 times the area of triangle P1,Q1,S1

and this is where we make use of the formula for the area of a 
triangle in determinant form.

Before we can apply the formula we must first find the Cartesian 
coordinates of P1, Q1 and S1.

If x = f(u,v) then a small increase dx in x is given by:

     dx = part(df/du).du + part(df/dv).dv

and if y = g(u,v) then a small increase dy in y is given by:

     dy = part(dg/du).du + part(dg/dv).dv

Now P1 coincides with P(x,y), therefore:

(a) P1 is the point (x,y)
(b) Q1 corresponds to Q, i.e. small changes from P

     dx = part(dx/du).du + part(dx/dv).dv

and  dy = part(dy/du).du + part(dy/dv).dv

But along P1Q1 v is constant and so dv = 0

Therefore dx = part(dx/du).du   and   dy = part(dy/du).du

     i.e. Q1 is the point (x+part(dx/du)du, y+part(dy/du)du)

Similarly for S1, since u is constant along P1S1, du = 0

and so S1 is the point(x+part(dx/dv)dv, y+part(dy/dv)dv).

We have then the coordinates of P1,Q1,S1 and can write down the 
determinant giving the area.

                 | 1           1                        1    |
     Area = (1/2)| x      x+part(dx/du)du     x+part(dx/dv)dv|
                 | y      y+part(dy/du)du     y+part(dy/dv)dv|

subtracting column (1) from columns (2) and (3) gives:

                 | 1          0                  0    |
     Area = (1/2)| x     part(dx/du)du   part(dx/dv)dv|
                 | y     part(dy/du)du   part(dy/dv)dv|

and this simplifies to:

     Area = (1/2)|part(dx/du)du   part(dx/dv)dv|
                 |part(dy/du)du   part(dy/dv)dv|

Then taking out the factor du from the first column and the factor dv 
from the second column, this becomes:

     Area = (1/2)|part(dx/du)    part(dx/dv)|du.dv
                 |part(dy/du)    part(dy/dv)|

The area of the approximate rectangle is twice the area of the 
triangle, and so:

     Area of rectangle = dA1 = |part(dx/du)  part(dx/dv)|du.dv
                               |part(dy/du)  part(dy/dv)|

For convenience this is often written as:

             part d(x,y)
     dA1  =  ----------- du.dv  =  J.du.dv
             part d(u,v)


where   J = |part(dx/du)   part(dx/dv)|
            |part(dy/du)   part(dy/dv)|

Note that in the transformation it is possible for the order of the 
points P1, Q1, R1, S1 to be reversed for that of P, Q, R, S and in 
this case dA1 may give a negative result when the determinant is 
evaluated. To ensure a positive element of area the result is given as 
the modulus of J.

The formula to remember is that when changing to a new coordinate 
system we have for x = f(u,v), y = g(u,v)

       INT.INT[F(x,y)dx.dy]
        area A

     = INT.INT[F{f(u,v),g(u,v)}.|J|.du.dv] 
        area A

We can check that this calculation of J agrees with the result for 
conversion from Cartesian to polar coordinates.

          x = r.cos(theta)                   y = r.sin(theta)

part(dx/dr) = cos(theta)     part(dx/d(theta)) = -r.sin(theta)

part(dy/dr) = sin(theta)     part(dy/d(theta)) = r.cos(theta)

and so    J = |cos(theta)    -r.sin(theta)|
              |sin(theta)     r.cos(theta)|

            =  r[cos^2(theta) + sin^2(theta)]

            =  r    as required

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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