Explicit proof of the Jacobian inverse

[Physgeek64]

Homework Statement

Given the transformations $x^2+y^2=2*r*cos(theta)$ and $x*y=r*sin(theta)$ prove the Jacobian explicitly

The question then goes on to ask how r and theta are related to the cylindrical coordinates rho and phi. I think $r=1/2*(x^2+y^2)$ and hence $r=1/2 rho$ but I am not really sure about this part I'm afraid, so haven't gotten very far

Homework Equations

$(partial(x,y)/partial(r,theta))*partial(r,theta)/partial(x,y))=1$

The Attempt at a Solution

So by multiplying the second expression by two and then squaring both and adding we get $r=1/2*(x^2+y^2)$ from which we can find the first two elements in the second Jacobian to be x and y respectively. By dividing the two transformations we can also get an expression for theta, which is relativity simple to differentiate, however when isolating x and y I seem to be going into pages of algebra, which leads me to think I have got the wrong approach, and there must be a simpler way since this is meant to be a relatively quick question.

Many thanks

 

[mighty2000]

for your post! I would like to clarify a few things before I begin my response.

Firstly, it seems like the transformations given are in polar coordinates, not cylindrical coordinates. The first equation is the equation of a circle in polar coordinates, while the second equation is the product of the polar coordinates. In cylindrical coordinates, the first equation would be $rho^2 = 2*r*cos(phi)$ and the second equation would be $rho*phi = r*sin(phi)$.

Assuming we are working in polar coordinates, the first step would be to rewrite the equations in terms of the variables r and theta. This can be done using the following equations:

$x = r*cos(theta)$
$y = r*sin(theta)$

Substituting these into the two given equations, we get:

$r^2*cos^2(theta) + r^2*sin^2(theta) = 2*r*cos(theta)$
$r^2*cos(theta)*r*sin(theta) = r*sin(theta)$

Simplifying, we get:

$r^2 = 2*r*cos(theta)$
$r = 1/2*cos(theta)$

Using these equations, we can now find the Jacobian. The Jacobian is defined as:

$J = partial(x,y)/partial(r,theta) = (partial(x)/partial(r))*(partial(y)/partial(theta)) - (partial(x)/partial(theta))*(partial(y)/partial(r))$

Substituting in the expressions for x and y in terms of r and theta, we get:

$J = (cos(theta))*(r*cos(theta)) - (-r*sin(theta))*(sin(theta))$
$J = r*cos^2(theta) + r*sin^2(theta)$
$J = r$

So the Jacobian is just equal to r. This means that the Jacobian is independent of theta, and only depends on the distance from the origin, r.

To answer the second part of the question, we can use the relationship between polar and cylindrical coordinates, which is:

$rho = r*sin(phi)$
$z = r*cos(phi)$

Substituting this into the equations we found for r and theta, we get:

$rho = (1/2)*cos(theta)*sin(phi)$
$z = (1/2)*cos(theta)*cos(phi)$

So we can see that the cylindrical coordinates are related to the polar coordinates by a factor of 1/