# Converted Cartesian coordinates to polar coordinates

• touqra
In summary, the Cartesian coordinates to polar coordinates conversion resulted in a wrong equation for the polar components. The Laplacian for the polar coordinates is not equivalent to the Laplacian for the Cartesian coordinates.
touqra
I don't know where have I gone wrong...
I converted Cartesian coordinates to polar coordinates:

$$\frac{\partial^2\Psi}{\partial x^2} +\frac{\partial^2\Psi}{\partial y^2}= \frac{1}{2}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2 }{\partial y^2})\Psi^2 - \Psi(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\Psi =\frac{\partial^2\Psi}{\partial r^2}+ \frac{1}{r^2}\frac{\partial^2\Psi}{\partial \Phi^2}$$

But on the left hand side (the Cartesian components) is just the Laplacian in 2D, but the final answer I got for the polar components is not equivalent to the Laplacian for polar coordinate system.
I'm missing the term $$\frac{1}{r}\frac{\partial\Psi}{\partial r}$$

Last edited:
You are misunderstanding something here.

$${\frac{\partial^2}{\partial x^2}\Psi^2 = \frac{\partial}{\partial x}\frac{\partial}{\partial x}\Psi^2 = \frac{\partial}{\partial x}( 2\Psi \frac{\partial \Psi}{\partial x}) = 2 ({\frac{\partial \Psi}{\partial x}})^2 + 2\Psi \frac{\partial^2 \Psi}{\partial^2 x}$$

That doesn't seem consistent with your (wrong) equation.

Last edited:
$$\frac{\partial^2\Psi}{\partial x^2} +\frac{\partial^2\Psi}{\partial y^2}= \frac{1}{2}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2 }{\partial y^2})\Psi^2 - \Psi(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\Psi=\frac{\partial^2\Psi}{\partial r^2}+ \frac{1}{r^2}\frac{\partial^2\Psi}{\partial \Phi^2}$$

Ooops, I typed the wrong stuffs. I'm sorry.

$$(\frac{\partial\Psi}{\partial x})^2 +(\frac{\partial\Psi}{\partial y})^2= \frac{1}{2}(\frac{\partial^2}{\partial x^2}+\frac{\partial^2 }{\partial y^2})\Psi^2 - \Psi(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})\Psi$$

Next, I plug in the Laplacian for the polar coordinates, essentially cylindrical coordinate, with z constant, and I end up with:
$$(\frac{\partial\Psi}{\partial x})^2 +(\frac{\partial\Psi}{\partial y})^2= (\frac{\partial\Psi}{\partial r})^2+ \frac{1}{r^2}(\frac{\partial\Psi}{\partial \Phi})^2$$

Next, I am required to get the Euler Lagrange equation for a system. The above is just the potential part. The time derivative kinetic is just $$\frac{1}{2}m\dot{\Psi}^2$$
Taking the Euler Lagrange for the Cartesian, I end up with an expression from the potential part:
$$\frac{\partial^2\Psi}{\partial x^2} +\frac{\partial^2\Psi}{\partial y^2}$$
And this is just a Laplacian.

But when taking the Euler Lagrange for the polar coordinates, I end up with an expression $$\frac{\partial^2\Psi}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2\Psi}{\partial \Phi^2}$$
and this is not equal to the Laplacian for the polar coordinates.
I am missing $$\frac{1}{r}\frac{\partial\Psi}{\partial r}$$

Last edited:

## 1. What is the difference between Cartesian and polar coordinates?

Cartesian coordinates, also known as rectangular coordinates, use two perpendicular axes (x and y) to locate points on a 2D plane. Polar coordinates, on the other hand, use a distance from the origin (r) and an angle (θ) to locate points on a 2D plane.

## 2. How do you convert Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x, y) to polar coordinates (r, θ), you can use the following formulas:
r = √(x² + y²)
θ = tan⁻¹(y/x)

## 3. What are the applications of polar coordinates?

Polar coordinates are often used in mathematics, physics, and engineering to represent circular or rotational motion, such as in polar graphs and polar equations. They are also commonly used in navigation and mapping, as well as in describing the position and movements of objects in polar coordinates systems.

## 4. Can you convert polar coordinates to Cartesian coordinates?

Yes, you can convert from polar coordinates (r, θ) to Cartesian coordinates (x, y) using the following formulas:
x = r cos(θ)
y = r sin(θ)

## 5. Are there any limitations to using polar coordinates?

While polar coordinates have many applications, they do have some limitations. They are not as intuitive as Cartesian coordinates and can be more difficult to visualize in 3D space. They also have limited use in certain mathematical calculations, such as finding the distance between two points or determining the slope of a line.

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