Polar to cartesian using e(theta) and e(r)

[ual8658]

Homework Statement

Homework Equations

NONE

The Attempt at a Solution

I'm trying to understand why the unit vector in the y direction is that formula. I get that e(theta) and e(r) are unit vectors used with polar coordinates that define direction and are perpendicular to each other always. What I don't get is their relationship with theta to get cartesian unit vectors.

 

[BvU]

What I don't get is their relationship with theta to get cartesian unit vectors

A simple rotation over the angle $\theta$. Your second picture shows it clearly for $\bf\hat\jmath$.
For $\bf \hat\imath$ you see $e_\theta$ points in the negative $\hat\imath$-direction so you get $\hat\imath = \cos\theta \; e_r - \sin\theta \; e_\theta$

 

[PeroK]

The Attempt at a Solution

I'm trying to understand why the unit vector in the y direction is that formula. I get that e(theta) and e(r) are unit vectors used with polar coordinates that define direction and are perpendicular to each other always. What I don't get is their relationship with theta to get cartesian unit vectors.

What do you think it should be?

 

[ual8658]

A simple rotation over the angle $\theta$. Your second picture shows it clearly for $\bf\hat\jmath$.
For $\bf \hat\imath$ you see $e_\theta$ points in the negative $\hat\imath$-direction so you get $\hat\imath = \cos\theta \; e_r - \sin\theta \; e_\theta$

I don't follow.

What do you think it should be?

I don't know what it should be. I originally thought the j vector was just the projection of e(theta) but that clearly cannot be because both j and e(theta) have length 1. Is there a general relationship that might explain how two perpendicular unit vectors in a plane can somehow form this relationship?

 

[PeroK]

I don't follow.
I don't know what it should be. I originally thought the j vector was just the projection of e(theta) but that clearly cannot be because both j and e(theta) have length 1. Is there a general relationship that might explain how two perpendicular unit vectors in a plane can somehow form this relationship?

Your diagram in post #1 only needs you to read off the sine and cosine components to get this relationship. However, another way is to note that (in Cartesian coordinates):

$j = (0, 1), \ \hat{r} = (\cos \theta, \sin \theta), \ \hat \theta = (-\sin \theta, \cos \theta)$

Then, you simply have to use a bit of vector algebra to express $j$ as a ,linear combination of $\hat{r}$ and $\hat \theta$.

 

[BvU]

Is there a general relationship that might explain how two perpendicular unit vectors in a plane can somehow form this relationship?

You can check it from the expressions I gave. A rotation preserves vector lengths

 

[Chestermiller]

Suppose that $\theta = 0$. What would j be equal to, $i_r$ or $i_{\theta}$?

Suppose that $\theta = 90\ degrees$. What would j be equal to, $i_r$ or $i_{\theta}$?

 

[ual8658]

Suppose that $\theta = 0$. What would j be equal to, $i_r$ or $i_{\theta}$?

Suppose that $\theta = 90\ degrees$. What would j be equal to, $i_r$ or $i_{\theta}$?

I follow this but I wanted the derivation for it.

Your diagram in post #1 only needs you to read off the sine and cosine components to get this relationship. However, another way is to note that (in Cartesian coordinates):

$j = (0, 1), \ \hat{r} = (\cos \theta, \sin \theta), \ \hat \theta = (-\sin \theta, \cos \theta)$

Then, you simply have to use a bit of vector algebra to express $j$ as a ,linear combination of $\hat{r}$ and $\hat \theta$.

Ok that first part makes sense. How can you at a glance determine that those two vectors r and theta would lineraly combine to form a vector of length 1?

 

[PeroK]

I follow this but I wanted the derivation for it.Ok that first part makes sense. How can you at a glance determine that those two vectors r and theta would lineraly combine to form a vector of length 1?

$\hat{r}$ and $\hat \theta$ are linearly independent so form a basis (in fact, they form an orthonormal basis). Any vector, therefore, can be expressed as a linear combination of them.

You can use the basic techniques of solving simultaneous equations to get the coefficients. In this case, we look for coefficients $a, b$ such that:

$a \hat{r} + b \hat \theta = j$

Hence

$a(\cos \theta, \sin \theta) + b(-\sin \theta, \cos \theta) = (0, 1)$

Which implies:

$a \cos \theta - b \sin \theta = 0$
$a \sin \theta + b \cos \theta = 1$

Which leads to the solution $a = \sin \theta, \ b = \cos \theta$.

 

[Chestermiller]

Just draw a right triangle anywhere in the domain with j as the hypotenuse, and the other two sides parallel to the local ir and itheta. Then determine the lengths of those other two sides.

 

[ual8658]

When you substitute with $cos(\alpha)=\sqrt{1-sin^2(\alpha)}$ , you then square both sides of the equation and get a quadratic equation in $u=sin^2(\alpha)$. (It is actually 4th power in $sin(\alpha)$ , but there is no $sin^3(\alpha)$ term and no $sin(\alpha)$ term=it is quadratic in $sin^2(\alpha)$.

$\hat{r}$ and $\hat \theta$ are linearly independent so form a basis (in fact, they form an orthonormal basis). Any vector, therefore, can be expressed as a linear combination of them.

You can use the basic techniques of solving simultaneous equations to get the coefficients. In this case, we look for coefficients $a, b$ such that:

$a \hat{r} + b \hat \theta = j$

Hence

$a(\cos \theta, \sin \theta) + b(-\sin \theta, \cos \theta) = (0, 1)$

Which implies:

$a \cos \theta - b \sin \theta = 0$
$a \sin \theta + b \cos \theta = 1$

Which leads to the solution $a = \sin \theta, \ b = \cos \theta$.

Now it makes sense. Thanks!

Just draw a right triangle anywhere in the domain with j as the hypotenuse, and the other two sides parallel to the local ir and itheta. Then determine the lengths of those other two sides.

I see it now. Thanks!