How Do Non-Perpendicular Unit Vectors Affect Coordinate Transformation?

[unscientific]

Homework Statement

The x-y coordinates are being transformed into the u-v coordinates.

Based on the diagram, u lies along x while v makes an angle α with x.

The Attempt at a Solution

The answer defined u and v weirdly..

Shouldn't

x = u

and

y = v sin α

??

 

[gabbagabbahey]

Shouldn't

x = u

and

y = v sin α

??

No. Consider the position vector $\mathbf{r}= x\mathbf{e}_x + y\mathbf{e}_y = u\mathbf{e}_u + v\mathbf{e}_v$:

You have $\mathbf{e}_u = \mathbf{e}_x$ since the two axes are parallel, but $\mathbf{e}_v$ has both a vertical and a horizontal component and is given by $\mathbf{e}_v = \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y$. Plugging this into the position vector definition gives $x\mathbf{e}_x + y\mathbf{e}_y= u \mathbf{e}_x + v( \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y)$, which gives you the relations in your image.

 

[unscientific]

Also, does the region of integration R change if we change the variables from (x,y) to (u,v)?


According to the answer, the region R → R', where R' is only σ/2∏ of the original R..

 

[gabbagabbahey]

According to the answer, the region R → R', where R' is only σ/2∏ of the original R..

No, that's not what they are saying. Read it again more carefully, what they are actually claiming is that $\int_{0}^{\infty} \int_{0}^{\infty} e^{-r^2} \left| \frac{\partial(x,y)}{\partial(u,v)} \right|dudv = \frac{\alpha}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-r^2}dxdy$

The integral on the left is only over positive $u$ & $y$, while the integral on the right is over all (R²) space.

 

[unscientific]

No, that's not what they are saying. Read it again more carefully, what they are actually claiming is that $\int_{0}^{\infty} \int_{0}^{\infty} e^{-r^2} \left| \frac{\partial(x,y)}{\partial(u,v)} \right|dudv = \frac{\alpha}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-r^2}dxdy$

The integral on the left is only over positive $u$ & $y$, while the integral on the right is over all (R²) space.

Yup, if you only consider x,y,u,v > 0,

it would be α/(∏/2) for ∫ 0 to infinity..

 

[unscientific]

No. Consider the position vector $\mathbf{r}= x\mathbf{e}_x + y\mathbf{e}_y = u\mathbf{e}_u + v\mathbf{e}_v$:

You have $\mathbf{e}_u = \mathbf{e}_x$ since the two axes are parallel, but $\mathbf{e}_v$ has both a vertical and a horizontal component and is given by $\mathbf{e}_v = \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y$. Plugging this into the position vector definition gives $x\mathbf{e}_x + y\mathbf{e}_y= u \mathbf{e}_x + v( \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y)$, which gives you the relations in your image.

I don't really understand what you mean...

My main problem here is why do they define u and v as such in the picture?

I thought u and v are defined when you drop a perpendicular line onto the axis?

And it's pretty clear that the lengths u, v they define are shorter than the ones in my picture..

 

[voko]

I thought u and v are defined when you drop a perpendicular line onto the axis?

That's the case when your coordinate lines are perpendicular. When they are not, you get what you see in this picture.

 

[unscientific]

That's the case when your coordinate lines are perpendicular. When they are not, you get what you see in this picture.

Are they defined this way?

 

[voko]

A coordinate system (on an plane) is defined by its origin and unit vectors $\vec{a}$ and $\vec{b}$. Any point $\vec{p} = u\vec{a} + v\vec{b}$. $u$ and $v$ are coordinates. Now if the coordinate unit vectors are not perpendicular, what do you get? Try it on a piece of paper.

 

[unscientific]

A coordinate system (on an plane) is defined by its origin and unit vectors $\vec{a}$ and $\vec{b}$. Any point $\vec{p} = u\vec{a} + v\vec{b}$. $u$ and $v$ are coordinates. Now if the coordinate unit vectors are not perpendicular, what do you get? Try it on a piece of paper.

Ah, using vectors everything seems much simpler now! Thank you!