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Coordinates transformation

  • #1
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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

??
 

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Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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6
Shouldn't

x = u

and

y = v sin α

??
No. Consider the position vector [itex]\mathbf{r}= x\mathbf{e}_x + y\mathbf{e}_y = u\mathbf{e}_u + v\mathbf{e}_v[/itex]:

You have [itex]\mathbf{e}_u = \mathbf{e}_x[/itex] since the two axes are parallel, but [itex]\mathbf{e}_v[/itex] has both a vertical and a horizontal component and is given by [itex]\mathbf{e}_v = \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y[/itex]. Plugging this into the position vector definition gives [itex]x\mathbf{e}_x + y\mathbf{e}_y= u \mathbf{e}_x + v( \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y)[/itex], which gives you the relations in your image.
 
  • #3
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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..
 

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  • #4
gabbagabbahey
Homework Helper
Gold Member
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6
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 [itex]\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[/itex]

The integral on the left is only over positive [itex]u[/itex] & [itex]y[/itex], while the integral on the right is over all (R2) space.
 
  • #5
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No, that's not what they are saying. Read it again more carefully, what they are actually claiming is that [itex]\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[/itex]

The integral on the left is only over positive [itex]u[/itex] & [itex]y[/itex], 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..
 
  • #6
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13
No. Consider the position vector [itex]\mathbf{r}= x\mathbf{e}_x + y\mathbf{e}_y = u\mathbf{e}_u + v\mathbf{e}_v[/itex]:

You have [itex]\mathbf{e}_u = \mathbf{e}_x[/itex] since the two axes are parallel, but [itex]\mathbf{e}_v[/itex] has both a vertical and a horizontal component and is given by [itex]\mathbf{e}_v = \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y[/itex]. Plugging this into the position vector definition gives [itex]x\mathbf{e}_x + y\mathbf{e}_y= u \mathbf{e}_x + v( \cos\alpha \mathbf{e}_x + \sin\alpha \mathbf{e}_y)[/itex], which gives you the relations in your image.
I dont 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..
 
  • #7
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390
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.
 
  • #8
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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?
 
  • #9
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390
A coordinate system (on an plane) is defined by its origin and unit vectors [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex]. Any point [itex]\vec{p} = u\vec{a} + v\vec{b}[/itex]. [itex]u[/itex] and [itex]v[/itex] are coordinates. Now if the coordinate unit vectors are not perpendicular, what do you get? Try it on a piece of paper.
 
  • #10
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A coordinate system (on an plane) is defined by its origin and unit vectors [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex]. Any point [itex]\vec{p} = u\vec{a} + v\vec{b}[/itex]. [itex]u[/itex] and [itex]v[/itex] 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! :smile:
 

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