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Post by Admin on Jan 17, 2018 12:08:12 GMT
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Post by mondo on Apr 9, 2022 20:18:39 GMT
Vasco,
thank you for posting your solution. I found it difficult to understand hence I would like to ask for some hints to fill the gap.
1. You say $\rho = Re^{2\theta - \frac{\pi}{2}}$ - why the angle is $\frac{\pi}{2} - 2\theta$ which is just our $2\theta$ minus 180 degrees? Shouldn't it be the angle $\rho$ makes with the real axis?
2. How does $i\rho = (R/r^2) \cdot z^2$ shows that vector field $z^2$ is tangent to the circle at $z$? What rule/relation is used here?
3. I feel like even if it shows the tangency of $z^2$ at $z$ it is not what this exercise asks for - they want us to show the tangency of vector field to the real axis. This is exactly what we can see on fig [1a] on page 450.
Thank you
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Gary
GaryVasco
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Post by Gary on Apr 10, 2022 7:01:00 GMT
Vasco, thank you for posting your solution. I found it difficult to understand hence I would like to ask for some hints to fill the gap. 1. You say $\rho = Re^{2\theta - \frac{\pi}{2}}$ - why the angle is $\frac{\pi}{2} - 2\theta$ which is just our $2\theta$ minus 180 degrees? Shouldn't it be the angle $\rho$ makes with the real axis? 2. How does $i\rho = (R/r^2) \cdot z^2$ shows that vector field $z^2$ is tangent to the circle at $z$? What rule/relation is used here? 3. I feel like even if it shows the tangency of $z^2$ at $z$ it is not what this exercise asks for - they want us to show the tangency of vector field to the real axis. This is exactly what we can see on fig [1a] on page 450. Thank you Mondo,
Here is how I understand it: 1. The two acute interior angles are equal, so the oblique angle is $2\theta$. The rest follows from knowing that the angle between orthogonal lines is $\pi/2$ and the angle about a point on a straight line is $\pi$.  2. Writing “i” is equivalent to multiplying by $e^{i \pi/2}$. It rotates an object by $\pi/2$. Then, since the ray $\rho$ has an angle of $(2\theta-(\pi/2))$, multiplying by $i$ rotates it back to $2\theta$, which is the same as the angle of $z^2$. The real value $(R/r^2)$ has no bearing on the angle of $z^2$. Since $i\rho$ is orthogonal to $\rho$, so is $z^2$, so the field $z^2$ anchored at $z$ must be tangent to the circle, because tangents are orthogonal to rays. 3. I share your confusion. Figure [1a] appears to show at least three circles centred at the origin. When z has traversed a half circle, the field vectors $z^2$ anchored at $z$ are tangent to the real line, but they are not tangent at the origin, and they would not touch the origin unless there were some reason to increase their length. Notice that Vasco and I have drawn circles centred away from the origin and tangent to the origin with the result that $z^2$ is tangent at the origin. But in my notes, I also drew field vectors according to the rules and obtained a figure like [1a]. We can infer the tangency of the vector field of $z^2$ on the circle in Vasco’s diagram to the real axis from the fact that it is tangent to the circle everywhere on the circle, which itself touches the origin. I think I would suggest that the question be modified to include the condition that z traverse circles that touch the origin. Gary |
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Post by Admin on Apr 10, 2022 12:22:19 GMT
Gary
If by 'question' in the last sentence above you mean the exercise, then it is already included that $z$ traverses circles that touch the origin.
Vasco
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Post by Admin on Apr 10, 2022 13:56:46 GMT
Vasco, thank you for posting your solution. I found it difficult to understand hence I would like to ask for some hints to fill the gap. 1. You say $\rho = Re^{2\theta - \frac{\pi}{2}}$ - why the angle is $\frac{\pi}{2} - 2\theta$ which is just our $2\theta$ minus 180 degrees? Shouldn't it be the angle $\rho$ makes with the real axis? 2. How does $i\rho = (R/r^2) \cdot z^2$ shows that vector field $z^2$ is tangent to the circle at $z$? What rule/relation is used here? 3. I feel like even if it shows the tangency of $z^2$ at $z$ it is not what this exercise asks for - they want us to show the tangency of vector field to the real axis. This is exactly what we can see on fig [1a] on page 450. Thank you Mondo 1. I agree with what Gary says in his reply to your question 1:the angle of the complex number $\rho$ is $2\theta-\pi/2$. $\pi/2$ is 90 degrees. 2. Multiplying a complex number by $i$ rotates it through a right angle. 3. My approach to this is to draw a circle which is tangent to the real axis and show that the vector field is tangent to the circle at z and so it must be a streamline of the vector field. In 1a on page 450 the streamlines are not circles centred at the origin, but circles tangent to the real axis at the origin. Look carefully at 1a and note the directions of the vectors tangent to the streamlines!Vasco |
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Gary
GaryVasco
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Post by Gary on Apr 10, 2022 15:09:59 GMT
Gary If by 'question' in the last sentence above you mean the exercise, then it is already included that $z$ traverses circles that touch the origin. Vasco Vasco,
Agreed. It is very clear. I don't know why I misread it.
Gary
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Gary
GaryVasco
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Post by Gary on Apr 10, 2022 15:19:43 GMT
Vasco,
Even knowing what I am looking for in [1a], I find it hard to see a line of dots in the shape of a circle, so I can see why it is misleading to refer back to this figure. But it is clear that the vectors rotate by a full circle as they collectively wend their way around the half plane. I think the conceptual problem I had with the exercise was trying to work from a set of vectors to a circle, rather than assuming the circle tangent to the origin and demonstrating that the vectors of $z^2$ are all tangent, thereby proving the circle is a streamline of the field.
Gary
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Post by Admin on Apr 10, 2022 15:38:04 GMT
Gary Have a look at this link using equation explorer. The problem is near the origin the vectors get very small. Play with the scaling factors and window size to see more clearly. linkVasco |
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Post by Admin on Apr 10, 2022 16:48:54 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Apr 10, 2022 19:30:03 GMT
Vasco,
When you overlay known streamlines on the field, they pick out (touch) their own vectors.
How did you write the expression for the vector field in EquationExplorer?
Gary
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Post by Admin on Apr 10, 2022 19:58:44 GMT
Vasco, When you overlay known streamlines on the field, they pick out (touch) their own vectors. How did you write the expression for the vector field in EquationExplorer? Gary
Gary $z^2=(x+iy)^2=x^2-y^2+2xyi$ So real part is $x^2-y^2$ and imaginary part is $2xy$. The EquationExplorer uses the unit vectors $i$ and $j$ in the real and imaginary directions and so we write $z^2$ as $(x^2-y^2)i+2xyj$ Vasco |
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Post by mondo on Apr 10, 2022 21:53:11 GMT
Thank you for answers, most things are clear now and I agree the proposed solution is fine. However I have one small comment to the graphical solution and specifically to the Figure 1. It appears to me the dashed line that aims to extend $z$ is wrong. I mean you extended the longest side of a triangle instead of $\rho$. Do you agree?
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Post by Admin on Apr 10, 2022 22:13:43 GMT
Thank you for answers, most things are clear now and I agree the proposed solution is fine. However I have one small comment to the graphical solution and specifically to the Figure 1. It appears to me the dashed line that aims to extend $z$ is wrong. I mean you extended the longest side of a triangle instead of $\rho$. Do you agree? Mondo No I don't agree. Why do you think it is wrong? Vasco |
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Post by mondo on Apr 10, 2022 22:37:52 GMT
Ok I think it is OK. Initially I thought you should measure the angle in respect to $z$ but this is really not needed. All that matters is to show that at the end $z$ and $z^2$ at a right angle which you did. Thank you  |
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Post by Admin on Apr 10, 2022 22:47:41 GMT
Ok I think it is OK. Initially I thought you should measure the angle in respect to $z$ but this is really not needed. All that matters is to show that at the end $z$ and $z^2$ at a right angle which you did. Thank you Mondo But the angle is with respect to $z$. $z$ is the vector from the origin to the point $z$, and the angle with the black arc is $\theta$. It's absolutely crucial to draw the dashed line where it is in order to do the proof. Vasco |
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