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Post by Admin on Oct 10, 2016 16:29:40 GMT
Gary
This suggested exercise is at the end of the second paragraph on page 303.
Refer to figure 22b on page 302.
Let the radius of $z$ be $r$ and the radius of $\widetilde{z}$ be $R$. Since the two right-angled triangles containing $ds$ and $d\widetilde{s}$ are similar we can write
$d\widetilde{s}/ds=R/r$
Since the two right-angled triangles containing $y$ and $\widetilde{y}$ are similar we can also write
$R/r=\widetilde{y}/y$ and so it follows that $d\widetilde{s}/ds=\widetilde{y}/y$ or $d\widetilde{s}/\widetilde{y}=ds/y$
So using (31) on page 298 we can write $d\widehat{\widetilde{s}}=d\widetilde{s}/\widetilde{y}=ds/y=d\widehat{s}$ and (37) follows from this result.
Vasco
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Post by salzburg99 on May 27, 2018 18:22:26 GMT
I don't understand the phrase: "the anticonformality of inversion implies that the image d(tilde-s) is also orthogonal to this radius" on page 303?
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Post by Admin on May 27, 2018 19:16:33 GMT
Hi
If a mapping is anticonformal (page 131), which inversion in a circle is, then the magnitude of the angles is preserved by the mapping but the direction of the angles is not preserved.
So because the angle between the radius and $ds$ is a right angle, the angle between the radius and $d\tilde{s}$ is also a right angle. Remember that $z$ is mapped to $\tilde{z}$ and $ds$ is mapped to $d\tilde{s}$ by inversion in $K$.
Vasco
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Post by salzburg99 on May 27, 2018 20:09:06 GMT
Thanks for your quick reply. I have a mental block on inversion in general I guess because I don't know what determines the length |d(tilde-s)|? In other words, how is ds mapped to d(tilde-s) by inversion? I get that z is mapped to tilde-z by inversion, but what about the tip of ds (the point not on the qz radius but on the other radius drawn in [22b])? Is the key that ds is infinitesimal? Because when I look at [2c] on page 126, the angle qab is not the same as q(tilde-a)(tilde-b).
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Gary
GaryVasco
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Post by Gary on May 28, 2018 1:36:52 GMT
Thanks for your quick reply. I have a mental block on inversion in general I guess because I don't know what determines the length |d(tilde-s)|? In other words, how is ds mapped to d(tilde-s) by inversion? I get that z is mapped to tilde-z by inversion, but what about the tip of ds (the point not on the qz radius but on the other radius drawn in [22b])? Is the key that ds is infinitesimal? Because when I look at [2c] on page 126, the angle qab is not the same as q(tilde-a)(tilde-b). salzburg99,
Please permit me to jump in. On p. 302, $ds_1$ is treated as an infinitesimal segment of $L$. On p. 303, para. 1, it appears that another $ds$ is also defined as an infinitesimal "line segment". This segment is orthogonal to $K$ at $z$. I think because of these definitions, it works to think of $ds$ as a series of points on the infinitesimal segment, not as a vector or single complex number. Then when one applies inversion to the points of $ds$, the result is $d\tilde{s}$ aligned as shown. One can say that $d\tilde{s}$ preserves the angle of $ds$ with $z$, because $d\tilde{s}$ is a line segment orthogonal to the radius at $\tilde{z}$. If one were to create an infinitesimal vector and call it $dz$ and apply inversion, then I think $d\tilde{z}$ would point in the opposite direction to $dz$. If $dz$ points in the direction of $iz$ at $z$, $d\tilde{z}$ points in the direction of $-iz$
I have to correct the above regarding the direction of $dz$. I found that plotting the inverse $\mathcal{I}_C$ of a small vector orthogonal to $z$ at $z$ does not preserve the angle magnitude. That is $\mathcal{I}_C (dz + z) - \mathcal{I}_C (z)$ is not orthogonal to $\mathcal{I}_C (z)$. I'm puzzled, too. Perhaps there was a calculation error. I changed this paragraph a few times. I hope I haven't confused the issue.
Regarding [c], p. 126, the corresponding angles have the same shade or coloring.
Gary
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Post by Admin on May 29, 2018 6:57:45 GMT
Thanks for your quick reply. I have a mental block on inversion in general I guess because I don't know what determines the length |d(tilde-s)|? In other words, how is ds mapped to d(tilde-s) by inversion? I get that z is mapped to tilde-z by inversion, but what about the tip of ds (the point not on the qz radius but on the other radius drawn in [22b])? Is the key that ds is infinitesimal? Because when I look at [2c] on page 126, the angle qab is not the same as q(tilde-a)(tilde-b). Hi Yes, the key is that $ds$ is infinitesimal. If you look at [3] on page 127 you can see that as $b$ approaches $a$, the line from $\tilde{b}$ to $\tilde{a}$ becomes more and more parallel to the line $L$ through $a$ and $b$, and in the limit the infinitesimal from $a$ to $b$ is mapped to the infinitesimal from $\tilde{a}$ to $\tilde{b}$. To be more precise, the situation in [22b] on page 302, is like the situation described in the paragraph under [3] on page 127, where the line $L$ intersects the circle $K$: The infinitesimal $ds$ is part of the line $L$ and the infinitesimal $d\tilde{s}$ is part of the circle that the line $L$ is mapped to by the inversion. Because it is infinitesimal we can treat $d\tilde{s}$ as a line segment along the tangent to the circle through $\tilde{z}$ with centre on $q\tilde{z}$. Vasco |
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Gary
GaryVasco
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Post by Gary on May 29, 2018 14:57:01 GMT
Thanks for your quick reply. I have a mental block on inversion in general I guess because I don't know what determines the length |d(tilde-s)|? In other words, how is ds mapped to d(tilde-s) by inversion? I get that z is mapped to tilde-z by inversion, but what about the tip of ds (the point not on the qz radius but on the other radius drawn in [22b])? Is the key that ds is infinitesimal? Because when I look at [2c] on page 126, the angle qab is not the same as q(tilde-a)(tilde-b). Hi Yes, the key is that $ds$ is infinitesimal. If you look at [3] on page 127 you can see that as $b$ approaches $a$, the line from $\tilde{b}$ to $\tilde{a}$ becomes more and more parallel to the line $L$ through $a$ and $b$, and in the limit the infinitesimal from $a$ to $b$ is mapped to the infinitesimal from $\tilde{a}$ to $\tilde{b}$. To be more precise, the situation in [22b] on page 302, is like the situation described in the paragraph under [3] on page 127, where the line $L$ intersects the circle $K$: The infinitesimal $ds$ is part of the line $L$ and the infinitesimal $d\tilde{s}$ is part of the circle that the line $L$ is mapped to by the inversion. Because it is infinitesimal we can treat $d\tilde{s}$ as a line segment along the tangent to the circle through $\tilde{z}$ with centre on $q\tilde{z}$. Vasco That also explains why $d\tilde{s}$ and $ds$ differ by a small, but perceptible angle when plotted as in [22], p. 302.
Gary
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Post by Admin on May 29, 2018 16:32:37 GMT
Gary
Good point. I guess it's a short arc of a circle, if you plot it. Beware of infinitesimals! Do you remember reading Needham's explanation of what he means by infinitesimal? I re-read it recently (pages 20-21).
Vasco
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Gary
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Post by Gary on May 29, 2018 21:31:22 GMT
Gary Good point. I guess it's a short arc of a circle, if you plot it. Beware of infinitesimals! Do you remember reading Needham's explanation of what he means by infinitesimal? I re-read it recently (pages 20-21). Vasco Vasco,
Thanks. I will look at it. Actually, I meant to write $dz$ and $d\tilde{z}$ rather than $ds$ and $d\tilde{s}$ when I remarked on the non-orthogonality of the infinitesimals to the ray.
Gary
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Post by Admin on May 29, 2018 22:20:44 GMT
Gary Good point. I guess it's a short arc of a circle, if you plot it. Beware of infinitesimals! Do you remember reading Needham's explanation of what he means by infinitesimal? I re-read it recently (pages 20-21). Vasco Vasco, Thanks. I will look at it. Actually, I meant to write $dz$ and $d\tilde{z}$ rather than $ds$ and $d\tilde{s}$ when I remarked on the non-orthogonality of the infinitesimals to the ray.
Gary
Gary I meant to write "...short chord of a circle..." above. Vasco |
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Gary
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Post by Gary on May 30, 2018 1:28:24 GMT
Vasco, Thanks. I will look at it. Actually, I meant to write $dz$ and $d\tilde{z}$ rather than $ds$ and $d\tilde{s}$ when I remarked on the non-orthogonality of the infinitesimals to the ray.
Gary
Gary I meant to write "...short chord of a circle..." above. Vasco Vasco,
Yes, I think that is more consistent.
Gary
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Post by Admin on May 30, 2018 5:31:44 GMT
Gary
I don't know why I wrote that, it should be "...arc of a circle...". We are using inversion and so lines are mapped to circles and line segments are mapped to arcs of circles. In this case infinitesimal line segments and arcs. Of course the arc and chord become indistinguishable in the limit.
Vasco
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Gary
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Post by Gary on May 30, 2018 16:16:37 GMT
Vasco,
Thinking of the infinitesimal $ds$ as the arc of a circle that does not pass through $q$, its inversion will be another infinitesimal arc outside K (by (9), p. 128). Both arcs are orthogonal to the ray. By the discussion of conformal and anticonformal transformations on pp. 128 to 129, one would expect that the magnitude of angle $abc$ = $\tilde{a}\tilde{b}\tilde{c}$ (Figure [4], p. 128). Is this the case?
Gary
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Post by Admin on May 30, 2018 17:59:53 GMT
Vasco, Thinking of the infinitesimal $ds$ as the arc of a circle that does not pass through $q$, its inversion will be another infinitesimal arc outside K (by (9), p. 128). Both arcs are orthogonal to the ray. By the discussion of conformal and anticonformal transformations on pp. 128 to 129, one would expect that the magnitude of angle $abc$ = $\tilde{a}\tilde{b}\tilde{c}$ (Figure [4], p. 128). Is this the case? Gary Gary No, because although in figure 4 on page 128 the arc $[bc]$ is mapped to the arc $[\tilde{b}\tilde{c}]$, the line $[bc]$ is not mapped to the line $[\tilde{b}\tilde{c}]$, but to an arc through $\tilde{b}$ and $\tilde{c}$, which is not shown in figure 4. The angle $abc$ will be equal to the angle between the line $q\tilde{a}$ and the tangent to the arc which is not shown. Don't forget that in figure 4 only the circles are mapped by the inversion to the other circle. The lines are drawn in afterwards and are not part of the mapping. Vasco |
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Gary
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Post by Gary on May 31, 2018 0:05:01 GMT
Vasco, Thinking of the infinitesimal $ds$ as the arc of a circle that does not pass through $q$, its inversion will be another infinitesimal arc outside K (by (9), p. 128). Both arcs are orthogonal to the ray. By the discussion of conformal and anticonformal transformations on pp. 128 to 129, one would expect that the magnitude of angle $abc$ = $\tilde{a}\tilde{b}\tilde{c}$ (Figure [4], p. 128). Is this the case? Gary Gary No, because although in figure 4 on page 128 the arc $[bc]$ is mapped to the arc $[\tilde{b}\tilde{c}]$, the line $[bc]$ is not mapped to the line $[\tilde{b}\tilde{c}]$, but to an arc through $\tilde{b}$ and $\tilde{c}$, which is not shown in figure 4. The angle $abc$ will be equal to the angle between the line $q\tilde{a}$ and the tangent to the arc which is not shown. Don't forget that in figure 4 only the circles are mapped by the inversion to the other circle. The lines are drawn in afterwards and are not part of the mapping. Vasco Vasco, Yes, it is as you say. By (7), if an extended chord of C does not pass through q, it must map to an arc of a circle that does pass through q. In the attached plot, the arc $\tilde{b}\tilde{c}$ is the inversion of the chord $bc$ in K. It is not apparent from the first plot, but when sufficiently extended, it must pass through q, as it would do in the second plot if the extension of chord bc were sufficiently long.
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