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Post by mondo on Dec 12, 2022 6:27:45 GMT
I have a few questions to Figure [6] and accompanying description. At the bottom of page 222 author says: "Since we have purely rotated $z$, its image will all move vertically up though a distance equal to the angle of rotation $\delta$. Now:
1. I understood that the left hand side of [6] shows us $z$ and the right hand side $log(z)$ - if so, why the above description I quoted, suggests the image is due to rotation by $\delta$? This is when author says, the image will move through a distance equal to the angle of rotation.
2. This aims to visually show this differentiation of $log(z)$ but I can't see how on page 223 author concludes the amplitwist is $\frac{1}{r}$ im terms of amplification and $-\theta$ in terms of a twist.
3. Also author says the image will rotate in the direction of the angle of rotation while on [6] we can see they are both opposite directions.
Thank you.
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Post by Admin on Dec 12, 2022 20:19:04 GMT
I have a few questions to Figure [6] and accompanying description. At the bottom of page 222 author says: "Since we have purely rotated $z$, its image will all move vertically up though a distance equal to the angle of rotation $\delta$. Now: In your quote the word 'image' should be 'images', and I changed $\sigma$ to $\delta$ as it should be. I think you have misunderstood this. Since we know that $\log z=\ln r+i(\theta+2m\pi)$ we can see that rotating $z$ keeps $r$ the same and so keeps the real part of $\log z$ the same ($=\log r$). So $\log z$ moves up a vertical line as its imaginary part varies. Since it is clear from the right hand diagram that the white arrow must be multiplied by $(1/r)e^{-i\theta}$ to obtain the black arrow, then by definition this is the amplitwist. I don't see where he says this. Vasco |
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Post by mondo on Dec 13, 2022 23:33:42 GMT
In your quote the word 'image' should be 'images', and I changed $\sigma$ to $\delta$ as it should be. Yes, thank you. I think you have misunderstood this. Since we know that $\log z=\ln r+i(\theta+2m\pi)$ we can see that rotating $z$ keeps $r$ the same and so keeps the real part of $\log z$ the same ($=\log r$). So $\log z$ moves up a vertical line as its imaginary part varies. I see this but I don't see how this rotation example relates to a differentiation which is the end goal. At the beginning of this subsection author says that in this example he will rely on examination of infinitesimal geometry (second to last paragraph of page 222) but I have a problem understanding how can we get from an infinitesimal rotation of $z$ and its image which is $log(z)$ to a derivative of $log(z)$? Since it is clear from the right hand diagram that the white arrow must be multiplied by $(1/r)e^{-i\theta}$ to obtain the black arrow, then by definition this is the amplitwist. Yes but how did he get right hand image from the left hand one? I don't see where he says this. |
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Post by Admin on Dec 14, 2022 13:12:27 GMT
Mondo
1. You ask "...but I have a problem understanding how can we get from an infinitesimal rotation of $z$ and its image which is $\log(z)$ to a derivative of $\log(z)$?"
Your question indicates to me that you do not fully understand what an amplitwist is. Please read sections III and IV of chapter 4 again and make sure you understand them. The amplitwist concept is the most important concept in this book.
You seem to have moved on to chapter 5 without fully understanding chapter 4.
2. "Yes but how did he get right hand image from the left hand one?"
Read and understand chapter 4 and especially that if $z$ is written as $re^{i\theta}$ then $\log z=\ln r+i(\theta+2m\pi)$, so that when $z$ rotates, its argument changes and consequently the imaginary part of $\log z$ changes.
3."End of page 222 - "Since we have purely rotated $z$, its image will all move in vertically up through a distance equal to the angle of rotation $\delta$".
This is perfectly correct in the book. The image moves vertically upwards, it does not rotate. $z$ rotates on the left. When $z$ rotates through $\delta$ then $\log z$ moves vertically upwards by $\delta$. This is the thin black arrow. Needham has drawn the white arrow representing the change in $z$ on the right hand side of [6] as well as the left, so that we can easily see what the amplitwist is.
You have misquoted the book and your question indicates that you have not understood chapter 4 and the amplitwist concept.
Vasco
PS the quote function seems OK to me. Please send me an example where it does not work and I'll see if I can work out what the problem is. Thanks
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Post by mondo on Dec 14, 2022 18:43:33 GMT
Vasco,
I think you are right, I thought I grasped amplitwist concept fully but now I see there are things that I missed. Especially in regards to graphical/visual method author uses to present amplitwist. That said I don't fully understand why on figure [5], page 194 the amplification is $2r$ and a twist is $\theta$. We mapped $z = re^{i\theta}$ to $z^2 = r^{2}e^{2i\theta}$ from this only we can see amplification is $r$ and a twist is $2$. This is because length of a point was extended by $r$ and its angle was multiplied by $2$. I think author in all these examples, links amplification with the concept of arc length but why? On figure [5] both black and white arrows were rotated upwards by $\theta$ as part of the $z^2$ mapping but their length should not be changed - the point to which they are attached $z$ was moved but this should not affect the length of vectors that originate from it.
So I don't see the arc length being involved in the move that both white and black arrow went through.
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Post by Admin on Dec 15, 2022 14:00:23 GMT
Mondo
We are looking at infinitesimal vectors emanating from a point. These infinitesimal vectors are represented by the black and white arrows in figure 5. At the end of subsection 2 of chapter 4 on page 193 we see that the infinitesimal vectors are multiplied by $2r$ and twisted by $\theta$ as illustrated in figure 5.
You are wrong when you say that the infinitesimal arrows don't change length when the transformation $z^2$ is applied. Of course they do, because for example the end points of the white arrow correspond to different points in the plane on the right of figure 5 than they do on the left of figure 5, and so their lengths must have changed.
Vasco
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Post by mondo on Dec 18, 2022 1:33:12 GMT
We are looking at infinitesimal vectors emanating from a point. These infinitesimal vectors are represented by the black and white arrows in figure 5. At the end of subsection 2 of chapter 4 on page 193 we see that the infinitesimal vectors are multiplied by $2r$ and twisted by $\theta$ as illustrated in figure 5. Yes and this is a result of Jacobian matrix. So are you saying the the the amplitwist can not be derived/proved from figure [5]? You are wrong when you say that the infinitesimal arrows don't change length when the transformation $z^2$ is applied. Of course they do, because for example the end points of the white arrow correspond to different points in the plane on the right of figure 5 than they do on the left of figure 5, and so their lengths must have changed. My conclusion was more in regards to what is shown on figure [5] only. From it I see there is a point $z$ from which two infinitesimal vector are drown, next upon $z^2$ mapping the white one (previously with length $\epsilon$) got a length of $2r\epsilon$. So as I asked above, looks like figure [5] is there just to visualize effect of Jacobian matrix (2) and not to derive amplitwist of $z^2$ mapping, right? UPDATE: I think the amplitwist values should be visible from the figure [5] but I can't see it. On the other hand I do understand figure [12] on page 230. Here upon $z^a$ mapping the infinitesimal arrow emanating from point $z$ was moved to the point at a distance $r^{a-1}$ further and twisted by $a$ hence the net effect is an arc length -> $ar^{a-1}$. This indeed proves the amplification (twist is obvious). But how should I understand figure [5]? |
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Post by mondo on Dec 22, 2022 6:54:41 GMT
Vasco, I still have some troubles figuring out why on figure [5] page 194 the infinitesimal vector is increased in length by $2r$. Can it be derived from the picture itself? Also what if these two vectors were on the other side of $r$? Then, rotating $r$ by $\theta$ would actually shrink them, right?
Thank you.
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Post by Admin on Dec 23, 2022 13:34:42 GMT
We are looking at infinitesimal vectors emanating from a point. These infinitesimal vectors are represented by the black and white arrows in figure 5. At the end of subsection 2 of chapter 4 on page 193 we see that the infinitesimal vectors are multiplied by $2r$ and twisted by $\theta$ as illustrated in figure 5. Yes and this is a result of Jacobian matrix. So are you saying the the the amplitwist can not be derived/proved from figure [5]? You are wrong when you say that the infinitesimal arrows don't change length when the transformation $z^2$ is applied. Of course they do, because for example the end points of the white arrow correspond to different points in the plane on the right of figure 5 than they do on the left of figure 5, and so their lengths must have changed. My conclusion was more in regards to what is shown on figure [5] only. From it I see there is a point $z$ from which two infinitesimal vector are drown, next upon $z^2$ mapping the white one (previously with length $\epsilon$) got a length of $2r\epsilon$. So as I asked above, looks like figure [5] is there just to visualize effect of Jacobian matrix (2) and not to derive amplitwist of $z^2$ mapping, right? UPDATE: I think the amplitwist values should be visible from the figure [5] but I can't see it. On the other hand I do understand figure [12] on page 230. Here upon $z^a$ mapping the infinitesimal arrow emanating from point $z$ was moved to the point at a distance $r^{a-1}$ further and twisted by $a$ hence the net effect is an arc length -> $ar^{a-1}$. This indeed proves the amplification (twist is obvious). But how should I understand figure [5]? Mondo If you understand figure [12] on page 230 then by setting a=2 you can see that this is exactly what happens in figure 5. Vasco |
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Post by Admin on Dec 23, 2022 13:48:55 GMT
Vasco, I still have some troubles figuring out why on figure [5] page 194 the infinitesimal vector is increased in length by $2r$. Can it be derived from the picture itself? Also what if these two vectors were on the other side of $r$? Then, rotating $r$ by $\theta$ would actually shrink them, right? Thank you. Mondo Yes we can see it from 5. Do you understand section 2 The Jacobian Matrix? If you do then reading 3 The Amplitwist Concept and referring to figure 5 you should be able to understand it. What do you mean exactly by "the other side of $r$"? Vasco |
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Post by mondo on Jan 7, 2023 22:48:15 GMT
Hi Vasco, sorry for late response! Vasco, I still have some troubles figuring out why on figure [5] page 194 the infinitesimal vector is increased in length by $2r$. Can it be derived from the picture itself? Also what if these two vectors were on the other side of $r$? Then, rotating $r$ by $\theta$ would actually shrink them, right? Thank you. Mondo Yes we can see it from 5. Do you understand section 2 The Jacobian Matrix? If you do then reading 3 The Amplitwist Concept and referring to figure 5 you should be able to understand it. Yes I do understand section 2 The Jacobian Matrix as well as 3 The Amplitwist Concept (beside this figure [5]). However, why do you ask if I understand them? If we agreed that the amplitwist can be seen on figure [5] then this figure alone (plus some basic geometric facts) should be enough to get the amplitwist. What do you mean exactly by "the other side of $r$"? On the left hand side of figure [5] we see the two infinitesimal vector are pointing in the direction of a real axis, what if they were flipped to the other side of $r$, pointing to imaginary axis? Would the amplitwist still be $2r$? If so, why? I feel like what I am missing is what happens to these infinitesimal vectors during the mapping $z -> z^2$ I can't get the movement that these vector perform during the mapping. In the case I mentioned, when they are originally on the other side of $r$ it looks like the mapping would lead to a shrinkage of them. |
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Post by Admin on Jan 8, 2023 10:01:01 GMT
Mondo
The reason I asked if you understand subsections 2 and 3 on pages 192-194 is because if you understood them it seems to me you would not be asking the questions that you are asking.
The essential thing to realise is that the Jacobean at the point $(r,\theta)$ is the same for all infinitesimal vectors emanating from this point.
This means that all infinitesimal vectors emanating from this point are multiplied by $2r$ and rotated anticlockwise through an angle $\theta$ by the mapping $z\mapsto z^2$. This is what we have shown in subsection 2.
So the amplitwist is the same for all infinitesimal vectors emanating from $(r,\theta)$
Vasco
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Post by mondo on Jan 8, 2023 17:13:23 GMT
Vasco, ok so you are confirming that one can't get the amplitwist just by figure [5] inspection. In other words, figure 5 is only to visualize the Jacobiam matrix. Is that right?
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Post by Admin on Jan 8, 2023 17:26:30 GMT
mondo
No. We can get the amplitwist from it.
Vasco
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Post by mondo on Jan 8, 2023 17:53:38 GMT
Vasco, I understand it from Jacobian point of view but can you explain what geometric relations are needed to show the length of the infinitesimal vector $\epsilon$ was increased to $2r\epsilon$?
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