|
|
Post by Admin on Jul 21, 2023 20:54:06 GMT
Mondo
Why do you say over $y$? This is just the derivative with respect to $z$.
I completely and very strongly disagree with what you have written above. If you write $f'$ then you are differentiating with respect to $z$.
It's clear to me that you do not understand what differentiation with respect to $z$ means and so you cannot possibly understand what an amplitwist is
Vasco
|
|
|
|
Post by Admin on Jul 21, 2023 21:02:29 GMT
Mondo
Try reading page 210 after (7) and (8) to the end of the page.
Vasco
|
|
|
|
Post by mondo on Jul 21, 2023 21:04:43 GMT
Why do you say over $y$? This is just the derivative with respect to $z$. Vasco But you also wrote $df/dz=f'=2z=2x+2yi=\partial_xu+i\partial_xv=\partial_yv-i\partial_yu$ Partials over $y$ are not part of differentiation in respect to $z$, which means at some point you must have calculated $\partial_y(f)$ and notice a relationship that you show above. |
|
|
|
Post by Admin on Jul 21, 2023 21:14:22 GMT
Mondo
You obviously do not follow what I have written. I never calculated $\partial_yf$.
Vasco
|
|
|
|
Post by Admin on Jul 21, 2023 21:20:54 GMT
Mondo
I calculated partial derivatives with respect to $x$ and $y$ for $u$ and $v$.
Vasco
|
|
|
|
Post by Admin on Jul 21, 2023 21:35:24 GMT
Mondo
If you disagree that the derivative of an analytic function is the same in all directions at a point $z$ then read the italics on page 197.
Vasco
|
|
|
|
Post by mondo on Jul 21, 2023 22:18:20 GMT
Mondo If you disagree that the derivative of an analytic function is the same in all directions at a point $z$ then read the italics on page 197. Vasco Yes but it says that if we have a function $f = z^2 = r^2e^{2i\theta}$ then every complex number with the same modulus ($r$ here) and originating from the same point (origin in this case) will have exactly the same amplification $2r$ and twist $\theta$. But it doesn't say that if we now switch to rectangular coordinates, then the derivative of that single complex number in the direction of $x$ and $y$ will be the same. |
|
|
|
Post by Admin on Jul 21, 2023 22:36:15 GMT
Mondo
Yes it does. Every direction.
Vasco
|
|
|
|
Post by Admin on Jul 21, 2023 22:39:12 GMT
Mondo
Look, if you think this way you are challenging the author. A very big deal indeed. You must get in touch with him immediately. He is still alive.
Vasco
|
|
|
|
Post by mondo on Jul 21, 2023 22:44:12 GMT
Now, that you brought amplitwist concept and page 197 my understanding of it is wiggling.. Let's say we have two complex numbers $A=[2,0]$ and $B=[0,2]$ so they point into real and imaginary axis respectively (same length, separated by $\pi/2$). Now, I would like to apply a mapping $f(z) = z^2$ so the first complex number $A$ will rotate through $\theta = 0$ but the second $B$, through angle $\pi/2$ so the angle of rotation is not the same and we can't say "complex numbers are twisted the same amount".
That's one thing, the second is $f(z) = z^2 = r^2e^{2i\theta}$ and if we calculate derivatives with respect to $r$, to see how much the length changes and then with respects to $\theta$ to see how much the rotation change we will get $2r$ and $2ie^{2i\theta}$. So twist is different when we differentiate with respect to $z$ and $\theta$
|
|
|
|
Post by Admin on Jul 21, 2023 22:56:19 GMT
Mondo
This is about infinitesimal complex numbers.
Vasco
|
|
|
|
Post by mondo on Jul 21, 2023 23:04:04 GMT
That's one thing, the second is $f(z) = z^2 = r^2e^{2i\theta}$ and if we calculate derivatives with respect to $r$, to see how much the length changes and then with respects to $\theta$ to see how much the rotation change we will get $2r$ and $2ie^{2i\theta}$. So twist is different when we differentiate with respect to $z$ and $\theta$ And what is wrong with this? PS: I calculated these derivatives by isolating magnitude and rotation of $z^2$ |
|
|
|
Post by mondo on Jul 22, 2023 5:00:22 GMT
I also reread the introduction of chapter 4, The Amplitwist and found one more thing that is confusing. On page 190 the first paragraphs talks about the conformality and concludes it $z^2$ is conformal everywhere except at two critical points $z=0$ and $z=\infty$ where angles are doubled, how should we understand that? Angles are always doubled by $z^2$ as we just showed, due to $(e^{i\theta})^2 = e^{2i\theta}$
|
|
|
|
Post by Admin on Jul 22, 2023 6:17:09 GMT
Mondo
I'll let you work it out. The book makes perfect sense to me.
Vasco
|
|
|
|
Post by Admin on Jul 22, 2023 8:17:48 GMT
That's one thing, the second is $f(z) = z^2 = r^2e^{2i\theta}$ and if we calculate derivatives with respect to $r$, to see how much the length changes and then with respects to $\theta$ to see how much the rotation change we will get $2r$ and $2ie^{2i\theta}$. So twist is different when we differentiate with respect to $z$ and $\theta$ And what is wrong with this? PS: I calculated these derivatives by isolating magnitude and rotation of $z^2$ Mondo We are talking about $f'(z)$. Vasco |
|
