|
|
Post by Admin on Jul 20, 2023 19:48:33 GMT
Mondo
That's not what we (I and the book) do, we calculate $\overline{\nabla}f=\overline{\nabla}(u+iv)$
Vasco
|
|
|
|
Post by Admin on Jul 20, 2023 20:00:25 GMT
Mondo
In my reply #10 the 5th line of algebra is just a rearrangement of the 4th line. What happens is dictated by the algebra not by any choices I make.
Vasco
|
|
|
|
Post by mondo on Jul 20, 2023 20:10:37 GMT
To me equations (7) and (8) on page 210 abuse the notation of $f'$. This is so called "Lagrange notation" and does great when we deal with a single variable functions. But here we have two variables $x$ and $y$ and immediately an ambiguity arise - as I wrote in my first post "$is f'$ a derivative in regards to $x$ or $y$"? The way I would do this exercise on page 518 is $\overline{\nabla}f = 2\frac{d}{dx}f = 2\frac{d}{dy}(-if)$
|
|
|
|
Post by Admin on Jul 20, 2023 20:29:07 GMT
Mondo
As I think I said earlier, $f'$ is the derivative with respect to $z$ not $x$ or $y$.
Vasco
|
|
|
|
Post by mondo on Jul 20, 2023 20:43:53 GMT
Mondo As I think I said earlier, $f'$ is the derivative with respect to $z$ not $x$ or $y$. Vasco If that were the case, then in your reply #10, instead of saying $\partial_x(u+iv) = f'$ you would do $\partial_z(u+iv) = f'$ $\overline{\nabla}=\partial_x-i\partial_y$ So $\overline{\nabla}f=(\partial_x-i\partial_y)(u+iv)$ $\overline{\nabla}f=\partial_xu+i\partial_xv-i\partial_yu+\partial_yv$ $\overline{\nabla}f=\partial_x(u+iv)+\partial_y(v-iu)=f'+f'=2f'$ |
|
|
|
Post by Admin on Jul 20, 2023 20:53:39 GMT
Mondo
Why would I? I am using $\overline{\nabla}$.
Vasco
|
|
|
|
Post by mondo on Jul 20, 2023 22:19:51 GMT
Mondo Why would I? I am using $\overline{\nabla}$. Vasco Ok and this part is fine. The problem starts to emerge when author associates these partial derivatives with $f'$. We have isolated this inconsistency - it all originates from page 210, I don't know how I "accepted" equations (7), (8) before. I still think it is abusing notation. When we say $f'$ we mean calculate the derivative of function $f$ in regards to some variable $z$, $x$, $y$, or whatever else. It also means, that regardless of which variable is is used we always differentiate the same function $f$. But what happens in equations (7) and (8)? In (7) we calculate a derivative of function $u+iv$ in regards to $x$ while in (8) we calculate derivative of function $v-iu$ in regards to $y$. This is clearly not the same, not only the variables of differentiation are different, more so, the functions are different! However, thanks to CR equations at the end these two differentiation results are the same but I say, just because the result is the same doesn't mean the input and operation made on it were the same. I think, it is a situation like: $f = 2+2$ and $g = 4 \cdot 1$ and we can say $f=g$ but having two different definitions of the same function $f$ would be weird. But this is exactly what author does in (7) and (8) redefined the same function. |
|
|
|
Post by Admin on Jul 21, 2023 6:09:55 GMT
Mondo
Chapter 4 and section VIII The Cauchy-Riemann Equations express the fundamental idea around which the whole book is built. You need to find your way through this again. The fundamental idea of the amplitwist is embodied in the equations on page 210.
We do not have two different definitions of the same function, but two expressions for the derivative of $f$ with respect to $z$.
This is connected to the fundamental fact that analytic functions have the same derivative in all directions at the point $z$.
Vasco
|
|
|
|
Post by Admin on Jul 21, 2023 18:07:01 GMT
Mondo
Take a simple example of an analytic function $f=z^2=x^2-y^2+2xyi=u+iv$
It follows that $u=x^2-y^2$ and $v=2xy$
Therefore
$df/dz=f'=2z=2x+2yi=\partial_xu+i\partial_xv=\partial_yv-i\partial_yu$
because $2x=\partial_xu=\partial_yv$ and $2y=\partial_xv=-\partial_yu$
Vasco
|
|
|
|
Post by mondo on Jul 21, 2023 19:46:08 GMT
This is not quite true right? The derivative of $f$ in regards to $x$ is $\partial_x{f}$ and this is not $\partial_y{f}$ but $-i\partial_y{f}$ (equation 8, page 210)
|
|
|
|
Post by Admin on Jul 21, 2023 19:57:02 GMT
Mondo
Yes it is completely correct, otherwise the whole idea on which the book is based would have to be thrown in the bin. This is the definition of an analytic function.
$f'$ is not equal to $\partial_xf$.
Vasco
|
|
|
|
Post by mondo on Jul 21, 2023 20:08:57 GMT
Mondo Yes it is completely correct, otherwise the whole idea on which the book is based would have to be thrown in the bin. This is the definition of an analytic function. $f'$ is not equal to $\partial_xf$. Vasco Wait wait wait, isn't that exactly what equation (7) says?  PS: In your Reply #23 you have a mistake in the first equation: it should be $x^2 - y^2$. |
|
|
|
Post by Admin on Jul 21, 2023 20:18:03 GMT
Mondo
Corrected reply 23 and yes sorry I thought you had written $\partial_xu$
Vasco
|
|
|
|
Post by Admin on Jul 21, 2023 20:27:32 GMT
Mondo
Those two expressions represent the same thing. One is not $f'$ in the $x$ direction and the other $f'$ in the $y$ direction!!
vasco
|
|
|
|
Post by mondo on Jul 21, 2023 20:43:16 GMT
Mondo Those two expressions represent the same thing. One is not $f'$ in the $x$ direction and the other $f'$ in the $y$ direction!! vasco But the one over $y$ is also multiplied by $-i$ so it is not a pure derivative of $\partial_y{f}$ as you suggest, otherwise it would be $\partial_y{u} + i\partial_y{v}$. I think it all makes sense from somebody who keeps equation (7) and (8) from page 210 in mind all the time. However if you take somebody who never read this book, and ask him to do calculate $f' = (x^2-y^2 + i2xy)'$ he will right away ask "in regards to which variable should I differentiate $f$?". And if you tell him, do it first over $x$ and then over $y$ you will get answers $\partial_x{f} = 2x + i2y$ and $\partial_y{f} = -2y + i2x$ and they are not equal! They only become equal if you take a direction into account and "map" $y$ onto $x$ by rotating it by $\pi/2$ clockwise which is to multiply it by $-i$. Hence $-i\partial_y{f} = -i(-2y + i2x) = 2x + i2y = \partial_x{f}$ |
|
