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Post by Admin on Jul 9, 2023 6:45:56 GMT
Mondo
In the book $u$ and $v$ are definitely scalars.
Vasco
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Post by Admin on Jul 9, 2023 7:01:11 GMT
Mondo
I would advise you also to read (i) at the bottom of page 481 and continued on page 482.
Vasco
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Post by mondo on Jul 9, 2023 7:01:51 GMT
Mondo If $u$ and $v$ were scalars it's completely untrue to say that $\partial_xu$ would always be zero If you write $u=[x,0]$ etc then they are vectors. So which is it? Vectors or scalars? Vasco I treat scalar a a constant i.e $6$ is a scalar while $6+6i$ is a complex number/vector. You probably mean a scalar function like $6x$ when you say a derivative of scalar is not always $0$. I can try to write down a proof for (13), but if we treat $u$ $v$ as a multiple of unit vector then do you agree with my idea here? |
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Post by mondo on Jul 9, 2023 7:16:13 GMT
Mondo I would advise you also to read (i) at the bottom of page 481 and continued on page 482. Vasco Yes and I do just this, I only flip $H(z)$, the image not the argument $z$ |
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Post by Admin on Jul 9, 2023 7:18:27 GMT
Mondo
A scalar is not a constant. When we use the word scalar we do not mean a constant. A scalar can be a constant but in general it is not.
I do not agree with your idea. $u$ and $v$ represent general scalar functions which may or may not be constant. but most importantly they are not vectors and so cannot be written as multiples of unit vectors.
I agree with you that if we write the vector $H$ as $u+iv$ then if $v=0$ we can think of $H$ as a vector pointing along the real axis. The complex notation breaks down a bit here I agree. We certainly cannot write things like $u+v$ as a vector.
Vasco
Let's see your proof! Go for it!
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Post by Admin on Jul 9, 2023 7:38:19 GMT
Mondo
If you want to think of $6$ as a vector along the real axis then I think we have various possibilities:
1. Use words to explain that you are considering it as a vector
2. write it as 6+0i
3. write it as (6,0) or [6,0] as you did
4 draw a diagram.
The main thing to remember is that we cannot add a vector and a scalar together without converting the scalar to a vector if we want the result to be a vector or without converting the vector to a scalar if we want the result to be a scalar.
If we define $u$ and $v$ to be scalars as in this situation then $u+v$ is a scalar and $u+iv$ is vector. In the case of $u+iv$ then we must think of this as the sum of the two vectors $u$ [now thought of as a vector $(u,0)$] and $iv$.
Vasco
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Post by Admin on Jul 9, 2023 7:50:37 GMT
Mondo If $u$ and $v$ were scalars it's completely untrue to say that $\partial_xu$ would always be zero If you write $u=[x,0]$ etc then they are vectors. So which is it? Vectors or scalars? Vasco I treat scalar a a constant i.e $6$ is a scalar while $6+6i$ is a complex number/vector. You probably mean a scalar function like $6x$ when you say a derivative of scalar is not always $0$. I can try to write down a proof for (13), but if we treat $u$ $v$ as a multiple of unit vector then do you agree with my idea here?Mondo If we think of $u$ and $v$ as multiples of unit vectors then they become vectors not scalars and so you are changing their definition. and they are vectors not scalars now. Also do you mean the same unit vector in each case? pointing in the same direction or different directions? Vasco |
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Post by mondo on Jul 9, 2023 8:11:26 GMT
I treat scalar a a constant i.e $6$ is a scalar while $6+6i$ is a complex number/vector. You probably mean a scalar function like $6x$ when you say a derivative of scalar is not always $0$. I can try to write down a proof for (13), but if we treat $u$ $v$ as a multiple of unit vector then do you agree with my idea here?Mondo If we think of $u$ and $v$ as multiples of unit vectors then they become vectors not scalars and so you are changing their definition. and they are vectors not scalars now. Also do you mean the same unit vector in each case? pointing in the same direction or different directions? Vasco Thank you for your answers Vasco. Yes, of course when I say "if we treat $u$ $v$ as a multiple of unit vectors" I do mean separate unit vectors in real and imaginary axis. So as I wrote I redefine $u = A[x,0]$ and $v = B[0,y]$ where $A,B$ are adequate multiplies. So now a vector $H(z) = u + iv$ can have a polya vector field $\overline{H} = u - iv$, yes exactly the same as normally, by definition of complex conjugate. The redefinition of $u,v$ is only to make them vectors so we can formally talk about vector addition, subtraction. If you look at the LHS of (13) this gives divergence and curl in respect to $H$ but now if you look at my diagram, and apply i.e $u-v$ instead of $u+v$ you will get the RHS of (13). Because the later gives $H(z)$ while the former $\overline{H}(z)$ So in essence this seems to be a "geometrical" way of getting this relation. |
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