Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Dec 12, 2015 21:40:50 GMT
Vasco,
I am puzzled by (23). Are the lines over f' and f'' symbols for the conjugates? If so, what is the motivation? I suspect there is an obvious derivation from the previous formula for k(\(\hat{\xi}\)), but I don't see it.
Gary
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Post by Admin on Dec 12, 2015 23:53:02 GMT
Vasco, I am puzzled by (23). Are the lines over f' and f'' symbols for the conjugates? If so, what is the motivation? I suspect there is an obvious derivation from the previous formula for k(\(\hat{\xi}\)), but I don't see it. Gary Gary Presumably you mean you are puzzled by (28) not (23)? Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 13, 2015 5:37:28 GMT
Vasco, I am puzzled by (23). Are the lines over f' and f'' symbols for the conjugates? If so, what is the motivation? I suspect there is an obvious derivation from the previous formula for k(\(\hat{\xi}\)), but I don't see it. Gary Gary Presumably you mean you are puzzled by (28) not (23)? Vasco Yes, (28). |
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Post by Admin on Dec 13, 2015 11:06:14 GMT
Vasco, I am puzzled by (23). Are the lines over f' and f'' symbols for the conjugates? If so, what is the motivation? I suspect there is an obvious derivation from the previous formula for k(\(\hat{\xi}\)), but I don't see it. Gary Gary Presumably you mean you are puzzled by (28) not (23)? Vasco Gary Yes the bars in (28) mean complex conjugate. On page 238 in the paragraph immediately preceding (28), where Needham writes: "It would appear that the most natural intrinsic quantity that can be abstracted from $k(\hat{\xi})$ is..." he means this in the sense that - if you read on you will see the justification for this, and in fact if you read on to (30) and beyond I think you will be able to see this. Look at the very last line on page 28 and you will see that the scalar product $\vec{a}\cdot \vec{b}$, where $\vec{a}$ and $\vec{b}$ are vectors in $\mathbb{R^2}$, referred to in the last sentence on page 238, can be written as Re$(\bar{a}b)$, which is needed to understand the derivation of (30). Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 13, 2015 23:04:01 GMT
Vasco, Thank you. That helped. But using the equations to plot a circle of curvature turned out to be a tricky exercise. The use of f = \(e^z\) seems to be both an advantage and a disadvantage. It's a simplification, because f' = f'' = \(e^z\), so f''/f' cancels out to 1 and plotting f on the slanted line should mean that \(\kappa\) = 0. But it caused me conceptual difficulties when trying to plot tangents orthogonal to f and curvature orthogonal to the tangent. The amplitwist provided the tangent and the curvature can be multiplied by a unit vector orthogonal to the tangent. Equation (30) on p. 239 is "neater and more intelligible" as Needham pointed out, but the scalar product requires extra steps to implement unless I am overlooking some quick way to calculate it using complex numbers, because you first have to extract the real and imaginary parts. Gary |
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Post by Admin on Dec 15, 2015 16:50:44 GMT
Gary
I am studying your attached file and I will get back to you as soon as I can. One comment: I'm not sure what you mean by "curvature orthogonal to the tangent", in your post above. The way I understand the curvature $\widetilde{\kappa}$ is that it it is a scalar quantity.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 15, 2015 21:27:23 GMT
Gary I am studying your attached file and I will get back to you as soon as I can. One comment: I'm not sure what you mean by "curvature orthogonal to the tangent", in your post above. The way I understand the curvature $\widetilde{\kappa}$ is that it it is a scalar quantity. Vasco Vasco, I knew that phrase might cause trouble, but I forgot to get back to it. Yes, $\widetilde{\kappa}$ is scalar. I think, since r = 1/k, where r is the radius of the circle of curvature, I just assumed that $\widetilde{\kappa}$ should be thought of as the length of a vector pointing from f(p) in the direction of the center of the circle of curvature, which should be the direction of f''(p) and orthogonal to the tangent at f(p). I revised the original attachment and added another. Gary |
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