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Post by Admin on Feb 25, 2017 12:15:41 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Feb 27, 2017 7:17:16 GMT
Vasco, I thought I should learn how to produce various loops, so I gave it a try. My drawing program produced wiggly loop curves, so I tried one (the one on the right) with Mathematica's BezierCurve[] function. It was pretty tedious and to match Needham's curve more closely would have taken quite a lot of time adjusting control points. I am wondering how you were able to do it.  |
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Post by Admin on Feb 27, 2017 8:47:41 GMT
Vasco, I thought I should learn how to produce various loops, so I gave it a try. My drawing program produced wiggly loop curves, so I tried one (the one on the right) with Mathematica's BezierCurve[] function. It was pretty tedious and to match Needham's curve more closely would have taken quite a lot of time adjusting control points. I am wondering how you were able to do it. View AttachmentGary Yes, they are a bit of a pain to draw. Mine are also not that close to Needham's originals. I don't think they need to be exact copies as long as they show the same characteristics as his diagrams I'm quite happy. The approach I take is to draw the curves by hand on graph paper and then identify some points on the curve and use them with a command called \psccurve (see below). Depending on the results I get, I then add extra points to them or subtract points from them until I'm happy with the result. The worst bit is placing and orienting the arrows. To use \pscurve the command is essentially \psccurve (x1,y1)(x2,y2)(x3,y3)(x4,y4)(x5,y5)(x6,y6) and you can add as many points as you like. I use a package called "ps-tricks" which has various options for drawing curves all based around a command called \pscurve. I used a variation of it called \psCcurve (my capitalisation), where the extra 'c' signifies a closed curve. You can read the details in the PSTricks manual in section 8 - to get there quickly do a pdf search for \pscurve after opening up the file. Here are two links - the first is a short intro to PSTricks which shows what it can do with examples, and the second is the full manual I refer to above. Introduction to PSTricksPSTricks manualVasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 27, 2017 22:34:33 GMT
Vasco, I thought I should learn how to produce various loops, so I gave it a try. My drawing program produced wiggly loop curves, so I tried one (the one on the right) with Mathematica's BezierCurve[] function. It was pretty tedious and to match Needham's curve more closely would have taken quite a lot of time adjusting control points. I am wondering how you were able to do it. Gary Yes, they are a bit of a pain to draw. Mine are also not that close to Needham's originals. I don't think they need to be exact copies as long as they show the same characteristics as his diagrams I'm quite happy. The approach I take is to draw the curves by hand on graph paper and then identify some points on the curve and use them with a command called \psccurve (see below). Depending on the results I get, I then add extra points to them or subtract points from them until I'm happy with the result. The worst bit is placing and orienting the arrows. To use \pscurve the command is essentially \psccurve (x1,y1)(x2,y2)(x3,y3)(x4,y4)(x5,y5)(x6,y6) and you can add as many points as you like. I use a package called "ps-tricks" which has various options for drawing curves all based around a command called \pscurve. I used a variation of it called \psCcurve (my capitalisation), where the extra 'c' signifies a closed curve. You can read the details in the PSTricks manual in section 8 - to get there quickly do a pdf search for \pscurve after opening up the file. Here are two links - the first is a short intro to PSTricks which shows what it can do with examples, and the second is the full manual I refer to above. Introduction to PSTricksPSTricks manualVasco Vasco, Thank you. PSTricks produces nice results. Rather than learning a new system, I think I will try to gain a better understanding of what Mathematica can do. I'm starting to get the hang of placing the points. With Bézier curves I've found that one could draw the curve, then draw some tangents from the curve to possible control points. I didn't have graph paper at hand, but that is how I do it in my mind. Mathematica has a FilledCurve[] function that has variable results. It completely filled the (d) curve which intersects itself three times, but it would not fill the interior loop of the (c) and (f) curves. The loop was filled with a disc plotted behind it. The arrows were drawn in, but could have been plotted. I notice that you still have the (e) curve labeled $\nu = 0$. Wouldn't $\nu = 1$ be more correct for the shaded region? nh.ch7.notes.p339.pdf (53.05 KB) Gary |
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Post by Admin on Feb 27, 2017 23:10:05 GMT
Gary
No, because I defined $\nu$ as the winding number of the loop about the origin, and in the diagram (e) the origin is outside the loop.
Having said that, there is no need for these captions to be there in order to answer the exercise. I think it would be better to remove them altogether from my answer.
Vasco
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Post by Admin on Feb 27, 2017 23:31:28 GMT
Gary
I have amended the diagram and accompanying text of my answer to make it less confusing, I hope! (see the post immediately before this one)
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 28, 2017 0:52:24 GMT
Yes, that works. But I think I will relabel them $\nu(L, p)$ rather than just $\nu$.
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Post by Admin on Feb 28, 2017 6:38:41 GMT
Yes, that works. But I think I will relabel them $\nu(L, p)$ rather than just $\nu$. Gary Each shaded diagram is meant to show the set of all points whose winding number is non-zero for that particular loop. That is: the set of all points $p$ for which $\nu(L,p)\neq 0$. In the case of (e), this set does not include the origin. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 28, 2017 16:33:13 GMT
Yes, that works. But I think I will relabel them $\nu(L, p)$ rather than just $\nu$. Gary Each shaded diagram is meant to show the set of all points whose winding number is non-zero for that particular loop. That is: the set of all points $p$ for which $\nu(L,p)\neq 0$. In the case of (e), this set does not include the origin. Vasco Vasco, I don't understand what I have written or drawn that would lead one to think that I thought that (e) includes the origin. I labeled it $\nu(L, p) = 1$ because the shaded points fall within a loop with a single wind. Also, I noticed that (c) should have been labeled $\nu(L, p) = \pm1$ (adding the $\pm$), because the loop segment on the RHS is clockwise for the points within it, while L is counterclockwise for the other sectors. Gary |
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Post by Admin on Feb 28, 2017 16:45:54 GMT
Gary Each shaded diagram is meant to show the set of all points whose winding number is non-zero for that particular loop. That is: the set of all points $p$ for which $\nu(L,p)\neq 0$. In the case of (e), this set does not include the origin. Vasco Vasco, I don't understand what I have written or drawn that would lead one to think that I thought that (e) includes the origin. I labeled it $\nu(L, p) = 1$ because the shaded points fall within a loop with a single wind. Gary Gary I wasn't commenting on what you had written, I was just trying to make it clear what my way of looking at the situation is. Vasco |
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Post by Admin on Feb 28, 2017 17:18:18 GMT
Gary
I assume you mean (d) not (c), but I don't agree with you about this because for any point $p$ not on the loop and letting $z$ be a point on the loop, we pivot the vector $z-p$ at $p$ and move $z$ round the whole loop, not just the sub-loop that the point $p$ is in.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Feb 28, 2017 20:41:20 GMT
Gary I assume you mean (d) not (c), but I don't agree with you about this because for any point $p$ not on the loop and letting $z$ be a point on the loop, we pivot the vector $z-p$ at $p$ and move $z$ round the whole loop, not just the sub-loop that the point $p$ is in. Vasco Vasco, Yes, (d). Here is how I see it. (Not say'n it's correct.): Unless you start from a particular point on the loop and rotate in a particular direction, a twisted loop like (d) has no intrinsic direction. If we select a point p in the right hand closed segment (Is there a name for this?) the loop is traveling clockwise relative to p. From the perspective of this point, that is the loop direction. The fact that it intersects and forms new enclosed regions does not affect the fact that it winds around p only once in the clockwise direction. Hence, $\nu(L,p) = -1$ OK. I thought I might have a basic misconception of the issue and I was trying to elicit the correct view. My view of it is the same as yours, I think. To test that, I would suggest that the location of the origin with respect to the interior or exterior of the loop is irrelevant to $\nu(L, p)$. Do you agree? Gary |
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Post by Admin on Mar 1, 2017 9:56:40 GMT
Gary
I misunderstood what you were saying in your original post. I agree that all the loops have no intrinsic direction - we can traverse them in whatever direction we like. That means that if we choose a point inside the small loop on the right of (d), we can go round the full loop in either direction and so $\nu$ for points inside any of the four small loops can be $\pm1$ depending on which direction we choose.
Looking at the suggested exercise again, it seems to me that it would be better to just illustrate the loops without any arrows, because the exercise tells us to shade the set $\nu(L,p)\neq0$, and this includes positive and negative winding numbers $\nu$. I will amend my document to remove the arrows and also the point at the origin, which only serves to confuse here.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Mar 1, 2017 18:37:29 GMT
Vasco,
We can go round in either direction, but since Needham has given L in (d) a particular direction I would retain it, in which case each set of p-points in the twisted loop has $\nu(L,p)$ that can be determined by following the segment of loop that surrounds it. So my label $\pm$ over (d) applies only to the group of p-point sets. For the right hand set and the left hand set, $\nu(L,p) = -1$. For the top and bottom sets, $\nu(L,p) = 1$. I should have put the numbers in the diagram and I will do so.
Gary
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Post by Admin on Mar 1, 2017 19:14:49 GMT
Maybe I didn't misunderstand you after all. In which case I would say that when the loop is traversed for a particular point $p$ we have to traverse the whole loop. So in diagram (d) for a point in the rightmost segment of the loop we cannot traverse just the segment of loop that surrounds it. Also, in figure 2 in the book on page 339, the values of $\nu$ in the little box on the right are derived from traversing the whole loop not just a segment of it for each of the sets $D_1,D_2,D_3,D_4$.
If we applied what you are proposing to figure 2 the value of $\nu$ for each of the four sets $D_1,D_2,D_3,D_4$ would be $\pm1$. Also as you traversed the segment containing the set $D_3$ you would sometimes be going with the arrows and sometimes against them.
Vasco
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