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Post by telemeter on Apr 15, 2020 8:51:23 GMT
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Post by Admin on Apr 16, 2020 10:21:20 GMT
telemeter
I have an answer for this exercise which I haven't yet posted. I will take a look at your answer and get back to you.
I have sent you a private message.
Vasco
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Post by Admin on Apr 16, 2020 11:21:42 GMT
telemeter
In your answer to part (i) you say "As an analytic mapping sends circles to circles...", but this is not so. It is correct however to say that an analytic mapping sends infinitesimal circles to infinitesimal circles.
See pages 197-8 of the book.
Vasco
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Post by telemeter on Apr 17, 2020 8:33:47 GMT
Vasco,
Thanks, a very good point, I was mixing in my Mobius transformations! I'm tempted to rely on the angles between field lines then.
telemeter
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Post by Admin on Apr 20, 2020 9:40:07 GMT
telemeter
I have looked at your answer and at mine and my answer is different as I find that the dipole moment is changed by the mapping.
Vasco
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Post by telemeter on Apr 20, 2020 20:39:09 GMT
Vasco,
I won't say I'm definitely correct but I had the following two lines of reasoning which made me comfortable...
A
1) A source maps to another source of the same strength. Therefore
2) A sink will map to another sink of the same strength. Therefore
3) A doublet with finite separation will map to a doublet of the same net strength
And just as a dipole is created by the limiting process of bringing a doublet's separation to zero in the z plane, you can also do this in the mapped, w, plane. If you are creating a dipole in the w and z planes from two sources of the same strength, and slowly bringing them together, should not the result be the same in both planes..as the rules of geometry are the same in each.
Thus, considering dipole construction, I see the fact that single charges are unchanged means that dipoles must also be unchanged.
B
A dipole will be amplitwisted in the mapping, as the angle and infinitesimal separation epsilon will change so d will go to f'*d. But this effect is countered by the "distortion of space" , leaving the new dipole moment to be d. Just considering d on its own (as if one could) that does change.
Do you see an error in the maths? Or is it that you think my statement that d' = f'd is unwarranted?
telemeter
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Post by Admin on Jul 1, 2020 13:04:22 GMT
telemeter The link in your first post in this thread no longer works. I have published my answer here. I do not agree with your conclusions about this exercise. If you still disagree with me I suggest you read the paragraph under figure 28 on page 548 of the book. Vasco |
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