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Post by Admin on Apr 10, 2018 23:20:29 GMT
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Post by telemeter on Mar 23, 2020 11:57:56 GMT
On Ex 1 part ii)
The answers attached are as ever comprehensive and detailed. I'm more of a physicist than a mathematician so will take the quick 'n dirty route wherever I can. In particular, part ii) of question 1 can be very involved...more than I can handle anyway!
If you want a quicker result (& more geometric approach) to part ii) without any differentiation at all, how about this:-
Let X = z bar = r*exp(-i*theta)
Let ds= element along streamlime; Let dp = element along the orthogonal equipotential
Let Rs = Radius of curvature of streamline at z; Let Rp = radius of curvature of equipotential at z
Then a quick diagram of these elements (in particular of the infinitesimals dr, ds and r*d(theta)) at z will convince you that:
r*d(theta)= ds* sin (2*theta) (1)
r*d(theta) = dp*cos(2*theta) (2)
dr = ds*cos(2*theta) (3)
dr = dp*sin(2*theta) (4)
Let the radii of curvature rotate by d(theta) then (ie same d(theta) at origin and centre of curvature...I think valid in the limit d(theta)=0)
Rs * d(theta) = ds (5)
Rp * d(theta) = dp (6)
To get the curvatures, substitute (5) and (6) into (1) and (2) respectively to get:
Rs = r/ sin (2*theta) (7) and Rp = -r/ cos( 2*theta) (8) The minus sign arising because Rp has to rotate oppositely to r to generate ds.
Considering Div....
d/ds = dr/ds*d/dr From (3) dr/ds = cos(2*theta) so,
d/ds (|X|) + |X|/Rp = 0 as required. (div zbar being zero, of course)
Considering curl...
d/dp = dr/dp*d/dr from (4) dr/dp = sin(2*theta) so,
-d/dp(|X|) + |X|/Rs = 0 as required (curl zbar being zero, of course).
Note that the expressions for Rs and Rp agree with the published answer if you remember your trig identities and set x= r*cos(theta); y=r*sin(theta).
The only query I have is around my use of d(theta) at the centres of curvature and the origin...my answer to any challenge here is that in the limit d(theta) goes to zero, my appraoch will be exact.
It looks a nice quick solution, does it not? Why o why did take many hours to find!?
telemeter
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Post by mondo on Aug 12, 2022 4:16:20 GMT
Vasco,
I have a questions to your proposed solution. You say "It is clear from this expression that the streamlines S are radial lines through the origin..". This is similar to what Needham said on page 496 right below equations for $\bar H$ "the streamlines of which are clockwise circles round the origin." I don't see it that clearly just by looking at $\bar H$ or in case of exercise $1$ on $X$. So what is the trick here to get the geometry of a streamline just by looking at a vector field equation? How can we tell that a streamlines of a vector field $\bar H = (y -x)$ is a circle round the origin while $X = \frac{1}{\bar z}$ has a radial streamlines?
Thank you.
Mondo
If $X=(1/\overline{z})=(1/r)e^{i\theta}$ then we can see immediately that the direction of $X$ is $\theta$, the same as the point $z=re^{i\theta}$ where $X$ originates. So it is clear that the streamlines are radial lines through the origin since the vector $X$ is tangent to the streamlines by definition.
As far as page 496 is concerned you have made the same mistake again $\overline{H}=-iz=y-ix$ NOT $y-x$. So $\overline{H}$ is $z$ rotated through an angle of $-\pi/2$ and so is tangent to an origin-centred circle - a streamline by definition.
Vasco
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Post by mondo on Aug 15, 2022 6:00:35 GMT
Sorry for late response on it. You answered by accidentally modifying my post so I did not get any notification. If $X=(1/\overline{z})=(1/r)e^{i\theta}$ then we can see immediately that the direction of $X$ is $\theta$, the same as the point $z=re^{i\theta}$ where $X$ originates. So it is clear that the streamlines are radial lines through the origin since the vector $X$ is tangent to the streamlines by definition. Yes, it is clear for me now as well  As far as page 496 is concerned you have made the same mistake again $\overline{H}=-iz=y-ix$ NOT $y-x$. Actually I did a different mistake - I poorly wrote my answer. When I wrote $\bar H = (y -x)$ I should have at least put a semicolon in between to indicate I mean a vector $\begin{pmatrix} y \\ -x \end{pmatrix}$. I am sorry about the confusion. So $\overline{H}$ is $z$ rotated through an angle of $-\pi/2$ and so is tangent to an origin-centred circle - a streamline by definition. I see the geometry now but I would like to clarify your refence to a streamline. I think it all boils down to the ability of visualizing the vector field. In this case a figure [8] from page 12 is handy (although the direction of rotation is different). So if we imagine only points $z$ located at a circle, $\bar H$ will be always tangent to that circle. Finally, if we know the relation between a vector field and a streamline we just found both. Do you agree Vasco? Thanks! |
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