Steve, same as elsewhere, when we get down to r=k,  STOP !!! HALTEN SIE!!!   You are entering an 11-dimensional zone !!!!!!!!!!!!
I keep having to pull him back from Sargasso Seas… Actually, why not somehow write in your URL in the very narrow space??  Tricia Mayo Butler does this in some filigree of her marvelous drawing/painting.


19 thoughts on “AT THE EDGE

  1. A related question we might revisit is the areas where |x| < 1 and |y| k. Should we try to let these areas back in? >>>This is how I started my LINE calcs. I think yesterday I demonstrated adequate expression for that whole range.


  2. Did a typo get into my question – I’ll try to restate. A related question we might revisit is the areas where |x| < 1 and |y| k?
    Huh??? I am allowed to, and have analyzed along y=k… there ais no restriction on x being in our physical theoretic realm, here.


  3. For some reason the comments are truncating my question. Sometimes r > k even if |y| < k. What I this means is whether more of the rectangular exclusion zone above/below the disk can be added into the plots?
    DON’T GO THERE, dangit. Lessee……….>calculating< I HAVE DECLARED r=k as the innermost radius, expressed by Kerr, OF OUR EXTERNAL VACUUM PHYSICS.


  4. FORGET YOUR ZONES.. In this algebra there is all kind of weird shit. Near x=1, and at close y, we get r = sqrt k. Put this in your pipe… Your first pic of the latest gorgeous series shows no ‘zones’ WE ARE TOUCHING A GOOD Q, since at points of x>0, the radius at y=k, is a little bit larger, yah??????????
    Let’s see, follow the distructions…


  5. I’d like to restate the question again in the event I’m not communicating it properly. If the exclusion zone is in fact r<k then it should look elliptical instead of rectangular in the plot. And there would be more real estate that gets plotted. Is this correct, or how is this incorrect?

    It seems that some points are inside the rectangle and outside the ellipse.


  6. This issue only becomes noticeable when we are zoomed in significantly. Right now I’m excluding based on the rectangle and the ellipse (a double test) when I think I should be testing only the ellipse? This seems simpler and more elegant.


  7. You will notice I have shined on this issue. BUSTED. Actually I looked at it yesterday, and I only need to know that there is not much further skating sideways, yah??
    Let us look at x=1/2. y=k. Easier to define a K = x^2 + y^2 – 1 AND HERE, it is -3/4 + k^2 . Now,
    2r^2 = K + sqrt[K^2 + 4y^2]. THE y-TERM IS SMALLER, except out at x≈1. EXPAND:
    , = K + |K| sqrt[ 1 + 4 (y/K)^2 ]= K + |K|[1+ 2(y/K)^2]. Did I do good? YOU GOTTA CHANGE THE SIGN ON K, cuz we do that with the sqrt WHEN K is negative. THIS IS THE ALGEBRAIC SWITCH.
    NOW WE HAVE EXPANDED THE ROOT, knowing the second term is ‘small’ . (K is negative NEAR THE DISK see???) r^2 = 0 + Ky^2/ K^2… CAREFUL WITH EXPONENTS !!!
    r^2 = y^2/K well whaddaya know?
    The definition of theta, is cos(theta) = y/r, so importantly, cos^2(theta) = K. ‘r’ is about the sqrt of |K| , or 0.86. π/6 radians… So, at x=1/2 and y ‘close’, the radius is only y/sqrt[|K|] = 1.18 k.
    DAMN, I STARTED FRESH AND NOW FEEL FRIED, SEE HOW IT GOES. OK look, it is helpful to do y=k, x=1, since we get to the infamous
    2r^2 = y^2 + sqrt[y^4 + 4y^2]. What’s the answer, I am toast. OK I CAN DO THIS, FOLLOW THE DISTRUCTIONS. y is ‘small’ so 2r^@ = y^2 + 2y.
    HERE THEN, AND ONLY HERE, r^2 ≈ y. Yeah. The sqrt of ‘k’ is larger than ‘k’, but still…………………….
    Except near the ring edge we can do the expansion with K >> y^2, yah? The one order of K comes out of the root, but there were two put under the 4y^2 term. WE ARE LEFT WITH HALF THAT TERM, or 2y^2/K = 2r^2 .


  8. Yes I did see you were looking at some cases with appropriate approximations. Thanks for the elaboration above.

    I do see sideways motion at the ellipse edge near the disk edge (WARNING new plot coming soon!!!). We can examine also your x=1/2 case in a bit.

    Liked by 1 person

  9. I’ll have the check the runs again to see if I’m attempting to test the square root of a negative number. Presently the code will terminate the line if this happens. Does this mean we can plot inside the ellipse if we switch the sign?
    DANGER, DANGER, Mr. whose-his-face I DO NO FIZZIKS in here. True madmen like Burinskii do !!!!!

    Liked by 1 person

  10. Or maybe I’ll have to become clearer on when to add the sqrt of the discriminant and when to subtract it. Are you delineating this in your discussion above? Perhaps I can generally add the sqrt(discriminant), except if 2r^2 would be < 0, then in that case I can subtract the sqrt(discriminant)?
    >>>SIMPLE, just look at the sign of x^2 + y^2 – 1. I did complete my discussion, despite fry-brain. INSIDE THE ROOT IT IS SQUARED, haaaaa. The detective will note outside the root is only the + sign.

    Liked by 1 person

  11. I can see that “x^2 + y^2 – 1” changes sign inside of a Cartesian circle of radius 1. So far the code seems to run OK if I employ the quadratic handling strategy mentioned immediately above.

    Perhaps sometime I can see if there’s a way to plot the situations where the sign is flipped by the code.
    >>> Mighta been my idiocy on y=k, but do explore into your coolly labelled FORBIDDEN ZONE, into y =k^2 at x=1. ‘k’ is a very small number, so its square is waaaay small.
    Actually it seems I have no problems here, because the cosine is going to zero, and the tangential field with it !!! See? Can I go out and play now???


  12. Thus the code when plotting these lines with theta within .07 degrees of the equator do not seem to hit any occasions of the flipping quadratic. This is the run zooming into the ring edge.
    >>>There they become dramatically nothing. [BIG HINT: this was planned.] Burinskii is no doubt choking on this tonight. This is his problem.

    Liked by 1 person

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