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Math Help - The crumpled sombrero

  1. #1
    Forum Admin topsquark's Avatar
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    The crumpled sombrero

    I have rather embarrassing request to make. See, I visualize badly in anything above 2 dimensions, so I have to work hard to figure out equations for 3D shapes. Additionally I am slightly "technologically challenged" as to 3D graphing software: the only one I've ever understood was Mathematica, which I no longer have a license for.

    So it comes down to this: I would like someone to graph a surface for me so I can double check that I have something of the correct form.

    What I am going for is this: The "base" of this shape is, in Physics, called the "sombrero" potential. (I've attached a cross-section of the graph below. It is named so for the rough similarity the shape has to the Mexican hat.) The basic form of this equation is z = -a \rho ^2 + b \rho ^4 where a and b are positive. (I'm using cylindrical coordinates.)

    Now, imagine a ball sitting at \rho = 0. It is in an unstable equilibrium. It will roll down the "sombrero" and eventually settle at the lowest point, of which there are an infinite number, due to the rotational symmetry.

    What I want to do is alter this function so that the ball can only roll down certain "channels": instead of having the same probability for ending up at any point on the minimum, I want it to have to choose one of several locations I have predetermined. So we must introduce some sort of sinusoidal variation into the sombrero.

    What I have come up with is this:
    z = (-a \rho ^2 + b \rho ^4) + c \rho ~ sin \left ( \frac{ 2\pi \theta}{k} \right )

    The value of k will be integral and will determine the number of "valleys" (absolute minima) the surface will contain.

    Since I need someone to graph this, let me give a sample graph to look at:
    z = (-6 \rho ^2 + \rho ^4) + \frac{1}{5} \rho ~ sin \left ( \frac{ \pi \theta}{2} \right )

    I would like to verify that this surface has the properties I am looking for (and that my poor deficient mind is coming up with as a solution.)

    Thank you for your trouble.

    -Dan

    PS If anyone has a "correct" Mathematical name for this surface I would be interested in knowing it. For the moment I'm calling it the "crumpled sombrero."
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-sombrero.jpg  
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  2. #2
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    Hello,

    instead of cylindrical coordinates I used the normal cartesian coordinate system.

    f(x,y)=9 \cdot e^{- \frac14 \cdot \sqrt{x^2+y^2}} \cdot \cos\left(\sqrt{x^2+x^2}  \right)
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-sombrero.jpg  
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  3. #3
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    Hi,

    next attempt: This sombrero has a flat base.

    f(x,y)=4 \cdot e^{-\frac1{10} \cdot \sqrt{x^2+y^2}} \cdot \left(1+\cos\left( \sqrt{x^2+y^2} \right) \right)
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-sombrero_flatbase.gif  
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    Forum Admin topsquark's Avatar
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    Hmmm... Thanks for the graphs. This isn't what I am looking for. Your graphs have radial symmetry about the z axis, which is not what I am looking for.

    Let's go for a "Close Encounters" approach.

    Say we have a lot of mashed potatoes. I am going to build my sombrero out of the mashed potatoes. Now, I'm going a little spastic due to radiation burns, so I'm going to take a spoon and cut out a channel from the top of the sombrero down to the rim in a straight line. (Call this channel the North side of the sombrero.) I'm going to do this three more times at the W, E, and S sides of the sombrero.

    That's the kind of shape I'm trying for. (It looks a little like a bundt cake.)

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    ...

    Say we have a lot of mashed potatoes. I am going to build my sombrero out of the mashed potatoes. Now, I'm going a little spastic due to radiation burns, so I'm going to take a spoon and cut out a channel from the top of the sombrero down to the rim in a straight line. ...
    Hi,

    must be a very nice hobby ... remeinds me of my son when he started to eat without any help

    Quote Originally Posted by topsquark View Post
    ...
    That's the kind of shape I'm trying for. (It looks a little like a bundt cake.)
    Hi, I've attached a graph which comes a little bit closer to your cake.

    I've used an epicycloid curve as boundary of the cross section:

    \left \{\begin{array}{l}x=\left(5 \cos\left(\frac14 t\right)-\frac12 \cos\left(\frac54 t\right)\right) \cdot s \\ y=\left(5 \sin\left(\frac14 t\right)-\frac12 \sin\left(\frac54 t\right)\right) \cdot s \\ z=\frac{4}{\frac14 s + 1} \cdot (1+\cos(s))\end{array}\right.
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-sombrero_epicykl.gif  
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  6. #6
    Forum Admin topsquark's Avatar
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    Thank you again. This seems to be a tougher problem than I had originally thought, so I appreciate the effort you are taking. (And I never would have thought to parametrize the surface.)

    I hate to be so demanding about it, but it's not quite what I am looking for. The "channels" I was talking about are there (down the central "cone" if you can call it that.) I need something extra, though. The cross-sections of this surface parallel to the xy plane shows that a ball could roll freely around the entire cross-section. That is to say, the minimum of the surface is level.

    What I need is something featuring those same channels such that a ball rolled down one of the channels will land in a "valley" and not come loose even if we shake the sombrero a bit. That is to say, the ball will land in a cup and not be able to freely move (far) along any direction.

    I'm explaining this badly, I know. Let's go back to that sombrero shape. The sombrero has a minimum that is circular. What I wish to do is make the "minimum" sinusoidal: 4 valleys and 4 hills where the minimum of the sombrero was.

    (sigh) I don't have time now. Perhaps it would be best if I used my rudimentary art skills to sketch what I'm after. I'll see what I can do along those lines later.

    Thanks again!

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    ...And I never would have thought to parametrize the surface.
    ...
    Hi, topsquark,

    to be honest I found this way by accident.
    Quote Originally Posted by topsquark View Post
    ...The "channels" I was talking about are there (down the central "cone" if you can call it that.) I need something extra, though. The cross-sections of this surface parallel to the xy plane shows that a ball could roll freely around the entire cross-section. That is to say, the minimum of the surface is level.

    What I need is something featuring those same channels such that a ball rolled down one of the channels will land in a "valley" and not come loose even if we shake the sombrero a bit. That is to say, the ball will land in a cup and not be able to freely move (far) along any direction....
    You actually explain your problem very well and if I understand you correctly you would like to get a kind of ditch the "hillsides" up and down and even through the "valley".

    I've changed the "active radius" of the outer circle. Now the valley is divided by a ridge so the movement of the ball is slightly limited but... yes I know it isn't quite that what you are looking for.

    By the way: I consider your problem as a challenge to me - and it makes me furious to experience that I can't help you because my math is so limited.
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-sombrero_epicykl2.gif  
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  8. #8
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    new model of sombrero

    Quote Originally Posted by topsquark View Post
    ...
    That's the kind of shape I'm trying for. (It looks a little like a bundt cake.)

    -Dan
    Hi, Dan,

    I've done a little bit of playing around and I found:

    \text{sombrero with ditches}=\left \{\begin{array}{l}x = s \cdot \cos(t) \\y = s \cdot \sin(t) \\ z = 3 \cdot e^{-0.1 \cdot s} \cdot (1+\cos(s))-\frac25 \cdot e^{1-\cos(4t)}\end{array} \right.
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-somb_all.jpg  
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    Forum Admin topsquark's Avatar
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    Sorry, earboth, I think I may have found it but apparently my last post got lost somehow.

    Could you try z = -\frac{3}{2} \rho ^2 + \frac{1}{2} \rho ^4  + \frac{1}{4} \rho sin(4 \theta)
    for me please? (I've tried to guess coefficients that will show things in a proper scale.)

    -Dan
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    Quote Originally Posted by topsquark View Post
    Sorry, earboth, I think I may have found it but apparently my last post got lost somehow.

    Could you try z = -\frac{3}{2} \rho ^2 + \frac{1}{2} \rho ^4  + \frac{1}{4} \rho sin(4 \theta)
    for me please? (I've tried to guess coefficients that will show things in a proper scale.)

    -Dan
    Hi, Dan,

    I've attached the graph of your equation.

    As you can see the graph is going up unlimited while I tried to get a "hill" with embankments and ditches.
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-somb_topsquark1.gif  
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    Quote Originally Posted by earboth View Post
    Hi, Dan,

    I've attached the graph of your equation.

    As you can see the graph is going up unlimited while I tried to get a "hill" with embankments and ditches.
    Actually that works for me. All I am interested in is the first set of "valleys." Pretty much all that matters to me is that the central peak has a (relatively) flat center and the structure of the valleys. (What I'm adding to the Physics model is the peaks and troughs where the sombrero potential would have its radially symmetric ring of minima values.

    _Dan
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    Quote Originally Posted by topsquark View Post
    Actually that works for me. All I am interested in is the first set of "valleys." Pretty much all that matters to me is that the central peak has a (relatively) flat center and the structure of the valleys. ...
    Hi, Dan,

    in this case you get a slightly enlarged graph so you have a closer and more accurate look on the area of interest.
    Attached Thumbnails Attached Thumbnails The crumpled sombrero-somb_topsquark2.gif  
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    Quote Originally Posted by earboth View Post
    Hi, Dan,

    in this case you get a slightly enlarged graph so you have a closer and more accurate look on the area of interest.
    Words cannot describe the feelings i experience when i look at graphs like these. Now THIS is a fantastic part of mathematics...
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    Quote Originally Posted by earboth View Post
    Hi, Dan,

    in this case you get a slightly enlarged graph so you have a closer and more accurate look on the area of interest.
    Oh that's gorgeous. And it is exactly what I was looking for.

    Let me explain a little of where I am going with this.

    In the GWS electroweak model, we take the potential for the (so far undiscovered) scalar Higg's particle to be -a \phi ^2 + b \phi^4, where \phi is a "radial" field in a bizarre functional space. This is the so-called sombrero potential.

    Now, as the Universe cooled after the Big Bang, the idea is that the Universe settled into a "false vacuum" at the top of the sombrero. This is an unstable equilibrium and as the Universe cools even more, the situation becomes very unstable. At some point, perhaps due to quantum fluctuation perhaps something else, the equilibrium is broken and the state of the Universe "rolls down" the sombrero into the circle of minimum values at the bottom of the sombrero.

    Which point on the sombrero it picks is completely arbitrary. The state of the Universe really doesn't depend on it, so long as the state is somewhere in the valley. The process of the state of the Universe moving from a false vacuum at the start and ending with a true vacuum at the end is called "broken symmetry." The equations for the state of the Universe were initially radially symmetric in the field variable \phi. When the Universe enters a true vacuum state, that symmetry no longer exists.

    This can all be put into a more solid Mathematical form using the principles of Quantum Field Theory. If you are interested, look up "symmetry breaking" and the "Goldstone Theorem."

    Now, when the Universe falls into the vacuum state, it "freezes" the state of the Universe. (Indeed, the context is literal: you can nearly instantly freeze a supercooled fluid.) This is actually the starting point of another kind of theory called "inflation" because when the Universe does this it releases a tremendous amount of energy, which goes into the expansion of the Universe.

    The problem I see with the freezing process is that not all regions of the Universe need fall into exactly the same state; all that is necessary is that all of the Universe falls into the same ground state potential. There's a level of arbitrariness here that I don't like. This is much the same thing as a heated magnet forming magnetic domains as it is quenched rapidly. The magnetic domains need not all point in the same direction. For the Universe it is not known how these domains would interact. Indeed, it is quite possible that they don't, which would mean that the universe that we see could potentially be just a small domain of the entire Universe.

    The whole point of this is that we really don't have any good idea of what the Higg's potential should look like. My thinking is that it makes more intuitive sense (to me anyway) that the Universe can only "roll" into a finite set of possible ground states. Since the Higg's potential works very well for the GWS electroweak model, I thought I'd modify the potential to select, say, only four possible ground states and see how it affects the Physics.

    Hence my pickiness on the structure of the surface representing the Higg's potential.

    -Dan
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