MisterStarshine
New member
- Joined
- Jan 1, 2016
- Messages
- 3
Hi.
I'm looking for a vector-field mapping function that has certain properties. Let us call the function F(x,y,z). I know the square of the magnitude of this function for all points. It is a rather messy function but I'll show it anyway.
\[\left( {x}^{2}+{y}^{2}+{z}^{2}\right) \,\left( {\left( x-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z-\frac{1}{\sqrt{3}}\right) }^{2}\right) \,\left( {\left( x+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z-\frac{1}{\sqrt{3}}\right) }^{2}\right) \,\left( {\left( x+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z+\frac{1}{\sqrt{3}}\right) }^{2}\right) \,\left( {\left( x-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z+\frac{1}{\sqrt{3}}\right) }^{2}\right) \]
The magnitude (and hence the vectorfield) is zero for the five points \[[0,0,0]\] \[[-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]\] \[[\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]\] \[[\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}]\] \[[-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}]\]
The knowledge of the magnitude alone doesn't lead to a specific answer for the vector field function but I know some more things. I happen to know how the function looks along the [1,1,1] direction and [-1,-1,1] and [-1,1,-1] and [1,-1,-1] directions.
In the [1,1,1] direction the following applies
\[{F}_{x}\left( \frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[{F}_{y}\left( \frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[{F}_{z}\left( \frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[\frac{\left( {x}^{2}-x\right) \,{\left( 3\,{x}^{2}+2\,x+3\right) }^{\frac{3}{2}}}{3}\]
The other directions have similar rules
\[{F}_{x}\left( -\frac{x}{\sqrt{3}},-\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[{F}_{y}\left( -\frac{x}{\sqrt{3}},-\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[-{F}_{z}\left( -\frac{x}{\sqrt{3}},-\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[\frac{\left( x-{x}^{2}\right) \,{\left( 3\,{x}^{2}+2\,x+3\right) }^{\frac{3}{2}}}{3}\]
and so on. Hopefully this is sufficient to determine the vector field I am searching for. I just can't wrap my head around the problem and figure out a method to determine an expression for the vector field. Thanks in advance.
Cheers
Tobbe
I'm looking for a vector-field mapping function that has certain properties. Let us call the function F(x,y,z). I know the square of the magnitude of this function for all points. It is a rather messy function but I'll show it anyway.
\[\left( {x}^{2}+{y}^{2}+{z}^{2}\right) \,\left( {\left( x-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z-\frac{1}{\sqrt{3}}\right) }^{2}\right) \,\left( {\left( x+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z-\frac{1}{\sqrt{3}}\right) }^{2}\right) \,\left( {\left( x+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z+\frac{1}{\sqrt{3}}\right) }^{2}\right) \,\left( {\left( x-\frac{1}{\sqrt{3}}\right) }^{2}+{\left( y+\frac{1}{\sqrt{3}}\right) }^{2}+{\left( z+\frac{1}{\sqrt{3}}\right) }^{2}\right) \]
The magnitude (and hence the vectorfield) is zero for the five points \[[0,0,0]\] \[[-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]\] \[[\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}]\] \[[\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}]\] \[[-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}]\]
The knowledge of the magnitude alone doesn't lead to a specific answer for the vector field function but I know some more things. I happen to know how the function looks along the [1,1,1] direction and [-1,-1,1] and [-1,1,-1] and [1,-1,-1] directions.
In the [1,1,1] direction the following applies
\[{F}_{x}\left( \frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[{F}_{y}\left( \frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[{F}_{z}\left( \frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[\frac{\left( {x}^{2}-x\right) \,{\left( 3\,{x}^{2}+2\,x+3\right) }^{\frac{3}{2}}}{3}\]
The other directions have similar rules
\[{F}_{x}\left( -\frac{x}{\sqrt{3}},-\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[{F}_{y}\left( -\frac{x}{\sqrt{3}},-\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[-{F}_{z}\left( -\frac{x}{\sqrt{3}},-\frac{x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right) \]\[=\]\[\frac{\left( x-{x}^{2}\right) \,{\left( 3\,{x}^{2}+2\,x+3\right) }^{\frac{3}{2}}}{3}\]
and so on. Hopefully this is sufficient to determine the vector field I am searching for. I just can't wrap my head around the problem and figure out a method to determine an expression for the vector field. Thanks in advance.
Cheers
Tobbe