Hausdorff measure and angle of a regular simplex/tetrahedron
Let $T subset mathbb{R^3}$ be a regular simplex/tetrahedron with boundary $partial{T}$ and edge lenght 1.
In which way can the angle between two different edges be computed?
I know the formula and the result of $frac{pi}{3}$, but I'm not sure if I can choose any vertex vectors. For example, I picked $v_1=(1,1,1),v_2=(1,-1,-1)$ and $v_3=(-1,1,-1)$ and I got this result. I don't see how these vectors describe the simplex appropriately.
So which vectors can be used to get this result?
Otherwise, how to compute the Hausdorff measure $mathcal{H^2}(partial T)$ of this simplex? It should be 6, since it has six sides, but how can it be shown with the properties of the Hausdorff measure?
geometry measure-theory
asked Jan 4 at 9:56
HertulHertul
425
add a comment |
Let $T subset mathbb{R^3}$ be a regular simplex/tetrahedron with boundary $partial{T}$ and edge lenght 1.
In which way can the angle between two different edges be computed?
I know the formula and the result of $frac{pi}{3}$, but I'm not sure if I can choose any vertex vectors. For example, I picked $v_1=(1,1,1),v_2=(1,-1,-1)$ and $v_3=(-1,1,-1)$ and I got this result. I don't see how these vectors describe the simplex appropriately.
So which vectors can be used to get this result?
Otherwise, how to compute the Hausdorff measure $mathcal{H^2}(partial T)$ of this simplex? It should be 6, since it has six sides, but how can it be shown with the properties of the Hausdorff measure?
geometry measure-theory
asked Jan 4 at 9:56
HertulHertul
425
add a comment |
Let $T subset mathbb{R^3}$ be a regular simplex/tetrahedron with boundary $partial{T}$ and edge lenght 1.
In which way can the angle between two different edges be computed?
I know the formula and the result of $frac{pi}{3}$, but I'm not sure if I can choose any vertex vectors. For example, I picked $v_1=(1,1,1),v_2=(1,-1,-1)$ and $v_3=(-1,1,-1)$ and I got this result. I don't see how these vectors describe the simplex appropriately.
So which vectors can be used to get this result?
Otherwise, how to compute the Hausdorff measure $mathcal{H^2}(partial T)$ of this simplex? It should be 6, since it has six sides, but how can it be shown with the properties of the Hausdorff measure?
geometry measure-theory
asked Jan 4 at 9:56
HertulHertul
425
Let $T subset mathbb{R^3}$ be a regular simplex/tetrahedron with boundary $partial{T}$ and edge lenght 1.
In which way can the angle between two different edges be computed?
I know the formula and the result of $frac{pi}{3}$, but I'm not sure if I can choose any vertex vectors. For example, I picked $v_1=(1,1,1),v_2=(1,-1,-1)$ and $v_3=(-1,1,-1)$ and I got this result. I don't see how these vectors describe the simplex appropriately.
So which vectors can be used to get this result?
Otherwise, how to compute the Hausdorff measure $mathcal{H^2}(partial T)$ of this simplex? It should be 6, since it has six sides, but how can it be shown with the properties of the Hausdorff measure?
geometry measure-theory
geometry measure-theory
asked Jan 4 at 9:56
HertulHertul
425
asked Jan 4 at 9:56
HertulHertul
425
asked Jan 4 at 9:56
HertulHertul
425
asked Jan 4 at 9:56
HertulHertul
425
asked Jan 4 at 9:56
HertulHertul
425
425
add a comment |
add a comment |
1 Answer
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First of all you use the (alternate) vertex vectors $v_i$ of a unit cube. Thus the edge vectors $w_i$ of the tetrahedron would be eg. $w_{21}=v_2-v_1$ and $w_{31}=v_3-v_1$. Note that those edge vectors will have the length $||w_i||=sqrt{2}$. And by means of the cosine form of the scalar product you could derive that already provided angle.
From $T subset mathbb{R^3}$ (full tetrahedron) it follows that $partial{T}$ would be the surface, i.e. the size of 4 times the area of the according triangle (each with side length $||w_i||=sqrt{2}$ for sure). And $mathcal{H^2}(partial T)$ thus would just describe that very value, esp. since the Hausdorff measure and the Lesbeque measure do coincide for integral dimension values. (Provided both exist.)
What you are after instead, if I understand you correctly, is rather the 1-dimensional Hausdorff measure of the edge skeleton. I tentatively would write that as $mathcal{H^1}(partialpartial T)$. And that value then clearly becomes $sum_i ||w_i||=6sqrt{2}$.
If you'd try to calculate $mathcal{H^2}(partialpartial T)$ instead, you surely become Zero. Just consider the square faces of the encasing cube, subdivide those into $n^2$ squares of side length $1/n$ each, then you'd get a cover of the contained tetrahedral edge by just the diagonal $n$ ones only. That is, you'd have to consider the limit $lim_{ntoinfty}n/n^2=0$.
--- rk
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
add a comment |
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First of all you use the (alternate) vertex vectors $v_i$ of a unit cube. Thus the edge vectors $w_i$ of the tetrahedron would be eg. $w_{21}=v_2-v_1$ and $w_{31}=v_3-v_1$. Note that those edge vectors will have the length $||w_i||=sqrt{2}$. And by means of the cosine form of the scalar product you could derive that already provided angle.
From $T subset mathbb{R^3}$ (full tetrahedron) it follows that $partial{T}$ would be the surface, i.e. the size of 4 times the area of the according triangle (each with side length $||w_i||=sqrt{2}$ for sure). And $mathcal{H^2}(partial T)$ thus would just describe that very value, esp. since the Hausdorff measure and the Lesbeque measure do coincide for integral dimension values. (Provided both exist.)
What you are after instead, if I understand you correctly, is rather the 1-dimensional Hausdorff measure of the edge skeleton. I tentatively would write that as $mathcal{H^1}(partialpartial T)$. And that value then clearly becomes $sum_i ||w_i||=6sqrt{2}$.
If you'd try to calculate $mathcal{H^2}(partialpartial T)$ instead, you surely become Zero. Just consider the square faces of the encasing cube, subdivide those into $n^2$ squares of side length $1/n$ each, then you'd get a cover of the contained tetrahedral edge by just the diagonal $n$ ones only. That is, you'd have to consider the limit $lim_{ntoinfty}n/n^2=0$.
--- rk
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
add a comment |
First of all you use the (alternate) vertex vectors $v_i$ of a unit cube. Thus the edge vectors $w_i$ of the tetrahedron would be eg. $w_{21}=v_2-v_1$ and $w_{31}=v_3-v_1$. Note that those edge vectors will have the length $||w_i||=sqrt{2}$. And by means of the cosine form of the scalar product you could derive that already provided angle.
From $T subset mathbb{R^3}$ (full tetrahedron) it follows that $partial{T}$ would be the surface, i.e. the size of 4 times the area of the according triangle (each with side length $||w_i||=sqrt{2}$ for sure). And $mathcal{H^2}(partial T)$ thus would just describe that very value, esp. since the Hausdorff measure and the Lesbeque measure do coincide for integral dimension values. (Provided both exist.)
What you are after instead, if I understand you correctly, is rather the 1-dimensional Hausdorff measure of the edge skeleton. I tentatively would write that as $mathcal{H^1}(partialpartial T)$. And that value then clearly becomes $sum_i ||w_i||=6sqrt{2}$.
If you'd try to calculate $mathcal{H^2}(partialpartial T)$ instead, you surely become Zero. Just consider the square faces of the encasing cube, subdivide those into $n^2$ squares of side length $1/n$ each, then you'd get a cover of the contained tetrahedral edge by just the diagonal $n$ ones only. That is, you'd have to consider the limit $lim_{ntoinfty}n/n^2=0$.
--- rk
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
add a comment |
First of all you use the (alternate) vertex vectors $v_i$ of a unit cube. Thus the edge vectors $w_i$ of the tetrahedron would be eg. $w_{21}=v_2-v_1$ and $w_{31}=v_3-v_1$. Note that those edge vectors will have the length $||w_i||=sqrt{2}$. And by means of the cosine form of the scalar product you could derive that already provided angle.
From $T subset mathbb{R^3}$ (full tetrahedron) it follows that $partial{T}$ would be the surface, i.e. the size of 4 times the area of the according triangle (each with side length $||w_i||=sqrt{2}$ for sure). And $mathcal{H^2}(partial T)$ thus would just describe that very value, esp. since the Hausdorff measure and the Lesbeque measure do coincide for integral dimension values. (Provided both exist.)
What you are after instead, if I understand you correctly, is rather the 1-dimensional Hausdorff measure of the edge skeleton. I tentatively would write that as $mathcal{H^1}(partialpartial T)$. And that value then clearly becomes $sum_i ||w_i||=6sqrt{2}$.
If you'd try to calculate $mathcal{H^2}(partialpartial T)$ instead, you surely become Zero. Just consider the square faces of the encasing cube, subdivide those into $n^2$ squares of side length $1/n$ each, then you'd get a cover of the contained tetrahedral edge by just the diagonal $n$ ones only. That is, you'd have to consider the limit $lim_{ntoinfty}n/n^2=0$.
--- rk
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
First of all you use the (alternate) vertex vectors $v_i$ of a unit cube. Thus the edge vectors $w_i$ of the tetrahedron would be eg. $w_{21}=v_2-v_1$ and $w_{31}=v_3-v_1$. Note that those edge vectors will have the length $||w_i||=sqrt{2}$. And by means of the cosine form of the scalar product you could derive that already provided angle.
From $T subset mathbb{R^3}$ (full tetrahedron) it follows that $partial{T}$ would be the surface, i.e. the size of 4 times the area of the according triangle (each with side length $||w_i||=sqrt{2}$ for sure). And $mathcal{H^2}(partial T)$ thus would just describe that very value, esp. since the Hausdorff measure and the Lesbeque measure do coincide for integral dimension values. (Provided both exist.)
What you are after instead, if I understand you correctly, is rather the 1-dimensional Hausdorff measure of the edge skeleton. I tentatively would write that as $mathcal{H^1}(partialpartial T)$. And that value then clearly becomes $sum_i ||w_i||=6sqrt{2}$.
If you'd try to calculate $mathcal{H^2}(partialpartial T)$ instead, you surely become Zero. Just consider the square faces of the encasing cube, subdivide those into $n^2$ squares of side length $1/n$ each, then you'd get a cover of the contained tetrahedral edge by just the diagonal $n$ ones only. That is, you'd have to consider the limit $lim_{ntoinfty}n/n^2=0$.
--- rk
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
answered Jan 4 at 11:55
Dr. Richard KlitzingDr. Richard Klitzing
1,40116
1,40116
add a comment |
add a comment |
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