Mixing
What does this correspond to geometrically?
$begingroup$
I recently was playing around with some maths and pondered the following:
Let us define
$$ Delta vec s_{12} = vec s_1 - vec s_2 $$
Squaring both sides:
$$ |Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2 $$
Choosing both $|s_1|=|s_2|$ and dividing by $|s_1|^2$:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} = 2(1 + cos(z_{12})) $$
Where $z_{12}$ is the angle. Multiplying both sides with $ 1 - cos(z_{12})$ we get:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} cos^2(frac{z_{12}}{2})= sin^2(z_{12}) $$
Taking $Delta s to ds $, $z_{12} to epsilon$ and multiplying $ d epsilon$ both sides and integrating around in a loop.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon $$
I was originally thinking of this in Hyperbolic or Euclidean geometry in $2$ dimensions. Is this correct? Can this be extended to $4$ dimensions? What does this correspond to geometrically?
geometry hyperbolic-geometry
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
$endgroup$
add a comment |
$begingroup$
I recently was playing around with some maths and pondered the following:
Let us define
$$ Delta vec s_{12} = vec s_1 - vec s_2 $$
Squaring both sides:
$$ |Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2 $$
Choosing both $|s_1|=|s_2|$ and dividing by $|s_1|^2$:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} = 2(1 + cos(z_{12})) $$
Where $z_{12}$ is the angle. Multiplying both sides with $ 1 - cos(z_{12})$ we get:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} cos^2(frac{z_{12}}{2})= sin^2(z_{12}) $$
Taking $Delta s to ds $, $z_{12} to epsilon$ and multiplying $ d epsilon$ both sides and integrating around in a loop.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon $$
I was originally thinking of this in Hyperbolic or Euclidean geometry in $2$ dimensions. Is this correct? Can this be extended to $4$ dimensions? What does this correspond to geometrically?
geometry hyperbolic-geometry
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
$endgroup$
1
$begingroup$
Your steps seem to be just restating $$cos theta = frac{ vec{s}_1 cdot vec{s}_2 }{lVertvec{s}_1 rVert , lVert vec{s}_2 rVert}$$That is the geometric definition of dot product, and works all Euclidean spaces (including the high-dimensional ones). As far as I understand, the last equality is just $pi = pi$ by definition. What am I missing here? I do have the feeling that there is an interesting question there, I just cannot see it.
$endgroup$
– Nominal Animal
Jan 21 at 15:34
$begingroup$
I was wondering if it worked with hyperbolic geometry (with imaginary $z$) and if there was a version of this in $4$ dimensions with signature $(-1,1,1,1)$?
$endgroup$
– More Anonymous
Jan 21 at 15:39
$begingroup$
Also how would u compute the L.H.S of the last equality?
$endgroup$
– More Anonymous
Jan 21 at 15:41
1
$begingroup$
You may need to restate the question, then, methinks. (Unfortunately, my own math-fu does not extend to non-Euclidean geometries.)
$endgroup$
– Nominal Animal
Jan 21 at 16:19
$begingroup$
Several mistakes in your development. Please proofread.
$endgroup$
– Yves Daoust
Jan 22 at 10:47
add a comment |
$begingroup$
I recently was playing around with some maths and pondered the following:
Let us define
$$ Delta vec s_{12} = vec s_1 - vec s_2 $$
Squaring both sides:
$$ |Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2 $$
Choosing both $|s_1|=|s_2|$ and dividing by $|s_1|^2$:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} = 2(1 + cos(z_{12})) $$
Where $z_{12}$ is the angle. Multiplying both sides with $ 1 - cos(z_{12})$ we get:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} cos^2(frac{z_{12}}{2})= sin^2(z_{12}) $$
Taking $Delta s to ds $, $z_{12} to epsilon$ and multiplying $ d epsilon$ both sides and integrating around in a loop.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon $$
I was originally thinking of this in Hyperbolic or Euclidean geometry in $2$ dimensions. Is this correct? Can this be extended to $4$ dimensions? What does this correspond to geometrically?
geometry hyperbolic-geometry
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
$endgroup$
I recently was playing around with some maths and pondered the following:
Let us define
$$ Delta vec s_{12} = vec s_1 - vec s_2 $$
Squaring both sides:
$$ |Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2 $$
Choosing both $|s_1|=|s_2|$ and dividing by $|s_1|^2$:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} = 2(1 + cos(z_{12})) $$
Where $z_{12}$ is the angle. Multiplying both sides with $ 1 - cos(z_{12})$ we get:
$$ frac{|Delta s_{12} |^2}{|s_1|^2} cos^2(frac{z_{12}}{2})= sin^2(z_{12}) $$
Taking $Delta s to ds $, $z_{12} to epsilon$ and multiplying $ d epsilon$ both sides and integrating around in a loop.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon $$
I was originally thinking of this in Hyperbolic or Euclidean geometry in $2$ dimensions. Is this correct? Can this be extended to $4$ dimensions? What does this correspond to geometrically?
geometry hyperbolic-geometry
geometry hyperbolic-geometry
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
asked Jan 21 at 15:12
More AnonymousMore Anonymous
36919
36919
1
$begingroup$
Your steps seem to be just restating $$cos theta = frac{ vec{s}_1 cdot vec{s}_2 }{lVertvec{s}_1 rVert , lVert vec{s}_2 rVert}$$That is the geometric definition of dot product, and works all Euclidean spaces (including the high-dimensional ones). As far as I understand, the last equality is just $pi = pi$ by definition. What am I missing here? I do have the feeling that there is an interesting question there, I just cannot see it.
$endgroup$
– Nominal Animal
Jan 21 at 15:34
$begingroup$
I was wondering if it worked with hyperbolic geometry (with imaginary $z$) and if there was a version of this in $4$ dimensions with signature $(-1,1,1,1)$?
$endgroup$
– More Anonymous
Jan 21 at 15:39
$begingroup$
Also how would u compute the L.H.S of the last equality?
$endgroup$
– More Anonymous
Jan 21 at 15:41
1
$begingroup$
You may need to restate the question, then, methinks. (Unfortunately, my own math-fu does not extend to non-Euclidean geometries.)
$endgroup$
– Nominal Animal
Jan 21 at 16:19
$begingroup$
Several mistakes in your development. Please proofread.
$endgroup$
– Yves Daoust
Jan 22 at 10:47
add a comment |
1
$begingroup$
Your steps seem to be just restating $$cos theta = frac{ vec{s}_1 cdot vec{s}_2 }{lVertvec{s}_1 rVert , lVert vec{s}_2 rVert}$$That is the geometric definition of dot product, and works all Euclidean spaces (including the high-dimensional ones). As far as I understand, the last equality is just $pi = pi$ by definition. What am I missing here? I do have the feeling that there is an interesting question there, I just cannot see it.
$endgroup$
– Nominal Animal
Jan 21 at 15:34
$begingroup$
I was wondering if it worked with hyperbolic geometry (with imaginary $z$) and if there was a version of this in $4$ dimensions with signature $(-1,1,1,1)$?
$endgroup$
– More Anonymous
Jan 21 at 15:39
$begingroup$
Also how would u compute the L.H.S of the last equality?
$endgroup$
– More Anonymous
Jan 21 at 15:41
1
$begingroup$
You may need to restate the question, then, methinks. (Unfortunately, my own math-fu does not extend to non-Euclidean geometries.)
$endgroup$
– Nominal Animal
Jan 21 at 16:19
$begingroup$
Several mistakes in your development. Please proofread.
$endgroup$
– Yves Daoust
Jan 22 at 10:47
1
1
$begingroup$
Your steps seem to be just restating $$cos theta = frac{ vec{s}_1 cdot vec{s}_2 }{lVertvec{s}_1 rVert , lVert vec{s}_2 rVert}$$That is the geometric definition of dot product, and works all Euclidean spaces (including the high-dimensional ones). As far as I understand, the last equality is just $pi = pi$ by definition. What am I missing here? I do have the feeling that there is an interesting question there, I just cannot see it.
$endgroup$
– Nominal Animal
Jan 21 at 15:34
$begingroup$
Your steps seem to be just restating $$cos theta = frac{ vec{s}_1 cdot vec{s}_2 }{lVertvec{s}_1 rVert , lVert vec{s}_2 rVert}$$That is the geometric definition of dot product, and works all Euclidean spaces (including the high-dimensional ones). As far as I understand, the last equality is just $pi = pi$ by definition. What am I missing here? I do have the feeling that there is an interesting question there, I just cannot see it.
$endgroup$
– Nominal Animal
Jan 21 at 15:34
$begingroup$
I was wondering if it worked with hyperbolic geometry (with imaginary $z$) and if there was a version of this in $4$ dimensions with signature $(-1,1,1,1)$?
$endgroup$
– More Anonymous
Jan 21 at 15:39
$begingroup$
I was wondering if it worked with hyperbolic geometry (with imaginary $z$) and if there was a version of this in $4$ dimensions with signature $(-1,1,1,1)$?
$endgroup$
– More Anonymous
Jan 21 at 15:39
$begingroup$
Also how would u compute the L.H.S of the last equality?
$endgroup$
– More Anonymous
Jan 21 at 15:41
$begingroup$
Also how would u compute the L.H.S of the last equality?
$endgroup$
– More Anonymous
Jan 21 at 15:41
1
1
$begingroup$
You may need to restate the question, then, methinks. (Unfortunately, my own math-fu does not extend to non-Euclidean geometries.)
$endgroup$
– Nominal Animal
Jan 21 at 16:19
$begingroup$
You may need to restate the question, then, methinks. (Unfortunately, my own math-fu does not extend to non-Euclidean geometries.)
$endgroup$
– Nominal Animal
Jan 21 at 16:19
$begingroup$
Several mistakes in your development. Please proofread.
$endgroup$
– Yves Daoust
Jan 22 at 10:47
$begingroup$
Several mistakes in your development. Please proofread.
$endgroup$
– Yves Daoust
Jan 22 at 10:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2$$
That should be
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 - 2 s_1 cdot s_2$$
Choosing both $|s_1| = |s_2|$
This means that your calculations only apply to a circle around the origin. It may seem like infinitesimally you can still apply it to other loops, but you are only picking up the contribution in the direction of $dtheta$, and dropping that in the direction of $dr$. So, no, it only holds for the circle.
Multiplying both sides with $1 - cos(z_{12})$
Because of the earlier sign change, this would be $1 + cos(z_{12})$. However, the next line has another error that cancels out the previous (perhaps the sign error was only when you copied it into here?)$$cos^2 frac theta 2 = frac{1 + cos theta}2ne frac{1 - cos theta}2$$ so the next formula is correct again.
Taking $Delta s to ds, z_{12} to epsilon$
The cosine of $z_{12}$ is the length of $Delta s$ divided by the radius. If $Delta s$ becomes infinitesimal, then $z_{12}to pi/2$ I.e. $epsilon = pi/2, depsilon = 0$, and your integrations are over a constant instead of a variable.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon$$
Even without the issue with $epsilon$ being constant, you should notice that the interior of the LHS contains an infinitesimal you are not integrating over, while the RHS does not. That alone is enough to tell you you've made some errors.
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
add a comment |
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$begingroup$
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2$$
That should be
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 - 2 s_1 cdot s_2$$
Choosing both $|s_1| = |s_2|$
This means that your calculations only apply to a circle around the origin. It may seem like infinitesimally you can still apply it to other loops, but you are only picking up the contribution in the direction of $dtheta$, and dropping that in the direction of $dr$. So, no, it only holds for the circle.
Multiplying both sides with $1 - cos(z_{12})$
Because of the earlier sign change, this would be $1 + cos(z_{12})$. However, the next line has another error that cancels out the previous (perhaps the sign error was only when you copied it into here?)$$cos^2 frac theta 2 = frac{1 + cos theta}2ne frac{1 - cos theta}2$$ so the next formula is correct again.
Taking $Delta s to ds, z_{12} to epsilon$
The cosine of $z_{12}$ is the length of $Delta s$ divided by the radius. If $Delta s$ becomes infinitesimal, then $z_{12}to pi/2$ I.e. $epsilon = pi/2, depsilon = 0$, and your integrations are over a constant instead of a variable.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon$$
Even without the issue with $epsilon$ being constant, you should notice that the interior of the LHS contains an infinitesimal you are not integrating over, while the RHS does not. That alone is enough to tell you you've made some errors.
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
add a comment |
$begingroup$
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2$$
That should be
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 - 2 s_1 cdot s_2$$
Choosing both $|s_1| = |s_2|$
This means that your calculations only apply to a circle around the origin. It may seem like infinitesimally you can still apply it to other loops, but you are only picking up the contribution in the direction of $dtheta$, and dropping that in the direction of $dr$. So, no, it only holds for the circle.
Multiplying both sides with $1 - cos(z_{12})$
Because of the earlier sign change, this would be $1 + cos(z_{12})$. However, the next line has another error that cancels out the previous (perhaps the sign error was only when you copied it into here?)$$cos^2 frac theta 2 = frac{1 + cos theta}2ne frac{1 - cos theta}2$$ so the next formula is correct again.
Taking $Delta s to ds, z_{12} to epsilon$
The cosine of $z_{12}$ is the length of $Delta s$ divided by the radius. If $Delta s$ becomes infinitesimal, then $z_{12}to pi/2$ I.e. $epsilon = pi/2, depsilon = 0$, and your integrations are over a constant instead of a variable.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon$$
Even without the issue with $epsilon$ being constant, you should notice that the interior of the LHS contains an infinitesimal you are not integrating over, while the RHS does not. That alone is enough to tell you you've made some errors.
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
add a comment |
$begingroup$
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2$$
That should be
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 - 2 s_1 cdot s_2$$
Choosing both $|s_1| = |s_2|$
This means that your calculations only apply to a circle around the origin. It may seem like infinitesimally you can still apply it to other loops, but you are only picking up the contribution in the direction of $dtheta$, and dropping that in the direction of $dr$. So, no, it only holds for the circle.
Multiplying both sides with $1 - cos(z_{12})$
Because of the earlier sign change, this would be $1 + cos(z_{12})$. However, the next line has another error that cancels out the previous (perhaps the sign error was only when you copied it into here?)$$cos^2 frac theta 2 = frac{1 + cos theta}2ne frac{1 - cos theta}2$$ so the next formula is correct again.
Taking $Delta s to ds, z_{12} to epsilon$
The cosine of $z_{12}$ is the length of $Delta s$ divided by the radius. If $Delta s$ becomes infinitesimal, then $z_{12}to pi/2$ I.e. $epsilon = pi/2, depsilon = 0$, and your integrations are over a constant instead of a variable.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon$$
Even without the issue with $epsilon$ being constant, you should notice that the interior of the LHS contains an infinitesimal you are not integrating over, while the RHS does not. That alone is enough to tell you you've made some errors.
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
$endgroup$
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 + 2 s_1 cdot s_2$$
That should be
$$|Delta s_{12} |^2 = |vec s_1|^2 + |vec s_2|^2 - 2 s_1 cdot s_2$$
Choosing both $|s_1| = |s_2|$
This means that your calculations only apply to a circle around the origin. It may seem like infinitesimally you can still apply it to other loops, but you are only picking up the contribution in the direction of $dtheta$, and dropping that in the direction of $dr$. So, no, it only holds for the circle.
Multiplying both sides with $1 - cos(z_{12})$
Because of the earlier sign change, this would be $1 + cos(z_{12})$. However, the next line has another error that cancels out the previous (perhaps the sign error was only when you copied it into here?)$$cos^2 frac theta 2 = frac{1 + cos theta}2ne frac{1 - cos theta}2$$ so the next formula is correct again.
Taking $Delta s to ds, z_{12} to epsilon$
The cosine of $z_{12}$ is the length of $Delta s$ divided by the radius. If $Delta s$ becomes infinitesimal, then $z_{12}to pi/2$ I.e. $epsilon = pi/2, depsilon = 0$, and your integrations are over a constant instead of a variable.
$$oint frac{|d s |^2}{|s|^2} cos^2(frac{epsilon}{2}) d epsilon = oint sin^2(epsilon) d epsilon$$
Even without the issue with $epsilon$ being constant, you should notice that the interior of the LHS contains an infinitesimal you are not integrating over, while the RHS does not. That alone is enough to tell you you've made some errors.
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
edited Jan 22 at 10:42
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
answered Jan 21 at 22:16
Paul SinclairPaul Sinclair
20.4k21443
20.4k21443
add a comment |
add a comment |
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$begingroup$
Your steps seem to be just restating $$cos theta = frac{ vec{s}_1 cdot vec{s}_2 }{lVertvec{s}_1 rVert , lVert vec{s}_2 rVert}$$That is the geometric definition of dot product, and works all Euclidean spaces (including the high-dimensional ones). As far as I understand, the last equality is just $pi = pi$ by definition. What am I missing here? I do have the feeling that there is an interesting question there, I just cannot see it.
$endgroup$
– Nominal Animal
Jan 21 at 15:34
$begingroup$
I was wondering if it worked with hyperbolic geometry (with imaginary $z$) and if there was a version of this in $4$ dimensions with signature $(-1,1,1,1)$?
$endgroup$
– More Anonymous
Jan 21 at 15:39
$begingroup$
Also how would u compute the L.H.S of the last equality?
$endgroup$
– More Anonymous
Jan 21 at 15:41
1
$begingroup$
You may need to restate the question, then, methinks. (Unfortunately, my own math-fu does not extend to non-Euclidean geometries.)
$endgroup$
– Nominal Animal
Jan 21 at 16:19
$begingroup$
Several mistakes in your development. Please proofread.
$endgroup$
– Yves Daoust
Jan 22 at 10:47