units in math, cross product
The cross product of two vectors has length equal to the area of the parallelogram they generate.
The cross product is also a vector and thus has dimensions. But the units of those dimensions are units of area, such as m2.
My question is - this vector can't be in the same vector space as the original vectors multiplied, right?
What does adding units to dimension do to a vector space?
cross-product
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
add a comment |
The cross product of two vectors has length equal to the area of the parallelogram they generate.
The cross product is also a vector and thus has dimensions. But the units of those dimensions are units of area, such as m2.
My question is - this vector can't be in the same vector space as the original vectors multiplied, right?
What does adding units to dimension do to a vector space?
cross-product
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
Probably related: en.wikipedia.org/wiki/Pseudovector
– lisyarus
Sep 24 '15 at 9:44
... or en.wikipedia.org/wiki/Bivector.
– Hans Lundmark
Sep 24 '15 at 9:59
A similar question: math.stackexchange.com/questions/705954/…
– Hans Lundmark
Sep 24 '15 at 10:01
Also seems helpful: mathoverflow.net/questions/4069/...
– Scorpion_God
Sep 24 '15 at 11:32
@JoonatanTalviste You might also find Muphrid's answer to this question informative.
– user137731
Oct 1 '15 at 16:59
add a comment |
The cross product of two vectors has length equal to the area of the parallelogram they generate.
The cross product is also a vector and thus has dimensions. But the units of those dimensions are units of area, such as m2.
My question is - this vector can't be in the same vector space as the original vectors multiplied, right?
What does adding units to dimension do to a vector space?
cross-product
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
The cross product of two vectors has length equal to the area of the parallelogram they generate.
The cross product is also a vector and thus has dimensions. But the units of those dimensions are units of area, such as m2.
My question is - this vector can't be in the same vector space as the original vectors multiplied, right?
What does adding units to dimension do to a vector space?
cross-product
cross-product
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
asked Sep 24 '15 at 9:34
Joonatan TalvisteJoonatan Talviste
61
61
Probably related: en.wikipedia.org/wiki/Pseudovector
– lisyarus
Sep 24 '15 at 9:44
... or en.wikipedia.org/wiki/Bivector.
– Hans Lundmark
Sep 24 '15 at 9:59
A similar question: math.stackexchange.com/questions/705954/…
– Hans Lundmark
Sep 24 '15 at 10:01
Also seems helpful: mathoverflow.net/questions/4069/...
– Scorpion_God
Sep 24 '15 at 11:32
@JoonatanTalviste You might also find Muphrid's answer to this question informative.
– user137731
Oct 1 '15 at 16:59
add a comment |
Probably related: en.wikipedia.org/wiki/Pseudovector
– lisyarus
Sep 24 '15 at 9:44
... or en.wikipedia.org/wiki/Bivector.
– Hans Lundmark
Sep 24 '15 at 9:59
A similar question: math.stackexchange.com/questions/705954/…
– Hans Lundmark
Sep 24 '15 at 10:01
Also seems helpful: mathoverflow.net/questions/4069/...
– Scorpion_God
Sep 24 '15 at 11:32
@JoonatanTalviste You might also find Muphrid's answer to this question informative.
– user137731
Oct 1 '15 at 16:59
Probably related: en.wikipedia.org/wiki/Pseudovector
– lisyarus
Sep 24 '15 at 9:44
Probably related: en.wikipedia.org/wiki/Pseudovector
– lisyarus
Sep 24 '15 at 9:44
... or en.wikipedia.org/wiki/Bivector.
– Hans Lundmark
Sep 24 '15 at 9:59
... or en.wikipedia.org/wiki/Bivector.
– Hans Lundmark
Sep 24 '15 at 9:59
A similar question: math.stackexchange.com/questions/705954/…
– Hans Lundmark
Sep 24 '15 at 10:01
A similar question: math.stackexchange.com/questions/705954/…
– Hans Lundmark
Sep 24 '15 at 10:01
Also seems helpful: mathoverflow.net/questions/4069/...
– Scorpion_God
Sep 24 '15 at 11:32
Also seems helpful: mathoverflow.net/questions/4069/...
– Scorpion_God
Sep 24 '15 at 11:32
@JoonatanTalviste You might also find Muphrid's answer to this question informative.
– user137731
Oct 1 '15 at 16:59
@JoonatanTalviste You might also find Muphrid's answer to this question informative.
– user137731
Oct 1 '15 at 16:59
add a comment |
1 Answer
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Yes -- using a vector to write the cross product is just a 3D convention to remind you it has 3 independent components. It's more naturally represented as a bivector or as a rank-2 tensor, but there's a duality between these and vectors in 3D.
Scale invariance/tensors in physics
The formal way of expressing your concern -- that the cross product has different "units" from other vectors in your space -- is that the cross product doesn't behave like a vector under scaling. Under scaling, a vector is supposed to scale as $vtolambda v$, because lengths scale that way, and areas scale as $ttolambda^2t$. The cross product scales as an area, so it makes a sucky vector.
This is sort of related to the definition of tensors in physics (specifically in relativity) -- a tensor is an object that transforms as a tensor under specific transformations, specifically Lorentz transformations (skews in the t-x/t-y/t-z plane and rotations in the other three, x-y/y-z/z-x). So scalars are invariant under Lorentz transformations, vectors transform as $Lambda_mu^{barmu}v^mu$, rank-2 tensors transform as $Lambda_mu^{barmu}Lambda_nu^{barnu}t^{munu}$, etc. So it's not enough to have the right number of components -- because this can be gamed with symmetries, like it is for the cross product (which has nine components, but only three independent one) -- you need to transform in the right way.
It's a bit more complicated with scaling, because different scalars transform differently (lengths scale like vectors, areas scale like rank-2 tensors), but the idea is the same -- the cross product is not a vector in the physics sense, although it is in the math sense. But nobody in math cares about the cross product anyway.
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
add a comment |
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Yes -- using a vector to write the cross product is just a 3D convention to remind you it has 3 independent components. It's more naturally represented as a bivector or as a rank-2 tensor, but there's a duality between these and vectors in 3D.
Scale invariance/tensors in physics
The formal way of expressing your concern -- that the cross product has different "units" from other vectors in your space -- is that the cross product doesn't behave like a vector under scaling. Under scaling, a vector is supposed to scale as $vtolambda v$, because lengths scale that way, and areas scale as $ttolambda^2t$. The cross product scales as an area, so it makes a sucky vector.
This is sort of related to the definition of tensors in physics (specifically in relativity) -- a tensor is an object that transforms as a tensor under specific transformations, specifically Lorentz transformations (skews in the t-x/t-y/t-z plane and rotations in the other three, x-y/y-z/z-x). So scalars are invariant under Lorentz transformations, vectors transform as $Lambda_mu^{barmu}v^mu$, rank-2 tensors transform as $Lambda_mu^{barmu}Lambda_nu^{barnu}t^{munu}$, etc. So it's not enough to have the right number of components -- because this can be gamed with symmetries, like it is for the cross product (which has nine components, but only three independent one) -- you need to transform in the right way.
It's a bit more complicated with scaling, because different scalars transform differently (lengths scale like vectors, areas scale like rank-2 tensors), but the idea is the same -- the cross product is not a vector in the physics sense, although it is in the math sense. But nobody in math cares about the cross product anyway.
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
add a comment |
Yes -- using a vector to write the cross product is just a 3D convention to remind you it has 3 independent components. It's more naturally represented as a bivector or as a rank-2 tensor, but there's a duality between these and vectors in 3D.
Scale invariance/tensors in physics
The formal way of expressing your concern -- that the cross product has different "units" from other vectors in your space -- is that the cross product doesn't behave like a vector under scaling. Under scaling, a vector is supposed to scale as $vtolambda v$, because lengths scale that way, and areas scale as $ttolambda^2t$. The cross product scales as an area, so it makes a sucky vector.
This is sort of related to the definition of tensors in physics (specifically in relativity) -- a tensor is an object that transforms as a tensor under specific transformations, specifically Lorentz transformations (skews in the t-x/t-y/t-z plane and rotations in the other three, x-y/y-z/z-x). So scalars are invariant under Lorentz transformations, vectors transform as $Lambda_mu^{barmu}v^mu$, rank-2 tensors transform as $Lambda_mu^{barmu}Lambda_nu^{barnu}t^{munu}$, etc. So it's not enough to have the right number of components -- because this can be gamed with symmetries, like it is for the cross product (which has nine components, but only three independent one) -- you need to transform in the right way.
It's a bit more complicated with scaling, because different scalars transform differently (lengths scale like vectors, areas scale like rank-2 tensors), but the idea is the same -- the cross product is not a vector in the physics sense, although it is in the math sense. But nobody in math cares about the cross product anyway.
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
add a comment |
Yes -- using a vector to write the cross product is just a 3D convention to remind you it has 3 independent components. It's more naturally represented as a bivector or as a rank-2 tensor, but there's a duality between these and vectors in 3D.
Scale invariance/tensors in physics
The formal way of expressing your concern -- that the cross product has different "units" from other vectors in your space -- is that the cross product doesn't behave like a vector under scaling. Under scaling, a vector is supposed to scale as $vtolambda v$, because lengths scale that way, and areas scale as $ttolambda^2t$. The cross product scales as an area, so it makes a sucky vector.
This is sort of related to the definition of tensors in physics (specifically in relativity) -- a tensor is an object that transforms as a tensor under specific transformations, specifically Lorentz transformations (skews in the t-x/t-y/t-z plane and rotations in the other three, x-y/y-z/z-x). So scalars are invariant under Lorentz transformations, vectors transform as $Lambda_mu^{barmu}v^mu$, rank-2 tensors transform as $Lambda_mu^{barmu}Lambda_nu^{barnu}t^{munu}$, etc. So it's not enough to have the right number of components -- because this can be gamed with symmetries, like it is for the cross product (which has nine components, but only three independent one) -- you need to transform in the right way.
It's a bit more complicated with scaling, because different scalars transform differently (lengths scale like vectors, areas scale like rank-2 tensors), but the idea is the same -- the cross product is not a vector in the physics sense, although it is in the math sense. But nobody in math cares about the cross product anyway.
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
Yes -- using a vector to write the cross product is just a 3D convention to remind you it has 3 independent components. It's more naturally represented as a bivector or as a rank-2 tensor, but there's a duality between these and vectors in 3D.
Scale invariance/tensors in physics
The formal way of expressing your concern -- that the cross product has different "units" from other vectors in your space -- is that the cross product doesn't behave like a vector under scaling. Under scaling, a vector is supposed to scale as $vtolambda v$, because lengths scale that way, and areas scale as $ttolambda^2t$. The cross product scales as an area, so it makes a sucky vector.
This is sort of related to the definition of tensors in physics (specifically in relativity) -- a tensor is an object that transforms as a tensor under specific transformations, specifically Lorentz transformations (skews in the t-x/t-y/t-z plane and rotations in the other three, x-y/y-z/z-x). So scalars are invariant under Lorentz transformations, vectors transform as $Lambda_mu^{barmu}v^mu$, rank-2 tensors transform as $Lambda_mu^{barmu}Lambda_nu^{barnu}t^{munu}$, etc. So it's not enough to have the right number of components -- because this can be gamed with symmetries, like it is for the cross product (which has nine components, but only three independent one) -- you need to transform in the right way.
It's a bit more complicated with scaling, because different scalars transform differently (lengths scale like vectors, areas scale like rank-2 tensors), but the idea is the same -- the cross product is not a vector in the physics sense, although it is in the math sense. But nobody in math cares about the cross product anyway.
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
edited Nov 30 '18 at 20:03
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
answered Jun 9 '18 at 15:36
Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir
864619
864619
add a comment |
add a comment |
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Probably related: en.wikipedia.org/wiki/Pseudovector
– lisyarus
Sep 24 '15 at 9:44
... or en.wikipedia.org/wiki/Bivector.
– Hans Lundmark
Sep 24 '15 at 9:59
A similar question: math.stackexchange.com/questions/705954/…
– Hans Lundmark
Sep 24 '15 at 10:01
Also seems helpful: mathoverflow.net/questions/4069/...
– Scorpion_God
Sep 24 '15 at 11:32
@JoonatanTalviste You might also find Muphrid's answer to this question informative.
– user137731
Oct 1 '15 at 16:59