What exactly are pseudovectors and pseudoscalars? And where could I read about them?
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I can't find good information on the internet.
In my mathematical physics class the definition of a vector was given as:
That object with magnitude and direction which doesn't change under inversions. Pseudovectors do change.
Scalars are that magnitude that don't change with inversions. Pseudoscalars do change.
Inversions were loosely defined as inverting all components of a vector.
The examples were given as:
Say $vecA=vecBtimesvecC$; under the inversion it becomes $vec-Btimesvec-C$ which is again, $vecA$ which then has to be a pseudovector.
Say $A=vecacdotvecbtimesvecc$ then under inversion $vec-acdotvec-btimesvec-c$, which is $-A$ and has to be a pseudoscalar.
I don't get it though. Why are inversions done on the individual vectors and not to the whole product?
What really is an inversion?
Where could I read more about this?
Thanks.
linear-algebra vector-analysis mathematical-physics
add a comment |Â
up vote
2
down vote
favorite
I can't find good information on the internet.
In my mathematical physics class the definition of a vector was given as:
That object with magnitude and direction which doesn't change under inversions. Pseudovectors do change.
Scalars are that magnitude that don't change with inversions. Pseudoscalars do change.
Inversions were loosely defined as inverting all components of a vector.
The examples were given as:
Say $vecA=vecBtimesvecC$; under the inversion it becomes $vec-Btimesvec-C$ which is again, $vecA$ which then has to be a pseudovector.
Say $A=vecacdotvecbtimesvecc$ then under inversion $vec-acdotvec-btimesvec-c$, which is $-A$ and has to be a pseudoscalar.
I don't get it though. Why are inversions done on the individual vectors and not to the whole product?
What really is an inversion?
Where could I read more about this?
Thanks.
linear-algebra vector-analysis mathematical-physics
Hmm. I'm sorry if this sounds dumb, but I guess I'm talking about the things usually referred to with these names in a physics class. I really don't know much about them apart from this. Pseudovectors would be stuff like angular velocity and torque, for example.
– ben ari
Jan 24 '15 at 1:34
I'm not sure which internet do you use, but my internet finds at least this, which lists at the bottom of the page quite a selection of references.
– Algebraic Pavel
Jan 24 '15 at 1:37
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I can't find good information on the internet.
In my mathematical physics class the definition of a vector was given as:
That object with magnitude and direction which doesn't change under inversions. Pseudovectors do change.
Scalars are that magnitude that don't change with inversions. Pseudoscalars do change.
Inversions were loosely defined as inverting all components of a vector.
The examples were given as:
Say $vecA=vecBtimesvecC$; under the inversion it becomes $vec-Btimesvec-C$ which is again, $vecA$ which then has to be a pseudovector.
Say $A=vecacdotvecbtimesvecc$ then under inversion $vec-acdotvec-btimesvec-c$, which is $-A$ and has to be a pseudoscalar.
I don't get it though. Why are inversions done on the individual vectors and not to the whole product?
What really is an inversion?
Where could I read more about this?
Thanks.
linear-algebra vector-analysis mathematical-physics
I can't find good information on the internet.
In my mathematical physics class the definition of a vector was given as:
That object with magnitude and direction which doesn't change under inversions. Pseudovectors do change.
Scalars are that magnitude that don't change with inversions. Pseudoscalars do change.
Inversions were loosely defined as inverting all components of a vector.
The examples were given as:
Say $vecA=vecBtimesvecC$; under the inversion it becomes $vec-Btimesvec-C$ which is again, $vecA$ which then has to be a pseudovector.
Say $A=vecacdotvecbtimesvecc$ then under inversion $vec-acdotvec-btimesvec-c$, which is $-A$ and has to be a pseudoscalar.
I don't get it though. Why are inversions done on the individual vectors and not to the whole product?
What really is an inversion?
Where could I read more about this?
Thanks.
linear-algebra vector-analysis mathematical-physics
edited Jan 24 '15 at 0:37
Omnomnomnom
121k784170
121k784170
asked Jan 24 '15 at 0:35
ben ari
626
626
Hmm. I'm sorry if this sounds dumb, but I guess I'm talking about the things usually referred to with these names in a physics class. I really don't know much about them apart from this. Pseudovectors would be stuff like angular velocity and torque, for example.
– ben ari
Jan 24 '15 at 1:34
I'm not sure which internet do you use, but my internet finds at least this, which lists at the bottom of the page quite a selection of references.
– Algebraic Pavel
Jan 24 '15 at 1:37
add a comment |Â
Hmm. I'm sorry if this sounds dumb, but I guess I'm talking about the things usually referred to with these names in a physics class. I really don't know much about them apart from this. Pseudovectors would be stuff like angular velocity and torque, for example.
– ben ari
Jan 24 '15 at 1:34
I'm not sure which internet do you use, but my internet finds at least this, which lists at the bottom of the page quite a selection of references.
– Algebraic Pavel
Jan 24 '15 at 1:37
Hmm. I'm sorry if this sounds dumb, but I guess I'm talking about the things usually referred to with these names in a physics class. I really don't know much about them apart from this. Pseudovectors would be stuff like angular velocity and torque, for example.
– ben ari
Jan 24 '15 at 1:34
Hmm. I'm sorry if this sounds dumb, but I guess I'm talking about the things usually referred to with these names in a physics class. I really don't know much about them apart from this. Pseudovectors would be stuff like angular velocity and torque, for example.
– ben ari
Jan 24 '15 at 1:34
I'm not sure which internet do you use, but my internet finds at least this, which lists at the bottom of the page quite a selection of references.
– Algebraic Pavel
Jan 24 '15 at 1:37
I'm not sure which internet do you use, but my internet finds at least this, which lists at the bottom of the page quite a selection of references.
– Algebraic Pavel
Jan 24 '15 at 1:37
add a comment |Â
2 Answers
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The concepts of pseudovectors and pseudoscalars arise from a clumsy attempt to make all geometric objects seem like vectors and scalars when they're not.
In 3d, pseudovectors and pseudoscalars are better understood as bivectors and trivectors instead.
Bivectors and trivectors in the 3d exterior algebra: direct representations of areas and volumes
Exterior algebra takes the vectors and scalars you know and builds from them bivectors and trivectors. The product needed for this is called the wedge product, and its properties are simple: if $a$, $b$, and $c$ are vectors, then
$$a wedge b = - b wedge a, quad (a wedge b) wedge c = a wedge (b wedge c)$$
So it's anticommutative (like the cross product) but also associative (unlike the cross product). $a wedge b$ is a bivector, and $a wedge b wedge c$ is a trivector.
Bivectors correspond directly to weighted, oriented planes or areas the way vectors correspond to weighted, oriented lines or directions. For instance, $hat x wedge hat y$ corresponds to the $xy$-plane. You can multiply this by a scalar, so for instance, $2 hat x wedge hat y$ also corresponds to the $xy$-plane, but with a different magnitude or weight. This is not different from regular vectors, as $2 hat x$ corresponds to the $x$-direction just as much as $hat x$ does. But having different magnitudes allows you to do addition and subtraction like usual.
In fact, bivectors form a vector space of their own, just as vectors form their own vector space. In 3d, there are three basis bivectors--$hat x wedge hat y, hat y wedge hat z, hat z wedge hat x$--and it's for this reason that many people identify bivectors with vectors that transform "differently" than regular vectors.
Do bivectors transform differently? Yes. The basic law of how linear maps ("matrices") act on a bivector is like so: if $T$ is a linear map, then
$$T(a wedge b) equiv T(a) wedge T(b)$$
Consider the case $T(a) = -a$, which is an inversion through the origin. All vectors point the opposite direction under inversion, but a bivector doesn't change:
$$T(a wedge b) = T(a) wedge T(b) = (-a) wedge (-b) = a wedge b$$
So this is part of what's different about bivectors compared to ordinary vectors.
Trivectors, too, are different from ordinary scalars. All trivectors can be written as a scalar multiple of $epsilon equiv hat x wedge hat y wedge hat z$, and we can interpret this as an oriented volume. If some trivector $tau = alpha epsilon$ for some positive $alpha$, then $tau$ is right-handed. If $alpha < 0$, then $tau$ is left-handed.
The definition above for linear maps acting on bivectors generalizes to trivectors, and you should see pretty quickly that $epsilon$ picks up a minus sign on inversion: inversion turns a right-handed volume into a left-handed one, and vice versa. This is called an orientation reversing transformation. (Though, note that bivectors were not reversed on inversion, and so whether a transformation reverses orientation can depend on what you're talking about.)
Relating bivectors and trivectors to common vectors and scalars: the clifford algebra
Exterior algebra doesn't have an operation to convert bivectors and trivectors to vectors and scalars, but the clifford algebra does. It defines a geometric product of vectors that incorporates the dot product. If $u,v, w$ are orthogonal vectors, then
$$uu = u cdot u, quad uv = -vu = u wedge v, quad (uv)w = u(vw)$$
With this in mind, we can write $epsilon = hat x hat y hat z$ under the geometric product, and we can use it with the geometric product to turn bivectors into vectors and trivectors into scalars. We can actually write an expression relating the cross product and triple scalar product to our clifford algebra stuff:
$$a times b = -epsilon (a wedge b), quad a cdot (b times c) = -epsilon(a wedge b wedge c)$$
(If you began to suspect $epsilon$ has to do with the Levi-Civita tensor, you'd be right! The components of $epsilon$ in some coordinate system are exactly those of the Levi-Civita tensor. And that's why I denote it $epsilon$.)
So, for any bivector, multiplying by $epsilon$ generates a corresponding vector--in fact, it is the normal vector to the plane represented by the bivector, and the convention adopted here ensures that the normal vector is related to the bivector by the usual right-hand rule.
Now, as we established, a bivector doesn't change under inversions, and so its normal vector doesn't change under inversions. This is what physicists often call "pseudovector" behavior, since all vectors should change under inversions. At first, this may seem paradoxical. The resolution to the paradox is simple: the transformation law of $epsilon$ has been ignored. Inversion makes $epsilon$ left-handed, but people kept using the right-hand rule to find the normal vector, even after transformation, which makes the normal vector seem different than regular vectors. The same explanation applies for pseudoscalars.
You can read further about exterior and clifford algebras on the internet in various places. You can do calculus with them either using the common formalism of "differential forms" or the calculus of clifford algebra, "geometric calculus." Bivectors and trivectors arise in many situations in physics, though most conventional texts ignore how quantities could be viewed this way. Angular momentum is a simple example of a bivector quantity, and electric flux through a surface can be viewed as a trivector. Though common problems seldom exploit these properties, they can be handy to remember when doing coordinate transformations.
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
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In the context of abstract mathematics, these are definitions of how to transform objects. More specifically, the rule is: if you invert your coordinate axis (taking $ hatx rightarrow - hatx $ etc.), then the components of a vector negate, the components of a pseudovector stay the same.
These definitions are motivated from objects in physics. It is a law/fact/assumption of physics whether a certain physical quantity is a pseudo vector/scalar or genuine vector/scalar. The laws of physics relating different quantities are then preserved under coordinate inversion - meaning precisely that if you transform each guy appearing in the equation by his transformation rule, the equation should stay the same. This is why in those examples, the inversions are done on individual vectors.
For example, it is a fact that position and velocity are vectors and angular momentum is a pseudovector. It had to be this way, because the quantities are related by $ vecL = vecr times m vecv $ . If $ vecL $ was defined to be a vector then upon coordinate inversion $ vecL = vecr times mvecv rightarrow - vecL = vecr times m vecv $, which changes the equation. With $ vecL $ defined as a pseudovector, there is not this problem.
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
The concepts of pseudovectors and pseudoscalars arise from a clumsy attempt to make all geometric objects seem like vectors and scalars when they're not.
In 3d, pseudovectors and pseudoscalars are better understood as bivectors and trivectors instead.
Bivectors and trivectors in the 3d exterior algebra: direct representations of areas and volumes
Exterior algebra takes the vectors and scalars you know and builds from them bivectors and trivectors. The product needed for this is called the wedge product, and its properties are simple: if $a$, $b$, and $c$ are vectors, then
$$a wedge b = - b wedge a, quad (a wedge b) wedge c = a wedge (b wedge c)$$
So it's anticommutative (like the cross product) but also associative (unlike the cross product). $a wedge b$ is a bivector, and $a wedge b wedge c$ is a trivector.
Bivectors correspond directly to weighted, oriented planes or areas the way vectors correspond to weighted, oriented lines or directions. For instance, $hat x wedge hat y$ corresponds to the $xy$-plane. You can multiply this by a scalar, so for instance, $2 hat x wedge hat y$ also corresponds to the $xy$-plane, but with a different magnitude or weight. This is not different from regular vectors, as $2 hat x$ corresponds to the $x$-direction just as much as $hat x$ does. But having different magnitudes allows you to do addition and subtraction like usual.
In fact, bivectors form a vector space of their own, just as vectors form their own vector space. In 3d, there are three basis bivectors--$hat x wedge hat y, hat y wedge hat z, hat z wedge hat x$--and it's for this reason that many people identify bivectors with vectors that transform "differently" than regular vectors.
Do bivectors transform differently? Yes. The basic law of how linear maps ("matrices") act on a bivector is like so: if $T$ is a linear map, then
$$T(a wedge b) equiv T(a) wedge T(b)$$
Consider the case $T(a) = -a$, which is an inversion through the origin. All vectors point the opposite direction under inversion, but a bivector doesn't change:
$$T(a wedge b) = T(a) wedge T(b) = (-a) wedge (-b) = a wedge b$$
So this is part of what's different about bivectors compared to ordinary vectors.
Trivectors, too, are different from ordinary scalars. All trivectors can be written as a scalar multiple of $epsilon equiv hat x wedge hat y wedge hat z$, and we can interpret this as an oriented volume. If some trivector $tau = alpha epsilon$ for some positive $alpha$, then $tau$ is right-handed. If $alpha < 0$, then $tau$ is left-handed.
The definition above for linear maps acting on bivectors generalizes to trivectors, and you should see pretty quickly that $epsilon$ picks up a minus sign on inversion: inversion turns a right-handed volume into a left-handed one, and vice versa. This is called an orientation reversing transformation. (Though, note that bivectors were not reversed on inversion, and so whether a transformation reverses orientation can depend on what you're talking about.)
Relating bivectors and trivectors to common vectors and scalars: the clifford algebra
Exterior algebra doesn't have an operation to convert bivectors and trivectors to vectors and scalars, but the clifford algebra does. It defines a geometric product of vectors that incorporates the dot product. If $u,v, w$ are orthogonal vectors, then
$$uu = u cdot u, quad uv = -vu = u wedge v, quad (uv)w = u(vw)$$
With this in mind, we can write $epsilon = hat x hat y hat z$ under the geometric product, and we can use it with the geometric product to turn bivectors into vectors and trivectors into scalars. We can actually write an expression relating the cross product and triple scalar product to our clifford algebra stuff:
$$a times b = -epsilon (a wedge b), quad a cdot (b times c) = -epsilon(a wedge b wedge c)$$
(If you began to suspect $epsilon$ has to do with the Levi-Civita tensor, you'd be right! The components of $epsilon$ in some coordinate system are exactly those of the Levi-Civita tensor. And that's why I denote it $epsilon$.)
So, for any bivector, multiplying by $epsilon$ generates a corresponding vector--in fact, it is the normal vector to the plane represented by the bivector, and the convention adopted here ensures that the normal vector is related to the bivector by the usual right-hand rule.
Now, as we established, a bivector doesn't change under inversions, and so its normal vector doesn't change under inversions. This is what physicists often call "pseudovector" behavior, since all vectors should change under inversions. At first, this may seem paradoxical. The resolution to the paradox is simple: the transformation law of $epsilon$ has been ignored. Inversion makes $epsilon$ left-handed, but people kept using the right-hand rule to find the normal vector, even after transformation, which makes the normal vector seem different than regular vectors. The same explanation applies for pseudoscalars.
You can read further about exterior and clifford algebras on the internet in various places. You can do calculus with them either using the common formalism of "differential forms" or the calculus of clifford algebra, "geometric calculus." Bivectors and trivectors arise in many situations in physics, though most conventional texts ignore how quantities could be viewed this way. Angular momentum is a simple example of a bivector quantity, and electric flux through a surface can be viewed as a trivector. Though common problems seldom exploit these properties, they can be handy to remember when doing coordinate transformations.
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
add a comment |Â
up vote
10
down vote
The concepts of pseudovectors and pseudoscalars arise from a clumsy attempt to make all geometric objects seem like vectors and scalars when they're not.
In 3d, pseudovectors and pseudoscalars are better understood as bivectors and trivectors instead.
Bivectors and trivectors in the 3d exterior algebra: direct representations of areas and volumes
Exterior algebra takes the vectors and scalars you know and builds from them bivectors and trivectors. The product needed for this is called the wedge product, and its properties are simple: if $a$, $b$, and $c$ are vectors, then
$$a wedge b = - b wedge a, quad (a wedge b) wedge c = a wedge (b wedge c)$$
So it's anticommutative (like the cross product) but also associative (unlike the cross product). $a wedge b$ is a bivector, and $a wedge b wedge c$ is a trivector.
Bivectors correspond directly to weighted, oriented planes or areas the way vectors correspond to weighted, oriented lines or directions. For instance, $hat x wedge hat y$ corresponds to the $xy$-plane. You can multiply this by a scalar, so for instance, $2 hat x wedge hat y$ also corresponds to the $xy$-plane, but with a different magnitude or weight. This is not different from regular vectors, as $2 hat x$ corresponds to the $x$-direction just as much as $hat x$ does. But having different magnitudes allows you to do addition and subtraction like usual.
In fact, bivectors form a vector space of their own, just as vectors form their own vector space. In 3d, there are three basis bivectors--$hat x wedge hat y, hat y wedge hat z, hat z wedge hat x$--and it's for this reason that many people identify bivectors with vectors that transform "differently" than regular vectors.
Do bivectors transform differently? Yes. The basic law of how linear maps ("matrices") act on a bivector is like so: if $T$ is a linear map, then
$$T(a wedge b) equiv T(a) wedge T(b)$$
Consider the case $T(a) = -a$, which is an inversion through the origin. All vectors point the opposite direction under inversion, but a bivector doesn't change:
$$T(a wedge b) = T(a) wedge T(b) = (-a) wedge (-b) = a wedge b$$
So this is part of what's different about bivectors compared to ordinary vectors.
Trivectors, too, are different from ordinary scalars. All trivectors can be written as a scalar multiple of $epsilon equiv hat x wedge hat y wedge hat z$, and we can interpret this as an oriented volume. If some trivector $tau = alpha epsilon$ for some positive $alpha$, then $tau$ is right-handed. If $alpha < 0$, then $tau$ is left-handed.
The definition above for linear maps acting on bivectors generalizes to trivectors, and you should see pretty quickly that $epsilon$ picks up a minus sign on inversion: inversion turns a right-handed volume into a left-handed one, and vice versa. This is called an orientation reversing transformation. (Though, note that bivectors were not reversed on inversion, and so whether a transformation reverses orientation can depend on what you're talking about.)
Relating bivectors and trivectors to common vectors and scalars: the clifford algebra
Exterior algebra doesn't have an operation to convert bivectors and trivectors to vectors and scalars, but the clifford algebra does. It defines a geometric product of vectors that incorporates the dot product. If $u,v, w$ are orthogonal vectors, then
$$uu = u cdot u, quad uv = -vu = u wedge v, quad (uv)w = u(vw)$$
With this in mind, we can write $epsilon = hat x hat y hat z$ under the geometric product, and we can use it with the geometric product to turn bivectors into vectors and trivectors into scalars. We can actually write an expression relating the cross product and triple scalar product to our clifford algebra stuff:
$$a times b = -epsilon (a wedge b), quad a cdot (b times c) = -epsilon(a wedge b wedge c)$$
(If you began to suspect $epsilon$ has to do with the Levi-Civita tensor, you'd be right! The components of $epsilon$ in some coordinate system are exactly those of the Levi-Civita tensor. And that's why I denote it $epsilon$.)
So, for any bivector, multiplying by $epsilon$ generates a corresponding vector--in fact, it is the normal vector to the plane represented by the bivector, and the convention adopted here ensures that the normal vector is related to the bivector by the usual right-hand rule.
Now, as we established, a bivector doesn't change under inversions, and so its normal vector doesn't change under inversions. This is what physicists often call "pseudovector" behavior, since all vectors should change under inversions. At first, this may seem paradoxical. The resolution to the paradox is simple: the transformation law of $epsilon$ has been ignored. Inversion makes $epsilon$ left-handed, but people kept using the right-hand rule to find the normal vector, even after transformation, which makes the normal vector seem different than regular vectors. The same explanation applies for pseudoscalars.
You can read further about exterior and clifford algebras on the internet in various places. You can do calculus with them either using the common formalism of "differential forms" or the calculus of clifford algebra, "geometric calculus." Bivectors and trivectors arise in many situations in physics, though most conventional texts ignore how quantities could be viewed this way. Angular momentum is a simple example of a bivector quantity, and electric flux through a surface can be viewed as a trivector. Though common problems seldom exploit these properties, they can be handy to remember when doing coordinate transformations.
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
add a comment |Â
up vote
10
down vote
up vote
10
down vote
The concepts of pseudovectors and pseudoscalars arise from a clumsy attempt to make all geometric objects seem like vectors and scalars when they're not.
In 3d, pseudovectors and pseudoscalars are better understood as bivectors and trivectors instead.
Bivectors and trivectors in the 3d exterior algebra: direct representations of areas and volumes
Exterior algebra takes the vectors and scalars you know and builds from them bivectors and trivectors. The product needed for this is called the wedge product, and its properties are simple: if $a$, $b$, and $c$ are vectors, then
$$a wedge b = - b wedge a, quad (a wedge b) wedge c = a wedge (b wedge c)$$
So it's anticommutative (like the cross product) but also associative (unlike the cross product). $a wedge b$ is a bivector, and $a wedge b wedge c$ is a trivector.
Bivectors correspond directly to weighted, oriented planes or areas the way vectors correspond to weighted, oriented lines or directions. For instance, $hat x wedge hat y$ corresponds to the $xy$-plane. You can multiply this by a scalar, so for instance, $2 hat x wedge hat y$ also corresponds to the $xy$-plane, but with a different magnitude or weight. This is not different from regular vectors, as $2 hat x$ corresponds to the $x$-direction just as much as $hat x$ does. But having different magnitudes allows you to do addition and subtraction like usual.
In fact, bivectors form a vector space of their own, just as vectors form their own vector space. In 3d, there are three basis bivectors--$hat x wedge hat y, hat y wedge hat z, hat z wedge hat x$--and it's for this reason that many people identify bivectors with vectors that transform "differently" than regular vectors.
Do bivectors transform differently? Yes. The basic law of how linear maps ("matrices") act on a bivector is like so: if $T$ is a linear map, then
$$T(a wedge b) equiv T(a) wedge T(b)$$
Consider the case $T(a) = -a$, which is an inversion through the origin. All vectors point the opposite direction under inversion, but a bivector doesn't change:
$$T(a wedge b) = T(a) wedge T(b) = (-a) wedge (-b) = a wedge b$$
So this is part of what's different about bivectors compared to ordinary vectors.
Trivectors, too, are different from ordinary scalars. All trivectors can be written as a scalar multiple of $epsilon equiv hat x wedge hat y wedge hat z$, and we can interpret this as an oriented volume. If some trivector $tau = alpha epsilon$ for some positive $alpha$, then $tau$ is right-handed. If $alpha < 0$, then $tau$ is left-handed.
The definition above for linear maps acting on bivectors generalizes to trivectors, and you should see pretty quickly that $epsilon$ picks up a minus sign on inversion: inversion turns a right-handed volume into a left-handed one, and vice versa. This is called an orientation reversing transformation. (Though, note that bivectors were not reversed on inversion, and so whether a transformation reverses orientation can depend on what you're talking about.)
Relating bivectors and trivectors to common vectors and scalars: the clifford algebra
Exterior algebra doesn't have an operation to convert bivectors and trivectors to vectors and scalars, but the clifford algebra does. It defines a geometric product of vectors that incorporates the dot product. If $u,v, w$ are orthogonal vectors, then
$$uu = u cdot u, quad uv = -vu = u wedge v, quad (uv)w = u(vw)$$
With this in mind, we can write $epsilon = hat x hat y hat z$ under the geometric product, and we can use it with the geometric product to turn bivectors into vectors and trivectors into scalars. We can actually write an expression relating the cross product and triple scalar product to our clifford algebra stuff:
$$a times b = -epsilon (a wedge b), quad a cdot (b times c) = -epsilon(a wedge b wedge c)$$
(If you began to suspect $epsilon$ has to do with the Levi-Civita tensor, you'd be right! The components of $epsilon$ in some coordinate system are exactly those of the Levi-Civita tensor. And that's why I denote it $epsilon$.)
So, for any bivector, multiplying by $epsilon$ generates a corresponding vector--in fact, it is the normal vector to the plane represented by the bivector, and the convention adopted here ensures that the normal vector is related to the bivector by the usual right-hand rule.
Now, as we established, a bivector doesn't change under inversions, and so its normal vector doesn't change under inversions. This is what physicists often call "pseudovector" behavior, since all vectors should change under inversions. At first, this may seem paradoxical. The resolution to the paradox is simple: the transformation law of $epsilon$ has been ignored. Inversion makes $epsilon$ left-handed, but people kept using the right-hand rule to find the normal vector, even after transformation, which makes the normal vector seem different than regular vectors. The same explanation applies for pseudoscalars.
You can read further about exterior and clifford algebras on the internet in various places. You can do calculus with them either using the common formalism of "differential forms" or the calculus of clifford algebra, "geometric calculus." Bivectors and trivectors arise in many situations in physics, though most conventional texts ignore how quantities could be viewed this way. Angular momentum is a simple example of a bivector quantity, and electric flux through a surface can be viewed as a trivector. Though common problems seldom exploit these properties, they can be handy to remember when doing coordinate transformations.
The concepts of pseudovectors and pseudoscalars arise from a clumsy attempt to make all geometric objects seem like vectors and scalars when they're not.
In 3d, pseudovectors and pseudoscalars are better understood as bivectors and trivectors instead.
Bivectors and trivectors in the 3d exterior algebra: direct representations of areas and volumes
Exterior algebra takes the vectors and scalars you know and builds from them bivectors and trivectors. The product needed for this is called the wedge product, and its properties are simple: if $a$, $b$, and $c$ are vectors, then
$$a wedge b = - b wedge a, quad (a wedge b) wedge c = a wedge (b wedge c)$$
So it's anticommutative (like the cross product) but also associative (unlike the cross product). $a wedge b$ is a bivector, and $a wedge b wedge c$ is a trivector.
Bivectors correspond directly to weighted, oriented planes or areas the way vectors correspond to weighted, oriented lines or directions. For instance, $hat x wedge hat y$ corresponds to the $xy$-plane. You can multiply this by a scalar, so for instance, $2 hat x wedge hat y$ also corresponds to the $xy$-plane, but with a different magnitude or weight. This is not different from regular vectors, as $2 hat x$ corresponds to the $x$-direction just as much as $hat x$ does. But having different magnitudes allows you to do addition and subtraction like usual.
In fact, bivectors form a vector space of their own, just as vectors form their own vector space. In 3d, there are three basis bivectors--$hat x wedge hat y, hat y wedge hat z, hat z wedge hat x$--and it's for this reason that many people identify bivectors with vectors that transform "differently" than regular vectors.
Do bivectors transform differently? Yes. The basic law of how linear maps ("matrices") act on a bivector is like so: if $T$ is a linear map, then
$$T(a wedge b) equiv T(a) wedge T(b)$$
Consider the case $T(a) = -a$, which is an inversion through the origin. All vectors point the opposite direction under inversion, but a bivector doesn't change:
$$T(a wedge b) = T(a) wedge T(b) = (-a) wedge (-b) = a wedge b$$
So this is part of what's different about bivectors compared to ordinary vectors.
Trivectors, too, are different from ordinary scalars. All trivectors can be written as a scalar multiple of $epsilon equiv hat x wedge hat y wedge hat z$, and we can interpret this as an oriented volume. If some trivector $tau = alpha epsilon$ for some positive $alpha$, then $tau$ is right-handed. If $alpha < 0$, then $tau$ is left-handed.
The definition above for linear maps acting on bivectors generalizes to trivectors, and you should see pretty quickly that $epsilon$ picks up a minus sign on inversion: inversion turns a right-handed volume into a left-handed one, and vice versa. This is called an orientation reversing transformation. (Though, note that bivectors were not reversed on inversion, and so whether a transformation reverses orientation can depend on what you're talking about.)
Relating bivectors and trivectors to common vectors and scalars: the clifford algebra
Exterior algebra doesn't have an operation to convert bivectors and trivectors to vectors and scalars, but the clifford algebra does. It defines a geometric product of vectors that incorporates the dot product. If $u,v, w$ are orthogonal vectors, then
$$uu = u cdot u, quad uv = -vu = u wedge v, quad (uv)w = u(vw)$$
With this in mind, we can write $epsilon = hat x hat y hat z$ under the geometric product, and we can use it with the geometric product to turn bivectors into vectors and trivectors into scalars. We can actually write an expression relating the cross product and triple scalar product to our clifford algebra stuff:
$$a times b = -epsilon (a wedge b), quad a cdot (b times c) = -epsilon(a wedge b wedge c)$$
(If you began to suspect $epsilon$ has to do with the Levi-Civita tensor, you'd be right! The components of $epsilon$ in some coordinate system are exactly those of the Levi-Civita tensor. And that's why I denote it $epsilon$.)
So, for any bivector, multiplying by $epsilon$ generates a corresponding vector--in fact, it is the normal vector to the plane represented by the bivector, and the convention adopted here ensures that the normal vector is related to the bivector by the usual right-hand rule.
Now, as we established, a bivector doesn't change under inversions, and so its normal vector doesn't change under inversions. This is what physicists often call "pseudovector" behavior, since all vectors should change under inversions. At first, this may seem paradoxical. The resolution to the paradox is simple: the transformation law of $epsilon$ has been ignored. Inversion makes $epsilon$ left-handed, but people kept using the right-hand rule to find the normal vector, even after transformation, which makes the normal vector seem different than regular vectors. The same explanation applies for pseudoscalars.
You can read further about exterior and clifford algebras on the internet in various places. You can do calculus with them either using the common formalism of "differential forms" or the calculus of clifford algebra, "geometric calculus." Bivectors and trivectors arise in many situations in physics, though most conventional texts ignore how quantities could be viewed this way. Angular momentum is a simple example of a bivector quantity, and electric flux through a surface can be viewed as a trivector. Though common problems seldom exploit these properties, they can be handy to remember when doing coordinate transformations.
answered Jan 24 '15 at 7:24
Muphrid
15.2k11439
15.2k11439
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
add a comment |Â
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
I never thought of electric flux as a trivector before. Gauss' Law, $iintvec Ecdothat n,dA=frac1varepsilon_0iiintrho,dV$, can be rewritten as $iintvec EwedgemathbfdA=frac1varepsilon_0iiintrho,mathbfdV$; this would imply that electric charge is also a trivector, not a scalar. But it can also be rewritten with $E$ as a timelike bivector, so charge would be a quadvector! This brings to mind CPT symmetry...
– mr_e_man
May 18 at 18:03
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
In terms of four-multivectors, I think Gauss' Law would be something like $iint FwedgemathbfdA=mu_0iiint JwedgemathbfdV=-cmu_0Q(hat e_1hat e_2hat e_3hat e_4)$. This is CP, CT, and PT, but not CPT invariant.
– mr_e_man
May 18 at 22:44
add a comment |Â
up vote
1
down vote
In the context of abstract mathematics, these are definitions of how to transform objects. More specifically, the rule is: if you invert your coordinate axis (taking $ hatx rightarrow - hatx $ etc.), then the components of a vector negate, the components of a pseudovector stay the same.
These definitions are motivated from objects in physics. It is a law/fact/assumption of physics whether a certain physical quantity is a pseudo vector/scalar or genuine vector/scalar. The laws of physics relating different quantities are then preserved under coordinate inversion - meaning precisely that if you transform each guy appearing in the equation by his transformation rule, the equation should stay the same. This is why in those examples, the inversions are done on individual vectors.
For example, it is a fact that position and velocity are vectors and angular momentum is a pseudovector. It had to be this way, because the quantities are related by $ vecL = vecr times m vecv $ . If $ vecL $ was defined to be a vector then upon coordinate inversion $ vecL = vecr times mvecv rightarrow - vecL = vecr times m vecv $, which changes the equation. With $ vecL $ defined as a pseudovector, there is not this problem.
add a comment |Â
up vote
1
down vote
In the context of abstract mathematics, these are definitions of how to transform objects. More specifically, the rule is: if you invert your coordinate axis (taking $ hatx rightarrow - hatx $ etc.), then the components of a vector negate, the components of a pseudovector stay the same.
These definitions are motivated from objects in physics. It is a law/fact/assumption of physics whether a certain physical quantity is a pseudo vector/scalar or genuine vector/scalar. The laws of physics relating different quantities are then preserved under coordinate inversion - meaning precisely that if you transform each guy appearing in the equation by his transformation rule, the equation should stay the same. This is why in those examples, the inversions are done on individual vectors.
For example, it is a fact that position and velocity are vectors and angular momentum is a pseudovector. It had to be this way, because the quantities are related by $ vecL = vecr times m vecv $ . If $ vecL $ was defined to be a vector then upon coordinate inversion $ vecL = vecr times mvecv rightarrow - vecL = vecr times m vecv $, which changes the equation. With $ vecL $ defined as a pseudovector, there is not this problem.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In the context of abstract mathematics, these are definitions of how to transform objects. More specifically, the rule is: if you invert your coordinate axis (taking $ hatx rightarrow - hatx $ etc.), then the components of a vector negate, the components of a pseudovector stay the same.
These definitions are motivated from objects in physics. It is a law/fact/assumption of physics whether a certain physical quantity is a pseudo vector/scalar or genuine vector/scalar. The laws of physics relating different quantities are then preserved under coordinate inversion - meaning precisely that if you transform each guy appearing in the equation by his transformation rule, the equation should stay the same. This is why in those examples, the inversions are done on individual vectors.
For example, it is a fact that position and velocity are vectors and angular momentum is a pseudovector. It had to be this way, because the quantities are related by $ vecL = vecr times m vecv $ . If $ vecL $ was defined to be a vector then upon coordinate inversion $ vecL = vecr times mvecv rightarrow - vecL = vecr times m vecv $, which changes the equation. With $ vecL $ defined as a pseudovector, there is not this problem.
In the context of abstract mathematics, these are definitions of how to transform objects. More specifically, the rule is: if you invert your coordinate axis (taking $ hatx rightarrow - hatx $ etc.), then the components of a vector negate, the components of a pseudovector stay the same.
These definitions are motivated from objects in physics. It is a law/fact/assumption of physics whether a certain physical quantity is a pseudo vector/scalar or genuine vector/scalar. The laws of physics relating different quantities are then preserved under coordinate inversion - meaning precisely that if you transform each guy appearing in the equation by his transformation rule, the equation should stay the same. This is why in those examples, the inversions are done on individual vectors.
For example, it is a fact that position and velocity are vectors and angular momentum is a pseudovector. It had to be this way, because the quantities are related by $ vecL = vecr times m vecv $ . If $ vecL $ was defined to be a vector then upon coordinate inversion $ vecL = vecr times mvecv rightarrow - vecL = vecr times m vecv $, which changes the equation. With $ vecL $ defined as a pseudovector, there is not this problem.
answered Jan 24 '15 at 1:54
user209126
1012
1012
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Hmm. I'm sorry if this sounds dumb, but I guess I'm talking about the things usually referred to with these names in a physics class. I really don't know much about them apart from this. Pseudovectors would be stuff like angular velocity and torque, for example.
– ben ari
Jan 24 '15 at 1:34
I'm not sure which internet do you use, but my internet finds at least this, which lists at the bottom of the page quite a selection of references.
– Algebraic Pavel
Jan 24 '15 at 1:37