Why does parity give raise to pseudovectors?

  • B
  • Thread starter Phylosopher
  • Start date
  • Tags
    Parity
In summary, the concept of pseudovectors arises from the parity transformation, which involves a mirror reflection followed by a rotation. This transformation only works in odd dimensions, and in general, there are two types of orthogonal transformations - proper rotations and improper rotations. The cross product of two vectors is a pseudovector, not a vector, because it transforms according to the determinant of the rotation matrix rather than the matrix itself. This can lead to confusion when dealing with objects such as the electric and magnetic fields, as they may behave like vectors or pseudovectors depending on the coordinate system. Ultimately, the distinction between vectors and pseudovectors lies in their transformation properties, rather than their contents.
  • #1
Phylosopher
139
26
TL;DR Summary
Why parity have vectors and pseudovectors? why not only vectors?
Summary: Why parity have vectors and pseudovectors? why not only vectors?

I am reading Griffiths "Introduction to elementary particle physics" Ed.1.

The book obviously is an undergraduate introduction.Thus, not much detail is presented, but I cannot get my head around pseudovectors (pseudoscalars as well).

Parity in the book is just an inversion, that is to say, a mirror reflection followed by a ##180^{\text{o}}## rotation. It is easy to show that the parity group is just ##\{P,I\}##, Where ##P## is the inversion matrix (or ##-I##). For a vector ##\text{v}\in R^{3}##:

$$P\text{v}=-\text{v}$$

This is how Griffiths explains it. He follows this by saying that for a cross product ##\text{w}=\text{u}\times\text{v}##, parity does not give ##-\text{w}##, because ##P## is distributive over cross products:

$$P(\text{w})=P(\text{u}\times\text{v})=P(\text{u})\times P(\text{v})=(-\text{u})\times (-\text{v})=\text{u}\times\text{v}=\text{w}$$

It looks to me that he is forcing ##P## to be a linear transformation over cross products. But ##\text{w}## is a vector and the solution must be ##-\text{w}##, one way or another. Unless, ##\text{w}## is not a vector (a pseudovector)**. But, in this sense, what he presented is not a cross product to begin with!

Can someone help me understand why parity have pseudo's?
**Around three years ago, I read a book about geomtric algebra (didn't complete it), it talked about bivectors and pseudovectors. Could this be the same thing? Because the explanation Griffiths presented is very incomplete in my opinion.
 
Physics news on Phys.org
  • #2
This is not specific to QM. The electric field is a vector, but the magnetic field is an axial vector.
 
  • Like
Likes ohwilleke and Phylosopher
  • #3
I always had a beef with pseudo-vectors, or rather how they are presented in the undergraduate curriculum. The parity transformation is usually presented as ##\vec x \to -\vec x##, but this only works in odd dimensions. In a more general sense, the parity transformations are about the two disjoint classes of orthogonal transformations, the group of proper rotations ##SO(n)## and the group of improper rotations, which are rotations + a reflection.

It then all comes down to representation theory. If you have a rotation ##A \in O(n)##, an object ##\vec v## in the fundamental representation transforms as ##\vec v \to A\vec v## when ##\vec x \to A\vec x##, i.e., it transforms as the position vector. However, this is far from the only representation of ##O(n)##. Another representation with the same dimension as the fundamental representation is obtained if you let ##A \to \operatorname{det}(A) A##. An object ##\vec w## that transforms according to this representations, i.e., ##\vec w \to \operatorname{det}(A)A \vec w## when ##\vec x \to A\vec x## is called a pseudo-vector (or axial vector). In particular, in three dimensions for the transformation ##\vec x \to -\vec x##, a pseudo-vector transforms according to ##\vec w \to - \operatorname{det}(-I) \vec w = \vec w## (where ##I## is the unit matrix), not ##\vec w \to - \vec w##.

The cross product of two vectors is a pseudo-vector, not a vector.
Phylosopher said:
But ww\text{w} is a vector and the solution must be −w−w-\text{w}, one way or another.
No, ##\vec w = \vec u \times \vec v## where ##\vec u## and ##\vec v## are vectors is a pseudo-vector. The cross product between two vectors is always a pseudo-vector because it transforms according to ##\operatorname{det}(A) A##, not according to ##A##.
 
  • Like
  • Informative
Likes ohwilleke, vanhees71, dextercioby and 2 others
  • #4
Orodruin said:
but this only works in odd dimensions

Well, if you're going to go down that path other issues show up. In 3-d angular momentum is an axial vector, but in 2-d it is a scalar, and in 4-d it's a 6-component skew-symmetric tensor.
 
  • Like
Likes ohwilleke
  • #5
Vanadium 50 said:
This is not specific to QM. The electric field is a vector, but the magnetic field is an axial vector.
Orodruin said:
No, →w=→u×→v\vec w = \vec u \times \vec v where →u\vec u and →v\vec v are vectors is a pseudo-vector. The cross product between two vectors is always a pseudo-vector because it transforms according to det(A)A\operatorname{det}(A) A, not according to AA.

This is very interesting. It is the first time someone states to me this.

In wiki, I found these two paragraphs:
Wikipedia said:
Using the cross product requires the handedness of the coordinate system to be taken into account (as explicit in the definition above). If a left-handed coordinate system is used, the direction of the vector n is given by the left-hand rule and points in the opposite direction.

This, however, creates a problem because transforming from one arbitrary reference system to another (e.g., a mirror image transformation from a right-handed to a left-handed coordinate system), should not change the direction of n. The problem is clarified by realizing that the cross product of two vectors is not a (true) vector, but rather a pseudovector.

How can I then distinguish pseudovectors from vectors if they live in the same realm. Pardon my ignorance, but if ##\vec{v}=\{1,0,0\}## and ##\vec{u}=\{0,1,0\}## are vectors that their cross product produce ##\vec{w}## ((##\vec{w}=\vec{v}\times\vec{u}=\{0,0,1\}##))

Then how can I judge, say ##\vec{s}=\{0,0,1\}##? Is ##\vec{s}## a vector or a pseudovector?
 
  • #6
It can be either. The contents of a vector don't determine whether it's a polar vector or an axial vector, its transformation properties do.

If I look at the electric field in a mirror, it points the same direction: a (polar) vector. If I look at the magnetic field, it points in the opposite direction: an axial vector.
 
  • Like
Likes Klystron, Phylosopher and Orodruin
  • #7
Vanadium 50 said:
Well, if you're going to go down that path other issues show up. In 3-d angular momentum is an axial vector, but in 2-d it is a scalar, and in 4-d it's a 6-component skew-symmetric tensor.
I was referring to pseudo-vectors in general, not necessarily angular momentum. Angular momentum generally is a 2-form, which in three dimensions happens to transform according to the pseudo-vector representation, but this does not hold in other dimensions (wrong dimension for the representation).
 
  • Like
Likes weirdoguy
  • #8
Vanadium 50 said:
It can be either. The contents of a vector don't determine whether it's a polar vector or an axial vector, its transformation properties do.
This is a very important lesson. A vector is not just a collection of three numbers, those are just the components of the vector in some given basis. A vector is an object that also implies particular transformation properties of the components, i.e., how the components in different bases relate to each other.

Vanadium 50 said:
If I look at the electric field in a mirror, it points the same direction: a (polar) vector. If I look at the magnetic field, it points in the opposite direction: an axial vector.
This is slightly misleading though and depends on which components you look at. If by "looking in a mirror" you mean doing a reflection such that the direction perpendicular to the mirror is flipped, then you would have (taking the third direction as the direction perpendicular to the mirror) ##\vec e_3 \to -\vec e_3##, ##\vec e_i \to \vec e_i## for ##i = 1## and ##2##. Under this transformation, a vector ##\vec v = v^i \vec e_i## would transform to ##v^1 \vec e_1 + v^2 \vec e_2 - v^3 \vec e_3## and to minus this if it were a pseudo-vector. Only if you do ##\vec x \to -\vec x## do you get ##\vec v \to -\vec v## for a proper vector and ##\vec v \to \vec v## for a pseudo vector. As a student, this had me stumped for a long time before I realized what was actually going on. This is partially why I find the "standard approach" of introducing parity as ##\vec x \to -\vec x## and at the same time talking about it as a "mirror" to be misleading.

Edit: Also note that a vector changes sign under ##\vec x \to -\vec x## whereas a pseudo-vector (axial vector) does not.
 
  • Like
Likes etotheipi
  • #9
Orodruin said:
This is a very important lesson.

Thank you,
Orodruin said:
This is slightly misleading though

I don't think so. (Obviously, because I wrote it) I think it's "mostly right" to associate a mirror with the parity operator, and the fact that E is even and B is odd is not just an example, it's a helpful example: B flips sign because the current loop that generates it flips sign. It is true that this is not completely precise, but this is a B-level thread after all.
 
  • #10
Vanadium 50 said:
I don't think so. (Obviously, because I wrote it) I think it's "mostly right" to associate a mirror with the parity operator, and the fact that E is even and B is odd is not just an example, it's a helpful example: B flips sign because the current loop that generates it flips sign. It is true that this is not completely precise, but this is a B-level thread after all.
No, this is wrong. In fact, note that the parity transformation that flips all spatial directions flips a vector (i.e., the electric field) but not an axial vector (i.e., the magnetic field is unchanged under this transformation). For a pure reflection (i.e., not flipping all spatial directions, just a mirror in a plane), it is true that, for a vector, the directions within the plane of reflection do not flip. However, the direction perpendicular to the plane of reflection does flip. This can be intuitively understood as the reflection makes the point under consideration lie on the other side of the charge and therefore the component must flip under the reflection.

For the pseudo-vector, e.g., magnetic field, it depends on how you orient the loop. The magnetic field in the centre of the loop of course points normal to the loop. If the loop lies planar parallel to the plane of reflection, then the pure reflection does nothing with the direction of the current and the magnetic field in the loop centre remains the same. If the loop lies orthogonal to the plane of reflection, then the pure reflection does change the direction of the current and the magnetic field in the loop centre flips sign. All of this is in perfect accordance with what I described above for the transformation properties of vectors and pseudo-vectors under pure reflections.

It is also an important lesson that a pure reflection does not just amount to possibly flipping a sign. It is not only the determinant, but also the transformation ##A## itself, that is important in the pseudo-vector representation ##\operatorname{det}(A) A##.
 
  • Like
Likes etotheipi and weirdoguy
  • #11
Well, I guess we disagree on what the appropriate degree of "wrongness" is for a B-level thread.
 
  • #12
Vanadium 50 said:
Well, I guess we disagree on what the appropriate degree of "wrongness" is for a B-level thread.
Well, first of all, I do not consider this a B-level thread. If the OP is advanced enough to read about vectors vs pseudo-vectors it is enough for me to consider him I-level. Second, the misconception that parity just "flips the sign" of vectors but not pseudo-vectors is very confusing to anyone who just starts thinking about looking at a wheel in a mirror. If the wheel is parallel to the plane of reflection, it will have the same angular momentum in the mirror world as in the "ordinary" world because both the wheel and the mirror wheel spin in the same direction. If it is orthogonal to the plane of reflection the angular momentum changes sign because the wheel and the mirror wheel now spin in the opposite directions. (This example is what first had me stumped and wondering what was going on.)

Also, note that it is vectors, not pseudo-vectors that flip sign under ##\vec x \to -\vec x##. There is no improper rotation that has ##\vec v \to \vec v## for all proper vectors ##\vec v##. Therefore I do think it is misleading, even at B-level, to state that the E-field remains the same after reflection.
 
  • Like
Likes weirdoguy
  • #13
I'll let you have the last word on this.
 
  • #14
Vanadium 50 said:
Well, I guess we disagree on what the appropriate degree of "wrongness" is for a B-level thread.
Only, because something is labeled as B-level, it's no excuse for inexact explanations. To the contrary, for a beginner it's very helpful to be correct, and @Orodruin did a great job!
 
  • Like
Likes weirdoguy
  • #15
Here the distinguishing feature of a reflection is that it is a transformation with a determinant of -1 as opposed to +1 .

Fortunately in 3D an inversion ##x \rightarrow -x## treats all directions equally and has determinant -1. Unlike a mirror refection, it is isotropic. That makes it a very useful and clear way to discuss parity and reflections is general.
 
Last edited by a moderator:

Related to Why does parity give raise to pseudovectors?

1. What is parity and how does it relate to pseudovectors?

Parity is a physical property that describes whether a system remains unchanged when its spatial coordinates are inverted. Pseudovectors are mathematical quantities that behave like vectors under rotations, but change sign under a parity transformation.

2. Why do pseudovectors arise from parity transformations?

Pseudovectors arise from parity transformations because they are defined as quantities that change sign under a parity transformation. This means that if a physical system is inverted, the direction of the pseudovector will also be inverted, resulting in a change in sign.

3. How does the concept of chirality relate to pseudovectors and parity?

Chirality is a property of a physical system that describes whether it is distinguishable from its mirror image. Pseudovectors are often associated with chirality because they are affected by parity transformations, which change the handedness of a system. This means that a system that is chiral will have a pseudovector associated with it.

4. What are some examples of pseudovectors in physics?

Some common examples of pseudovectors in physics include angular momentum, magnetic moment, and torque. These quantities change sign under a parity transformation, making them pseudovectors. Another example is the Coriolis force, which is a pseudovector that arises in rotating reference frames.

5. How do pseudovectors affect our understanding of physical systems?

Pseudovectors play a crucial role in our understanding of physical systems because they help us describe and analyze phenomena that involve chirality or rotation. They also help us understand the effects of parity transformations on physical quantities and can provide insights into the underlying symmetries of a system.

Similar threads

Replies
1
Views
1K
Replies
3
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
2K
  • Special and General Relativity
Replies
4
Views
946
  • Linear and Abstract Algebra
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
Replies
9
Views
820
Replies
2
Views
974
Back
Top