Talk:Pseudovector

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Contents 1 More on pseudovectors 2 Microprocessors 3 too technical 4 the conspiracy of pseudo vectors

[edit] More on pseudovectors

The cross product of two polar vectors or two pseudovectors will be a pseudovector while cross product of a pseudovector and a polar vector is again a polar vector. Moreover, dot product of a pseudovector with a polarvector gives a pseudoscalar while dot product of two polar vectors or two pseudovectors is a scalar.

[edit] Microprocessors

Shouldn't there at least be a mention of the microprocessor application of pseudo-vectors? (John2kx 21:05, 27 April 2007 (UTC))

[edit] too technical

This article may be more technical than it needs to be. Please help improve the article by making the article accessible to the widest possible audience.

I have no idea what this article is talking about; can it be explained in a manner comprehensible to somebody who majored in a subject other than mathematics? As it stands now, the article is of use only to those who already understand the topic. 69.140.159.215 (talk) 03:22, 10 January 2008 (UTC)

I think you are right. Furthermore, I want a more technical definition of what this is all about. The "transforms like" is very vague. I sounds like it has a life of its own and "transforms" out of free will! —Bromskloss (talk) 16:11, 18 January 2008 (UTC)

I tried to clarify the lead. However, we can expect people to read up on what a vector is, and how vectors transform. I've opted to put links to the relevant articles in the lead in so far as they were not already present. Before you learn to run, first learn to walk. It's useless to memorise the differences between vectors and pseudovectors if you don't have a general idea of how vectors behave in the first place. Shinobu (talk) 00:47, 28 November 2008 (UTC)

[edit] the conspiracy of pseudo vectors

This article needs a LOT of work. It absolutely confuses me what the author means by 'inversion':

do you invert all vectors including the given ones a and b and the basis vectors? do you invert only the vectors a and b but not the basis vectors? do you invert the basis vectors but keep the same a and b?

are you talking about active or passive transformation? when you say a becomes -a after inversion do you mean it is the opposite vector geometricaly i. e. the components of a in the old basis flip or you mean it has the opposite components in the flipped vector basis which compensates and its the same geometrical vector?

does the geometrical, right screw, definition of vector product changes when you do 'inversion' to a left screw?

I haven't seen a single place where those questions were addressed systematically and I've been asking myself for years if the right screw for vector product turns into left screw in left coordinate system ... All textbooks just gloss over the problem and never explain what exactly they mean by 'inversion'. How can you talk about operation without clearly defining it? I have the feeling nobody in the world actually knows it, they just repeat like parrots 'vector' and 'pseudo vector'... —Preceding unsigned comment added by 128.135.235.188 (talk) 02:59, 14 January 2009 (UTC)

You're probably reading physics textbooks. It sounds like you may enjoy looking into the mathematical physics literature, where things like active versus passive transformations are treated carefully and rigorously. Anyway, I happen to prefer the active-transformation approach, in which case the following are not changed: (Direction of the +x axis, direction of the +y axis, direction of the +z axis), and the following are inverted: (a, b, any other vector). The right-hand rule is always the right-hand rule. If you simultaneously invert the coordinate system, vectors, and definition of right-hand rule, then you haven't done anything at all, of course.

Did you read the example about the car wheels in the article?

Anyway, I'm not implying that the article doesn't need improvement. :-) --Steve (talk) 05:00, 14 January 2009 (UTC)

It's me again from the pseudo vector conspiracy.

The article talks about 'coordinate inversion'. To me that implies inversion of the coordinate system axes but the article should be more explicit about what is being inverted and not leave that to the reader to guess. Also the article says that x goes to -x and y to -y. These are in bold face implying they are vectors probably the basis vectors. Again that is not specified implicitly, the reader has to guess the meaning of the symbols. So a goes to -a, b goes to -b, and a x b remains the same (according to right screw rule) which is the opposite of what a vector would do so it's a pseudo vector. This is the active transformation view. But wait a minute, active view is when you invert all vectors except the basis vectors (the coordinate axes). So what is the meaning then of 'x goes to -x' .....

Now let's consider the passive transformation i.e. inversion only of the basis vectors (coordinate axes) and all other vectors remaining the same. Then a remains a, b remains b, a x b remains a x b, and any other vector remains the same. So in this case a x b behaves like any other vector, it remains the same geometrically. Where is the 'pseudo-vector' in this case?

The article should be rewritten so that these questions are so clearly addressed that the above confusion does not arise at all. And a clear definition of what is meant by inversion cannot be substituted with the usual confusing and not clarifying 'examples from the real world' about wheels, magnets, mirror worlds and whatever.

I personally have NEVER seen a clear address of the above question in ANY book, in neither mathematics nor physics. If such book exists, a reference should be put pointing to it for the readers that want to understand the stuff better than some car wheel example. If the article is not rewritten it simply propagates the usual confusion found in thousands of textbooks and is simply of no value to Wikipedia. —Preceding unsigned comment added by 75.22.195.143 (talk) 09:22, 14 January 2009 (UTC)

I guess I only understood this stuff well later on in my physics education--after courses in GR and QFT. I can't think of a book off-hand that addresses this topic well at an intro-physics level. I vaguely remember Feynman's Lectures on Physics discussing this, but I don't have it with me and I could be wrong.

Does the new section I wrote help? I put in the "Examples" section to show how you can do things from the ground up. (I'm not implying that the introduction section doesn't also need improvement.)

To answer your specific question, your argument is correct: If you do a passive transformation, you have to switch to the left-hand rule. Sorry that I implied otherwise! More specifically, if you have right-hand coordinate axes you need the right-hand rule, and if you have left-hand coordinate axes you need the left-hand rule. This has to be the case, because it must always be true that (1,0,0) X (0,1,0) = (0,0,1), i.e. x-hat cross y-hat equals z-hat. (When in doubt, forget about left hands and right hands, and just use the algebraic definition, which never changes.) So the a vector doesn't move, but its coordinates change from (a_x,a_y,a_z) to (-a_x,-a_y,-a_z), and ditto for b, and if you algebraically compute the cross product of a and b with their new coordinates, and then plot the result using the new coordinate axes, you'll see that a x b has in fact switched directions. Again, this is why I prefer the active-transformation approach. :-) --Steve (talk) 21:30, 14 January 2009 (UTC)

Feynman discusses the definition of vectors as things that transform like the coordinates (i.e. are contravariant), but he doesn't discuss pseudovectors as far as I can tell. Jackson's Classical Electrodynamics and Arfken & Weber's Mathematical Methods for Physicists both have good discussions of the subject (and are cited as references in our article), especially the latter. Wikipedia's article on covariance and contravariance of vectors is, unfortunately, rather incomprehensible IMO (partly because it starts too generally, with linear functionals and curved manifolds). —Steven G. Johnson (talk) 21:47, 14 January 2009 (UTC)

There seems to be an enormous amount of confusion about this topic. Vectors and vector cross products are defined so that they are independent of coordinate system. Whether you use a right handed or left handed coordinate system has no bearing on the use of the right hand rule to determine the direction of the cross product of two vectors. It is conventional to always use a right hand rule to determine the direction of the cross product of two vectors. In a right handed basis i x j =k. In a left handed basis i x j=-k. Since the directions of both vectors and pseudovectors (cross products) are fixed independently of coordinate system, they must transform in the same way under a passive rotation regardless of whether it is proper or improper. The distinction between vectors and pseudovectors arise only for active improper rotations.Jcpaks3 (talk) 23:17, 27 June 2009 (UTC)

Jcpaks3, if you use the definition

then (1,0,0) x (0,1,0) = (0,0,1) always. So in a left-handed coordinate system, you need to use the left-hand rule.

I'm sure other definitions of the cross product are possible, but this one is pretty standard as far as I can tell. Also, with this definition, everything is the same whether you take the point of view of active or passive transformations. With the definition you like, they're quite different, which I view as a sign that your definition is not a great definition. After all, many (probably most) physicists only use passive transformations, and yet have no problem defining and discussing pseudovectors. --Steve (talk) 02:24, 28 June 2009 (UTC)

The definition you quote is not a fundamental definition of cross product and only applies to a right handed coordinate system. The Levi Civita symbol is a tensor density, not a tensor. If you transform to a left-handed basis set, the relation you use acquires an additional factor of det(R) where R is the orthogonal transformation matrix. The cross product of two vectors (pseudo vector) is independent of coordinate system. If A and B are two known vectors, you don't need a coordinate system to determine A x B. By convention, we have made the choice of using a right hand rule to determine the direction of A x B. The determinant that we use to find the cross product in elementary physics (and where your "definition" comes from) implicitly assumes a right hand coordinate system. For example, in classical physics the electric field (E) is a vector and the magnetic field (B) is a pseudo vector (whose direction is determined by a small bar magnet at the point in question, B is parallel to the magnet and directed from the south to the north pole). It is easy to generate static electric and magnetic fields at the same point that are parallel. Simply changing to a different coordinate system does nothing to E and B. They must remain parallel. This is obviously true under a proper or improper rotation. Therefore E and B must transform in the same way under any passive orthogonal transformation. It is only under an active (physical system changes, coordinate system remains the same) orthogonal transformation that E and B transform differently.Jcpaks3 (talk) 17:36, 28 June 2009 (UTC)

Hmm, I checked all the textbooks I have on hand, and none discusses pseudovectors (or cross-products) from the passive-transformation approach. Do you have a textbook that does this? Or says that it's impossible? If so, that would be a nice addition to the article.

You're using the definition "cross product always follows the right-hand-rule, even in a left-handed coordinate system". In which case, as long as Maxwell's equations don't change, E and B obviously "transform" the same way. I'm using the definition "cross product follows the right-hand rule in a right-handed coordinate system and left-hand rule in a left-handed coordinate system". In which case a passive parity inversion would obviously change the direction of B but not E. It would also switch all the "right-hand rules" with "left-hand rules" in magnetism, and given a bar magnet it would switch which pole was labeled "north" and which was labeled "south". The physics all works out, despite the changing direction of B, and that includes your bar magnet example.

I would call my transformation "passive coordinate inversion". You would call it "passive coordinate inversion plus changing the meaning of the cross product". I don't know which is the more common thing to call it. Both are probably acceptable. But can we agree that, whatever you call it, this is a way to define and discuss pseudovectors in a (basically) passive-transformation point-of-view? --Steve (talk) 19:02, 28 June 2009 (UTC)

I am a retired physics professor who taught an intermediate E&M class last year. The text I used had a discussion of passive transformations of vectors and then introduced the notion of pseudo vectors in a problem. I emailed the author and tried to convince him, with limited success, that vectors and pseudo vectors had to transform the same way under passive coordinate rotations, proper or improper. You and I seem to disagree about how the cross product is defined. I think that you want to choose a coordinate system first and then define the cross product according to whether the coordinate system is right or left handed. My interpretation of cross product is as follows. Two vectors determine a plane. The cross product of the two vectors is defined to be perpendicular to that plane. The only ambiguity is which direction to give the cross product. This ambiguity is resolved by using the right hand rule. This makes the definition independent of any coordinate system. The magnetic field is a pseudo vector. If we define its direction by the bar magnet that I mentioned previously, then the magnetic force on a moving charge is q(v x B) if you use a right hand rule for the cross product. This vector equation should be the same in any coordinate system, and it will be if the right rule is always used to determine the cross product. The right hand rule has absolutely nothing to do with the coordinate system one chooses to use. Consistency requires that the direction of A x B is always determined by using a right hand rule. I think that the confusion arises when we attempt to find the components of the cross product in a particular coordinate system. The fact that an improper rotation (or left handed coordinate system) introduces an extra negative sign seems to be the source of confusion. This sign change is necessary for a pseudo vector to remain the same under a passive coordinate change. It would be a strange world if inverting a coordinate system changed the direction of a magnetic field (and hence the poles of a bar magnetic).

I have not been able to find a source that discusses this issue in detail. In fact the sources that I have found seem to say that cross products transform differently from vectors under passive improper rotations. I don't believe that this is the case for the reasons I have stated. That is why I found your article in the first place. I noticed that one of your earlier responders also said that vectors and pseudo vectors transform the same under improper rotations. Since there seems to be a lot of confusion about this, I think it would be great if you were able to shed some light on it or convince me that I am wrong.Jcpaks3 (talk) 20:27, 28 June 2009 (UTC)

I agree with essentially everything you're saying. Insofar as you can define things in a coordinate-independent way, passive transformations are meaningless, obviously. In fact, I had a physics professor who said "active transformations are superior in every way to passive transformations"...not coincidentally, his main research interest was proving physics theorems in a coordinate-independent way. Anyway, how about this:

Pseudovectors are usually discussed using active transformations. An alternate approach, more along the lines of "passive transformations", is to keep the universe fixed, but switch "right-hand rule" with "left-hand rule" and vice-versa everywhere in physics, in particular in the definition of the cross-product. Any polar vector (e.g., a translation vector) would be unchanged, but pseudovectors (e.g., the magnetic field vector at a point) would switch signs. Quantities defined via right-hand rules would likewise switch, such as the "north pole" and "south pole" of a magnet.

(I left out links and references, but have some.) Would you agree with this? I tried to skirt the issue of how the cross-product is fundamentally defined. --Steve (talk) 22:34, 28 June 2009 (UTC)

There are times when active transformations have advantages, but in relativity passive transformations (to my knowledge) are almost exclusively studied.

I don't understand your desire to skirt the fundamental definition of the cross product. I also think that using a left hand rule in a left handed coordinate system is inconsistent. For example, suppose that the velocity of a charged particle is along the x-axis and the magnetic field is along the y-axis in a right handed coordinate system. If the charge is positive, the force is along the z-axis. Now consider a passive inversion: i'=-1, j'=-j, k'=-k. The force doesn't change. It is given by

f=q(vi)x(Bj)=qvbk =q(-vi')x(-Bj').

According to your prescription, the last term would be evaluated using a left hand rule giving i'xj'=k'=-k and f=vBk=-vBk. Using a right hand rule gives i'xj'=-k which is consistent. Unless I have done something wrong, this is clearly inconsistent.Jcpaks3 (talk) 15:33, 29 June 2009 (UTC)

When you switch to left-hand rule, you need to change the sign of the magnetic field (like any other pseudovector).

What I'm trying to get at is Feynman Volume 1 Chapter 52: "...if some demon were to sneak into all the physics laboratories and replace the word "right" for "left" in every book in which "right-hand rules" are given, and instead we were to use all "left-hand rules," uniformly, then it should make no difference whatever in the physical laws." (Except phenomena involving the weak force, of course.) If you read the chapter, you'll see he gives examples with magnets, it all works out.

I don't know how the cross product is "supposed" to be defined in a left-handed coordinate system. Instead I propose saying explicitly whether we're using the left-hand rule or right-hand rule. If we do that, everything is clear and explicit, and the conventional definition, whatever it is, doesn't come up. --Steve (talk) 15:53, 29 June 2009 (UTC)

The right hand rule is just a convention to determine which of two possible normals to a plane to use to define a cross product. It is obvious that a left hand rule would work just as well. If we kept the same form for the Biot-Savart law, but adopted a left hand rule the direction of the magnetic field (and other pseudo vectors) would change. But if you let your choice of whether to use a right or left hand rule depend on whether you use a right or left handed coordinate system, things would be pretty chaotic. I don't think you really want to advocate a formalism where the direction of the earth's magnetic field depends on whether you use a right or left handed coordinate system.Jcpaks3 (talk) 19:53, 29 June 2009 (UTC)

Good, we're in agreement. I'm very happy to forget about the handedness of the coordinate system. We can just talk about right-hand-rule versus left-hand-rule, without thinking about coordinates. Would you agree with the addition of this text (or something like it) to the article?...

Pseudovectors are usually discussed using active transformations. An alternate approach, more along the lines of "passive transformations", is to keep the universe fixed, but switch "right-hand rule" with "left-hand rule" and vice-versa everywhere in physics, in particular in the definition of the cross-product. Any polar vector (e.g., a translation vector) would be unchanged, but pseudovectors (e.g., the magnetic field vector at a point) would switch signs.

--Steve (talk) 23:18, 29 June 2009 (UTC)

The choice of right hand vs left hand rule is simply a matter of convention. I am not aware that this is really a point that anyone is worried about. Feynman's comments are perfectly obvious. We also made a choice in assuming that the velocity vector points in the direction a particle is moving rather than the opposite direction. Although it would be unnatural, we could reformulate mechanics if we change the conventional direction of velocity and get a perfectly consistent formalism. In quantum mechanics we choose the commutator of x and p to be i (in natural units). We could have chosen it to be -i and developed an equivalent theory. We make choices like this all the time. The important thing is to be consistent once a choice is made. That is why you can't switch between right and left hand rules depending on the orientation of an arbitrary coordinate system.

The thing that concerns me is that there seems to be some confusion as to whether cross products transform differently from vectors under both passive and active improper rotations. The sources that I have found indicate that cross products do transform differently from vectors in both cases. I don't believe that because I don't believe the earth's magnetic field changes direction under a coordinate inversion. I think that it is only under active transformations that vectors and pseudo vectors transform differently. This is what I was hoping you could discuss.Jcpaks3 (talk) 14:54, 30 June 2009 (UTC)

(1) Pseudovectors acquire an extra sign-flip under active improper rotations.

(2) Pseudovectors acquire an extra sign-flip under the "convention-change" from right-hand-rule to left-hand-rule.

I'm saying that (1) is the active-transformation definition of pseudovectors and (2) is the closest thing there is to a passive-transformation definition of pseudovectors. It's not literally a passive transformation, but it's reminiscent of passive transformations, insofar as you're not actively transforming the system. Plus, some would argue that (2) is a passive transformation to a left-handed coordinate system. Personally, I don't know about that last sentence, since it depends on how you define cross-product, and I've seen both possible definitions in textbooks. So I would duck that controversy by not saying that (2) is a passive-transformation statement, but instead just saying that (2) is "more along the lines of a 'passive transformation' approach", or something like that. --Steve (talk) 15:48, 30 June 2009 (UTC)

I think that would help. You seem like a reasonable guy. It has been nice talking to you.Jcpaks3 (talk) 20:46, 30 June 2009 (UTC)