This is perhaps more a maths question, but it's about a piece of maths

[Philip Wood]

This is perhaps more a maths question, but it's about a piece of maths that physicists use a lot, and I'm interested in physicists' views. The question is, perhaps, rather subtle (though I hope not pedantic), and will take a bit of explaining. Sorry.

We're told that magnetic flux density and angular momentum are pseudovectors. Home in on flux density, B. Suppose we have a current-carrying loop, and another current carrying loop, which is a reflection in a plane of the first one. In each case the B (say at the centre of the loop) is related to the sense of the current by a right hand rule. This means that if the coil's plane is parallel to the reflection plane, then the B produced by the loops is normal to the reflecting plane and in the same direction - not opposite directions. An ordinary vector normal to the reflecting plane would be reversed by the reflecting plane. Conversely, the field due to a loop with its plane normal to the reflecting plane would produce a B parallel to the plane, but the B due to the reflected loop would be antiparallel to the first B. An ordinary vector parallel to the reflecting plane would not be reversed by the reflecting plane. There's a nice diagram in wiki under pseudovector - an excellent article in anycase.

Now the direction of B due to a square loop is that of c = axb in which a and b are are vectors representing successive adjacent sides directed according to the current.

The reflection behaviour is captured by the equation
R(axb) = -R(a)xR(b).
R is the operation of reflection in the plane. In words, the reflected cross product (vector product) of a and b is in the opposite direction to the cross product of the reflection of a and the reflection of b.

Now here's my question... Why is this behaviour regarded as a feature of the 'output vector', c, of the cross-product operation, rather than as a feature of the operation itself? It seems to me that c calculated for the unreflected system is just a vector, which reflects (see left hand side of equation above) as a normal (polar) vector reflects.

R(a)xR(b) is surely not sensibly regarded as the transformation (by reflection) of axb, since it has different vector 'inputs': R(a) and R(b) rather than a and b. [It's a bit like claiming that when we produce a real number c from the operation axb=c, then c isn't an ordinary real number because if we multiply it by -1 we get a different result from when we input -a and -b, instead of a and b, into axb=c.]

To sum up: why do we speak of pseudovectors and vectors rather than simply drawing attention to the reflection quirk of the vector product operation, as summed up in the equation above? [Or, to return to the current loop, why do we speak of B not obeying the same reflection rules as an ordinary vector, when we're not actually comparing B's due to the same current loop, but B's due to a current loop and its reflection?]

 

[vanhees71]

I guess, I've not understood your question properly. I can only try to explain what physicist understand under vectors and pseudovectors.

Of course, first of all vectors have in physics the same meaning as in mathematics, namely being elements of a vector space over a number field.

Your particular question is related to vectors in physical space within Newtonian physics or special relativity. In both space-time models, an inertial observer describes, after fixing an arbitrary point as the origin, the physical space as a Euclidean three-dimensional real vector space (more precisely the space is an Euclidean affine point space, but for our purposes, it's convenient to fix an arbitrary origin and describe each point in space by the corresponding position vector).

Now, for physics symmetries play an eminently important role, and after fixing the origin, we have rotations around this point as a symmetry of the space. This is described by the special orthogonal group SO(3), which leaves the Euclidean scalar product of any pair of vectors invariant.

Now physical quantities with a well-defined behavior under rotations are particularly nice objects to describe phenomena since all physical laws must obey the fundamental symmetry of space-time to be consistent with the underlying space-time model. Thus, any physical quantity which doesn't change under rotations is called a scalar, any vector quantity transforms under rotations like the position vectors. From this one can build multi-linear forms which transform by the Kronecker product of rotations. These can be further classified by irreducible representations of the SO(3), and so on.

Now, one can also take into account the discrete symmetries of space-time. Here we are interested in the space reflection. In physics it's more costumary to consider the parity transformation which maps any position vector $$\vec{x}$$ into $$P(\vec{x})=-\vec{x}$$. Now $$P=\mathrm{diag}(-1,-1,-1)$$ is orthogonal but has determinant -1. Any quantity that transforms under rotations and the parity transformation as the postition vectors is called a polar vector.

Now the vector product of two polar vectors transforms under rotations as a usual vector, but under the parity transformation, we have

$$P(\vec{a}) \times P(\vec{b})=(-\vec{a}) \times (-\vec{b})=+\vec{a} \times \vec{b}.$$

Thus under parity auch a vector quantity doesn't behave like a polar vector but just stays unchanged under the parity transformation. Such a quantity we call an axial vector or pseudovector.

Examples are obviously angular momenta since $$\vec{L}=\vec{x} \times \vec{p}$$ since $$\vec{x}$$ and $$\vec{p}$$ are polar vectors.

Further the force on a particle is also a polar vector. The force on a charge in an electromagnetic field is given by the Lorentz force

$$\vec{F}=q (\vec{E} + \frac{\vec{v}}{c} \times \vec{B}).$$

This implies that $$\vec{E}$$ must be a polar and $$\vec{B}$$ must be an axial vector to be consistent with $$\vec{F}$$ being a polar vector.

It's easy to prove that the complete set of Maxwell's Equations together with the equations of motion of charged particles is form invariant under parity transformations, i.e., electromagnetic interactions obey not only the continuous rotational symmetry (group SO(3)) but also under parity and thus under the larger group, O(3).

Of course the full symmetry group of classical electromagnetism is the Poincare group including paritiy and time reversal, i.e., it is not only invariant under the continuous part of the Poincare group (i.e., the proper orthochronous Poincare group) but under the whole Poincare group.

I hope from this example the meaning of "pseudovectors" in physics has become a bit clearer.

 

[Philip Wood]

Many thanks for taking the trouble to write such a full and informative reply. I hope you enjoyed writing it as much as I enjoyed reading it. We don't differ over anything except one issue of interpretation - and I expect this is due to some mental block of mine...

Whereas I considered the operation of reflection in a plane, you considered the parity change operation. This is NOT the issue in question, because both are improper rotations whose matrices have determinant -1, and the same point about vector products emerges from your equation:
P(a)xP(b) = (-a)x(-b) = axb
as emerges from mine:
R(a)xR(b) = -R(axb).

For me, either equation (rather than a distinction between pseudovector and polar vector) is the proper end-point of this line of reasoning. Either equation points to a distinctive feature of the vector product operation, setting it aside from, say, vector addition, for which:
P(a)+P(b) = (-a)+(-b) = -{a+b} .

Why do I baulk at the pseudovector/polar vector distinction? It's usually stated that these objects transform in different ways under reflection or parity change. Now, the so-called pseudovector is the 'output' of the vector product operation, i.e. c, given by c=axb. Surely if we subject c to a parity change it will become -c, just like a polar vector?

But we're asked to base the pseudovector nature of c on the fact that if we subject the inputs, a and b, to the vector product to a parity change (i.e. if we reverse both a and b) then c does not reverse direction. But subjecting a and b to a parity change is not at all the same thing as subjecting c to a parity change.

c, I'm arguing, is just an ordinary, polar vector, and it's weird to make it bear the burden (by calling it a pseudovector) of the (undisputed) distinctive parity behaviour of the vector product operation.















P(a)+P(b) = (-a)+(-b) = -{a+b}

 

[BruceW]

This is a good question.
So putting the question in my own words: while c is being defined as cross product of a and b, the reflection of a and b will not cause c to change. But if we do a reflection of c itself, then it does change.
So maybe the point physicists are trying to make with this pseudovector thing is that we will always be reflecting a and b (never reflecting c itself).
I now wonder why this should be true. Maybe it is because a and b are the physical vectors, and c is only a mathematical quantity that is calculated from the physical vectors. (So reflection should only happen on a and b because they are the physical ones)?
This is just me guessing, if anyone knows the answer, I'd like to find out.

 

[Philip Wood]

BruceW. Yes, I think you see what I'm driving at. I was beginning to worry that I'd gone out of my mind. Turning to your interesting suggestion for resolving the difficulty, I'm not sure that I'd want to withhold physical status from magnetic field (a so-called pseudovector).

 

[The_Duck]

Another example: Angular momentum is also a pseudovector. Classically angular momentum is, like the magnetic field, the result of a cross product r x p, but there is quantum mechanical spin angular momentum that is some intrinsic physical thing that is not obviously the result of a cross product operation.

One reason to insist on a distinction between polar and axial vectors is to make clear that under normal circumstances they should be kept apart from each other. It usually doesn't make sense to add a polar vector to an axial vector, for instance. If you do add them, you expect weird things to happen under reflections, as the sum of a polar and axial vector does not behave simply under parity. This is how parity violation occurs in the theory of the weak force.

 

[Philip Wood]

Agree. Your point that "quantum mechanical spin angular momentum [...] is some intrinsic physical thing that is not obviously the result of a cross product operation" is a very interesting one, and I shall, in my dull-witted way, ponder its implications.

I was going to tone down my objection to the polar vector / pseudovector distinction by saying the distinction is useful, but shouldn't be defined in terms of the (identical) transformation properties of the vectors themselves. Instead, what matters about them, I claimed, is their emergence from cross-product operations, whose parity behaviour is where the gist of the distinction lies. But you've made me have second thoughts about the primacy of the cross-product. Another aspect to which I haven't paid enough attention is the participation of pseudovectors as INPUTS to the cross product.

Many thanks.

 

[Studiot]

Hello Philip, I have attached a simple calculation based on the three unit vectors i. j, k to show the difference between vectors and pseudovectors and how this arises.

In general if you use symbols in compact (bold) form for vectors you will not be able to tell the difference.

Also a vector crossed with a vector yields a psuedovector, whereas a vector crossed with a psuedovector yields a vector.

Both vectors and pseudovectors are sets of three quantities. Their difference is how they react to transformations.

To see this consider a simple reflection in the xy plane on the true vector k and the cross product (i x j), which is a pseudovector. Applying the transformation, T to both yields interesting results.

It is worth also working out the effect of the parity transformation, P (all -1s on lead diagonal).

What this is telling us in physical terms is that the cross product of, say, given current and motion will produce the same real field direction, regardless of coordinate system, which is of course what we want in physics.

go well

 

[atyy]

B is a pseudovector because its direction depends on what we believe to be a choice of convention about using the right-hand rule to define the cross-product.

Since it is our arbitrary choice, the choice should not show up in the physics. So it is a pseudovector. What is observed is the acceleration of a charged particle, not B.

The Lorentz force law gives: acceleration ~ velocity X B

The B field is generated by a current souce: B ~position X current

So acceleration ~ velocity X B X position (order is confused, but you can figure that out). You will find that the double cross product produces a vector acceleration whose direction is independent of our choice of right hand rule in the cross-product convention.

However, it turns out that there are phenomena in which the universe is handed.

 

[Philip Wood]

Studiot. Agree with your calculations except that you've written T(ixj) = T(i)xT(j), whereas your very own calculations show that T(ixj) = -T(i)xT(j).

 

[Studiot]

except that you've written T(ixj) = T(i)xT(j), whereas your very own calculations show that T(ixj) = -T(i)xT(j).

I'm sorry, how do you make that out?

The real situation is that the cross product of two vectors does not yield a vector it yields a skew-symmetric second order cartesian tensor.
Discussed in thse terms you can see why the properties are attributed to the output, not to the cross product process.
Are you comfortable with tensors?

I'm having no little difficulty with the new programming on this site at the moment, but I will try to keep an eye on this thread.

 

[BruceW]

I read the wikipedia article on cross product, and it said that "The vector cross product can also be expressed as the product of a skew-symmetric matrix and a vector."
So the cross product of two vectors would yield a vector. (If I'm wrong, sorry, please do point it out).
After reading the article, I feel like I do understand the concept of a pseudovector now:
When you calculate the cross-product in the left-hand co-ord system, you get the negative of what you would calulate in a right-hand co-ord system.
So the output of the cross-product depends on which handed-ness you use.
A reflection causes a change in handed-ness of the sytem.
Therefore we have to define the reflection of a pseudovector to be -1 times what the reflection of a normal vector would be!
So if $\vec{a} = i$ and $\vec{b} = j$ and of course $\vec{c} = k$ then:
$$\vec{a} \wedge \vec{b} = \vec{c}$$
Then doing a reflection in z direction to both sides gives:
k = -1 times -k
(since the left side contains no k's, so it is not affected, and the right side contains k, so it is times -1, but then times -1 again because it is a pseudovector.)
Or, doing a reflection in x direction gives:
$$-i \wedge j = -k$$
since the right contains no i's, but we defined $\vec{c}$ to be a psudovector, so it gets multiplied by -1
So in conclusion, we have to define the reflection of a pseudovector to be -1 times what the normal reflection of that vector would be. And this keeps the equations consistent.

 

[D H]

So the cross product of two vectors would yield a vector. (If I'm wrong, sorry, please do point it out).

As has been pointed out, the cross product of two vectors is a pseudovector, not a vector. Pseudovectors share many of the same properties as do vectors: You can add them vectorially and you can scale them by a scalar. However, they do not transform like vectors upon reflection.

Studiot mentioned that a better view is that the cross product of two vectors is a skew-symmetric rank 2 tensor, or a bivector.

Rhetorical question: How many distinct values are needed to define a skew-symmetric matrix? The answer is is that it depends on the dimensionality of the matrix. For a 2x2 matrix, only one parameter is needed. 24 are needed for a 4x4 matrix. In general, N*(N-1)/2 parameters are needed, where N is the dimensionality. There is one very special case where N*(N-1)/2=N, and this is N=3. Pluck the right three elements in the right order out of the six non-zero elements of a 3x3 skew symmetric matrix and you have something that looks a lot like a vector. That does not mean that it *is* a vector.

 

[vanhees71]

The cross product is indeed a skew-symmetric bilinear map $V \times \vec{V} \rightarrow V$, where $V$ is isomorphic to $\mathbb{R}^3$.

The behavior of the vector product under reflections on a plane is also easily seen by first defining the corresponding linear operator. Let $\vec{n}$ the normal vector of the plane. Then the component $\vec{n} (\vec{n} \cdot \vec{x}$ wrt. to the normal direction flips sign, while the rest stays as it is, i.e., you have

$$R_\vec{n} \vec{x}=\vec{x}-2 \vec{n} (\vec{n} \cdot \vec{x}).$$

Then you have for two vectors

$$(R_{\vec{n}} \vec{a}) \times (R_{\vec{n}} \vec{b})=-R (\vec{a} \times \vec{b}).$$

This can also be seen by the fact that reflection on a plane is a parity transformation followed by a rotation with 180 degrees around $\vec{n}$. The general transformation rule of a vector product for an arbitrary O(3) transformation thus reads

$$(O \vec{a}) \times (O \vec{b})=\det O O(\vec{a} \times \vec{b}).$$

 

[Studiot]

This is perhaps more a maths question, but it's about a piece of maths that physicists use a lot, and I'm interested in physicists' views. The question is, perhaps, rather subtle (though I hope not pedantic), and will take a bit of explaining. Sorry.

I'm mindful that Philip is seeking physical motivations behind the cross product.

With the cartesian system I described, i, j are basis vectors for the vector space spanned by them. They are generators for the xy plane which is the vector space they span.

All the other rules of (mathematical) vector algebra lead to another vector within this space, or to number (scalar) or coefficient.

The cross product is different in that it leads to an object that is outside the vector space and not a member of it.

Alternatively we need to expand the vector space to three dimensions to accommodate it.

This situation occurs elsewhere in applied maths and led Hamilton to develop quaternions.
It also occurs where we sum to infinity a series of continuous functions to yield a discontinuous one in Fourier analysis.

So there is indeed a strong link between physics and maths, although the maths seems a bit disconnected at first.

go well

 

[BruceW]

Interesting stuff. So you could say that the cross-product is either: 1) a vector that doesn't obey the normal rules. Or, 2) a 2nd order anti-symmetric tensor.
Choice 2) would be the more general way, and choice 1) is a specific case. Is all this about right?
And so I guess that the reason we're taught that the cross-product is a vector is because its a simpler way to learn it the first time, and because to begin with, we're not doing problems that deal with reflections of the system.
It seems that a lot of physics is taught this way - specific examples first, leading up to general cases. And it looks like I've got a lot more maths to learn :)

 

[Studiot]

1) a vector that doesn't obey the normal rules.

It's not as bad as it sounds. Pseudovectors follow most of the normal rules that matter.

Most of the transformations of note are rotations - often called 'proper rotations'. Both vectors and psuedovectors tranform the same for these.

It is in the so called 'improper rotations' - parity and reflections, if you conside them separately, that they differ.

In summary:

A vector transforms the same as the coordinate axes under both proper and improper rotations., which means it transforms the same as the position vector.

A pseudovector transforms the same as the coordinate axes under proper rotations but is invariant under improper rotations.

Incidentally there are also scalars and pseudoscalars.

A scalar is invariant under both proper and improper rotations.

A pseudoscalar is invariant under proper rotations, but changes sign under improper rotations.

go well

 

[Philip Wood]

Studiot: I'm sorry about my abruptness in reply to your post 8. Tell me where I'm wrong, but surely T(i)xT(j) = (-i)x(-j) = (-1)(-1)ixj=k, whereas T(ixj) = T(k) = -k.
This assumed that i, j, k are a right-handed set, but the same result, T(i)xT(j) = -T(ixj) would hold for a left-handed set. Have I missed something?

Thank you very much for raising the point about the skew-symmetric tensor, which, I agree, is an excellent animal. I'll think about its implications for my embarrassing personal problem (the interpretation of pseudovectors).

 

[atyy]

Isn't it simply that in doing the cross-product, you choose the right hand rule. But the observed phenomena in electrodynamics don't depend on such a choice. So vectors that are the result of a cross-product are named "pseudovectors" just to help us remember that they cannot be directly observed.

 

[Studiot]

Studiot: I'm sorry about my abruptness in reply to your post 8.

Nothing to be sorry for, you had a doubt (though I wasn't sure what it was) and you stated it politely.

The point I was trying to show was that if we state i x j = k then we are stating that the result (output) of a cross product is a vector - in this case k.

I was trying to show, non symbolically, that this can't be true since although the output behaves like k in some respects, it does not do so in all respects.

So although it looks like a bear and growls like a bear, it does not smell like a bear.

So it is not a bear!

The next bit was to show, again non symbolically, why by operating with the transform in the inputs to the cross product, or if you like swapping the order of operators (xproduct and reflection).

The reason for asking about tensors is that if you use the Einstein tensor index notation this all falls out quite naturally.

Does this explain it better?

 

[atyy]

Another way to motivate the language that Studiot and D H have been advocating is to think that the cross product forms an oriented area VXW, while an oriented volume is U.(VXW). Naively we would like the volume to be a generalization of area "UXVXW", which is wrong if we use the cross product. There is a different operator, the wedge product, which does make the oriented volume U^V^W. It is in that language that the wedge product takes two "one forms" and outputs a "two form".

Maxwell's equations can be formulated using differential forms. Stokes's theorem, which relates the differential and integral forms of Maxwell's equations, has its most general formulation in that language.
http://sophia.dtp.fmph.uniba.sk/~fecko/regensburg.html

If an N dimensional space has a metric, the Hodge star is a linear map between p forms and N-p forms. In 3D, this is the relationship between vectors and pseudovectors.
http://unapologetic.wordpress.com/2009/11/10/the-cross-product-and-pseudovectors/
http://www.damtp.cam.ac.uk/research/gr/members/gibbons/gwgPartIII_DGeometry2011-1.pdf

 

[Studiot]

Thank you for those links, atyy. Most of my resources are paper based so I am always pleased to note good weblinks. At least I've (or my wife has) managed to throw out the stone tablets as too bulky.

go well