# Can vectors in physics be represented by complex numbers and can they be divided? [closed]

Below is attached for reference, but the question is simply about whether vectors used in physics in a vector space can be represented by complex numbers and whether they can be divided.

In abstract algebra, a field is an algebraic structure with notions of addition, subtraction, multiplication, and division, satisfying certain axioms. The most commonly used fields are the field of real numbers, the field of complex numbers, and the field of rational numbers, but there are also finite fields, fields of functions, various algebraic number fields, p-adic fields, and so forth.

Any field may be used as the scalars for a vector space, which is the standard general context for linear algebra. The theory of field extensions (including Galois theory) involves the roots of polynomials with coefficients in a field; among other results, this theory leads to impossibility proofs for the classical problems of angle trisection and squaring the circle with a compass and straightedge, as well as a proof of the Abel–Ruffini theorem on the algebraic insolubility of quintic equations. In modern mathematics, the theory of fields (or field theory) plays an essential role in number theory and algebraic geometry.

In mathematics and physics, a scalar field associates a scalar value to every point in a space. The scalar may either be a mathematical number, or a physical quantity. Scalar fields are required to be coordinate-independent, meaning that any two observers using the same units will agree on the value of the scalar field at the same point in space (or spacetime). Examples used in physics include the temperature distribution throughout space, the pressure distribution in a fluid, and spin-zero quantum fields, such as the Higgs field. These fields are the subject of scalar field theory.

In mathematics, an algebra over a field is a vector space equipped with a bilinear vector product. That is to say, it is an algebraic structure consisting of a vector space together with an operation, usually called multiplication, that combines any two vectors to form a third vector; to qualify as an algebra, this multiplication must satisfy certain compatibility axioms with the given vector space structure, such as distributivity. In other words, an algebra over a field is a set together with operations of multiplication, addition, and scalar multiplication by elements of the field.

A vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied ("scaled") by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex numbers, rational numbers, or even more general fields instead. The operations of vector addition and scalar multiplication have to satisfy certain requirements, called axioms,... An example of a vector space is that of Euclidean vectors which are often used to represent physical quantities such as forces: any two forces (of the same type) can be added to yield a third, and the multiplication of a force vector by a real factor is another force vector. In the same vein, but in more geometric parlance, vectors representing displacements in the plane or in three-dimensional space also form vector spaces.

In classical mechanics as in physics, the field is not real, but merely a model describing the effects of gravity. The field can be determined using Newton's law of universal gravitation. Determined in this way, the gravitational field around a single particle is a vector field consisting at every point of a vector pointing directly towards the particle. The magnitude of the field at every point is calculated applying the universal law, and represents the force per unit mass on any object at that point in space. The field around multiple particles is merely the vector sum of the fields around each individual particle. An object in such a field will experience a force that equals the vector sum of the forces it would feel in these individual fields.

Because the force field is conservative, there is a scalar potential energy per unit mass at each point in space associated with the force fields, this is called gravitational potential.

Gauss' law for gravity is mathematically equivalent to Newton's law of universal gravitation, but is stated directly as vector calculus properties of the gravitational field.

• This sounds like more of a question for math.SE, but I'll answer it anyway, because it's interesting, and has relations to physics! – Noldorin Jan 24 '11 at 2:46
• You already asked this question and it was closed! – MBN Jan 24 '11 at 5:04
• I can't vote to close for some mysterious bug - but this is clearly ot – Sklivvz Jan 24 '11 at 9:40

Can vectors in physics be represented by complex numbers?

Absolutely. There exists a direct isomorphism between the 2D Euclidean vector space and the Argand plane, for a start.

In fact, it is possible to talk of mathematical objects called quaternions and use quaternion algebra analogously to vector algebra. Historically quaternions were used to represent geometrical operations and transformations in 3D space - in the days before vector algebra. (They still are used, especially in areas such as computer graphics where they offer one or two advantages over the simpler world of vectors.) In any case, the relationship to vector algebra is a very close one.

Can vectors in physics be divided?

In general, no, vector-vector division is not a well-defined operation. At least, not within the bounds of linear algebra. i.e. There exist none or multiple solutions to the equation $\vec{y} = \mathbb{A} \vec{x}$. (See the Wolfram page.)

Saying this, the concept of vector division has an interesting relationship with your first question. If we map vectors to complex numbers (or quaternions in > two dimensions), we can use complex division or quaternion algebra respectively to define an analogous "vector division" operation. Note that quaternions can in fact be extended to higher dimensions, which allows for interesting possibilities. This study of this falls under the area of Clifford algebras.

(Note: vector-scalar division is of course well-defined, as is point-wise division of equal-length vectors, but I presume you are not referring to that.)

• So, the answer to the first question is no, unless we are considering vectors from a 2d real vector space! – MBN Jan 24 '11 at 4:14
• @MBN: it works more generally. There are quaternions in 4d, octonions in 8d, sedenions in 16d... But with each level you lose some structure (commutativity with quaternions and associativity with octonions) and finally sedenions aren't even a division algebra so after that you can't reasonably divide anymore. – Marek Jan 24 '11 at 9:00
• @Noldorin: what do you mean quaternions can be extended? You mean in the sense of finding the most structure in a vector space of given dimension, which leads to octonions and sedenions or in in the sense of Clifford algebras (of which quaternions are a special example)? – Marek Jan 24 '11 at 9:03
• Noldorin, I know, but these are not complex numbers, and his questions seems to be about complex numbers. – MBN Jan 24 '11 at 13:15
• @MBN: You're being really, really anal there. They're very closely related, and clearly relevant to the question (as are Clifford algebras). – Noldorin Jan 24 '11 at 17:32

In a certain sense, 3-d vectors can be divided in physics. To define the division, you begin by defining multiplication using the cross product. I.e. if $\vec{u} \times \vec{v} = \vec{w}$ then $\vec{w}/\vec{v} = \vec{u}$. This is a definition of division that is non commutative. It is also not well defined (as written) because you can always add multiples of $\vec{v}$ to $\vec{u}$ and still get a product of $\vec{w}$. That is, $\vec{u} \times \vec{v} = \vec{w}$ implies that $(\vec{u}+k\vec{v})\times\vec{v} = \vec{w}$. To make the division well defined you can require that $\vec{u}\cdot\vec{v}=0$ by subtracting off any component of $\vec{u}$ in the $\vec{v}$ direction.

Before you have division, you might want to have inverses. That is, for a vector $\vec{u}$, we need to define a vector $\vec{v}$ such that $\vec{u}\vec{v} = 1$. In the usual approach to physics this is nonsense because "1" is not a vector. But it is possible to make sense out of this and a group of physicists have done just that. To get a "1" from a vector product you use the dot product. Then the inverse of a real vector is just the same vector, divided by its squared magnitude.

In actual practice, it's probably clear that you need both the dot product and the cross product to fully define vector multiplication. The discussion I've given here is intended only to whet the interest of the reader, an adequate understanding will require a lot of work. For example, in the first paragraph, w is always perpendicular to both u and v; how do you define division when w and u aren't perpendicular?

Not very many physicists are interested in this sort of thing. It's studied by groups led by David Hestenes and makes up a part of what is called the "Geometric Algebra". As you can see from the above, it requires that vectors be generalized. For example, since addition works with vectors, one must allow constructions such as the sum of a vector and a scalar such as $\vec{u} + 3$.

As you would expect, the geometric algebra also implies a geometric calculus (which includes all the usual vector calculus relations you're familiar with such as Stoke's theorem). To learn more about it will likely require a great deal of time. To start you can click around on his website:
http://geocalc.clas.asu.edu/html/Intro.html

The geometric algebra is used in standard physics under a different name; the "Dirac algebra" or "gamma matrices" (so important to our modeling of the electron) are the geometric algebra in 3+1 dimensions. In this algebra, the usual spacetime or 4-vectors are generalized to include scalars, tensors, psuedo-vectors, and a pseudo-scalar.

• Nice answer mentioning geometric algebra, Carl. However I am wondering if someone is actually doing calculations with it (exception being the guys from QFT who cannot represent anything without it). I already own a book but do you think its worth the effort as an everyday physicist to learn it? Greets – Robert Filter Jul 18 '11 at 12:35
• Feynman said something to the effect that it's useful for a theoretical physicist to have a lot of tools in his toolbox. I don't think many people are using GA, but it's the most elegant method. That's why I think it's a great tool to use. But GA is not popular. So I published a paper that uses GA without mentioning it: arxiv.org/abs/1006.3114 – Carl Brannen Jul 19 '11 at 0:04
• So how long did it take you to acutally be able to calculate something with it that would have been not possible before? Don't get me wrong, I admire the beauty of GA but I dont know if it can pay of on a small time scale. Greets – Robert Filter Jul 19 '11 at 6:24
• Dr. Hestenes told me that his grad students had to study the material for a year (maybe two) before they were useful in it. As far as stuff that had not been possible; that may not exist as there always seems to be different ways of skinning cats. – Carl Brannen Jul 19 '11 at 21:14
• The usual approach to the unification of the forces of elementary particles is by assuming that the methods used up to now continue all the way down. So one looks for deeper symmetries. The search fails and so one broadens it to "broken" symmetries and renormalization. In my point of view, the whole field is a combination of intellectual cargo cultism (a naive understanding of how mathematics works) and glorified curve fitting. – Carl Brannen Jul 19 '11 at 21:26