Fields

[QuarkHead]

Fields are obviously of interest to mathematicians, but they seem to be very important in applications, especially physics and engineering.
However, in mathematics, the definition of a field depends on the context. Let's see........

In abstract algebra a field is defined as a set equipped with a pair of closed binary operations, usually called $$+$$ and $$\times$$, for each of which there is an inverse, respectively $$-x$$ and $$\frac{1}{x}$$ and an identity, respectively $$0$$ and $$1$$ such that $$x+(-x)=0$$ and $$x \times \frac{1}{x} = 1$$.

This is elementary. Examples of such fields are the Real Numbers $$\mathbb{R}$$ and the Complex Numbers $$\mathbb{C}$$. Do not run away with the idea that these sorts of fields are not important in physics - they are used all the time.

However in linear algebra, a field is defined somewhat differently. Here a field may be thought of as the assignment to every point in some chosen space a mathematical object - scalar, vector, tensor, whatever. In the foregoing, I chose not to call it a definition, as a moment's thought tells you that it depends entirely on the infamous Axiom of Choice, which mathematicians try (rarely successfully) to do without. Here's a better one.......

Consider a manifold (don't be deterred by the M-word - the humble 2-plane is a manifold, albeit not an interesting one). Anyway let's say that at every point on our manifold there exists a tangent vector space - this is always a given.

Now the set-theoretic union of all these vector spaces is called a vector bundle (technically it's a disjoint union - never mind about that). Then given some important constrains (which I'll come to shortly) a cross-section, or simply section, of our vector bundle defines a tangent vector field - at each point there exists a tangent vector

Note first that we can have ("choose") any section we like, hence the Axiom of Choice is not altogether avoided.

Now we will demand that from our vector bundle there is a projection "down" to our manifold. Calling our bundle as $$B$$ and our manifold as $$M$$ we may write $$\pi:B \to M, \pi(b)=m (b\in B, m\in M)$$. The inverse image of our projection $$\pi^{-1}$$ is called a fibre (or fiber if you prefer) over $$m \in M$$.

But these so-called fibres may be "tangled", so we finally require that, for the section $$\sigma:M \to B$$ that $$(\pi \circ \sigma)m=m$$ ie the composite function (as always reading right to left) is an identity.

Now we have field we can be proud of

 

[exchemist]

Fields are obviously of interest to mathematicians, but they seem to be very important in applications, especially physics and engineering.
However, in mathematics, the definition of a field depends on the context. Let's see........

In abstract algebra a field is defined as a set equipped with a pair of closed binary operations, usually called $$+$$ and $$\times$$, for each of which there is an inverse, respectively $$-x$$ and $$\frac{1}{x}$$ and an identity, respectively $$0$$ and $$1$$ such that $$x+(-x)=0$$ and $$x \times \frac{1}{x} = 1$$.

This is elementary. Examples of such fields are the Real Numbers $$\mathbb{R}$$ and the Complex Numbers $$\mathbb{C}$$. Do not run away with the idea that these sorts of fields are not important in physics - they are used all the time.

However in linear algebra, a field is defined somewhat differently. Here a field may be thought of as the assignment to every point in some chosen space a mathematical object - scalar, vector, tensor, whatever. In the foregoing, I chose not to call it a definition, as a moment's thought tells you that it depends entirely on the infamous Axiom of Choice, which mathematicians try (rarely successfully) to do without. Here's a better one.......

Consider a manifold (don't be deterred by the M-word - the humble 2-plane is a manifold, albeit not an interesting one). Anyway let's say that at every point on our manifold there exists a tangent vector space - this is always a given.

Now the set-theoretic union of all these vector spaces is called a vector bundle (technically it's a disjoint union - never mind about that). Then given some important constrains (which I'll come to shortly) a cross-section, or simply section, of our vector bundle defines a tangent vector field - at each point there exists a tangent vector

Note first that we can have ("choose") any section we like, hence the Axiom of Choice is not altogether avoided.

Now we will demand that from our vector bundle there is a projection "down" to our manifold. Calling our bundle as $$B$$ and our manifold as $$M$$ we may write $$\pi:B \to M, \pi(b)=m (b\in B, m\in M)$$. The inverse image of our projection $$\pi^{-1}$$ is called a fibre (or fiber if you prefer) over $$m \in M$$.

But these so-called fibres may be "tangled", so we finally require that, for the section $$\sigma:M \to B$$ that $$(\pi \circ \sigma)m=m$$ ie the composite function (as always reading right to left) is an identity.

Now we have field we can be proud of

I'm not sure what the point of discussion is here, but a question occurs to me. Michael Faraday was, I believe, the first to apply the term "field" to physics, for electric and magnetic action-at-a-distance influences. Do you know whether he used the word "field" as a result of the mathematical definition, or did the mathematical definition arise from generalising his term? It would be useful to know when the mathematical concept of a field was first defined.

 

[mathman]

I suspect the term "field" in physics is based on everyday usage, like a field of flowers or a field for sports or a field of wheat, etc. A gravitational field is thus a part of space where there is gravity. In mathematics fields have well defined properties.

 

[arfa brane]

In abstract algebra I believe there are three kinds of 'algebraic structures': groups, rings, and fields. But you find that a field is a commutative ring with unity (a multiplicative identity), with every nonzero element a unit.

Roughly, a field is a ring in which division is defined, or where the cancellation rule holds.

But what do I know?

 

[exchemist]

Rhetorical?

 

[QuarkHead]

es. Do you know whether he [Faraday] used the word "field" as a result of the mathematical definition, or did the mathematical definition arise from generalising his term? .

I don't know. I do know that when Maxwell wrote down his Electromagnetic Theory, he first used quaternions (you what?). It was Oliver Heavyside who translated them into the vector field format we now all know. And Heavyside was himself an engineer! Good gracious, an engineer!

You might find the following amusing (I do). Suppose we call the set of all vector fields over our manifold $$M$$ as $$\mathcal{X}(M)$$. And further suppose that $$\alpha,\, \beta \in \mathbb{R}$$. Then, for any $$X,\,Y \in \mathcal{X}(M)$$ it is easy to show that $$X+Y \in \mathcal{X}M$$ and that $$\alpha X +\beta Y \in \mathcal{X}(M)$$.

In other words the set of all vector fields over a manifold is itself a vector space. I love such curiosities that arise in mathematics

 

[exchemist]

I don't know. I do know that when Maxwell wrote down his Electromagnetic Theory, he first used quaternions (you what?). It was Oliver Heavyside who translated them into the vector field format we now all know. And Heavyside was himself an engineer! Good gracious, an engineer!

You might find the following amusing (I do). Suppose we call the set of all vector fields over our manifold $$M$$ as $$\mathcal{X}(M)$$. And further suppose that $$\alpha,\, \beta \in \mathbb{R}$$. Then, for any $$X,\,Y \in \mathcal{X}(M)$$ it is easy to show that $$X+Y \in \mathcal{X}M$$ and that $$\alpha X +\beta Y \in \mathcal{X}(M)$$.

In other words the set of all vector fields over a manifold is itself a vector space. I love such curiosities that arise in mathematics

Er, hoho?

 

[arfa brane]

What is the difference between a mathematical field and a physical field like the one Maxwell's equations describe? What do the equations 'describe', exactly?

That depends on a difference between the two kinds of fields, I think. To at least a first approximation, say.

/hoho

p.s. my guess is there's a relation between certain symmetries, some aren't physical and some are.

 

[arfa brane]

I think you will find that a physical field like an electric or magnetic field (or some combination), is quite a different thing than a mathematical field.

As G. t'Hooft points out, a physical field's degrees of freedom are the same as those of an 'exchange particle'; so for instance a photon's polarisation states encapsulate the d.o.f. of the electromagnetic field in three dimensions. The spin states of electrons encapsulate the d.o.f. of a field of fermions, etc.

Also, particles emit or generate fields; this isn't something that numbers do (or vectors or tensors).
Mathematical fields are things that depend on algebraic relations between numbers; $$ \mathbb R $$ is a field 'by itself'. In physics you need the underlying space $$ \mathbb R^3 \times \mathbb R$$.