*The Joshua Tree*, by U2, out of curiosity. A few weeks later, I gave it away!

Anyway, this isn't about U2, it's about U(1) and

*nominalism*. U(1) is the Lie group of rotations $R_{\theta}$ about an axis, parametrized by some angle $\theta \in [0, 2\pi]$. So, you might rotate a coffee cup around a vertical axis through the middle of cup, by a certain angle $\theta$. Indeed, if the cup is quite "symmetrical", then the result always "looks the same": roughly, this is because the

*set*$X$ of points occupied by the material of the coffee cup remains invariant when every point $x \in X$ is rotated by the rotation $R_{\theta}$: in symbols, $R_{\theta}[X] = X$. Each of these rotations is parametrized by an angle $\theta$; the rotations can be composed, there is an "identity" rotation (i.e., angle $= 0$) and each rotation has an inverse. So, the set of rotations forms a group.

Well, that's

*physicsese*. In mathematicsese, U(n) is the unitary group of $n \times n$ matrices, and U(1) is the case where $n = 1$. The elements of U(1) are identified with the complex numbers $e^{i \theta}$, and group multiplication is simply complex multiplication. The identity is $1$ and the inverse of $e^{i \theta}$ is $e^{-i \theta}$. U(1) is Abelian because multiplication of complex numbers, $z_1, z_2$, is commuative: in this case, $e^{i \theta_1}e^{i \theta_2} = e^{i (\theta_1 + \theta_2)} = e^{i \theta_2}e^{i \theta_1}$. The connection between rotations and with complex numbers comes from the 2-dimensional representation of $\mathbb{C}$ as the plane. If $z \in \mathbb{C}$, then $e^{i \theta} z$ is the result of "rotating" $z$ by angle $\theta$ (about the origin).

*Nominalism*is the philosophical doctrine that there are no abstract entities, and,

*a fortiori*, no numbers, sets, functions, groups, manifolds, Hilbert spaces and so on. Consequently, as frequently pointed out by Quine and Putnam, nominalism is

*inconsistent with science*. For example, the following is a true statement of physics:

U(1) is the gauge group of the electromagnetic field $A_{\mu}$.If nominalism is true, then there are no groups. If there are no groups, then this statement of physics is false.

A very brief explanation of gauge theory: Particles of matter are described by some field $\phi$; and, if they are charged, then $\phi$ is a complex field, so $\phi(x) \in \mathbb{C}$, at each point $x$ in spacetime. Associated with the field is a quantity called the Lagrangian (written $\mathcal{L}(\phi, \partial_{\mu} \phi)$) which, via an "action principle", implies the equations of motion for the field: e.g., a certain Lagrangian for the field $\phi$ implies that $\phi$ satisfies the Klein-Gordon equation:

$(\partial^{\mu} \partial_{\mu} + m^2)\phi = 0$.Suppose one multiplies the field $\phi$ by a constant phase factor $e^{i \theta}$ (which is a kind of rotation of each complex number $\phi(x)$ specifying the field). That is,

$\phi^{\prime}(x) = e^{i \theta}\phi(x)$.One can check that leaves the Lagrangian $\mathcal{L}$ invariant. We say that the Lagrangian $\mathcal{L}$ has a global U(1) symmetry (= a global gauge invariance; = a global gauge symmetry).

However, what if $\phi$ is subjected to a

*local*gauge transformation, a local phase rotation, i.e.,

$\phi^{\prime}(x) = e^{i \theta(x)}\phi(x)$,where the parameter $\theta$ can be non-constant function, varying from point to point? It turns out that the Lagrangian is not invariant.

However, this invariance can be restored by introducing a new compensating gauge field, $A_{\mu}$, and a modified Lagrangian $\mathcal{L}^{\ast}$, so that under a gauge transformation $e^{i \theta}$, the field $A_{\mu}$ transforms to

$A_{\mu}^{\prime}(x) = A_{\mu}(x) + (\partial_{\mu} \theta)(x)$.The resulting field $A_{\mu}$ is called a gauge field: the U(1) gauge field, which "couples" with the (current of the) original charged matter field $\phi$ exactly as the electromagnetic field does: $A_{\mu}$ gives rise to the electromagnetic field. More exactly, $A_{\mu}$ is the electromagnetic potential: the electromagnetic field $F_{\mu \nu}$ itself is the exterior derivative, $\frac{\partial A_{\mu}}{\partial x^{\nu}} - \frac{\partial A_{\nu}}{\partial x^{\mu}}$. And the field $F_{\mu \nu}$ is gauge-invariant.

Here is a really brilliant exposition of the ideas of gauge theories, and much more (co-ordinate systems, fibre bundles and whatnot), by Terence Tao.

[Update (Aug 30th), minor changes.]

Great post. One problem: how does one accidentally do something out of curiosity?

ReplyDeleteAdmittedly, it's quite incoherent! Still, I blame Bono for my incoherence.

ReplyDelete