Gauge fixing?

Discussion in 'Physics & Math' started by QuarkHead, Jan 1, 2017.

1. QuarkHeadRemedial Math StudentValued Senior Member

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What is it?

What follows is my intuitive understanding of gauge theories, though I do have a lot of the relevant geometric mathematics to hand if it would interest anyone - from a purely mathematical viewpoint it is quite nice.

So, a physical theory is said to be gauge invariant - loosely a gauge theory - if it takes the same form under local continuous symmetry transformations. Since continuous symmetry transformations are given by some Lie group $G$ - presumably specific to the physical theory at hand - I picture the situation as follows......

Suppose an n-manifold $M$. As our theory ranges over this manifold, at each point $m \in M$ it (our theory) "encounters" a Lie group $G$ and may "choose" there an element $g \in G$ and retain the same form.

Since at each $m \in M$ the choice of $g \in G$ is arbitrary (is it?), there exist multiple manifold-wide "collections" of such choices, each of which is called a gauge.

So what is meant be "fixing a gauge"? (assuming the above is not gibberish even at its admittedly superficial level)

3. karenmanskerHSIRIBanned

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'fixing a gauge?' . . . changing the parameters so that it fits the data/model, perhaps?

5. karenmanskerHSIRIBanned

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'fixing a gauge' . . . . not unlike 'renormalization' to get rid of unwanted infinte terms in equations . . . .

7. SchmelzerValued Senior Member

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If you have a gauge field, you have some forces you can really measure, $F_{mn}(x,t)$. Say, for the EM field they describe the electric force $E_i = F_{0i}$ and the magnetic force $H_i = \varepsilon_{ijk} F_{jk}$.

But there is a much more elegant and simpler description of the EM field using, instead of these six fields, only four fields, the gauge potentials $A_n(x,t)$. The EM forces may be simply computed if one knows these fields, by simple formulas $F_{mn} = \partial_m A_n - \partial_n A_m$. And the equation for the $A_n(x,t)$ is much easier than the Maxwell equations for the forces, it is simply the wave equation $\square A_n = 0$.

But can one somehow derive these equations $\square A_n = 0$ from the Maxwell equations? From point of view of Popper's philosophy of science, not really a meaningful question. All what matters is that we have a simple theory for the fields $\square A_n = 0$, and can derive from these equations the Maxwell equations. The very question is, essentially, a result of the misguided empiricist program that we somehow would have to derive the fundamental theory from observations. But, once all we can measure are the $F_{mn}(x,t)$, and observe if the Maxwell equations hold for them, the empiricist, positivist program prescribes that we would have to derive the equation $\square A_n = 0$ for the potentials from the Maxwell equation.

But this is an unsolvable problem, because of gauge freedom. Use a completely arbitrary function $\omega(x,t)$ and modify the potentials by the rule $A_n \to A_n + \partial_n \omega$ and these modified potentials define the same EM field $F_{mn} = \partial_m A_n - \partial_n A_m$, simply because the term which depends on $\omega(x,t)$ cancels. So, the question "which $\omega(x,t)$ is the correct one" has no solution in the positivist program to derive equations from observations.

Classical scientists would not have cared at all about this. Because they have a simple preferred choice, namely the one with the simplest equation $\square A_n = 0$. So, for the classical scientist the "gauge fixing" is not a problem at all, and done by choosing the $\square A_n = 0$ as the equation for the gauge potentials.

In classical theory, this gauge freedom is quite uninteresting, because it does not lead to any physical consequences. The situation is more confusing in the quantum part, because in this case one has to consider not only classical solutions of $\square A_n = 0$, but also oscillations around them. This is also not really problematic for the classically thinking scientist. In this case, he has to consider all fields $A_n(x,t)$ as different possible field configurations. But what if two such field configurations differ only by $A_n \to A_n + \partial_n \omega$ for some $\omega(x,t)$? The classically thinking scientist answers "so what?" and sees no reason to care. The positivist thinks differently. Once we cannot distinguish the two fields by observation, they have to be really the same field. Philosophical nonsense, but popular.

The problem is that a quantum field theory which would be based on the fields $F_{mn}(x,t)$ and the Maxwell equations would be different from one based on the $A_n(x,t)$ and the equation $\square A_n = 0$. A way to illustrate how such difference may appear the following: Whatever the quantum oscillations of $A_n(x,t)$, some of the Maxwell equations hold exactly - they are simply mathematical consequences of $F_{mn} = \partial_m A_n - \partial_n A_m$. Instead, in a quantum theory with the $F_{mn}(x,t)$ as fundamental fields these tautological equations would not hold exactly too, they would be only classical approximations, so there would be also oscillations around them. Mainstream quantum gauge field theory is even a different, even more complicate mess, where some $A_n(x,t)$ exists, but fields connected by $A_n \to A_n + \partial_n \omega$ for some $\omega(x,t)$ are nonetheless somehow identified. And this identification also leads to some further modifications of the theory.

8. rpennerFully WiredStaff Member

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Wrong.
Off-base and wrong.

Gauge-fixing is breaking a theoretical symmetry (called gauge freedom) so that calculations can proceed without leaving the extra terms used to represent the symmetry in every stage.

An analogy is making a choice of coordinate frame to carry SR calculations on N object so that every object has one velocity, momentum, energy instead of N-1 relative velocities, momenta, energies. Making a choice of state of motion which is to be considered at rest breaks SR symmetry where no inertial motion is special, but makes calculations easier and we can still recover the original.

But rather than the three degrees of freedom of the choice of a standard of rest, a gauge freedom is a symmetry with an infinite number of degrees of freedom.

In QED, the electric and magnetic fields take a backseat to the four potential $A^{\mu} = \begin{pmatrix} \phi /c \\ \vec{A} \end{pmatrix}$ , which the Aharonov–Bohm effect, says is real ... up to a gauge transformation. Take any smooth scalar function of space and time: $\psi$ and use its four-gradient $\partial^{\mu} \psi = \begin{pmatrix} \frac{1}{c} \frac{d \; }{dt} \psi \\ - \nabla \psi \end{pmatrix}$ to make an altered four potential. $\color{blue}{A^{\mu}} = A^{\mu} + \partial^{\mu} \psi$.

Then
$\color{blue}{\phi} = \phi + \frac{d \; }{dt} \psi \\ \color{blue}{\vec{A}} = \vec{A} - \nabla \psi$
$\color{blue}{\Delta \theta} = \frac{q}{\hbar} \left( \int_{a}^{b} \color{blue}{\vec{A}} \cdot d \vec{x} - \int_{a}^{b} \color{blue}{\vec{A}} \cdot d \vec{x}' \right) = \frac{q}{\hbar} \left( \int_{a}^{b} \vec{A} \cdot d\vec{x} +\psi(a) - \psi(b) - \int_{a}^{b} \vec{A} \cdot d\vec{x}' +\psi(b) - \psi(a) \right) = \Delta \theta$
With the last line showing that the phase difference for two paths between the same two points (a, b) does not depend on the choice of $\psi$.
Likewise, for classical EM,
$\color{blue}{\vec{E}} = - \nabla \color{blue}{\phi} - \frac{d \; }{dt} \color{blue}{\vec{A}} = - \nabla \phi - \nabla \frac{d \; }{dt} \psi - \frac{d \; }{dt} \vec{A} + \frac{d \; }{dt} \nabla \psi = - \nabla \phi - \frac{d \; }{dt} \vec{A} = \vec{E} \\ \color{blue}{\vec{B}} = \nabla \times \color{blue}{\vec{A}} = \nabla \times \vec{A} - \nabla \times \nabla \psi = \nabla \times \vec{A} = \vec{B}$

So since any choice of $\psi$ gives the same physical meaning (gauge freedom), we are free to choose one (gauge fixing) that makes working with $\color{blue}{A^{\mu}}$ "nice."

Last edited: Jan 1, 2017
9. karenmanskerHSIRIBanned

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Thanks for quantifying what I qualified

10. arfa branecall me arfValued Senior Member

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The photon is a gauge particle. In quantum mechanics it's about the phase of a field and how you can't determine a global phase, you can only determine differences in phase (interference exists because of the phase-differences in the field).

The gauge is a shift in this phase, for instance if you are using electrons in a double slit setup, you can alter the local phase with a shielded magnet and change the interference pattern. But wait, you can get the same change with a half-wave plate over one slit (or its electron wave equivalent) which shifts the phase of half the electrons. You can't tell which choice was made if you only see the results.

11. QuarkHeadRemedial Math StudentValued Senior Member

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I thank you both for your efforts. Regarding Schmelzer's post, I had no real problems (except I was unable to render some of his LaTex), but was slightly puzzled by the introduction of "philosophy". Never mind. Regarding this......
OK, you are assuming commutativity. As in, the symmetry group is $U(1)$ as expected for any EM field theory
I had a slight problem here. You write $\nabla \times \vec{A}- \nabla \times \nabla \psi= \nabla \times\vec{A}$ which as far as I know is $\nabla \times \vec{A}-\nabla^2 \psi = \nabla \times \vec{A}$ only for the scalar field $\psi$ where according to Laplace $\nabla^2 \psi = 0$ for some scalar field $\psi$.

Leave it with me for a while - I will get there!

12. rpennerFully WiredStaff Member

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Because the curl of the gradient of any scalar field is always the zero vector field (because partial derivatives commute).

$\nabla \times \nabla \psi = \nabla \times \begin{pmatrix} \frac{\partial \;}{\partial x} \psi \\ \frac{\partial \;}{\partial y} \psi \\ \frac{\partial \;}{\partial z} \psi \end {pmatrix} = \begin{pmatrix} \frac{\partial^2 \; \; }{\partial y \partial z} \psi - \frac{\partial ^2 \; \; }{\partial z \partial y} \psi \\ \frac{\partial^2 \; \; }{\partial x \partial z} \psi - \frac{\partial ^2 \; \; }{\partial z \partial x} \psi \\ \frac{\partial^2 \; \; }{\partial x \partial y} \psi - \frac{\partial ^2 \; \; }{\partial y \partial x} \psi \end {pmatrix} = \vec{0}$.

So there's no relation between the physics content of $\phi$ and $\vec{A}$ and the gauge freedom represented by $\psi$.

Also, please note there is no requirement that a gauge freedom have an infinite number of degrees of freedom.

13. arfa branecall me arfValued Senior Member

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An illustration

Gauge fixing of a twisted cylinder. (Note: the line is on the surface of the cylinder, not inside it.)

By looking at a cylindrical rod can one tell whether it is twisted? If the rod is perfectly cylindrical, then the circular symmetry of the cross section makes it impossible to tell whether or not it is twisted. However, if there were a straight line drawn along the length of the rod, then one could easily say whether or not there is a twist by looking at the state of the line.

Drawing a line is gauge fixing. Drawing the line spoils the gauge symmetry, i.e., the circular symmetry U(1) of the cross section at each point of the rod. The line is the equivalent of a gauge function; it need not be straight. Almost any line is a valid gauge fixing, i.e., there is a large gauge freedom. To tell whether the rod is twisted, you need to first know the gauge. Physical quantities, such as the energy of the torsion, do not depend on the gauge, i.e., are gauge invariant.

Thanks, Wikipedia!

14. QuarkHeadRemedial Math StudentValued Senior Member

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OK, the answer to mine own question (after some help here) is simpler that I had thought.

So....

Consider a manifold $M$ and a theory that is invariant at each point $m \in M$ under the action of some symmetry group $G \ni g$.

Then my theory will remain "true" at any point $m \in M$ irrespective of my choice of $g \in G$ at that point. So, again at that point, choose some particular $g \in G$ to represent the symmetry of my theory, and do the same at each point $m \in M$.

This is called a "fixed gauge" if and only if no arbitrary "collection" of such choices do not intersect.

Of course it is not quite as simple as that, due to the fact that the transition from $m \mapsto m' \in M$ may be complicated by the fact that my manifold may not be flat in the sense of Euclid.

Again, if this is nuts, please pull no punches.......

15. arfa branecall me arfValued Senior Member

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Suppose the cylinder in the wiki example is the bundle of directions over a hemisphere, or just the bundle over the equator . . .

If you twist the cylinder (and bend all the fibers) so that points on the equator are mapped to the base space z degrees out of "phase", is that a valid choice of gauge?

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16. arfa branecall me arfValued Senior Member

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I'll post those attached files, I didn't after realising they had little to do with my actual question.
I took screenshots of an article from SciAm, author: G. t'Hooft.