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16

In normal usage, a gauge is a particular choice, or specification, of vector and scalar potentials $\mathbf A$ and $\phi$ which will generate a given set of physical force fields $\mathbf E$ and $\mathbf B$. More specifically, a physical situation is specified by the electric and magnetic fields, $\mathbf E$ and $\mathbf B$. A set of potentials $\mathbf A$ ...


15

Just a quick complaint about naming first: Maxwell's equations are written in terms of the electric and magnetic fields, which are physical degrees of freedom. You're instead talking about the Euler-Lagrange equations of the Maxwell Lagrangian. At the classical level, if we want to write electromagnetism in terms of a four-potential, that potential has to ...


13

This is not the case. The Aharanov-Bohm effect yields an observable of the form $$\oint_{\mathcal C} A,$$ where $\mathcal{C}$ is some circuit. This is however gauge invariant. A way to see this is by noting that it can be written in terms of the field strength $$\int_\Sigma F,$$ via Stokes' theorem. In here we chose a surface $\Sigma$ whose boundary is $\...


12

Setting the einbein to $1$ corresponds to a diffeomorphism of the metric, as the einbein is given by $e_{\tau\tau}=\sqrt{g_{\tau\tau}}$, which can be easily deduced from the fact that a the vielbein is given as the transformation coefficients from the coordinate basis to a non-coordinate basis. Hence, the dimensionality of the einbein depends on that of the ...


12

Although I think this is a good question - the finding of meaning and relationships between physics notions is always worthwhile - "physical meaning" is not a good choice of words here. This is because gauge invariance is a redundancy in the mathematical description of a system; it means that we can partition the solutions of the description into equivalence ...


11

If $\phi$ is non-zero, fixing the phase of $\phi$ is a perfectly valid gauge condition. It's used frequently in Standard Model calcuations involving the Higgs field, where it goes by the name unitarity gauge. This is a nice gauge in some ways, because it makes manifest the fact that there's a massive vector field in the system. Edit: Some caution is ...


11

Comment to the question (v1): It seems OP is conflating, on one hand, a gauge transformation $$ \tilde{A}_{\mu} ~=~ A_{\mu} +d_{\mu}\Lambda $$ with, on the other hand, a gauge-fixing condition, i.e. choosing a gauge, such e.g., Lorenz gauge, Coulomb gauge, axial gauge, temporal gauge, etc. A gauge transformation can e.g. go between two gauge-fixing ...


11

You are misunderstanding what a gauge theory is if you think we shouldn't get rid of the gauge symmetry at some point. A gauge symmetry is not like other symmetries, it does not relate configurations of the dynamical variables that are physically distinct - instead, it relates configuration of the dynamical variables which are physically indistinguishable. ...


11

The gauge $\phi = A_0 = 0$ is called Weyl gauge or temporal gauge. This gauge is incomplete, as one can see from the definition of a gauge transformation, $$A_\mu \to A_\mu + \partial_\mu \alpha(x).$$ We can still perform any gauge transformation with gauge parameter $\alpha$ independent of $t$, as this keeps $A_0$ the same. To remove some of the residual ...


10

Contrary to popular belief, it is not necessary to choose a gauge to quantize a gauge theory. It is just convenient, since the non-gauge-fixing approaches are often difficult to implement for all but the simplest cases. Gauge theories are, in the Hamiltonian picture, certain kinds of constrained Hamiltonian systems. Dirac's canonical quantization procedure, ...


10

It's important to note that none of this is specific to quantum mechanics, and that you get exactly the same structure for the corresponding problem within classical hamiltonian mechanics. There, you have the hamiltonian $$ H=\frac{1}{2m}p_x^2 + \frac{1}{2m}\left(p_y -\frac{eB}{c}x \right)^2, $$ which naturally conserves $p_y$ since $y$ does not appear in $...


10

It is natural to generalize to an Abelian $p$-form gauge field $$A~=~\frac{1}{p!} A_{\mu_1\mu_2\ldots\mu_p} \mathrm{d}x^{\mu_1}\wedge\ldots\wedge \mathrm{d}x^{\mu_p}\tag{1}$$ with $\begin{pmatrix} D \cr p \end{pmatrix}$ real component fields $A_{\mu_1\mu_2\ldots\mu_p}$ in a $D$-dimensional spacetime. I) Massless case: There is a gauge symmetry $$ \...


9

I would like to know how exactly the equations of motion in the Lorenz gauge removes the second degree of freedom. In the Lorenz 'gauge', we have $$\Box A^{\mu} = \mu_0j^{\mu}$$ If $A^{\mu}$ is a solution, then so is $A^{\mu} + N\epsilon^{\mu}e^{-ik\cdot x}$ if $$\Box (N\epsilon^{\mu}e^{-ik\cdot x}) = 0$$ Consistency with the Lorenz condition $$\...


9

The fields are not affected by this gauge transformations, and only those quantities have physical meaning that are invariant under the transformations. So essentially the physics is the same in these two gauges. But if you want some of the implications of choosing one of these gauges, so you can choose the appropriate one, the following may be of help: ...


9

First, the gauge invariance means that the solutions $A_\mu(x^\alpha)$ are not unique. For every solution, the gauge transformations of it are solutions, too. That may be a problem because sometimes we want to have specific values of $A_\mu(x^\alpha)$ that answer a physical question. Second, we sometimes gauge fix because the equations simplify. For example,...


8

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2 $$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined $$D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu k^\...


8

You are correct that there are gauge degrees of freedom in the solution for $A_\mu$ - precisely the ordinary gauge transformations. But $A$ is not physical, the electromagnetic field strength $F$ is. There are no gauge degrees of freedom in $F$, and as an equation for $F$ Maxwell's equations are physical. The polarization of the classical $A_\mu$ has nothing ...


7

I) The un-gauge-fixed QED Lagrangian density reads $$ {\cal L}_0~:=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \bar{\psi}(i\gamma^{\mu}D_{\mu} -m)\psi.\tag{1}$$ The gauge-fixed QED Lagrangian density in the $R_{\xi}$-gauge reads $$ {\cal L}~=~ {\cal L}_0 +{\cal L}_{FP}-\frac{1}{2\xi}\chi^2 , \tag{2}$$ where the Faddeev-Popov term is $$ {\cal L}_{FP}~=~ -d_{\...


7

In this answer, we summarize the results. The analysis itself can be found in textbooks, see e.g. Refs. 1 & 2. $\downarrow$ Table 1: Massless spin $j$ field in $D$ spacetime dimensions. $$\begin{array}{ccc} \text{Massless}^1 & \text{Off-shell DOF}^2 & \text{On-shell DOF}^3 \cr j=0 & 1 & 1 \cr j=\frac{1}{2} & n & \frac{n}{2} ...


7

A gauge symmetry means that the equations of motion do not uniquely determine the evolution of all the configuration variables, i.e., that the Euler-Lagrange system is under-determined. The canonical example is the case of classical electrodynamics, where the equations of motion are \begin{equation} (\partial^2\delta^\mu_\nu-\partial^\mu\partial_\nu)A_\mu=0\...


7

The two-dimensional Polyakov action for a string with worldsheet $\Sigma$ and worldsheet metric $h_{ab}$ $$ \frac{T}{2}\int_\Sigma \sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu$$ has full conformal symmetry under the Virasoro algebra and under Weyl transformations1 , which can be seen as gauge degrees of freedom. It follows that we can always ...


7

The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...


7

Yes, all those things are correct. The equivalence class of potentials that are related by gauge transformation is called a gauge orbit, since it is an orbit for the action of the group of gauge transformations on the space of potentials. Choosing/fixing a gauge means picking out particular representants $A$ from each gauge orbit according to a rule ...


7

Answer to the first question You can fix the gauge doing the transformation $A^\mu \rightarrow {A^\prime}^\mu = A^\mu + \partial^{\,\mu} f$ and choosing an appropriate function $f$. You cannot fix it by impositing whatever condition on ${A^\prime}^\mu$ you want. You have to guarantee there is a function $f$ which can be used so as your chosen condition on ${...


6

The terms "Landau gauge" and "Feynman gauge" (among others) were introduced by Bruno Zumino. I accidentally learned about it an hour ago from David Derbes http://motls.blogspot.com/2014/06/bruno-zumino-1923-2014.html?m=1 in this blog post about a sad event, Bruno Zumino's death a week ago. David Derbes wrote: I met Bruno Zumino at the Scottish ...


6

For clarity let's work with a Lorentzian signature. Our $g$ is a metric for a 2 dimensional Lorentzian manifold $M$. It is well-known that any two dimensional pseudo-Riemannian manifold is conformally flat, that is $$g = e^{2\omega}\eta$$ Where $\eta$ is the flat 2D Minkovski metric. Your Lightcone gauge example You didn't define your $x^\pm$s but I ...


6

Continuous symmetries of the action of a system which are global, that is, do not depend on where they act, give rise through Noether's theorem to conserved quantities. For example, a translation in time $t \to t+\epsilon$ for $\epsilon \in \mathbb{R}$ is a global transformation, and leads to energy conservation. On the other hand, if an action is invariant ...


6

When you introduce a new field to make the Lagrangian gauge invariant, then you are at liberty to choose the transformation behaviour of the new field such that the Lagrangian becomes gauge invariant. If $\sigma\mapsto\sigma+\alpha$ leads to an invariant Lagrangian, then you are free to choose $\sigma$ as a field transforming such. In the situation you ...


6

The four-potential is not an observable because it is not invariant under a change of gauge. And no predictions of any physical theory are dependent on the choice of gauge, so the four-potential is not observable. What is gauge invariant and observable is the integral of the four potential around a loop, and that is what is observed in the AB effect. ...


6

Ok, so lets start with the basics, the answer we are expecting is given by: $$\vec B= \frac{\mu_0I}{2\pi r} \hat e_\theta$$ Which is from Ampere's law. From this we can kind of backwards engineer, to show that: $$\vec A=-\frac{\mu_0I}{2\pi} \ln(r) \hat e_z$$ would work as the potential. The reason I don't think your method works is because you are forcing ...


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