New answers tagged gauge-theory
1
The answer to the question in the title is no.
Everything comes to the idea of the swampland in the string theory context. Not every effective QFT can be UV-completed into a string theory, in particular that appears to be the case for $\mathcal{N}=2$ theories in four dimensions.
One of the swampland conjectures explicitly locate $\mathcal{N}=2,d=4$ ...
3
Short answer: Read Weinberg, Vol I, Section 2.5.
Longer answer, hopefully covering more of what Weinberg takes for granted:
In relativistic physics, the Hilbert space is a representation of the Poincare group; rotations, translations, and boosts act on the state vectors linearly, turning one state into another. Representation theory is useful here ...
3
In the case of linearised GR, one of the main reasons for choosing our gauge fixing condition is just sheer convenience. If we're doing a metric perturbation of $g_{\mu \nu} = \eta_{\mu \nu} + h _{\mu \nu} $, then our corresponding action that we get from perturbing our Ricci scalar is the Fierz-Pauli action which is second order in $h $ (since the first ...
0
In the first point you're wrong in the normalisation condition which is $$\text{Tr}(T^aT^b) = \frac{1}{2}\delta^{ab}$$ and it couldn't be $\delta_{ij}$ since you're tracing over the indices $$\text{Tr}(T^aT^b) =T^a_{ij}T^b_{ji}$$ With this the first result is trivial.
The second point just comes from the definition of the covariant derivative $D_\mu = \...
1
How am I meant to differentiate the $F_{\mu\nu}$ on the first term of (2)? I don't understand how doing so will give me $F_{\mu\nu}$. (as explained by Lubos in the link)
The Lagrangian should be thought as a function of $4+16=20$ variables in this case, denoted as $y_\mu$ and $y_{\mu,\nu}$, i.e. $$ \mathcal L=\mathcal L(\{y_{\mu}\},\{y_{\mu,\nu}\}). $$ ...
2
Consider the expression "$\partial^\mu A^\nu$" as a single object – call it "$X^{\mu\nu}$" for example. I will discuss this point later. Then $F^{\mu\nu} = X^{\mu\nu}- X^{\nu\mu}$.
We now rewrite the derivative by disambiguating indices (you used same letters for free and summed indices), raising some of them, and writing the sums explicitly:
$$
\begin{...
3
Why the path integral needs gauge-fixing is e.g. discussed in this, this, this & this Phys.SE posts.
The currently most general quantization scheme is Batalin-Vilkovisky (BV) quantization. Within the BV formalism the gauge-fixing fermion is more or less arbitrary as long as it is Grassmann-odd, a Lorentz scalar, has ghost number -1, and certain rank ...
3
We are looking for a vector field $A_{\mu}(x)$ which has spin 1 particle excitations, and does NOT require gauge invariance to describe it. Let's figure this out systematically, although I won't go through the gory details (references are below). First of all, the vector field is in the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group, so ...
4
According to Sredicki's QFT pg. 120:
Theories with spin-one fields are renormalizable for $d = 4$ if and only if the spin-one fields are associated with a gauge symmetry.
So I guess that means that you can have low-energy effective field theories of spin-1 bosons without gauge symmetry, but not UV-complete theories.
3
Is every vector boson a gauge boson? The answer is no. The Kalb-Ramond field in string theory is an example of an antisymmetric two-index tensor $B_{\mu \nu}$ ($\mu , \nu$ $\in$ $\{1, ... ,d\}$) whose components $B_{\mu i}$ (where $i$ are indices on a compactification space) behave as vector bosons after dimensional reduction and (strictly speaking) such ...
0
The third term is a gauge fixing term. At this point in time, the Wikipedia article on gauge fixing has has a nice section on it, here: https://en.wikipedia.org/wiki/Gauge_fixing#R.CE.BE_gauges These express the so-called $R_ξ$ gauges - with $\xi=1$ called the Feynman–'t Hooft gauge, and the $\lim\xi\to0$ being the Landau gauge, the limit taken after ...
5
The set of matrices $\sigma^{\mu \nu } = \gamma^\mu \gamma^\nu - \gamma^\nu \gamma^\mu$ is defined in such a way that $\sigma^{\mu \nu } = - \sigma^{\nu \mu } $. Therefore, inside the parenthesis you really have the field strength tensor $F^{\mu \nu}$.
8
$A_{\mu}$ is not gauge invariant, and $\partial_{\mu} A_{\nu}$ also isn't.
But its antisymmetric part is:
$$
\frac{1}{2} \left(\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}\right) = \frac{1}{2} F_{\mu \nu}.
$$
Since in your expression you multiply by an antisymmetric $\sigma^{\mu \nu}$, you're allowed to anti-symmetrize the tensor $\partial_{\mu} A_{\nu}...
0
By the star I'm assuming you mean group product.
Yes you would be able to still have confinement. In the standard model, as far as I can tell, we REQUIRE confinement due to experimental observations.
Confinement basically means any particle made of quarks need to be in a trivial representation of SU(3). So, if we had an SU(2) strong force instead, then we'...
2
Absolutely! Even if someone rejects string theory, D-branes are still a very useful and geometrical ways to construct (and predict!) a plethora of new objects in QFT. Everthing from perturbativate states like fermions, mesons, baryons to everthing non-perturbative like instantons, monopoles, dyons, votices, kinks, textures, condensates, cosmic strings, brane ...
1
$\newcommand{\D}{\mathrm{D}}\newcommand{\Tr}{\mathrm{Tr}}$
In the usual treatment, $\theta$ is a coupling constant, i.e. a fixed parameter and not a field. Therefore shifting $\theta$ to $\theta' \color{gray}{(\neq \theta+2\pi n, \ n\in\mathbb{Z})}$ definitely changes the physics. Your question is equivalent to asking whether changing the parameters of a ...
0
You cannot arrive at the Aharonov Bohm effect using the field tensor. It is zero in the domain of interaction. You need the potential. Hence the potential is the physical quantity.
More generally, in Lagrangian mechanics the potential plays the role of the coordinate. It is not possible to derive Maxwell's equations from an action principle using the ...
2
There is no definition of "more fundamental" so no answer that is unequivocally correct here.
I note that most answers have nevertheless opted for the potentials over the fields. But one should notice that this is in part owing to a feeling that Lagrangians offer the royal route to the physics. So it depends a bit on how one likes to construct ones field ...
0
I don't have Henneaux's lectures on hand, but I assume the context is classical (not quantum) gauge theory.
The paradox comes from thinking that the gauge-transform parameter $\Lambda$ is not allowed to depend on the gauge field $A$ that is being transformed, but that is incorrect. Within any contractible patch, we can take the quantities $A_a$ (and also $\...
11
Since you've already received a good answer to the full question, I'll just add a bit more detail on the analogue of the Aharanov-Bohm effect for gravity. As noted in the existing answer and elsewhere, any analogy between gravity and gauge theory can't be perfect because the graviton has spin $2$.
On the other hand, we don't need a perfect analogy if we ...
15
Review of the electromagnetic case
In the EM case, the Aharonov-Bohm effect can be deduced like this. The lagrangian for a non-relativistic charged particle is
$$
L\sim \dot{\mathbf{x}}^2/2+\dot{\mathbf{x}}\cdot\mathbf{A},
\tag{1}
$$
where the gauge field $\mathbf{A}$ is related to the magnetic field $\mathbf{B}$ by
$$
\mathbf{B}=\nabla\times\mathbf{A}.
...
0
Re-posting my previous downvoted/deleted answer:
--> Assuming that the first component of $\mathbf{G}$ was zero amounts to a "choice of gauge." <---
My answer was, and is, factually correct, and I hoped, helpful. Maybe I can add a bit:
When you say that "the first component ... was zero," let me assume that by "component" you mean vector component (...
0
Can you write it as a total divergence? If so, then you can use the divergence theorem to argue that it goes to 0 at the surface, which you can take to infinity.
1
Sterile neutrinos do not have any gauge interaction (electroweak or strong) but only gravitational.
0
I thought a little about this question and I'm still not a 100% sure how to answer it. However, I think the crucial point about this assumption is that (with the assumption) the angular derivative of $\phi$ at "infinity" is always proportional to the action of a Lie algebra element of $\mathfrak{g}$, i.e. $$\partial_\theta \phi= i \lambda^a T_a \phi$$ is ...
1
On one hand, Dirac-type and Wu-Yang-type magnetic monopoles usually refer to different mathematical descriptions of the same$^1$ underlying class of physical phenomenon in various space dimensions and with various gauge group:
A Dirac-type monopole uses singular Dirac-strings/delta-distributions and globally defined singular gauge fields on $\mathbb{R}^d$. ...
2
In the Lagrangian of the path integral, a popular class of gauge-fixing terms is of the form$^1$
$${\cal L}_{GF}~=~-\frac{\chi^2}{2\xi}.$$
The word "gauge" is here confusingly used in 2 ways:
The gauge-fixing function $\chi$, e.g. Lorenz gauge $\chi=\partial_{\mu}A^{\mu}$, Coulomb gauge $\chi=\vec{\nabla}\cdot \vec{A}$, etc.
The gauge parameter $\xi>0$, ...
0
Weak force is a force because it can also cause a push and a pull. It is just not the main part of the weak force. The main part of the weak force is the ability for quarks and lepton to change flavors. Force part of the weak force is for the most part enforced by the Z boson.
1
I am not sure if this answers OP's question, but here's something.
I think the proper way to think about Poincaré gauge theory is to first imagine the Poincaré symmetry to be a purely internal one. The following can be formalized precisely using principal fibre bundles and Cartan geometry, but let us use a local formalism here.
The idea is that if we work ...
5
We have that $D_\mu=\partial_\mu-iqA_\mu$. Evaluating the commutator directly on some function $f$:
\begin{align}
[D_\mu,D_\nu]f&=D_\mu D_\nu f-D_\nu D_\mu f\\
&=(\partial_\mu-iqA_\mu)(\partial_\nu-iqA_\nu)f-(\partial_\nu-iqA_\nu)(\partial_\mu-iqA_\mu)f\\
&=\partial_\mu\partial_\nu f-iqA_\mu\partial_\nu f-iq\partial_\mu (A_\nu f)-q^2A_\mu A_\nu ...
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