Skip to main content
Planned maintenance impacting Stack Overflow and all Stack Exchange sites is scheduled for Monday, September 16, 2024, 5:00 PM-10:00 PM EDT (Monday, September 16, 21:00 UTC- Tuesday, September 17, 2:00 UTC). The email/password authentication method will be unavailable for logging in and registering. Read more here

New answers tagged

0 votes

Does Cauchy stress tensor act on $SO(3)$?

Cauchy’s theorem, under standard physical laws of continuous bodies proves that the stress vector $s$ at $p$ referred to the normal unit $n$ has the form $$s(p,n)_k = \sigma(p)_{kj} n_j.$$ Above, I ...
Valter Moretti's user avatar
1 vote

Vector/axial current and vector/axial transformation

They are not just related, they are identical, joined at the hip. I'll illustrate by the original legendary 2-flavor chiral model, with the chiral group $\text{SU}(2)_L \times \text{SU}(2)_R$. This ...
Cosmas Zachos's user avatar
0 votes

Holonomy group of Schwarzschild spacetime, other interesting examples?

Firstly, it is important to note that the holonomy groups are non-local in nature, even in trivial topology. Consider, for example, a Lorentzian manifold $(\mathcal M,g)$ such that the Riemann ...
Lukas Nullmeier's user avatar
0 votes

Can we do better than "a spinor is something that transforms like a spinor"?

What you are asking about are what are called "coordinate free descriptions". A tensor can be defined transformatively or to use the correct term, covariantly, once we define a basis in the ...
Mozibur Ullah's user avatar
1 vote

Projective representations and reduction of half-integer spin representations under $C_{\infty v}$

I have a partial answer to some of the questions posed. First off, by analyzing the characters of the action of group $G \equiv \{R^z_\phi : \phi \in [0,4\pi)\} \cup \{\Sigma^z_\phi : \phi \in [0,4\pi)...
creillyucla's user avatar
2 votes

On JD Jackson's derivation of Matrix Representations of Lorentz Tranformations

The "sines and cosines" comes from the algebraic properties of the generator. The Lie algebra of the rotation group $SO(3)$ is given (as can be checked by differentiating the $SO(3)$ ...
Lourenco Entrudo's user avatar
1 vote
Accepted

Group of time translations for finite-dimensional quantum systems and incommensurable eigenvalues

Not a group theory person, but here are my thoughts on the problem. For general $e_1$ and $e_2$, when is the function $U\left(t\right)$ one-to-one? If $U\left(t_{1}\right)=U\left(t_{2}\right)$ then $\...
Yuli's user avatar
  • 96
0 votes

Physical intuition of spin connection and spinor bundles?

A spinor bundle is defined via the associated bundle construction. Lets first make a couple of preliminary definitions: A rotation bundle is defined as a principal bundle whose gauge structure group ...
Mozibur Ullah's user avatar
6 votes

Showing that a generator exponentiates to a $\mathbb{R}$ group

From now on $\{\cdot,\cdot\}$ denotes the Lie commutator of smooth vector fields on a smooth manifold. As far as I understand, you have a vector field $X$ on $TM$, where $M$ is the Schwarzschild ...
Valter Moretti's user avatar
1 vote

How can one says that a particle IS a representation of some group?

I think when someone says “a particle IS a representation of some group” in the context of quantum field theory it would always be more precise to say, “at any point in space time, the state space of ...
Jagerber48's user avatar
  • 14.9k
1 vote

How can one says that a particle IS a representation of some group?

But how else do you define a particle? Something that has a definite position or momentum, definite mass, and definite spin? Just like, erm, a representation of the Poincare group? (By the way, a ...
T.P. Ho's user avatar
  • 442
0 votes

How can one says that a particle IS a representation of some group?

Saying that particles are irreducible representation of Poincare group is like saying that energy is a number, not a full story. Also note that a particle is a tensor field.
Alien from future's user avatar
1 vote
Accepted

Cartesian tensor transformation and representations of $SO(3)$

The question is very unclear so maybe what the OP means is that any matrix $$ M=\pmatrix{M_{xx}&M_{xy}&M_{xz}\\ M_{yx}&M_{yy}&M_{yz}\\ M_{zx}&M_{zy}&M_zz} $$ can be written in ...
ZeroTheHero's user avatar
  • 46.9k
1 vote

Recovering the electromagnetic field from the $(1,0) \oplus (0,1)$ representation

The $(1,0)\oplus (0,1)$ representation is the $2$-covariant anti-symmetric tensor representation $$F = \frac{1}{2}F_{\mu\nu}dx^\mu\wedge dx^\nu.$$ Moreover, the $F_{\mu\nu}$ are already the components ...
Gold's user avatar
  • 36.7k
0 votes

Recovering the electromagnetic field from the $(1,0) \oplus (0,1)$ representation

After some calculations I found what I wanted. We can take the change-of-basis matrix $P$ to be \begin{gather} P = \frac{1}{2} \begin{bmatrix} 1 & 0 &0 &-i & 0 & 0 \\ 0 & 1 &...
Thomas Bastos's user avatar
3 votes

Recovering the electromagnetic field from the $(1,0) \oplus (0,1)$ representation

The $(1,0)$ representation is $F_{(\alpha\beta)}$ and the $(0,1)$ representation is $F_{({\dot \alpha}{\dot \beta})}$. The field strength in spacetime is then given by $$ F_{\mu\nu} = (\sigma_{\mu\nu})...
Prahar's user avatar
  • 27k
1 vote
Accepted

Group actions confusion

This should be familiar to you from the first example of generators of transformations we usually encounter: The bracket between position $q$ and momentum $p$ is constant as $\{q,p\} = 1$, and ...
ACuriousMind's user avatar
  • 127k
6 votes
Accepted

Why is the gauge field $A_\mu$ real for $\mathrm{U}(N)$ symmetry?

The first point is that the Lie group $G=U(N)$ [which consists of unitary $N\times N$ matrices] is a real Lie group, which is perhaps best explained as that the tangent spaces, or equivalently, the ...
Qmechanic's user avatar
  • 209k
3 votes

Why is the gauge field $A_\mu$ real for $\mathrm{U}(N)$ symmetry?

First, I question the importance of this in practice. Because if you have an action where the $A_\mu^a$ are initially real and you decide that you don't like this, you can always change this by ...
Connor Behan's user avatar
  • 8,431

Top 50 recent answers are included