The Poincaré group, named after Henri Poincaré (1906), was first defined by Hermann Minkowski (1908) as the group of Minkowski spacetime isometries. It is a ten-dimensional non-abelian Lie group, which is of importance as a model in our understanding the most basic fundamentals of physics. For example, in one way of rigorously defining exactly what a subatomic particle is, Sheldon Lee Glashow has expressed that "Particles are at a very minimum described by irreducible representations of the Poincaré group."
Given $$M_{\rho \sigma} = i (x^{\rho} \partial_{\sigma} - x^{\sigma} \partial_{\rho})$$
and $$W^{\mu} = \frac{1}{2} \epsilon^{\mu \nu \rho \sigma} P_{\nu} M_{\rho \sigma}$$
Why does ##W^{\mu}## pick up only the spin part of the total angular momentum?
I just want to make sure I understand this correctly.
For an infinite-dimensional representation, the generators of translation can be written as ##i \frac{\partial}{\partial_{\mu}}= i \partial_{\mu}##, where the generators of the Lorentz group can be written as ##i (x^{\mu}\partial_{\nu} -...
So, all of the generators of the Poincare group are associated with pretty well-known physical quantities. Time translation is associated with energy, space translations with momentum, rotations with angular momentum, and boosts... well, boosts are generated by the "generators of boosts". Do...
Using differential expressions for the generator, verify the commutator expression for ##[J_{\mu\nu},P_{\rho}]=i(\eta_{\mu\rho}P_{\nu}-\eta_{\nu\rho}P_{\mu})## in Poincare group
Generator of translation: ##P_{\rho}=-i\partial_{\rho}##
Generator of rotation...
I see that the first four equations are definitions. The problem is about the dimensions of the quotient.
Why does the set Kx forms a six dimensional Lie algebra?
I want to make certain that my proof is correct:
Since ## P^2 = P_\nu P^\nu=P^\nu P_\nu ##, then ## [P^2,P_\mu]=[P^\nu P_\nu,P_\mu]=P^\nu[P_\nu,P_\mu]+[P^\nu,P_\mu]P_\nu=[P^\nu,P_\mu]P_\nu=g^{\nu\alpha}[P_\alpha,P_\mu]P_\nu=0 ##, since ## g^{\nu\alpha} ## is just a number, I can bring it...
Homework Statement
Show that for
$$W^\mu = -\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}M^{\nu\rho}P^{\sigma},$$
where ##M^{\mu\nu}## satisfies the commutation relations of the Lorentz group and ##\Psi## is a bispinor that transforms according to the ##(\frac{1}{2},0)\oplus(0,\frac{1}{2})##...
Hello guys,
In 90% of the papers I've read about diferent ways to achieve generalizations of the Proca action I've found there's a common condition that has to be satisfied, i.e: The number of degrees of freedom allowed to be propagated by the theory has to be three at most (two if the fields...
Homework Statement
Does ##x_p\partial_v\partial_u-x_v\partial_p\partial_u=0##
Homework Equations
I need this to be true to show a poincare algebra commutator.
We have just shown that ##[P_u, P_v] =0 ##, i.e. simply because partial derivatives commute.
Where ##P_u=\partial_u##
The Attempt...
Hey guys,
as this is a basic QFT question, I wasn't sure to put it in the relativity or quantum section. Since this question specifically is about manipulating tensor expressions, i figured here would be appropriate.
My question is about equating coefficients in tensor expressions...
Homework Statement
Consider the Klein-Gordan action. Show that the Noether charges of the Poincare Group generate the Poincare Algebra in the Poisson brackets. There will be 10 generators.Homework Equations
\{ A,B \}=\frac{\delta A}{\delta \phi}\frac{\delta B}{\delta \pi}-\frac{\delta...
Hi,
I am working through Maggiore's QFT book and have a small problem that is really bothering me.
It involves finding the representation of the Poincare algebra generators on fields. I always end up with a minus sign for my representation of a translation on fields compared to Maggiore...
Hi,
Can someone please explain the following statement on page 62 of Weinberg's Vol 1 on QFT:
(I understand the part for P ~ mv, so the "quote" is slightly distorted, intentionally).
Also how is
?
Thanks in advance!
I am facing a dilemma which leaves me quite puzzled. I hope someone can straighten this out.
The short version is : when we use representations of the charges as differential operators to calculate their commutators, we always get -1 times the correct result. So, doe sthat mean that we...
Homework Statement
Find the commutators [P^\sigma,J^{\mu \nu}]
The answer is part of the Poincare algebra
[P^\sigma,J^{\mu \nu}]=i(g^{\mu \sigma}P^\nu-g^{\nu \sigma}P^\mu)
If someone can convince me that \partial_i T^{0\mu} = 0, (i.e. the energy-momentum tensor has no explicit spatial...