All Questions
7
questions
2
votes
0
answers
75
views
How do Maxwell's equations follow from the action of Lorentz generators on field strength?
Following Warren Siegel's book on Field theory (pg. 223), one might derive the action of Lorentz generators $S_{ab}$ on an antisymmetric 2-tensor field strength $F_{cd}$ which arises for example in ...
- 785
6
votes
1
answer
300
views
Inverse of anti-symmetric rank 4 tensors?
I am trying to find an inverse of a tensor of the form
$$M_{\mu\nu\rho\sigma}$$ such that $M$ is anti-symmetric in the $(\mu, \nu)$ exchange and $(\rho, \sigma)$ exchange. The inverse should be such ...
- 1,200
0
votes
1
answer
83
views
How can I calculate the square of Pauli-Lubanski vector in a rest frame?
recently I've been trying to demonstrate that, $$\textbf{W}^2 = -m^2\textbf{S}^2$$ in a rest frame, with $W_{\mu}$ defined as $$W_{\mu} = \dfrac{1}{2}\varepsilon_{\mu\alpha\beta\gamma}M^{\alpha\beta}p^...
- 1
0
votes
1
answer
91
views
Which is correct definition between $J^i\equiv \frac{1}{2}\epsilon^i_{~jk}J^{jk}$ and $J^i\equiv \frac{1}{2}\epsilon^{ijk}J^{jk}$?
The generators of the Lorentz group are denoted by $J^{\mu\nu}$ (suppose they are defined in terms of raised indices, as shown). From this, in my opinion, the angular momentum generators $J^i$'s and $...
- 11.8k
4
votes
1
answer
876
views
Levi-Civita tensor and the Lorentz group generators in the vector representation
In the vector representation of the Lorentz group its generators are given by -
$$(J^{\mu\nu})_{\alpha\beta} = i(\delta^\mu_\alpha\delta^\nu_\beta-\delta^\mu_\beta\delta^\nu_\alpha)$$
It can be ...
- 1,407
3
votes
1
answer
1k
views
Why is the Mixed Faraday Tensor a matrix in the algebra so(1,3)?
The mixed Faraday tensor $F^\mu{}_\nu$ explicitly in natural units is:
$$(F^\mu{}_\nu)=\left(\begin{array}{cccc}0&E_x&E_y&E_z\\E_x&0&B_z&-B_y\\E_y&-B_z&0&B_x\\E_z&...
- 88.5k
2
votes
1
answer
170
views
Representations of Lorentz algebra
It is well known that the Lorentz algebra can be written as two $SU(2)$ algebras. By defining
$$N_i=\frac{1}{2}(J_i+iK_i), \qquad
N^{\dagger}_i=\frac{1}{2}(J_i-iK_i)$$
we have
$[N_i,N_j]=i\...
- 1,654