Generators of Lorentz Lie Algebra

In summary, the generators of the Lorentz Lie algebra relations obey a commutator relation that is satisfied for any representation of the Lie algebra. However, when trying to prove this, it is important to not fully expand the commutator using the particular form of the generators, but only the first term in the expansion.
  • #1
latentcorpse
1,444
0
So the generators of the Lorentz Lie algebra relations obey

[itex][M^{\rho \sigma}, M^{\tau \mu}] = g^{\sigma \tau} M^{\rho \mu} - g^{\sigma \mu} M^{\rho \tau} + g^{\rho \mu} M^{\sigma \tau} - g^{\rho \tau} M^{\sigma \mu}[/itex]

where [itex](M^{\rho \sigma})^\mu{}_\nu = g^{\rho \mu} \delta^\sigma{}_\nu - g^{\sigma \mu} \delta^\rho{}_\nu[/itex]

Now the commutator relation above will be satisfied for any representation of the LIe algebra since a representation of a Lie algebra is just a set of matrices such that the Lie bracket is given by the commutator.

So anyways, I was trying to prove this and just got totally bogged down in the algebra:

[itex][M^{\rho \sigma}, M^{\tau \mu}] \\
= (M^{\rho \sigma})^\lambda{}_\kappa ( M^{\tau \mu})^\kappa{}_\lambda - (M^{\tau \mu})^\lambda{}_\kappa (M^{\rho \sigma})^\kappa{}_\lambda \\
=(g^{\rho \lambda} \delta^\sigma{}_\kappa - g^{\sigma \lambda} \delta^\rho{}_\kappa)(g^{\tau \kappa} \delta^\mu{}_\lambda - g^{\mu \kappa} \delta^\tau{}_\lambda) -(g^{\tau \lambda} \delta^\mu{}_\kappa - g^{\mu \lambda} \delta^\tau{}_\kappa)(g^{\rho \kappa} \delta^\sigma{}_\lambda - g^{\sigma \kappa} \delta^\rho{}_\lambda)[/itex]

but expanding this out isn't going to give me any M terms like I need in the answer. Any ideas where I've messed up?

Thanks.
 
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  • #2
You shouldn't fully expand the commutator using the particular form of M, but only the first of the M's in the expanded commutator.
 
  • #3
bigubau said:
You shouldn't fully expand the commutator using the particular form of M, but only the first of the M's in the expanded commutator.

ok. so i get to

[itex]g^{\rho \sigma} (M^{\tau \mu})^\sigma{}_\lambda - g^{\sigma \lambda} ( M^{\tau \mu})^\rho{}_\lambda - g^{\tau \lambda} ( M^{\rho \sigma})^\mu{}_\lambda + g^{\mu \lambda} ( M^{\rho \sigma})^\tau{}_\lambda[/itex]

but don't know how to finish it off...
 

1. What is the Lorentz Lie algebra?

The Lorentz Lie algebra is a mathematical structure that describes the symmetries of special relativity. It consists of a set of generators, or operators, that represent transformations between reference frames in which the laws of physics remain the same.

2. How is the Lorentz Lie algebra related to the Lorentz group?

The Lorentz Lie algebra is the set of all possible commutators of the generators, which form the basis for the Lorentz group. This means that the Lie algebra provides a way to represent the group in a more compact and algebraic form.

3. What are the generators of the Lorentz Lie algebra?

The generators of the Lorentz Lie algebra are the six operators corresponding to the three rotations and three boosts in special relativity. They are usually denoted by J and K, respectively, and can be represented as matrices or differential operators.

4. How do the generators of the Lorentz Lie algebra act on physical quantities?

The generators of the Lorentz Lie algebra act as infinitesimal transformations on physical quantities, such as position and momentum. They are used to describe how these quantities change under different reference frames and are essential for understanding the laws of physics in special relativity.

5. What are some practical applications of the Lorentz Lie algebra?

The Lorentz Lie algebra has numerous applications in theoretical physics, particularly in the study of special relativity and quantum field theory. It is also used in the development of mathematical models for physical phenomena, such as particle interactions and black hole thermodynamics.

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