Pesking & Schroeder Eqn 3.18 (Lorentz algebra)

In summary: J's?).However, if we multiply them as tensors, then I am right and 3.17 is ok.Let me explain. If we multiply J_{\alpha\beta} (a matrix with 2 indices) by J_{\gamma\delta} (a matrix with 2 indices), we get a matrix with 4 indices, let's call it M_{\alpha\beta\gamma\delta}.In component notation, this is just a sum over an index, like this:M_{\alpha\beta\gamma\delta}=\sum_\mu J^{\mu}_{\;\;\;\;\alpha\beta}J^{\mu}_{\;\;\;\;\gamma\delta}.It looks
  • #1
DivergentMind
4
0
I am trying to show that (for 4x4 matrices) the representation given by equation 3.18 (Peskin and Schroeder, page 39):
[tex]
(J^{\mu\nu})_{\alpha\beta}
=i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})
[/tex]
implies the commutation relations in 3.17:
[tex]
[J^{\mu\nu},J^{\rho\sigma}]
=i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})
[/tex]
For some reason I cannot even get this to work for the [tex](\mu,\nu,\rho,\sigma)=(0,1,1,2)[/tex] component:
[tex]
[J^{01},J^{12}]_{\alpha\beta}
=J^{01}_{\alpha\gamma}J^{12}_{\gamma\beta}
-J^{12}_{\alpha\gamma}J^{01}_{\gamma\beta}
=i(\delta^0_{\alpha}\delta^1_{\gamma}-\delta^0_{\gamma}\delta^1_{\alpha})
i(\delta^1_{\gamma}\delta^2_{\beta}-\delta^1_{\beta}\delta^2_{\gamma})
-i(\delta^1_{\alpha}\delta^2_{\gamma}-\delta^1_{\gamma}\delta^2_{\alpha})
i(\delta^0_{\gamma}\delta^1_{\beta}-\delta^0_{\beta}\delta^1_{\gamma})
[/tex]
Now, sums like [tex]\delta^1_{\gamma}\delta^2_{\gamma}[/tex] vanish, whereas [tex]\delta^1_{\gamma}\delta^1_{\gamma}[/tex] is 1, so we get:
[tex]
[J^{01},J^{12}]_{\alpha\beta}=
-\delta^0_{\alpha}\delta^2_{\beta}
+\delta^0_{\beta}\delta^2_{\alpha}
=i(J^{02})_{\alpha\beta}
[/tex]
On the other hand, the right hand side of 3.17 was supposed to give us:
[tex]
i(g^{11}J^{02}-g^{01}J^{12}-g^{12}J^{01}+g^{02}J^{11})_{\alpha\beta}
=i((-1)J^{02}-0-0+0)_{\alpha\beta}=-i(J^{02})_{\alpha\beta}
[/tex]
ugh...
I suspect I messed up with the metric at some point, but I don't see where.
 
Last edited:
Physics news on Phys.org
  • #2
There's something indeed dubious with your calculation, since there's no summation over \gamma anywhere (actually both \gamma's are downstairs, so they can't be summed over). So try to rewrite this again and sum correctly with the metric.
 
  • #3
OK, changed this to
[tex]
[J^{01},J^{12}]_{\alpha\beta}
=(J^{01})_{\alpha\gamma}(J^{12})^{\gamma}_{\beta}
-(J^{12})_{\alpha}^{\gamma}(J^{01})_{\gamma\beta}
=(J^{01})_{\alpha\gamma}g^{\gamma\delta}(J^{12})_{\delta\beta}
-(J^{12})_{\alpha\delta}g^{\delta\gamma}(J^{01})_{\\gamma\beta}
[/tex]
and this works. Thanks.
 
  • #4
DivergentMind said:
OK, changed this to
[tex]
[J^{01},J^{12}]_{\alpha\beta}
=(J^{01})_{\alpha\gamma}(J^{12})^{\gamma}_{\beta}
-(J^{12})_{\alpha}^{\gamma}(J^{01})_{\gamma\beta}
=(J^{01})_{\alpha\gamma}g^{\gamma\delta}(J^{12})_{\delta\beta}
-(J^{12})_{\alpha\delta}g^{\delta\gamma}(J^{01})_{\\gamma\beta}
[/tex]
and this works. Thanks.
This time you're starting with an equality that's false. If you're denoting the component on row [itex]\alpha[/itex] column [itex]\beta[/itex] of the matrix [itex]J^{\mu\nu}[/itex] by [itex](J^{\mu\nu})_{\alpha\beta}[/itex], you can't just change that to [itex](J^{\mu\nu})^\alpha_\beta[/itex] in the middle of a calculation.



DivergentMind said:
For some reason I cannot even get this to work for the [tex](\mu,\nu,\rho,\sigma)=(0,1,1,2)[/tex] component:
What is
[tex]i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})
[/tex]when the only two indices with the same values are [itex]\nu[/itex] and [itex]\rho[/itex], and that value is 1?
 
  • #5
Fredrik said:
This time you're starting with an equality that's false. If you're denoting the component on row [itex]\alpha[/itex] column [itex]\beta[/itex] of the matrix [itex]J^{\mu\nu}[/itex] by [itex](J^{\mu\nu})_{\alpha\beta}[/itex], you can't just change that to [itex](J^{\mu\nu})^\alpha_\beta[/itex] in the middle of a calculation.

i think dexter is implying that the J's are tensors, not matrices, and therefore the product of two Js is their contractions, and hence it is ok that one of the contracted indices is up and one is down (for otherwise, you cannot sum them)

What is
[tex]i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})
[/tex]when the only two indices with the same values are [itex]\nu[/itex] and [itex]\rho[/itex], and that value is 1?

the answer to this is at the last line of my first post:
[itex]
-iJ^{\mu\sigma}
[/itex]
(but when [itex]\mu[/itex] or [itex]\sigma[/itex] are also 1, then you get 0 of course)
 
Last edited:
  • #6
DivergentMind said:
I am trying to show that (for 4x4 matrices) the representation given by equation 3.18 (Peskin and Schroeder, page 39):
[tex]
(J^{\mu\nu})_{\alpha\beta}
=i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})
[/tex]
implies the commutation relations in 3.17:
[tex]
[J^{\mu\nu},J^{\rho\sigma}]
=i(g^{\nu\rho}J^{\mu\sigma}-g^{\mu\rho}J^{\nu\sigma}-g^{\nu\sigma}J^{\mu\rho}+g^{\mu\sigma}J^{\nu\rho})
[/tex]
I tried to do this calculation myself. I'm assuming that you meant that the first formula defines the 16 matrices [itex]J^{\mu\nu}[/itex] by, for each one of those matrices, specifying its component on row [itex]\alpha[/itex], column [itex]\beta[/itex], for all [itex]\alpha,\beta[/itex].

The result I'm getting is [tex][J^{\mu\nu},J^{\rho\sigma}]_{\alpha\beta}= i\delta^\nu_\gamma\delta^\rho_\gamma(J^{\mu\sigma})_{\alpha\beta}+\cdots[/tex] (Yes, I meant that there's a sum over all values of [itex]\gamma[/itex] even though they're both downstairs). Clearly each term of [itex]\delta^\nu_\gamma\delta^\rho_\gamma[/itex] is =0 when [itex]\rho\neq\nu[/itex]. When [itex]\rho=\nu[/itex], one of the terms is 1 and the others are 0, so they add up to 1. So the result is never -1, and that means it can't be true that [itex]\delta^\nu_\gamma\delta^\rho_\gamma=g^{\nu\rho}[/itex] for all [itex]\nu,\rho[/itex].

I'm not ruling out the possibility that I made some silly mistake, but right now I think the result we're supposed to get looks wrong. Edit: The result isn't wrong, but maybe the definition of the matrices is.
 
Last edited:
  • #7
Fredrik said:
I tried to do this calculation myself. I'm assuming that you meant that the first formula defines the 16 matrices [itex]J^{\mu\nu}[/itex] by, for each one of those matrices, specifying its component on row [itex]\alpha[/itex], column [itex]\beta[/itex], for all [itex]\alpha,\beta[/itex].
right, this is what i meant
The result I'm getting is [tex][J^{\mu\nu},J^{\rho\sigma}]_{\alpha\beta}= i\delta^\nu_\gamma\delta^\rho_\gamma(J^{\mu\nu})_{\alpha\beta}+\cdots[/tex] (Yes, I meant that there's a sum over all values of [itex]\gamma[/itex] even though they're both downstairs). Clearly each term of [itex]\delta^\nu_\gamma\delta^\rho_\gamma[/itex] is =0 when [itex]\rho\neq\nu[/itex]. When [itex]\rho=\nu[/itex], one of the terms is 1 and the others are 0, so they add up to 1. So the result is never -1, and that means it can't be true that [itex]\delta^\nu_\gamma\delta^\rho_\gamma=g^{\nu\rho}[/itex] for all [itex]\nu,\rho[/itex].

I'm not ruling out the possibility that I made some silly mistake, but right now I think the result we're supposed to get looks wrong.

yes, this is the same problem i was facing. i tried to demonstrate it for specific indices, in my first post.

i think it comes down to how we multiply one J by the other
Edit: if we multiply J's as matrices, then you are right and there is probably something wrong with formula 3.17 (the definition of the matrices)
 
Last edited:
  • #8
I think I got it. For each [itex]\mu,\nu[/itex], the formula [itex](J^{\mu\nu})_{\alpha\beta}
=i(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}-\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})[/itex] defines 16 numbers, but those numbers aren't the [itex]\alpha\beta[/itex] components (row [itex]\alpha[/itex], column [itex]\beta[/itex]) of the matrix [itex]J^{\mu\nu}[/itex]. Instead [itex]J^{\mu\nu}[/itex] is defined as the matrix with [itex]\alpha\beta[/itex] component [itex](J^{\mu\nu})^\alpha{}_\beta =g^{\alpha\gamma}(J^{\mu\nu})_{\gamma\beta}[/itex]. Similarly, the [itex]\alpha\beta[/itex] component of the commutator is [itex][J^{\mu\nu},J^{\rho\sigma}]^\alpha{}_\beta[/itex], not [itex][J^{\mu\nu},J^{\rho\sigma}]_{\alpha\beta}[/itex].
 
Last edited:

1. What is the Pesking & Schroeder Eqn 3.18?

The Pesking & Schroeder Eqn 3.18, also known as the Lorentz algebra, is a mathematical equation used in the study of special relativity. It describes the algebraic relationships between the components of four-vectors, which are quantities used to represent space and time in the theory of relativity.

2. What does the Lorentz algebra represent?

The Lorentz algebra represents the symmetries and transformations of space and time in special relativity. It is based on the work of Hendrik Lorentz, who first described these transformations in the late 19th century.

3. How is the Lorentz algebra used in physics?

The Lorentz algebra is used in physics to calculate and predict the behavior of particles moving at high speeds, close to the speed of light. It is also used in the development of theories of quantum mechanics and particle physics.

4. What are some applications of the Lorentz algebra?

The Lorentz algebra has many practical applications in fields such as astrophysics, nuclear physics, and engineering. It is used in the design of particle accelerators, the study of cosmic rays, and the development of GPS technology.

5. Are there any limitations to the Lorentz algebra?

While the Lorentz algebra is a powerful tool for understanding the behavior of particles at high speeds, it is limited to the realm of special relativity. It does not take into account the effects of gravity or the curvature of spacetime, which are described by general relativity.

Similar threads

Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
359
  • Quantum Physics
Replies
6
Views
962
Replies
24
Views
2K
Replies
1
Views
816
  • Special and General Relativity
Replies
1
Views
708
  • Special and General Relativity
Replies
1
Views
547
  • Quantum Physics
Replies
1
Views
630
  • Special and General Relativity
Replies
5
Views
1K
  • Quantum Physics
Replies
6
Views
750
Back
Top