- #1
Pacopag
- 197
- 4
Homework Statement
Find the commutators [tex][P^\sigma,J^{\mu \nu}][/tex]
The answer is part of the Poincare algebra
[tex][P^\sigma,J^{\mu \nu}]=i(g^{\mu \sigma}P^\nu-g^{\nu \sigma}P^\mu)[/tex]
If someone can convince me that [tex]\partial_i T^{0\mu} = 0[/tex], (i.e. the energy-momentum tensor has no explicit spatial dependence) then I got it.
But I'll still post my solution below.
Homework Equations
[tex]P^\mu = \int d^3x T^{0\mu}[/tex]
[tex]J^{\nu \sigma}=\int d^3x(x^\nu T^{0\sigma}-x^\sigma T^{0\nu})[/tex]
The Attempt at a Solution
An unsatisfactory explanation of why [tex]\partial_i T^{0\mu} = 0[/tex] is that [tex]T^{\mu \nu}[/tex] depends on the fields, so there is no [tex]explicit[/tex] spacetime dependence, but the fields in turn depend on spacetime. You'll see where this comes in below.
Since [tex]J^{\nu \sigma}=-J^{\sigma \nu}[/tex], we only have to consider the commutators
[tex][P^0,J^{0i}][/tex], [tex][P^0,J^{ij}][/tex], [tex][P^i,J^{0j}][/tex], [tex][P^i,J^{jk}][/tex]
Let's take [tex][P^i,J^{0j}][/tex] as an example. If I can get this one, then I can get them all.
[tex][P^i,J^{0j}] = \left[P^i, \int d^3x(x^0 T^{0j}-x^j T^{00}) \right][/tex]
[tex]=x^0[P^i,P^j]-[P^i,\int d^3x x^j T^{00}][/tex]
The first term is zero since momenta commute. For the second term, since [tex]P^i[/tex] is the generator of translation in the i-direction, then as in quantum mechanics we get
[tex][P^i,J^{0j}] = -i \partial^{'}_i \int d^3x x^j T^{00}[/tex]
[tex]=-i \left( \delta_{ij} \int d^3x T^{00} + \int d^3x x^j \partial_j T^{00}\right)[/tex]
[tex]=-i \left( \delta_{ij} P^0 + \int d^3x x^j \partial_j T^{00}\right)[/tex]
If the second term vanishes, then we get the right answer. There are a couple of ways this can happen:
(1) [tex]\partial_i T^{0\mu} = 0[/tex], which is what I'd really like to show.
(2) [tex]T^{00}[/tex] is for some reason a spatially odd function (i.e. [tex]T^{00}(x)=-T^{00}(-x)[/tex].
The latter seems completely unlikely due to the whole Lorentz invariance thing. It just seems wierd.
The (1) reason seems better to me. Of course, we do know that [tex]T^{\mu \nu}[/tex] is conserved, but in the sense that [tex]\partial_\mu T^{\mu \nu} = 0[/tex]. But I don't see how this tells us that individual partial derivatives are zero.