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