(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant 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]

3. 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.

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# Homework Help: Deriving the Poincare algebra in scalar field theory

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