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Why has the trace of the energy-momentum tensor to vanish for conserved scaling currents to exist?

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In this paper, the authors say that the trace of the energy-momentum tensor has to vanish to allow for the existence of conserved dilatation or scaling currents, as defined on p 10, Eq(22)

$$ \Theta^{\mu} = x_{\nu} \Theta^{\mu\nu} + \Sigma^\mu.$$

$\Theta^{\mu\nu}$ is the energy-momentum tensor and $\Sigma^\mu = d_{\phi}\pi_i^{\mu}\phi^i$ is the internal part.

This fact is just mentioned and I dont understand why this has to be the case, so can somebody explain it to me?

asked Oct 19, 2012 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
A related question by OP: physics.stackexchange.com/q/35693/2451

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Qmechanic

1 Answer

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Without the internal part:

The divergence $$\nabla_\mu (x_\nu \Theta^{\mu\nu}) = x_\nu \nabla_\nu \Theta^{\mu\nu} + \frac12(\nabla_\mu x_\nu + \nabla_\nu x_\mu) \Theta^{\mu\nu} $$ where I used that $\Theta^{\mu\nu}$ is symmetric. Recalling that the energy-momentum tensor is divergence free, the first term drops out. Assuming that $x^\nu$ generates a dilation/scaling symmetry (and not a bona fide symmetry), we know that its deformation $$ \nabla_\mu x_\nu + \nabla_\nu x_\mu \propto \mathcal{L}_x g_{\mu\nu} \propto g_{\mu\nu} $$ where $\mathcal{L}$ is the Lie derivative. (In the case $x^\nu$ generates a symmetry the term vanishes from Killing's equation.)

Hence in this case for the current to be conserved (that is, divergence free), we need that $g_{\mu\nu} \Theta^{\mu\nu} = 0$; that is, the energy momentum tensor is tracefree.

This post imported from StackExchange Physics at 2014-03-12 15:45 (UCT), posted by SE-user Willie Wong
answered Oct 19, 2012 by Willie Wong (570 points) [ no revision ]

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