- #1

provolus

- 18

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"for a theory involving only scalar fields, the energy-momentum tensor is always symmetric". But I've some doubts on how to demonstrate it. So I started with a particular example.

Given a density of Lagrangian [tex]L[/tex] for a scalar field [tex]\phi[/tex] (our field is in Minkowsky space with metric tensor [tex] \eta^{\mu\nu}[/tex] +---) of the form:

[tex]L=\frac{1}{2}(\partial_{\mu}\phi\partial^{\mu}\phi-m^2\phi^2})-\frac{\lambda}{4!}\phi^4[/tex]

with real constant [tex]\lambda[/tex] and [tex]m[/tex], according to the definition of energy-momentum tensor [tex] T^{\mu\nu}[/tex], that is

[tex]T^{\mu\nu}=\frac{\partial L}{\partial(\partial_{\mu}\phi_i)}\partial^{\nu}\phi_i-\eta^{\mu\nu}L} [/tex]

(with latin index

*i*running on space coordinates) for this [tex]L[/tex], substituing into the last formula, I should get (dropping the index

*i*since [tex]\phi[/tex] is a scalar):

[tex]T^{\mu\nu}=\frac{1}{2}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\phi\partial^{\mu}\phi)-m^2\phi^2-\frac{\lambda}{4!}\phi^4[/tex]

Arrived at this point my first question is: is this [tex]T^{\mu\nu}[/tex] symmetric?

Trying to answer by myself but I'm new in tensor computing and I know that it would be a joke with a few of QFT and GR knowledge...

given that [tex]\partial^{\mu}\phi[/tex] is a vector, and then [tex]\partial^{\mu}\phi\partial^{\nu}\phi [/tex] is a dot product (but where are the contracted indices?), so a scalar, it should be

[tex]\frac{1}{2}(\partial^{\mu}\phi\partial^{\nu}\phi-\partial^{\nu}\phi\partial^{\mu}\phi)=0[/tex]

so with all scalars my tensor should be symmetric, isn't?

But, generally speaking, my second question is why a theory of only scalar field the energy-momentum tensor is always symmetric?

Trying to answer by myself I think that being the Langrangian Lorentz invariant, in certain way this is reflected to the tensor... but I'm not sure...

can someone be so kind to explain to help me in answering these two questions?