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It's a question on Sean Carroll's Spacetime and Geometry, where we are supposed to prove that the energy momentum tensor of scalar field theory satisfies Weak Energy Condition (WEC). The energy momentum tensor is


and the condition for WEC is

$$T_{\mu\nu} U^\mu U^\nu \geq 0,$$

where $U^\mu$ is an arbitrary non-spacelike vector(=timelike or null).

But how can this be proved when there are no known properties about the scalar field variable $\phi$ and potential $V(\phi)$?

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You derived the stress-energy tensor above from a lagrangian by varying with respect to $g^{\mu \nu}$. What happens when you vay with respect to $\phi$? – Jerry Schirmer Jan 8 '12 at 13:28
@Jerry Schirmer: If you mean the Euler-Lagrangian equation for field variables, sadly that gives only differential, not algebraic properties of $V(\phi)$. – Siyuan Ren Jan 8 '12 at 14:25
up vote 2 down vote accepted

I don't have the book, so can't check out his assumptions, so this might not quite answer your question, since you're asking about arbitrary 4 vectors $U^{\mu}$, but I'll offer it in case some of it is useful. In proving the weak energy condition (which is part way to proving the dominant energy condition), the 4 vectors in question are timelike. If this is the case, I might try the following:

Assume a signature (- + + + )

Starting with $$T_{\mu\nu}=\nabla_{\mu}\phi\nabla_{\nu}\phi-\frac{1}{2}g_{\mu\nu}\left(\nabla_{\lambda}\phi\nabla^{\lambda}\phi+V(\phi)\right)$$

if $U^{\mu}$ is timelike and future pointing, then at any given point we can work in an orthonormal frame for which the components are $U^{\mu}=(1,0,0,0)$

If we then demonstrate the positivity of $T_{\mu\nu}U^{\mu}U^{\nu}$ in that frame, then it will hold in any frame since it's a scalar.

So, plugging in the components of $U^{\mu}$, we get



So provided $V(\phi)$ is positive and $\phi$ is a real field (which it surely is otherwise you'd have had complex conjugates in the energy momentum tensor), then in that frame, at that point $T_{\mu\nu}U^{\mu}U^{\nu}$ is positive.

But this is just the weak energy condition you'd have to work a bit harder to prove the dominant energy condition.

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You can always boost to make the vector U be the time axis (when it is strictly spacelike), and then the expression for the energy density can be written down explicitly

$$ T_{00} = {1\over 2} \dot{\phi}^2 + {1\over 2} |\nabla\phi|^2 + V(\phi) $$

This expression is found by substituting 0 for the indices, and the explicit Minkowski form o the metric tensor. All contributions are manifestly positive definite (assuming V is bounded below). the first contribution is the field kinetic energy, the second is the field gradient potenetial energy, and the third is the field value potential energy. Proving the inequality for null vectors can be done by taking a limit of the vector becoming null.

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Well, if the vector is null, the expression is explicitly the square of a real number. – Jerry Schirmer Jan 10 '12 at 13:50
@JerrySchirmer: How is that? – Siyuan Ren Mar 10 '12 at 0:17
Because, for all null vectors, $\ell^{a}\ell^{b}g_{ab}=0$. So, we have $T_{ab}\ell^{a}\ell^{b}=\ell^{a}\ell^{b}\nabla_{a}\phi \nabla_{b}\phi = \left(\ell^{a}\nabla_{a}\phi\right)^{2}$. – Jerry Schirmer Mar 12 '12 at 18:38

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