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 $V(\phi)$?

  • $\begingroup$ 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 $\phi$? $\endgroup$ Jan 8, 2012 at 13:28
  • $\begingroup$ @Jerry Schirmer: If you mean the Euler-Lagrangian equation for field variables, sadly that gives only differential, not algebraic properties of $V(\phi)$. $\endgroup$
    – Siyuan Ren
    Jan 8, 2012 at 14:25
  • $\begingroup$ This answer may be useful for the full calculation: physics.stackexchange.com/questions/283488/… $\endgroup$
    – Rexcirus
    Aug 17, 2022 at 16:09

2 Answers 2


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.


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.

  • $\begingroup$ Well, if the vector is null, the expression is explicitly the square of a real number. $\endgroup$ Jan 10, 2012 at 13:50
  • $\begingroup$ @JerrySchirmer: How is that? $\endgroup$
    – Siyuan Ren
    Mar 10, 2012 at 0:17
  • 1
    $\begingroup$ Because, for all null vectors, $\ell^{a}\ell^{b}g_{ab}=0$. So, we have $T_{ab}\ell^{a}\ell^{b}=\ell^{a}\ell^{b}\nabla_{a}\phi \nabla_{b}\phi = \left(\ell^{a}\nabla_{a}\phi\right)^{2}$. $\endgroup$ Mar 12, 2012 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.