# Is there an invariant way to derive the energy-momentum tensor?

On a (pseudo-)Riemannian manifold $$(M, g)$$ I can define the following action for any $$\phi \in C ^{\infty}(M)$$: $$\mathcal{S}(\phi) = \int_M g(\text{grad }\phi, \text{grad }\phi) \mathrm{d} V.$$ According to the Wikipedia page, the energy-momentum tensor associated with this action is given by the variation through a ''spacetime translation''. I take this to mean that given a vector field $$X$$ on $$M$$ with flow $$\psi$$, the composition $$\Phi(t) := \phi \circ \psi_t$$ gives a variation of $$\phi$$ as $$t$$ varies from some small $$-\epsilon$$ to $$\epsilon$$ (at least if $$X$$ is compactly supported). Then for any $$v \in T_p M$$ \begin{align*} g_p(\text{grad } \phi \circ \psi_t|_p, v) &= \mathrm{d}( \phi \circ \psi_t)|_p (v) = v( \phi \circ \psi_t) \\ &= \mathrm{d} \psi_t|_p(v) \phi = \mathrm{d} \phi|_{\psi_t(p)}(\mathrm{d} \psi_t |_p(v)) \\ &= g_{\psi_t(p)}(\text{grad } \phi |_{\psi_t(p)}, \mathrm{d} \psi_t|_p(v)) \\ &= g_p((\mathrm{d} \psi_t) ^{\top}|_{\psi_t(p)}\text{grad } \phi |_{\psi_t(p)}, v), \end{align*} where $$\mathrm{d} \psi_t|_{\psi_t(p)} : T_{\psi_t(p)} M \to T_pM$$ is the adjoint of $$\mathrm{d} \psi_t|_p : T_p M \to T_{\psi_t(p)} M$$ with respect to $$g$$. Thus, by diffeomorphism invariance of integrals: \begin{align*} \mathcal{S}(\Phi(t)) &= \int_M g_p(\text{grad }\phi \circ \psi_t|_p, \text{grad }\phi \circ \psi_t|_p) \mathrm{d} V \\ &= \int_M g_p((\mathrm{d} \psi_t) ^{\top}|_{\psi_t(p)}\text{grad } \phi |_{\psi_t(p)}, (\mathrm{d} \psi_t) ^{\top}|_{\psi_t(p)}\text{grad } \phi |_{\psi_t(p)}) \mathrm{d} V \\ &= \int_M g_{\psi_{-t}(p)}((\mathrm{d} \psi_t) ^{\top}\text{grad } \phi |_p, (\mathrm{d} \psi_t) ^{\top}\text{grad } \phi |_p) \psi_{-t} ^* \mathrm{d} V \end{align*} Upon taking the derivative with respect to $$t$$ and using the Leibniz rule I see that \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \Big|_{t = 0} \mathcal{S}(\Phi(t)) &= \int_M \frac{\mathrm{d}}{\mathrm{d} t} \Big|_{t = 0}g_{\psi_{-t}(p)}((\mathrm{d} \psi_t) ^{\top}\text{grad } \phi |_p, (\mathrm{d} \psi_t) ^{\top}\text{grad } \phi |_p) \mathrm{d} V \\ &- \int_M g_p(\text{grad } \phi|_p, \text{grad } \phi|_p) \text{div}(X) \mathrm{d} V. \end{align*} If I compare this with the the contraction of the actual energy-momentum tensor against the deformation tensor $$\pi ^{\alpha \beta} = \nabla ^{\alpha} X ^{\beta} + \nabla ^{\beta} X ^{\alpha}$$ I get \begin{align*} T_{\alpha \beta} \pi ^{\alpha \beta} &= \nabla_{\alpha} \phi \nabla_{\beta} \phi (\nabla ^{\alpha} X ^{\beta} + \nabla ^{\beta} X ^{\alpha}) - \frac{1}{2} g_{\alpha \beta} \nabla ^{\gamma} \phi \nabla_{\gamma} \phi (\nabla ^{\alpha} X ^{\beta} + \nabla ^{\beta} X ^{\alpha}) \\ &= 2g(\nabla_{\text{grad } \phi} X, \text{grad }\phi) - g(\text{grad }\phi, \text{grad }\phi) \text{div}(X). \end{align*} The divergence terms match, so all I have to do now is show that \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \Big|_{t = 0}g_{\psi_{-t}(p)}((\mathrm{d} \psi_t) ^{\top}\text{grad } \phi |_p, (\mathrm{d} \psi_t) ^{\top}\text{grad } \phi |_p) &= 2 g(\nabla_{\text{grad }\phi} X, \text{grad }\phi) \\ &= \mathcal{L}_X g( \text{grad }\phi, \text{grad }\phi). \end{align*} But this does not seem right. The left hand side is almost $$-\mathcal{L}_X g(\text{grad }\phi, \text{grad }\phi)$$, in fact if $$X$$ is Killing this is the case, so I am not sure if the variation I am using is the correct one, or what Wikipedia means by ''the energy tensor is the Noether current of spacetime translations''. I have seen derivations by using coordinates, but is there a way to proceed invariantly similar to what I am trying to do above?

• Where are you using coordinates in your approach? Commented Feb 28, 2023 at 4:30
• @Gibbs I'm not. I want to know if computing this without coordinates is feasible. Commented Feb 28, 2023 at 5:01
• Sometimes people call that integral $\mathcal{S}(\phi)$ the "Dirichlet energy" of $\phi$. Critical points of the Dirichlet energy, at least in $\Bbb R^n$, are harmonic functions (see Example 34, which unfortunately used coordinates). Commented Mar 2, 2023 at 7:26

Computing \begin{align*} \mathrm{d}(\phi \circ \psi_t)|_p(Y) &= Y|_p \phi \circ \psi_t = \mathrm{d}\psi_t|_p(Y) \phi = \mathrm{d} \phi|_{\psi_t(p)} (\mathrm{d} \psi_t|_p(Y)) = (\psi_t ^* \mathrm{d} \phi)_p(Y) \end{align*} yields \begin{align*} g_p ^{-1}(\mathrm{d}(\phi \circ \psi_t), \mathrm{d}(\phi \circ \psi_t)) &= g_p ^{-1}(\psi_t ^* \mathrm{d} \phi, \psi_t ^* \mathrm{d} \phi). \end{align*} Using diffeomorphism invariance \begin{align*} \mathcal{S}(\Phi(t)) = \int_M g ^{-1}_p(\mathrm{d} (\phi \circ \psi_t), \mathrm{d} ( \phi \circ \psi_t)) \mathrm{d}V &= \int_M g ^{-1}_{\psi_{-t}(p)}((\psi_{-t} ^{-1}) ^* \mathrm{d} \phi, (\psi_{-t} ^{-1}) ^* \mathrm{d} \phi) \phi_{-t} ^*\mathrm{d}V \\ &= \int_M (\psi_{-t} ^* g ^{-1})_p(\mathrm{d} \phi, \mathrm{d} \phi) \phi_{-t} ^* \mathrm{d}V \end{align*} and taking derivatives we have \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \Big|_{t = 0} \mathcal{S}(\Phi(t)) &= \int_M \mathcal{L}_{-X} g ^{-1}(\mathrm{d} \phi ,\mathrm{d}\phi) \mathrm{d} V + \int_M g ^{-1}(\mathrm{d} \phi, \mathrm{d} \phi) \mathcal{L}_{-X} \mathrm{d} V \\ &= -\int_M (\mathcal{L}_X g ^{-1}(\mathrm{d} \phi, \mathrm{d}\phi) + g ^{-1}(\mathrm{d}\phi, \mathrm{d} \phi) \text{div}(X)) \mathrm{d} V. \end{align*} But the Lie derivative of a musical isomorphism is \begin{align*} \mathcal{L}_X Y ^{\flat}(Z) &= X Y ^{\flat}(Z) - Y ^{\flat}([X, Z]) = X g(Y, Z) - g(Y, [X, Z]) \\ &= g(\nabla_X Y, Z) + g(Y, \nabla_X Z - [X, Z]) \\ &= g(\nabla_X Y, Z) + g(Y, \nabla_Z X) =: (\nabla_XY) ^{\flat}(Z) + \theta_{Y}(Z), \end{align*} whence \begin{align*} \mathcal{L}_X g ^{-1}(Y ^{\flat}, Z ^{\flat}) &= X g ^{-1}(Y ^{\flat}, Z ^{\flat}) - g ^{-1}(\mathcal{L}_X Y ^{\flat} , Z ^{\flat}) - g ^{-1}(Y ^{\flat}, \mathcal{L}_X Z ^{\flat} ) \\ &= X g (Y , Z) - g ^{-1}((\nabla_X Y) ^{\flat} , Z ^{\flat}) - g ^{-1}(Y ^{\flat}, (\nabla_X Z) ^{\flat} ) \\ &- g ^{-1}(\theta_{Y}, Z ^{\flat}) - g ^{-1}(Y ^{\flat}, \theta_{Z}) \\ &= - g(\theta_{Y} ^{\sharp}, Z) - g (Y , \theta_{Z} ^{\sharp}) \\ &= - \theta_Y(Z) - \theta_Z(Y) \\ &= -g(Y, \nabla_Z X) - g(Z, \nabla_YX) = - \mathcal{L}_X(Y, Z). \end{align*} We can then write $$\delta S = \int_M (\mathcal{L}_X g (\text{grad }\phi, \text{grad }\phi) - g(\text{grad }\phi, \text{grad }\phi) \text{div}(X)) \mathrm{d} V,$$ as required.