# Energy-momentum tensor transformation [closed]

I've been trying to find how the energy-momentum tensor changes if we add a total derivative to the lagrangian: $$L\to L+\mathrm d_\mu X^\mu.\tag{1}$$

From the answer key: $$T^{\mu\nu}\to T^{\mu\nu}+\partial_\lambda F^{\lambda\mu\nu}\tag{2}$$ Where $$F^{\lambda\mu\nu}=\frac{\partial X^\lambda}{\partial(\partial_\mu\phi)}\partial^\nu\phi -\frac{\partial X^\mu}{\partial(\partial_\lambda\phi)}\partial^\nu\phi.\tag{3}$$ Is the above answer right? I'm not able to get this result! any help?

## closed as off-topic by AccidentalFourierTransform, DilithiumMatrix, ACuriousMind♦, Asher, CuriousOneMay 25 '16 at 21:56

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• $\uparrow$ Which book? – Qmechanic May 25 '16 at 11:17
• actually my friend gave me this result I'm not able to find it in any text book – jane May 25 '16 at 11:20

## 1 Answer

1. Let us assume that the new and old Lagrangian density $${\cal L} \quad\to\quad {\cal L} +\sum_{\mu=0}^3 d_{\mu} {\cal X}^{\mu}\tag{A}$$ does not depend explicitly on the spacetime point $x$. Then Noether's 1st theorem states that the canonical stress-energy-momentum (SEM) tensor density $${\cal T}^{\mu}{}_{\nu} ~:=~ \frac{\partial{\cal L}}{\partial( \partial_{\mu}\phi)}\partial_{\nu}\phi- \delta^{\mu}_{\nu}{\cal L}\tag{B}$$ is conserved on-shell $$d_{\mu}{\cal T}^{\mu}{}_{\nu}~\approx~0. \tag{C}$$ (In this answer we use calligraphic letters to denote densities.)

2. Note that OP's formula (3) suggests that the vector density ${\cal X}^{\mu}(\phi(x),\partial\phi(x))$ depends on spacetime derivatives $\partial\phi(x)$ of the field $\phi(x)$. This would imply that the Lagrangian density (A) depends on higher spacetime derivatives $\partial^2\phi(x)$. This in turn goes against common physics lore, cf. e.g. this Phys.SE post and links therein.

3. Let us for simplicity assume that the vector density ${\cal X}^{\mu}(\phi(x))$ does not depends on spacetime derivatives $\partial\phi(x)$. Then $$d_{\nu} {\cal X}^{\mu}~=~\frac{\partial {\cal X}^{\mu}(\phi)}{\partial\phi}\partial_{\nu}\phi.\tag{D}$$ And hence the canonical SEM tensor density (B) changes as $${\cal T}^{\mu}{}_{\nu}\quad\to\quad {\cal T}^{\mu}{}_{\nu}+ d_{\nu} {\cal X}^{\mu} -\delta^{\mu}_{\nu} d_{\lambda} {\cal X}^{\lambda} ~=~{\cal T}^{\mu}{}_{\nu}+d_{\lambda} {\cal F}^{\lambda\mu}{}_{\nu}, \tag{E}$$ where the improvement term is $${\cal F}^{\lambda\mu}{}_{\nu}~=~ \delta^{\lambda}_{\nu}{\cal X}^{\mu}-\delta^{\mu}_{\nu}{\cal X}^{\lambda}~=~ - {\cal F}^{\mu\lambda}{}_{\nu} \tag{F}.$$

4. In particular, the energy density $${\cal T}^0{}_0\quad\to\quad {\cal T}^0{}_0 - \sum_{i=1}^3 d_i {\cal X}^i \tag{G}$$ changes with a spatial 3-divergence, so that the energy $${\cal E} ~=~\pm \int_V \! d^3x ~ {\cal T}^0{}_0 \tag{H}$$ is unchanged under normal circumstances by the divergence theorem. Here we have used the Minkowski sign convention $(\pm,\mp,\mp,\mp)$.