Proceeding like here, let us consider $N$ independent scalar fields which satisfy the Euler-Lagrange equations of motion and are denoted by $\phi^{(i)}(x) \ ( i = 1,...,N)$, and are extended in a region $\Omega$ in a $D$-dimensional model spacetime $\mathcal{M}_D$. Now consider the classical Lagrangian density, $\mathcal{L}(\phi^{(i)}, \partial_\mu \phi^{(i)}, x^\mu)$. We apply the following infintesimal fixed-boundary transformation to $\mathcal{M}_D$.

\begin{align*} x &\to \widetilde{x}^\mu \equiv x^\mu + \delta x^\mu (x), \tag{1} \\ \text{and the fields transform as: }\ \phi^{(i)}(x) &\to \widetilde{\phi}^{(i)}(\widetilde{x}) \equiv \phi^{(i)} (x) + \delta\phi^{(i)} (x). \tag{2} \\ \end{align*}

Accordingly, up to first order in the variation, the Lagrangian density is given by: $$ \boxed{ \delta \mathcal{L} = \partial_\mu \Big( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}(\delta\phi^{(i)} - \partial_\nu \phi^{(i)} \delta x^\nu) + \mathcal{L} \delta x^\mu \Big) - \mathcal{L} \partial_\mu (\delta x^\mu) } \,. \tag{3} $$

Now, following the conventional route to finding the energy-momentum tensor, consider that $\mathcal{L}$ does not explicitly depend on the spacetime coordinates and that the variation $\delta\phi^{(i)}$ is induced by the infinitesimal translation $\delta x^\mu \equiv \epsilon \ a^\mu(x)$ for some infinitesimal number $\epsilon > 0$, under which $$ \widetilde{\phi}^{(i)}(\widetilde{x}) = \phi^{(i)} (x + \epsilon \ a)=\phi^{(i)}(x) + \epsilon \ a^\mu \partial_\mu\phi^{(i)}(x) + \mathcal{O}(\epsilon^2) \,. \tag{4}$$

The variation $\delta\phi^{(i)}$ to first order in $\epsilon$, is therefore given by $$ \delta\phi^{(i)}=\epsilon \ a^\mu \partial_\mu\phi^{(i)}=\partial_\mu\phi^{(i)}\delta x^\mu \tag{5}$$

which means in conjunction with (3) that

$$ \delta \mathcal{L} = \partial_\mu \mathcal{L} \delta x^\mu=\epsilon \ a^\mu \partial_\mu\mathcal{L} = \epsilon \Big(\frac{\partial}{\partial \epsilon} \mathcal{L}\big(\widetilde{\phi}^{(i)}(\widetilde{x}), \widetilde{\partial}_\mu \widetilde{\phi}^{(i)}(\widetilde{x})\big)\Big)_{\epsilon=0}\,. \tag{6}$$

Eqn.(6) is tautological and does not give us any new information. I do not see how I can proceed to find the energy-momentum tensor following this route.


1 Answer 1


You are missing an important piece of assumption here, which is that you have to consider an active transformation of the fields in order to achieve your goal of finding the stress-energy tensor. In your notation, that is to say

$$\widetilde{\phi}^{(i)}(\widetilde{x}) = \phi^{(i)} (x) \,. \tag7$$

That means $ \, \delta \phi^{(i)} = 0 $, implying that the conserved Noether current is

$$ J^\mu = -\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )} \partial_\nu \phi^{(i)} \delta x^\nu + \mathcal{L} \delta x^\mu + F^\mu = {\Theta^\mu}_\nu \delta x^\nu + F^\mu \tag8$$


$$ {\Theta^\mu}_\nu \equiv -\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )} \partial_\nu \phi^{(i)} + {\delta^\mu}_\nu \mathcal{L} \tag9$$

with the property that

$$ (\partial_\mu {\Theta^\mu}_\nu)\delta x^\nu + {\Theta^\mu}_\nu \partial_\mu (\delta x^\nu) + \partial_\mu F^\mu = 0$$ $$\Rightarrow (\partial_\mu {\Theta^\mu}_\nu)\delta x^\nu + \partial_\mu F^\mu = 0 \,. \tag{10}$$

Note we have used that $\delta x^\nu$ is a constant shift in space-time coordinates.

This quantity $ {\Theta^\mu}_\nu $ is a possible candidate for the stress-energy tensor ${T^\mu}_\nu $. However, recall that an important feature of the latter is that it is symmetric. We can use the flexibility in the choice of the auxiliary field $F^\mu$ to construct the symmetric tensor we require. Pick the following field

$$ F^\mu \equiv {\Theta_\nu}^\mu \delta x^\nu \tag{11} $$

and define the stress-energy tensor by

$$ T^{\mu\nu} \equiv \Theta^{\mu\nu} + \Theta^{\nu\mu} \,.\tag{12} $$

Eqn.(10) now implies that

$$ \boxed{\partial_\mu T^{\mu\nu} = 0} \tag{13}$$


  1. For eqn.(11) to reflect a valid choice, one needs to assume proper asymptotic fall-off conditions for the fields, their derivatives and the Lagrangian. This is because the constant space-time shift $\delta x^\mu$ does not vanish at infinity, so that $\Theta^{\mu\nu}$ should compensate for this.

  2. You considered a passive transformation and that did not work out well because of very good reasons. The stress-energy tensor of a system is a physical quantity that remains conserved if there is a symmetry of the system with respect to space-time translations (recall from mechanics that time translation symmetry relates to energy conservation and space translation symmetry yields momentum conservation). Therefore, you need to physically transform the field configuration only. The transformed field configuration should relate to the original one by a space-time translation. This is different from keeping the actual field configuration fixed and transforming your coordinate system and observing how the fixed physical fields appear from your new coordinate system. That is not what you want to do. The physical principle is that your fields have a certain symmetry and therefore you necessarily need to transform your fields to probe the nature of the symmetry. Hence, active transformations are required.

  3. These notes are a great compilation of the fundamental physical principles required to understand the stress-momentum tensor.


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