How does Weinberg conclude that momentum and angular momentum are unperturbed by interaction terms?

Here's an argument to backup Weinberg's "assumption." In particular, I will show that the energy $P_0$ is changed by the presence of a potential, while the linear momentum $P_i$ and angular momentum $L_i$ are not, for a scalar field theory with a potential. I'll also consider a case below where $P_i$ is affected by the presence of a peculiar kind of interaction.

We can use the following definition to compute stress-energy tensor: \begin{equation} T_{\mu\nu} = -\frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu}}\Big|_{g_{\mu\nu}=\eta_{\mu\nu}} \end{equation}

Using the stress energy tensor, we can compute

• the energy, $P=T_{00}$,
• the linear momentum, $P_i = T_{0i}$,
• and the angular momentum, $L_i = \frac{1}{2} \epsilon_{ijk} (x^j T^{0k} - x^k T^{0j})$

Let's consider a scalar field with a potential. Let's write the action for this. I'm intentionally writing it in a way where we can take functional derivatives with respect to $g^{\mu\nu}$ easily, but in the end we are interested in Minkowski space, $g_{\mu\nu}=\eta_{\mu\nu}$. I'm using a mostly plus metric convention, $\eta_{\mu\nu}={\rm diag}(-1,1,1,1)$. \begin{equation} S = \int {\rm d}^4 x \sqrt{-g} \left(-\frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi - V(\phi)\right) \end{equation} Taking a variation $g_{\mu\nu}\rightarrow g_{\mu\nu}+\delta g_{\mu\nu}$, we obtain \begin{equation} \delta S = \int {\rm d}^4 x \sqrt{-g} \left( -\frac{1}{2} \partial_\mu \phi \partial_\nu \phi + \frac{1}{4} g_{\mu\nu} (\partial \phi)^2+\frac{1}{2} g_{\mu\nu} V(\phi)\right) \delta g^{\mu\nu} \end{equation} so \begin{equation} T_{\mu\nu} = -\frac{2}{\sqrt{-g}} \frac{\delta S}{\delta g^{\mu\nu}}\Big|_{g_{\mu\nu}=\eta_{\mu\nu}} = \partial_\mu \phi \partial_\nu \phi - \frac{1}{2} \eta_{\mu\nu}(\partial_\alpha \phi)^2 - \eta_{\mu\nu} V(\phi) \end{equation}

Now let's consider different cases to verify Weinberg's "assumption".

• The energy is $P=T_{00}=\frac{1}{2} \dot{\phi}^2 + \frac{1}{2} (\partial_i \phi)^2 + V(\phi)$, and depends on the potential, where a dot refers to a time derivative and $\partial_i$ is a spatial derivative.
• The linear momentum is $P_i = T_{0i} = \frac{1}{2} \dot{\phi}\partial_i \phi$, and does not depend on the potential.
• The angular momentum is $L_i = \frac{1}{2} \epsilon_{ijk} (x^j P^k - x^k P^j)$ also does not depend on the potential, since the linear momentum does not.

Finally, in the interest of seeing a situation where this assumption can break down, we can consider a case where interactions will affect the linear momentum and angular momentum operators:

\begin{equation} S = \int {\rm d}^4 x \sqrt{-g} \left(-\frac{1}{2} g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi + \frac{c}{\Lambda^3} (\partial \phi)^2 \square \phi \right) \end{equation} where $c$ is a dimensionless coupling parameter and $\Lambda$ is a scale with units of energy, and this interaction is a so-called "cubic Galileon" interaction.

If we work out the stress energy tensor, we find there is a new term (please double check the signs and coefficients if you actually use this somewhere): \begin{equation} T_{\mu\nu} = T_{\mu\nu}^{(c=0)} -\frac{2c}{\Lambda^3} \left[ \left(\partial_\mu \phi \partial_\nu \phi - \frac{1}{2} \eta_{\mu\nu}(\partial_\alpha \phi)^2\right) \square \phi + (\partial \phi)^2 \partial_\mu \partial_\nu \phi \right] \end{equation} where $T^{\mu\nu}_{(c=0)}$ is the stress energy tensor with $c=0$.

The last term will lead to a correction to the linear momentum \begin{equation} P_i = T_{0i} = P_i^{(c=0)} - \frac{2c}{\Lambda^3} \left(-\dot{\phi}^2 + (\partial_i \phi)^2\right) \partial_i \dot{\phi} \end{equation} Similarly, the angular momentum will also receive a correction.