# Stress-energy tensor for a massive particle in the electromagnetic field

I want to write down the expression for the stress-energy tensor for a massive particle in the electromagnetic field such that it can be split into two parts for the massive term and the field term, respectively: $$\begin{equation*} T^{\mu\nu} = T^{\mu\nu}_{(m)} + T^{\mu\nu}_{(f)} \,. \end{equation*}$$ Given the definition of the stress energy tensor $$\begin{equation*} T^{\mu\nu} = \sum_s \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi_s)} \partial^\nu \phi_s - g^{\mu\nu} \mathcal{L} \end{equation*}$$ and the Dirac-Lagrangian for a massive particle in the EM-field $$\begin{equation*} \mathcal{L} = \bar{\phi}(i\gamma^\mu D_\mu - \tfrac{mc}{\hbar})\phi - \tfrac{1}{4\mu_0} F_{\mu\nu} F^{\mu\nu} \,, \end{equation*}$$ where $$D_\mu = \partial_\mu + i\tfrac{e}{\hbar}A_\mu$$ is the covariant derivative, $$F^{\mu\nu}$$ the EM-field tensor and for $$\phi_s$$ we have $$\phi_1 = \phi \,, \ \phi_2 = \bar{\phi} = \phi^\dagger \gamma^0$$, I get \begin{align*} T^{\mu\nu} &= \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial^\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})} \partial^\nu \bar{\phi}- g^{\mu\nu} \mathcal{L} \\ &= \bar{\phi} i\gamma^\mu \partial^\nu \phi - g^{\mu\nu} \left[ \bar{\phi}(i\gamma^\alpha D_\alpha - \tfrac{mc}{\hbar})\phi - \tfrac{1}{4\mu_0} F_{\alpha\beta} F^{\alpha\beta} \right] \,. \end{align*} Now first I am not sure if this result is correct since this topic is new to me and I can't find a result to compare it with. Second, if this is the correct result, I am not sure which terms belong to the massive and which to the field term of the tensor. I guess the first term should be part of the massive term and the last term containing the EM-field tensor should belong to the field term, but I am uncertain what to do with the term containing the covariant derivative. Any input or help would be greatly appreciated!

Edit: notation

• Analogy: the pressure of a mixture of two non-interacting gases is the sum of partial pressures. This is not true for interactiong gases (note that pressure can be seen as the "average diagonal" of the energy-momentum tensor). Message: for interacting fields you can not split the energy momentum such that each part depends only on a single field. Oct 16, 2023 at 16:16
• Related: physics.stackexchange.com/q/119838/2451 and links therein. Oct 16, 2023 at 16:28
• @Quillo this is a good analogy, thank you. If I understand correctly, since the charged massive particles interact with the EM-field the separation is not possible. But is the result itself the correct one? I find that for the Dirac Lagrangian without the EM-field some terms cancel such that the term multiplied with the metric vanishes and if something like this can be done here. Oct 17, 2023 at 10:32
• @PascalS. I didn't check the algebra. True that rarely people write down the full energy-momentum. What conventions are you using? (Metric, gammas...) Are you following the notation of a reference book? Maybe to check your algebra Cadabra could be useful: physics.stackexchange.com/q/222/226902 Oct 17, 2023 at 10:34
• @Quillo I just follow the notation of the lecture I took on this topic. As for the metric and gammas I use (+---) and the Dirac representation. Oct 17, 2023 at 11:12

I am quite sure I have found the correct result now. My mistake was to not include the contribution of $$A_\mu$$ to the stress-energy tensor. I must have overlooked it since it is not explicitly written in the Lagrangian, but of course it is implicitly part of $$D_\mu$$ and $$F_{\mu\nu}$$. So actually we have a Lagrangian $$\mathcal{L} = \mathcal{L}(\phi,\bar{\phi},A_\mu,...)$$ such that $$\begin{equation*} T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial^\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})} \partial^\nu \bar{\phi} + \frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\sigma)} \partial^\nu A_\sigma - g^{\mu\nu} \mathcal{L} \,. \end{equation*}$$ Let's calculate the terms separately. The first two terms give us the well known result from the Dirac Lagrangian: $$\begin{equation*} \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial^\nu \phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \bar{\phi})} \partial^\nu \bar{\phi} = \bar{\phi} i \gamma^\mu \partial^\nu \phi = T^{\mu\nu}_{\mathrm{Dirac}} \,. \end{equation*}$$ The next term also gives a known result for the EM-field: $$\begin{equation*} \frac{\partial \mathcal{L}}{\partial (\partial_\mu A_\sigma)} \partial^\nu A_\sigma = - \tfrac{1}{\mu_0} F^{\mu\sigma} \partial^\nu A_\sigma \,. \end{equation*}$$ For the last term the contribution from the Dirac Lagrangian vanishes because the spinor field $$\phi$$ has to fulfill Dirac's equation such that only the contribution from the EM-Lagrangian is left: $$\begin{equation*} g^{\mu\nu} \mathcal{L} = - \tfrac{1}{4\mu_0} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} \,. \end{equation*}$$ In total we now have $$\begin{equation*} T^{\mu\nu} = \bar{\phi} i \gamma^\mu \partial^\nu \phi - \tfrac{1}{\mu_0} F^{\mu\sigma} \partial^\nu A_\sigma + \tfrac{1}{4\mu_0} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} \,. \end{equation*}$$ This stress-energy tensor is not symmetric but can be symmetrized by the addition of divergence-free fields to the Lagrangian. For the Dirac contribution we add $$\begin{equation*} f^\mu = -\tfrac{i}{2} \bar{\phi} \gamma^\mu \phi \end{equation*}$$ and for the EM contribution we choose $$\begin{equation*} C^{\mu\nu} = F^{\mu\sigma} \partial_\sigma A^\nu \end{equation*}$$ such that $$\begin{equation*} \mathcal{L} \rightarrow \mathcal{L} + \partial_\mu f^\mu + \partial_\mu C^{\mu\nu} \,. \end{equation*}$$ For further information see Energy-Momentum Tensor for the Electromagnetic Feild and Energy-momentum tensor of transformed Dirac Lagrangian. Finally we have $$\begin{equation*} T^{\mu\nu} = \underbrace{\tfrac{i}{2} \left( \bar{\phi} \gamma^\mu \partial^\nu \phi - \phi \gamma^\mu \partial^\nu \bar{\phi} \right)}_{=T^{\mu\nu}_{(m)}(\phi,\bar{\phi})} - \underbrace{g_{\sigma\lambda} F^{\mu\sigma} F^{\lambda \nu} + \tfrac{1}{4\mu_0} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta}}_{=T^{\mu\nu}_{(f)}(A)} \,. \end{equation*}$$ I hope I have done everything correctly and am open to further discussion.