I can find a lot of references which treat the derivation of Maxwell equations and the associated Energy-Stress Tensor from the action principle. But I cannot find any information on the Energy-Stress Tensor for electromagnetic fields with sources (classical treatment). Is it not possible or something like that?

The Lagrange density is obviously given by: $$\mathcal{L}=\frac{1}{c}A_\mu j^\mu+\frac{1}{16\pi}F^{\mu\nu}F_{\mu\nu}$$ where $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$

  • $\begingroup$ Hint: Use the definition of the metric/Hilbert SEM tensor. $\endgroup$
    – Qmechanic
    Nov 9, 2018 at 17:33
  • $\begingroup$ Do you mean you wish to derive Maxwell's equations with an external source without the means of an action principle? $\endgroup$ Nov 9, 2018 at 17:37
  • $\begingroup$ No. Actually I just want the Energy Stress Tensor for Electromagnetism with sources derived from the action. $\endgroup$
    – eeqesri
    Nov 9, 2018 at 18:06
  • $\begingroup$ Refer to 2.8 (page 83) of T. Padmanabhan's "Gravitation: Foundations and Frontiers" for a full treatment (or better yet, try it yourself!). $\endgroup$ Nov 9, 2018 at 18:39
  • $\begingroup$ Use Noether’s theorem. $\endgroup$
    – my2cts
    Aug 28, 2023 at 10:48

2 Answers 2


For an external source $j^\mu$, we could define the stress-energy tensor in the usual gauge-invariant way as $$ T^{\mu\nu}(x)\sim \frac{1}{\sqrt{|g|}}\,\frac{\delta S}{\delta g_{\mu\nu}(x)} \tag{1} $$ using the action $$ S\sim\int d^4x\ \sqrt{|g|} \left(F_{\mu\nu}F^{\mu\nu} +A_\mu j^\mu \right). $$ with a generic metric field $g_{\mu\nu}$. (I'm not worrying about the coefficients here because those details aren't important to the question.) To keep the action gauge-invariant, the external source should satisfy $$ \partial_\mu \, \sqrt{|g|}\, j^\mu=0. \tag{2} $$ If the current is due to another dynamic field rather than being imposed externally, then we can use the same approach by including that other dynamic field in the action. For example, we could consider the system with Lagrangian $$ L\sim F_{\mu\nu}F^{\mu\nu}+ (D_\mu\phi)^*(D^\mu\phi), $$ where $\phi$ is a scalar field and $D_\mu\sim\partial_\mu+iA_\mu$. Then we can use equation (1) again to derive the stress-energy tensor, which will now depend on both fields $A_\mu$ and $\phi$. This is the stress-energy tensor that belongs in the usual Einstein field equation $R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R\sim T_{\mu\nu}$, and it is conserved in the sense that $\nabla_\mu T^{\mu\nu}=0$. To relate the scalar field $\phi$ to the current $j^\mu$, we can write the equation of motion for the gauge field $A_\mu$ as $$ \frac{\delta S}{\delta A_\mu}=0, $$ which can be written in the form $$ \partial_\mu F^{\mu\nu}\sim j^\nu $$ (in the simplest case of a flat metric). This defines the current $j^\nu$ in terms of the scalar field $\phi$.

We could also define the stress-energy tensor by appealing to Noether's theorem, but then we're left with the extra step of figuring out how to make it gauge-invariant without disrupting its conservation. I used the metric-based definition here because it comes out gauge-invariant automatically — as long as the external current, if any, satisfies equation (2).


The approach outlined above used an arbitrary (variable) metric. This is necessary in order to define the variation of the action with respect to the metric. After the variation is computed, we can set the metric to be anything we want, such as the Minkowski metric.

But what's the justification for this? If we only care about the flat-spacetime version, then why should we need to temporarily consider arbitrary metrics?

The usual motives for considering the stress-energy tensor are (1) it is conserved ($\nabla_a T^{ab}=0$), and (2) it shows up in the Einstein field equation. If we're not doing general relativity, then motive #2 doesn't apply, but motive #1 still applies. We can think of the flat-space model is just one member of a family of models with different background metrics, and this whole set of models is invariant under diffeomorphisms even though the individual models (each with a specific metric) are not. This "collective" version of diffeomorphism invariance is sufficient for deriving the conservation law $\nabla_a T^{ab}=0$, as long as we start with an action that is (collectively) invariant under diffeomorphisms. This conservation law holds with any background metric, including flat spacetime. The generality of this result justifies thinking of $T^{ab}$ as something that every model "has", just like the generality of Noether's theorem justifies thinking of those conserved quantities as things that every model "has" (if enough symmetry is present).

But then why is the $T^{ab}$ we get from the metric-varying recipe consistent with the $T^{ab}$ we get from Noether's theorem in flat spacetime? Noether's therem seems unrelated to the metric-varying recipe. I don't have a clear understanding of this connection yet, but maybe it's related to the fact that when we use Noether's theorem to define $T^{ab}$ as a conserved quantity associated with the symmetries of flat spacetime, we are still relying on a mathematical symmetry — not the full diffeomorphism group, but part of it. I would like to understand this connection more clearly.

  • $\begingroup$ thank you for your answer. I have one doubt though, What if I am in flat space? Then the metric would just be the Minkowski metric. How to take the variation then as you indicated in (1). $\endgroup$
    – eeqesri
    Nov 9, 2018 at 18:42
  • $\begingroup$ @user139383 I added an appendix to my answer to try to address your comment. The basic idea is that we need to think of the metric as variable in order to calculate the variation of the action, but after that calculation is done, we can choose the metric to be whatever we want. The conservation law $\nabla_a T^{ab}=0$ follows automatically from a kind of "collective" diffeomorphism invariance of the action (explained in the appendix), and the resulting conservation law holds for an arbitrary metric, including the flat metric. $\endgroup$ Nov 9, 2018 at 20:12

Since the "Lagrangian density" is actually a density, technically it has a factor of $\sqrt{|g|}$ in it, where $g$ is the determinant of the metric. To make it consistent with your definition, we'll assume the metric $g_{μν}$ is constant, that it has a determinant $g = 1$, that the inverse of the metric has components $g^{μν}$. Your Lagrangian density is: $$\begin{align} 𝔏 &= \sqrt{|g|}\left(\frac 1 c A_μ j^μ + \frac 1 {16π} F_{μν} g^{μρ} g^{νσ} F_{ρσ}\right) \\ &= \frac 1 c A_μ j^μ + \frac 1 {16π} F_{μν} F^{μν} \end{align}$$ where you're using the convention $$F^{μν} = g^{μρ} g^{νσ} F_{ρσ}.$$

The canonical stress tensor density associated with a Lagrangian density $𝔏(q,v)$ that is a function of a field with components $q^A$ and gradients $v^A_μ = ∂_μ q^A$ is given by $$𝔓^ρ_ν = \frac{∂𝔏}{∂v^A_ρ} v^A_ν - δ^ρ_ν 𝔏.$$ Here, we use $𝔓$ to denote the tensor density, reserving $𝔗$ for the adjusted tensor density defined later below. When this is applied to the electromagnetic field, the role of the $q$'s is played by the potential $A_μ$, with $()_μ$ playing the role of $()^A$. Thus, when applied to the potential, $𝔓$ takes on the form: $$𝔓^ρ_ν = \frac{∂𝔏}{∂\left(∂_ρ A_μ\right)} ∂_ν A_μ - δ^ρ_ν 𝔏.$$

The response fields are densities defined by the respective derivatives, $$𝔊^{μν} = -\frac{∂𝔏}{∂F_{μν}} = -\frac1{4π} F^{μν}, \hspace 1em 𝔍^μ = \frac{∂𝔏}{∂A_μ} = \frac 1 c j^μ,$$ with the way you have the Lagrangian density set up, and satisfy the Euler-Lagrange equations: $$∂_ν 𝔊^{μν} = 𝔍^μ.$$ Note the double-counting for $𝔊^{μν}$ in the tabulation of the derivative with respect to $F_{μν}$, that occurs because $F_{μν} = -F_{νμ}$. So, your Euler-Lagrange equations are: $$-\frac1{4π} ∂_ν F^{μν} = \frac 1 c j^μ.$$ These are actually tensor densities involved here, since there's an implied factor of $\sqrt{|g|}$.

For consistency, the continuity equations for the source $$∂_μ 𝔍^μ = 0$$ must hold, since they are implied by the Euler-Lagrange equations. This means your source must satisfy the continuity equation $∂_μ j^μ = 0$.

In terms of the response field, we have $$\frac{∂𝔏}{∂\left(∂_ρ A_μ\right)} = -𝔊^{ρμ},$$ thus: $$𝔓^ρ_ν = -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏.$$

The stress tensor density $𝔗$ is an adjustment of $𝔓$ defined by: $$𝔗^ρ_ν = 𝔓^ρ_ν + ∂_μ 𝔭^{ρμ}_ν,$$ for a suitably-defined tensor density $𝔭$ that is anti-symmetric in its upper indices $𝔭^{ρμ}_ν = -𝔭^{μρ}_ν$. The anti-symmetry ensures that that continuity equation applies equally to both stress tensors, by ensuring that they have the same divergence: $$∂_ρ 𝔗^ρ_ν = ∂_ρ 𝔓^ρ_ν.$$ The relevant adjustment is termed the Belinfante correction and is given by: $$𝔭^{ρμ}_ν = 𝔊^{ρμ} A_ν.$$ Although it is not the defining requirement that it do so, when it's combined with the canonical stress tensor it yields a stress tensor where the functional dependence on the gradient of $A$ reduces to a functional dependence on the field strength $F$.

When evaluated with the source term $j$, using the Euler-Lagrange equation on the response field, we get: $$\begin{align} 𝔗^ρ_ν &= 𝔓^ρ_ν + ∂_μ 𝔭^{ρμ}_ν \\ &= -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏 + ∂_μ \left(𝔊^{ρμ} A_ν\right) \\ &= -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏 + ∂_μ 𝔊^{ρμ} A_ν + 𝔊^{ρμ} ∂_μ A_ν \\ &= 𝔊^{ρμ} F_{μν} + 𝔍^ρ A_ν - δ^ρ_ν 𝔏, \end{align}$$ using your equation $F_{μν} = ∂_μ A_ν - ∂_ν A_μ$ for the field strength. Applied to your Lagrangian density and field strengths, this yields the following: $$𝔗^ρ_ν = -\frac1{4π} F^{ρμ} F_{μν} + \frac 1 c j^ρ A_ν - δ^ρ_ν \left(\frac 1 c A_μ j^μ + \frac 1 {16π} F_{μσ} F^{μσ}\right).$$

This is not the whole story. The Belinfante correction is supposed to give you a symmetric stress tensor, which (here) means $g_{μρ} 𝔗^ρ_ν = g_{νρ} 𝔗^ρ_μ$. But, actually, you don't exact conservation laws at all because with an external source $j$, there is explicit coordinate-dependence for the Lagrangian density $𝔏$ through the external source $j$ expressed as a function of the coordinates.


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