# Coupling Hilbert action to a matter action

When coupling a matter action to the Hilbert action in general relativity, why does one simply add the two actions and not add a coupling action? In electromagnetism, a coupling term is needed between the matter action and the field action, so I’m not sure why such a term is not needed in general relativity. For example, say I want to write the action for two interacting charges in purely electromagnetic theory. Then I can write the field action coupled to two point charges along with the matter action which describes how the charges interact with the field. Then, to do the same thing in general relativity, I think I would just need to add a matter action which gives the right energy momentum tensor to the Hilbert action, but no coupling term.

• It's the same concept. The matter-field coupling for electromagnetism is not arbitrary but enters only through the gauge covariant derivative. So you do "simply add" the matter action in EM too, but using the covariant derivative. Commented Jan 19, 2021 at 6:02
• May be it is worth adding it is not that you "have" to, that is just the simplest case, so tends to draw more attention. But there are plenty of models with explicit coupling terms such as $R\phi$ Commented Jan 19, 2021 at 8:08

… why does one simply add the two actions and not add a coupling action?

One does add coupling action. When we replace the partial derivatives that are present in the matter action in flat spacetime in Cartesian coordinates with covariant derivatives and integrate using volume form of the curved spacetime (this approach is called minimal coupling), this is precisely the introduction of coupling.

This is most easily seen if we consider variation of the matter action with slight variation of metric: $$g_{μν}=g^{(0)}_{μν}+δg_{μν}$$, which has the form: $$\delta S_\text{matter}=\frac 12 \int T^{μν} δg_{μν} \sqrt{-g}\,d^4x,$$ where $$T_{μν}$$ is the stress–energy tensor. If we consider background spacetime having Minkowski metric $$g^{(0)}_{μν}=\eta_{μν}$$ then this variation of the matter action has the interpretation of coupling action for a “field” $$δg_{μν}$$ with matter via the “current” $$T^{μν}$$.

At the level of perturbation theory this is fully analogous to the coupling action of electromagnetism: $$S_\text{int}= - \int j^μ A_μ \sqrt{-g}\,d^4x.$$ But outside weak gravitational fields we must consider that GR is a nonlinear theory, and separation of sources and fields created by them is generally not possible.

Going in the opposite direction, we could “hide” coupling action of electromagnetism by introducing gauge covariant derivative for charged matter fields: $$\partial_μ \to D_μ = \partial_μ -i e A_μ.$$ By replacing all derivatives in kinetic terms with gauge covariant derivatives the action for charged matter would now contain interaction term.

Generally one should consider gauge invariance of electromagnetic theory and diffeomorphism invariance of GR not as just redundancy in description but as a builtin mechanism for introducing couplings between matter and EM or gravitational fields.

For example, say I want to write the action for two interacting charges in purely electromagnetic theory. Then I can write the field action coupled to two point charges along with the matter action which describes how the charges interact with the field. Then, to do the same thing in general relativity, I think I would just need to add a matter action which gives the right energy momentum tensor to the Hilbert action, but no coupling term.

In GR the notion of point particle is internally inconsistent: if we try to cram finite mass into sufficiently small volume this mass becomes a black hole, so while it is certainly possible to write equations of motion for test point particles consideration of the fields that they are creating requires some sort of approximation scheme.

So expanding the action within the perturbation theory the part where stress–energy tensor appears would be precisely the coupling term.