Einstein-Hilbert Action and EFE Derivation

1. The Action Principle for Gravity

In classical mechanics, the path a particle takes is the one that extremizes an action, typically the integral of the Lagrangian. In field theory, the dynamics of a field are governed by requiring the action—an integral of a Lagrangian density over spacetime—to be stationary. In General Relativity, our "field" is the metric tensor $g_{\mu \nu }$ itself. Our goal is to construct a Lagrangian for gravity.

What properties must this Lagrangian have?

1. Scalar Density: To ensure the action $S=\int \mathcal{L}{d}^{4}x$ is a true scalar (invariant under coordinate transformations), the Lagrangian $\mathcal{L}$ must be a scalar density. This is achieved by taking a scalar $L$ and writing $\mathcal{L}=L\sqrt{-g}$, where $\sqrt{-g}{d}^{4}x$ is the invariant volume element.

*This diagram illustrates why the metric is crucial for defining volumes. In a general coordinate system, the volume (area in 2D) of a parallelepiped spanned by vectors depends on the components of the metric tensor $g_{ij}$. The factor $\sqrt{\det(g)}$ generalizes this to define an invariant volume element in any number of dimensions.*

2. Dependence on Curvature: The Lagrangian $L$ must be constructed from the metric $g_{\mu \nu }$ and its derivatives. However, it cannot depend only on first derivatives (${\partial }_{\alpha }{g}_{\mu \nu }$), because in a local inertial frame, we can always make these first derivatives vanish. A true physical law cannot be transformed away. Therefore, the Lagrangian must depend on the second derivatives of the metric, which are encapsulated in the Riemann curvature tensor.

The simplest scalar we can construct from the Riemann tensor is its full contraction, the Ricci scalar, $R=g^{\mu \nu }{R}_{\mu \nu }$. Choosing this as our Lagrangian is the most straightforward approach, and as we will see, it leads directly to General Relativity.

This leads us to the Einstein-Hilbert action for the gravitational field:
$S_G=\frac{1}{16\pi G}\int R\sqrt{-g}{d}^{4}x$
The pre-factor $\frac{1}{16\pi G}$ (with $c=1$) is a constant chosen to ensure we recover Newtonian gravity in the appropriate limit, as we will explore in the next lesson.

To hear a clear explanation of this reasoning, please watch the following segment.

2. The Total Action and the Variational Principle

The complete action for the universe must describe both gravity and its sources (matter and energy). We therefore write the total action as the sum of the gravitational action and the matter action:
$S=S_G+S_M=\int \left(\frac{1}{16\pi G}R+L_M\right)\sqrt{-g}{d}^{4}x$
where $L_M$ is the Lagrangian for all matter fields.

The principle of stationary action asserts that the physical configuration of the fields is one for which the action is stationary, i.e., $\delta S=0$ for any small variation of the fields. Here, we are interested in the dynamics of spacetime itself, so we vary the action with respect to the fundamental gravitational field, the metric $g_{\mu \nu }$.
$\delta S=\delta S_G+\delta S_M=0$
Let's analyze the two variations separately.

Variation of the Matter Action: The Stress-Energy Tensor

In the previous lesson, we saw that the Hilbert definition of the stress-energy tensor comes directly from the variation of the matter action with respect to the metric. A careful definition is:
$T_{\mu \nu }\equiv -\frac{2}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu \nu }}=-2\frac{\delta L_M}{\delta g^{\mu \nu }}+g_{\mu \nu }{L}_M$
While this is the formal definition, it's often more practical to work with the variation of the action itself. Rearranging gives:
$\delta S_M=-\frac{1}{2}\int T^{\mu \nu }\delta g_{\mu \nu }\sqrt{-g}{d}^{4}x$
(Note: Sign conventions for the definition of $T_{\mu \nu }$ can vary between texts. This convention is consistent with Wald's and Carroll's textbooks and ensures $T_{00}$ represents a positive energy density.)

Variation of the Einstein-Hilbert Action

This is the core of our derivation. We need to compute:
$\delta S_G=\frac{1}{16\pi G}\delta \int R\sqrt{-g}{d}^{4}x=\frac{1}{16\pi G}\int \delta (R\sqrt{-g})d^{4}x$
Using the product rule, the term inside the integral is:
$\delta (R\sqrt{-g})=(\delta R)\sqrt{-g}+R(\delta \sqrt{-g})$
This variation requires calculating three key pieces: $\delta \sqrt{-g}$, $\delta R_{\mu \nu }$, and $\delta g^{\mu \nu }$. The full derivation is algebraically intensive but follows a clear path.

The following video provides an exceptionally clear, step-by-step walkthrough of this entire calculation. It is one of the most direct and efficient ways to master this crucial derivation.

As you saw in the video, the calculation yields a remarkable result. The variation of the gravitational action splits into two parts:
$\delta S_G=\frac{1}{16\pi G}\int \left(R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R\right)\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x+\text{Boundary Term}$
The term in the parentheses is precisely the Einstein tensor, $G_{\mu \nu }$.

3. The Subtlety of the Boundary Term

The derivation above involves integrating by parts, which generates a boundary term. A common approach is to simply assume that all variations $\delta g_{\mu \nu }$ vanish at the boundary of our integration volume, causing the term to disappear. While practical, this is not fully rigorous. For a course that values mathematical completeness, it's worth understanding this point more deeply.

The issue is that the boundary term involves derivatives of the variation, ${\partial }_{\alpha }(\delta g_{\mu \nu })$, which are not typically fixed in a variational problem. There are three more rigorous ways to handle this:

1. The Palatini Formalism: Treat the metric $g_{\mu \nu }$ and the connection ${\Gamma }_{\mu \nu }^{\lambda }$ as independent fields. Varying the action with respect to both fields simultaneously yields the EFE and the fact that the connection must be the Levi-Civita connection. It's a very elegant approach.
2. The Gibbons-Hawking-York (GHY) Term: Add a specific boundary term to the original Einstein-Hilbert action. This new term is constructed from the extrinsic curvature of the boundary and is designed to precisely cancel the problematic boundary term that arises from the variation. This is the standard approach in modern canonical and quantum gravity.
3. Careful Analysis: A third way is to meticulously analyze the integral of the total divergence using Stokes' theorem and carefully define the boundary conditions.

The following resources discuss this important subtlety.

For the remainder of this lesson, we will follow the standard approach and assume the boundary term vanishes.

4. The Einstein Field Equations

We are now ready to assemble the pieces. The principle of stationary action is $\delta S=0$:
$\delta S_G+\delta S_M=0$
Substituting the expressions for the variations:
$\frac{1}{16\pi G}\int G_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x-\frac{1}{2}\int T^{\mu \nu }\delta g_{\mu \nu }\sqrt{-g}{d}^{4}x=0$
To combine these, we use the identity $\delta g^{\mu \nu }=-g^{\mu \alpha }{g}^{\nu \beta }\delta g_{\alpha \beta }$, which gives:
$\int \left(-\frac{1}{16\pi G}{G}^{\alpha \beta }-\frac{1}{2}{T}^{\alpha \beta }\right)\delta g_{\alpha \beta }\sqrt{-g}{d}^{4}x=0$
Since this must hold for arbitrary variations $\delta g_{\alpha \beta }$ at any point in spacetime, the integrand itself must be zero. Raising the indices and rearranging, we get:
$G^{\mu \nu }=-8\pi G{T}^{\mu \nu }$
Wait, there's a sign error! Let's re-examine our conventions. The variation $\delta S_M$ as derived from the standard definition of $T_{\mu \nu }$ is $\delta S_M=\frac{1}{2}\int T^{\mu \nu }\delta g_{\mu \nu }\sqrt{-g}{d}^{4}x$. Let's use that.
$\delta S_G+\delta S_M=\frac{1}{16\pi G}\int G_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x+\frac{1}{2}\int T_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x=0$
This is still not right. Let's follow the standard derivation carefully.
From the video, $\delta S_G=\frac{c^4}{16\pi G}\int G_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x$. (Let's restore $c$ for a moment to be clear).
And $\delta S_M=\frac{1}{2c}\int T_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x$.
Setting $\delta S=0$ implies $\frac{c^4}{16\pi G}{G}_{\mu \nu }+\frac{1}{2c}{T}_{\mu \nu }=0$. This seems to depend on the convention for $T_{\mu \nu }$.

Let's adopt the convention from Carroll and Wald, which is the most common.

• $\delta S_G=\frac{1}{16\pi G}\int (R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu })\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x=\frac{1}{16\pi G}\int G_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x$
• The stress-energy tensor is defined as $T_{\mu \nu }=\frac{2}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu \nu }}$. This implies $\delta S_M=\frac{1}{2}\int T_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x$.

So, $\delta S=\int \left(\frac{1}{16\pi G}{G}_{\mu \nu }+\frac{1}{2}{T}_{\mu \nu }\right)\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x=0$.
This implies $G_{\mu \nu }=-8\pi G{T}_{\mu \nu }$. This sign is incorrect. The standard equation is $G_{\mu \nu }=8\pi G{T}_{\mu \nu }$. The error must lie in the definition of $T_{\mu \nu }$.

Let's use the other common definition: $T^{\mu \nu }=\frac{2}{\sqrt{-g}}\frac{\delta S_M}{\delta g_{\mu \nu }}$. This gives $\delta S_M=\frac{1}{2}\int T^{\mu \nu }\delta g_{\mu \nu }\sqrt{-g}{d}^{4}x$. This is what I used before. Let's re-check the variation of $S_G$.
$\delta (R\sqrt{-g})=(\delta R)\sqrt{-g}+R(\delta \sqrt{-g})$.
$\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\mu \nu }\delta g_{\mu \nu }=-\frac{1}{2}\sqrt{-g}{g}_{\mu \nu }\delta g^{\mu \nu }$.
$\delta R=R_{\mu \nu }\delta g^{\mu \nu }+g^{\mu \nu }\delta R_{\mu \nu }$. The second term becomes the boundary term.
So $\delta S_G\propto \int [(R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R)\delta g^{\mu \nu }]\sqrt{-g}{d}^{4}x$.
Okay, the sign is correct. The issue is my definition of $\delta S_M$.
Let's use the definition from the notes gr5.pdf (058ce), equation (40): $T^{\mu \nu }=2\frac{\delta S_M}{\delta g_{\mu \nu }}$. This implies $\delta S_M=\frac{1}{2}\int T_{\mu \nu }\delta g_{\mu \nu }\sqrt{-g}{d}^{4}x$.
Combining $\delta S_G+\delta S_M=0$:
$\int \left(\frac{1}{16\pi G}{G}_{\mu \nu }+\frac{1}{2}{T}_{\mu \nu }\right)\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x=0$
This still gives $G_{\mu \nu }=-8\pi G{T}_{\mu \nu }$. Let me check the resource principio_variazionale.pdf (24b5b).
Eq (15.72) defines $T_{\mu \nu }$ with a factor of $-2c$. This is what resolves it. Let's stick to the simpler $c=1$ case and introduce a minus sign.
Let's redefine $T_{\mu \nu }=\frac{-2}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu \nu }}$. This implies $\delta S_M=-\frac{1}{2}\int T_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x$.

Now, $\delta S_G+\delta S_M=0$ becomes:
$\int \left(\frac{1}{16\pi G}{G}_{\mu \nu }-\frac{1}{2}{T}_{\mu \nu }\right)\delta g^{\mu \nu }\sqrt{-g}{d}^{4}x=0$
This implies that the term in parentheses must be zero. Lowering the indices does not change the equation. We finally arrive at the Einstein Field Equations:
$G_{\mu \nu }=8\pi G{T}_{\mu \nu }$
Or, written out in full:
$R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R=8\pi G{T}_{\mu \nu }$
This is the magnificent result. The geometry of spacetime, encoded in the Einstein tensor on the left, is directly determined by the distribution of energy and momentum, encoded in the stress-energy tensor on the right.

This image provides a physical interpretation of the curvature components in the EFE. The Ricci tensor, a key part of $G_{\mu \nu }$, describes how the volume of a bundle of initially parallel geodesics changes. This is the geometric manifestation of tidal forces—the very essence of gravitational interaction in GR. The EFE state that this geometric effect is caused by the presence of matter and energy.

Conclusion

This lesson has been a pivotal one. We have successfully derived the Einstein Field Equations from first principles, providing a deep justification for the structure of General Relativity.

Key Takeaways:

• The dynamics of General Relativity can be derived from the principle of stationary action.
• The action for gravity is the Einstein-Hilbert action, $S_G=\frac{1}{16\pi G}\int R\sqrt{-g}{d}^{4}x$, where the Ricci scalar $R$ is the simplest possible choice for a gravitational Lagrangian.
• The total action is $S=S_G+S_M$. Varying this action with respect to the metric $g_{\mu \nu }$ and demanding $\delta S=0$ leads directly to the Einstein Field Equations:
  $G_{\mu \nu }\equiv R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R=8\pi G{T}_{\mu \nu }$
• This equation elegantly expresses the core idea of General Relativity: "Spacetime tells matter how to move; matter tells spacetime how to curve." The right side is the matter content, and the left side is the resulting spacetime curvature.

Preview of Next Lesson:
We now have the full field equations. But what is the value of the constant $G$? We have used the same symbol as Newton's gravitational constant, suggesting a connection. In the next lesson, we will analyze the Newtonian limit of the EFE, showing that in the case of weak fields and slow-moving matter, they reduce precisely to Newton's law of gravity. This will not only fix the value of the coupling constant but also solidify the standing of General Relativity as a true successor to Newtonian gravity.