Spacetime and Relativity Summary

Fundamental Postulates and the Lorentz Transformation

Galilean relativity assumes time is invariant across different inertial reference frames. Special Relativity, however, assumes the speed of light $c$ is invariant in all inertial reference frames. It also assumes that spacetime is flat and isotropic, and the laws of physics are identical everywhere, implying no reference frame is absolute or special.

From these postulates, we derive the Lorentz transformation, which defines the coordinate mapping from a rest frame to a moving reference frame. For a frame moving in the $x$-direction with velocity $v$:

$$\begin{bmatrix}ct'\\x'\end{bmatrix}=\begin{bmatrix}\gamma&-\gamma\beta\\-\gamma\beta&\gamma\end{bmatrix}\begin{bmatrix}ct\\x\end{bmatrix}$$

where $\beta=v/c$ and $\gamma=1/\sqrt{1-\beta^2}$.

By symmetry, the inverse transformation flips the sign of $\beta$:

$$\begin{bmatrix}ct'\\x'\end{bmatrix}=\begin{bmatrix}\gamma&+\gamma\beta\\+\gamma\beta&\gamma\end{bmatrix}\begin{bmatrix}ct\\x\end{bmatrix}$$

Transformations for basis vectors and components are complementary. Components transform contravariantly, while basis vectors transform covariantly. An object's rest frame always measures the maximum spatial length (proper length $L_0$) and the minimum elapsed time interval between two co-located events (proper time $\tau$).

Deriving the Lorentz Transformation

Because the transformation must be linear, the equation relating position $x'$ in $S'$ to coordinates $(x, t)$ in $S$ must take the form:

$$x' = A x + B t$$

Let's look at the origin of frame $S'$. By definition, its coordinate in $S'$ is always $x'=0$. In frame $S$, this origin is moving at velocity $v$, so its position is $x=vt$. Substituting these into the linear equation gives:

$$0 = A(vt) + Bt \implies B = -Av$$

Substituting $B$ back into our linear equation, we get:

$$x' = A(x - vt)$$

By the Principle of Relativity, neither frame is special. From the perspective of $S'$, frame $S$ is moving in the negative $x'$-direction at velocity $-v$. Because space is isotropic, the scaling factor $A$ must be identical for both frames. We will call this constant $\gamma$.

This gives us a symmetric pair of transformation equations:

$$x' = \gamma(x - vt)$$

$$x = \gamma(x' + vt')$$

To find $\gamma$, we invoke the second postulate. Imagine a flash of light is emitted at the exact moment the origins coincide ($t=t'=0$).

Because the speed of light is $c$ in both frames, the position of the light wavefront along the $x$-axis must satisfy: In frame $S$:

$$x = ct$$

In frame $S'$:

$$x' = ct'$$

Substitute these expressions for $x$ and $x'$ into our symmetric transformation equations:

$$ct' = \gamma(ct - vt) = \gamma t(c - v)$$

$$ct = \gamma(ct' + vt') = \gamma t'(c + v)$$

Now, multiply these two equations together:

$$c^2 t t' = \gamma^2 t t' (c - v)(c + v)$$

Divide both sides by $t t'$ (assuming $t>0$) and expand the right side:

$$c^2 = \gamma^2 (c^2 - v^2)$$

Solve for $\gamma^2$:

$$\gamma^2 = \frac{c^2}{c^2 - v^2} = \frac{1}{1 - v^2/c^2}$$

Taking the positive root (since axes are aligned in the same direction), we obtain the Lorentz factor:

$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

We have the spatial transformation $x'$ and the factor $\gamma$. We now need to find how time transforms ($t'$). We start with the inverse spatial equation:

$$x = \gamma(x' + vt')$$

Substitute the expression for $x'$ ($x' = \gamma(x - vt)$) into this equation:

$$x = \gamma(\gamma(x - vt) + vt')$$

Expand and isolate the $t'$ term:

$$x = \gamma^2 x - \gamma^2 vt + \gamma v t'$$

$$\gamma v t' = x - \gamma^2 x + \gamma^2 vt$$

$$t' = \gamma t + \frac{x(1 - \gamma^2)}{\gamma v}$$

To simplify the fraction on the right, we express $(1 - \gamma^2)$ in terms of $v$ and $c$:

$$1 - \gamma^2 = 1 - \frac{1}{1 - v^2/c^2} = \frac{1 - v^2/c^2 - 1}{1 - v^2/c^2} = \frac{-v^2/c^2}{1 - v^2/c^2} = -\frac{v^2}{c^2}\gamma^2$$

Substitute this back into our equation for $t'$:

$$t' = \gamma t + \frac{x(-\frac{v^2}{c^2}\gamma^2)}{\gamma v}$$

$$t' = \gamma t - \gamma \frac{vx}{c^2}$$

$$t' = \gamma\left(t - \frac{vx}{c^2}\right)$$

Spacetime Diagrams and Causality

A coordinate frame can be visualized on a Minkowski diagram with $ct$ on the vertical axis and $x$ on the horizontal, forming a square coordinate grid. Transforming to a moving reference frame shifts these axes into a rhombus, bringing the $ct'$ and $x'$ axes closer to the 45° line like a pair of scissors.

Vectors symmetric about this 45° line are Minkowski orthogonal; their slopes relative to the spatial axis are reciprocals. This geometry ensures the spacetime area of the coordinate grid remains constant under transformation, and the 45° line strictly bisects the coordinate axes. This enforces the invariance of the speed of light (1 unit of distance per 1 unit of time).

The 45° line represents the light cone:

Timelike Events: Worldlines falling strictly within the light cone (closer to the $ct$ axis) represent timelike intervals. These events are causally connected. An inertial reference frame exists where the two events occur at the same spatial location.
Spacelike Events: Worldlines falling outside the light cone (closer to the $x$ axis) represent spacelike intervals. An inertial reference frame exists where these events occur simultaneously at different spatial locations.

Because no worldline can cross the light cone (as nothing propagates faster than $c$), a timelike, causal relationship can never become spacelike under any transformation. Cause and effect cannot be reversed. However, for spacelike intervals, changing the reference frame shifts the line of simultaneity. This means two events that are simultaneous in one frame may occur in different chronological orders in other reference frames. This relativity of simultaneity resolves seeming paradoxes, such as Bell’s spaceship paradox.

4-Vectors and Invariants

Using the Lorentz transformation, the spacetime interval squared $s^2=(ct)^2-x^2$ is invariant under coordinate changes. This geometry is encoded in the Minkowski metric tensor. Using the $(+, -, -, -)$ signature convention, the diagonal elements are $(1, -1, -1, -1)$, and all off-diagonal elements are 0.

Because proper time $\tau$ is an invariant scalar, differentiating the spacetime position 4-vector by $\tau$ yields the 4-velocity $U$:

$$U=\frac{d}{d\tau}(ct, x, y, z)=\gamma(c, u_x, u_y, u_z)$$

The invariant magnitude squared of the 4-velocity is constant: $U\cdot U=c^2$.

Multiplying $U$ by the invariant mass $m$ yields the 4-momentum $P$:

$$P=mU=(E/c, p_x, p_y, p_z)$$

The dot product $P\cdot P$ is invariant. Expanding this dot product yields:

$$P\cdot P=(E/c)^2-p^2=(mc)^2$$

Rearranging this gives Einstein’s energy-momentum relation:

$$E^2=(mc^2)^2+(pc)^2$$

When a particle is at rest ($p=0$), this reduces to $E=mc^2$. For massless particles traveling at $c$, proper time $\tau$ is undefined, invariant mass $m=0$, and momentum is derived via quantum relations ($p=h/\lambda$) or $E=pc$.

Differentiating 4-momentum or 4-velocity with respect to $\tau$ yields 4-force and 4-acceleration, respectively. Unlike 4-velocity, the spatial components of 4-acceleration and 4-force are not strictly parallel to their classical 3-vector counterparts.

Total 4-momentum $P$ is conserved in isolated systems. Because invariant mass $m$ is dictated by the magnitude of the total 4-momentum rather than a simple sum of scalars, the total invariant mass of a system can change during processes such as inelastic collisions, even while $P$ remains rigorously conserved.

Note that though the 4-position is invariant under coordinate change, it does vary as thr world line evolves with proper time. However, the 4-velocity, and therefore the 4-momentum, are conserved, which means they are unchanged not just under coordinate transformations, but also as the world line evolves.

Hyperbolic Formulation

The Lorentz transformation naturally maps to hyperbolic geometry, satisfying the constant hyperbolic curve $(ct)^2-x^2=\text{constant}$:

$$\begin{bmatrix}ct'\\x'\end{bmatrix}=\begin{bmatrix}\cosh\phi&-\sinh\phi\\-\sinh\phi&\cosh\phi\end{bmatrix}\begin{bmatrix}ct\\x\end{bmatrix}$$

Here, $\phi$ is the rapidity, defined by the relation $\tanh\phi=v/c=\beta$. This hyperbolic formulation allows relative velocities to be calculated via the simple addition of rapidities: $\phi_{AC}=\phi_{AB}+\phi_{BC}$, bypassing the non-linear velocity addition formula.

This formulation also means that under constant proper acceleration $a$, the rapidity $\phi$ changes linearly with respect to proper time $\tau$ according to $d\phi/d\tau=a/c$. Thus, the worldline of an object under constant proper acceleration (the acceleration measured by an accelerometer in the object's rest frame) can be evaluated by integrating the 4-velocity vector. This integration yields a worldline where the Lorentz transformation is applicable at every individual point to map to a Momentarily Comoving Inertial Frame (MCIF), but the global worldline traces a hyperbola.

Let us define the hyperbola of a worldline under constant proper acceleration $a$ such that it intersects the $ct=0$ axis of an inertial frame at the spatial coordinate $x=c^2/a$ (let us define this constant distance as $D$). This worldline will asymptotically approach, but never cross, the light cone emanating from the origin. Thus, the frame's coordinate velocity approaches $c$ but never reaches it. At the same time, any light worldline originating from the origin $(0,0)$ after a certain coordinate time will never intersect the object's worldline. From the perspective of the accelerating frame, light from these regions is infinitely redshifted because the coordinate speed of light drops to zero at the horizon (where the metric component $g_{00}$ becomes zero). This is analogous to what happens at the event horizon of a black hole, where an object falling toward the black hole appears to freeze at the horizon for a distant inertial observer.

The non-inertial frame's local basis vectors are $\vec{e}_t$, which is the tangent to the hyperbola at a given event (parallel to the 4-velocity), and the Minkowski orthogonal spatial vector $\vec{e}_x$. Because the global Minkowski 4-position vector $\vec{S}$ (originating from the focal center of the hyperbola) is orthogonal to the hyperbola's tangent, its projection onto the local time axis is exactly zero. Consequently, from the point of view of the non-inertial frame's instantaneous basis, the position vector from the origin is given entirely by its spatial component: $\vec{S}=\tilde{x}\vec{e}_x$, where $\tilde{x}$ is the invariant coordinate distance along the instantaneous $\vec{e}_x$ direction (specifically, $\tilde{x}=c^2/a$). Furthermore, because the 4-velocity has a strictly constant invariant magnitude, its derivative (the 4-acceleration) must be Minkowski orthogonal to it. Therefore, while the 4-velocity is aligned with $\vec{e}_t$, the 4-acceleration is entirely aligned with $\vec{e}_x$.

Consider a scenario where two reference frames begin accelerating at the same time and with the same proper acceleration from the perspective of an inertial frame (Bell's Spaceship Paradox). In the inertial frame, the coordinate distance between them remains a constant $L$ throughout their worldlines. However, from the perspective of the rear non-inertial frame's MCIF, the proper distance between them continually grows. Because the rear frame's line of simultaneity (aligned with its $\vec{e}_x$ axis) tilts forward into the future as it accelerates, it intersects the forward frame's worldline at a progressively later proper time. Consequently, in the rear frame's instantaneous rest frame, the forward frame always appears to have accelerated for a longer duration, reaching a higher velocity and thus stretching the distance between them.

These hyperbolic trajectories form the basis of the Rindler coordinate system $(T, X)$, which maps spacetime for uniformly accelerating observers. For a family of hyperbolas intersecting the inertial axis at distances $D, 2D, 3D$, the associated proper accelerations are $c^2/D, c^2/2D, c^2/3D$.

The relationship transforming the Rindler coordinates $(T, X)$ (where the reference observer with acceleration $a$ is at $X=0$) to the global inertial coordinates $(t, x)$ is given by:

$$ct=\left(X+\frac{c^2}{a}\right)\sinh\left(\frac{aT}{c}\right)$$

$$x=\left(X+\frac{c^2}{a}\right)\cosh\left(\frac{aT}{c}\right)$$

In this non-inertial coordinate system, governed by the Rindler metric, inertial worldlines (straight lines in Minkowski space) map to curved trajectories, and light worldlines asymptotically approach the Rindler horizon at $X=-c^2/a$ as coordinate time $T \to \infty$. This demonstrates geometrically how the light cone boundary acts as an absolute horizon for the accelerating observer.

In the global inertial (Minkowski) frame, the accelerating object's spatial coordinate $x$ changes continuously with time $t$ as it traces its hyperbolic worldline. It is in constant motion, and its coordinate velocity $dx/dt$ asymptotically approaches $c$.

When we switch to the Rindler coordinate system $(T, X)$, we are mathematically forcing the coordinate grid to accelerate alongside the object. Depending on how you define the origin of the Rindler chart, the accelerating object is assigned a permanently fixed spatial coordinate:

Shifted Origin: The object is defined to sit exactly at $X=0$.
Horizon Origin: The object is defined to sit exactly at $X=c^2/a$.

In either convention, the spatial coordinate $X$ of the accelerating object is a constant constant scalar.

NOte that since the time basis in the Rindler coordinates is the tangent of the hyperbloic curve, which also matches the velocity vector, it means the spatial velocity of the object is 0 in that coordinate system, and therefore at rest:

$$\frac{dX}{dT}=0$$

While the object is kinematically at rest ($dX/dT=0$), the system remains a non-inertial frame. The physical reality of the acceleration is transferred from the object's motion into the geometry of the spacetime metric itself.

In the Rindler frame, an onboard accelerometer still measures a proper acceleration $a$. However, because the object is at rest in these coordinates, this measurement is not interpreted as kinematic acceleration, but rather as the object resisting a fictitious gravitational field. The spatial dependence of the Rindler metric time component, $g_{00}=(1+aX/c^2)^2$, acts mathematically identical to a uniform gravitational potential pulling everything toward the Rindler horizon. The object remains at rest at $X=0$ because it is "supported" against this fictitious field, much like a book resting stationary on a table in Earth's gravity.

Einstein Field Equations

Let us look at what the Riemann and Ricci tensors actually mean from the perspective of curvature. Take the Riemann tensor component $R^\rho_{\mu\nu\sigma}$. If you look at this equation in terms of the relative acceleration of the separation vector between geodesics, you get:

$$A^\rho = -R^\rho_{\mu\nu\sigma} V^\mu X^\nu V^\sigma$$

$A^\rho$ is the acceleration of the separation vector along the $\rho$ direction.
$V^\mu$ and $V^\sigma$ are the components of the 4-velocity of the reference frame in the $\mu$ and $\sigma$ directions.
$X^\nu$ is the separation vector between two adjacent geodesics in the $\nu$ direction.

The Riemann Tensor Components

A bunch of Riemann components will be strictly $0$ from mathematical symmetries, for example, $R^0_{000} = 0$.
$R^0_{101}$ represents an analysis of a beam of particles with a bulk velocity strictly along the $x$-axis (the 2nd and 4th indices are both $1$). We take a separation vector between a reference particle and a second particle infinitesimally close to it in the same frame, separated only by a time delay $X^0$ (the 3rd index). This component dictates how that time delay accelerates in the time direction $A^0$ (the 1st index). Physically, this represents the longitudinal stretching of the time interval (time dilation) between particles seperated by time but at the same spatial location such as between pulses in a beam.
$R^0_{102}$ evaluates how the time interval between two consecutive particles stretches or shrinks when the entire beam possesses a simultaneous velocity in both the $x$ and $y$ directions (y comes from the 4th index being 2).
$R^1_{010}$ evaluates the change in acceleration along the $x$-axis of a spatial separation between particles along the $x$-axis, when the particles have a velocity strictly in time (they are stationary relative to the source).
$R^\rho_{0\nu 0}$ represents the standard Newtonian tidal forces on a stationary swarm of objects, such as a cluster of particles dropped from rest above the Earth. Such tidal forces are the gravitational equivalent of the electric field $\vec{E}$, causing straightforward stretching and squeezing.
$R^\rho_{0\nu i}$ represents the tidal forces on an object possessing spatial velocity. These cross-terms manifest as relativistic effects like gravitomagnetism or frame-dragging.

The Ricci Tensor

$R_{11}$ is the trace (contraction) of the Riemann components with the same first and third index. It evaluates the spatial and temporal divergence when the particle swarm has strict $x$-velocity i.e. acceleration along the direction of the spatial seperation, when the particle has x velocity. So for eg. R^2_121 would be the y acceleration along the y direction. So R_11 becomes the contraction in the transverse plane (yz-plane or fixed x) and the longitudinal time dilation, as the particles travel along the x-axis. (Remember that R^1_111 is 0 so we don't take it into account in the summation)
$R_{12}$ represents the shear-stress equivalent of curvature. It dictates how motion along the $x$-axis and $y$-axis couple together to shear the transverse cross-section of the particle beam.
$R_{00}$ evaluates the pure 3D volume convergence or divergence of a stationary swarm of test particles dropped from rest. This is equivalent to the Energy density. Yes, several specific names from continuum mechanics and the gravito-electromagnetic analogy map directly to these tensor components. By categorizing them with these formal names, you directly connect the abstract geometry to the physical properties of the matter sourcing the gravitational field.
$R_{01}, R_{02}, R_{03}$ represent Momentum Density / Energy Flux. This specific Ricci component dictates the rotation of the local inertial frame, which is the direct source of Frame-Dragging (the Lense-Thirring effect).
$R_{11}, R_{22}, R_{33}$ represent Normal Stress / Principal Pressure.
$R_{12}, R_{13}, R_{23}$ represent shear stress and evaluate how motion along one spatial axis couples with momentum in another. They map identically to the mechanical shear stress within the matter field (e.g., viscous fluid layers sliding past one another).

The Ricci Scalar

The Ricci scalar $R$ is the complete trace of the Ricci tensor across all time and spatial dimensions. It represents an invariant measure of the total scalar volume deviation of a 4D hyper-volume compared to flat Euclidean space.

The Stress-Energy-Momentum Tensor

The Stress-Energy-Momentum Tensor (often denoted as $T_{\mu\nu}$ or $T^{\mu\nu}$) is a fundamental ten-component symmetric quantity that describes the density and flux of energy and momentum in spacetime. It is the "source" term on the right-hand side of the Einstein Field Equations:

$$G_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

This dictates how matter and energy curve spacetime. It acts as the bridge between the geometry of spacetime (curvature) and the matter/energy content within it. It quantifies the "stuff" that causes spacetime to curve, and conversely, the curvature dictates how that "stuff" moves.

First, let us understand what momentum flux means. Flux is the quantity of something crossing a boundary over a unit time period. For example, we keep an $x$-coordinate constant (defining a $y$-$z$ plane) and count the object crossing it as time varies. We can generalize this to boundaries of constant $y$ and $z$. We can also generalize this to a boundary of constant time $t$, where we measure "flux" over a constant $t$ within a unit of $x, y, z$ volume. This temporal flux is essentially spatial density.

If we look at momentum flux across a boundary of constant $t$, this becomes momentum density (which is proportional to energy flux). If we look at momentum flux across a boundary of constant $x$, it evaluates the change of momentum over time. Using the product rule, $\frac{d(mv)}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt} = ma + v\frac{dm}{dt}$.

The right part of this sum ($v\frac{dm}{dt}$) represents how bulk mass flow transports momentum across point $x$ (advective flux).
The left part of the sum ($ma$) is the force being applied on the body of particles or fluid (stress flux). Both these mechanisms contribute to the total momentum flux.

The tensor in Einstein's equation is a generalization of this. $T^{\mu\nu}$ is the flux of the $p^\mu$ component of 4-momentum across a volume boundary of constant $x^\nu$ (equivalently, a surface with a normal vector in the $\nu$ direction). The tensor is a $4 \times 4$ matrix with 16 components, but because it is symmetric ($T^{\mu\nu} = T^{\nu\mu}$), it has exactly 10 independent components.

Physical Meaning of Each Element $T^{\mu\nu}$:

The $T^{00}$ component is the flux of $p^0$ (energy) across a surface of fixed $t$ (a spatial volume). This is energy density.
The $T^{0i}$ components represent the flux of energy across fixed spatial coordinates $x, y, z$. This is energy flux.
The $T^{i0}$ components represent the flux of spatial momentum across fixed $t$. This is momentum density. (Due to symmetry, energy flux equals momentum density).
The pure spatial diagonals ($T^{11}, T^{22}, T^{33}$) evaluate spatial momentum along axis $i$ across a fixed boundary $i$. This represents $\frac{\text{Force}}{\text{Area}}$, which is normal stress or pressure.
The spatial off-diagonals (e.g., $T^{12}$) are cross-terms accounting for shear stress.

Index	Meaning	Physical Interpretation	Analogy
$T^{00}$	Energy Density	Energy per unit volume at a specific point in spacetime.	Mass density (assuming rest mass dominates).
$T^{0i}$ or $T^{i0}$	Energy Flux / Momentum Density	$T^{0i}$: Energy flowing through unit area per unit time. $T^{i0}$: Momentum per unit volume.	Bulk fluid flowing in one direction carrying momentum.
$T^{ii}$ (Diagonal)	Normal Stress / Pressure	Flux of spatial momentum in the $i$-direction across a surface oriented in the $i$-direction.	Pressure exerted outward by a fluid or gas.
$T^{ij}$ (Off-Diagonal, $i \neq j$)	Shear Stress	Flux of spatial momentum in the $i$-direction across a surface oriented in the $j$-direction.	Friction or shear forces between layers of a fluid or solid.

We can tie these back to the left side of Einstein's Field Equation's involving the Ricci Tensor.

$R_{00}$ evaluates the pure 3D volume convergence or divergence of a stationary swarm of test particles dropped from rest. This is equivalent to the Energy density.
$R_{01}, R_{02}, R_{03}$ represent Momentum Density / Energy Flux. This specific Ricci component dictates the rotation of the local inertial frame, which is the direct source of Frame-Dragging (the Lense-Thirring effect).
$R_{11}, R_{22}, R_{33}$ represent Normal Stress / Principal Pressure.
$R_{12}, R_{13}, R_{23}$ represent shear stress and evaluate how motion along one spatial axis couples with momentum in another. They map identically to the mechanical shear stress within the matter field (e.g., viscous fluid layers sliding past one another).

We can take the 4-divergence of the energy-momentum tensor:

$$\nabla_\mu T^{\mu\nu} = 0$$

If we assume flat isotropic spacetime (Special Relativity), the covariant derivative reduces to a partial derivative. For the time component ($\nu = 0$), this expands to:

$$\frac{\partial T^{00}}{\partial (ct)} + \frac{\partial T^{10}}{\partial x} + \frac{\partial T^{20}}{\partial y} + \frac{\partial T^{30}}{\partial z} = 0$$

This translates to the time derivative of energy density plus the spatial divergence of energy flux equating to zero. This is the continuity equation for energy conservation.

In General Relativity, we cannot claim global energy conservation due to the dynamic nature of spacetime curvature. Instead of the partial derivative, we take the covariant derivative, which incorporates the geometric curvature via Christoffel symbols ($\Gamma^\mu_{\alpha\beta}$):

$$\nabla_\alpha T^{\mu\nu} = \partial_\alpha T^{\mu\nu} + \Gamma^\mu_{\alpha\beta} T^{\beta\nu} + \Gamma^\nu_{\alpha\beta} T^{\mu\beta} = 0$$

Setting this covariant divergence to zero mathematically enforces the local conservation of energy and momentum.

Stress-Energy Tensor of Stationary Dust

Consider a cloud of stationary dust particles that exert no pressure on each other. If $p^\mu$ is the 4-momentum of a single particle and $N^\nu$ is the 4-number flux of the dust cloud, the energy-momentum tensor is the tensor product:

$$T^{\mu\nu} = p^\mu \otimes N^\nu$$

The number flux is $N^\nu = n U^\nu$, where $n$ is the rest number density and $U^\nu$ is the 4-velocity. The 4-momentum is $p^\mu = m U^\mu$. Therefore:

$$T^{\mu\nu} = m n U^\mu U^\nu = \rho U^\mu U^\nu$$

where $\rho = mn$ is the mass density. Since the dust particles are stationary, only the time component of the 4-velocity is non-zero, with a value $U^0 = c$. Consequently, the tensor only possesses a single non-zero component:

$$T^{00} = \rho c^2$$

Einstein Field Equation

It would be great if we could just set the Ricci tensor as equal to the Stress-Energy tensor. Unfortunately the covariant divergence of the Ricci tensor is not zero whereas the covariant divergence of the Stress-Energy tensor is zero.

However, we can fix this by adding in the Ricci scalar. The standard formulation of the Einstein Field Equations, explicitly expanding the Einstein tensor $G_{\mu\nu}$, is written as:

$$R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

Divergence and the Cosmological Constant

The addition of the cosmological constant $\Lambda$ is mathematically justified by analyzing the covariant divergence of the tensors involved in the field equations.

The Einstein Tensor ($G_{\mu\nu}$): The covariant divergence of the geometric side of the equation vanishes identically as a fundamental requirement of Riemannian geometry, specifically due to the contracted Bianchi identities:

$$\nabla^\mu \left( R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} \right) = \nabla^\mu G_{\mu\nu} = 0$$

The Stress-Energy Tensor ($T_{\mu\nu}$): To enforce the local conservation of energy and momentum, the covariant divergence of the matter side of the equation must also be zero:

$$\nabla^\mu T_{\mu\nu} = 0$$

The Metric Tensor ($g_{\mu\nu}$): In General Relativity, the spacetime manifold utilizes the Levi-Civita connection, which is defined by metric compatibility. This mandates that the covariant derivative of the metric tensor is strictly zero everywhere:

$$\nabla^\mu g_{\mu\nu} = 0$$

Because the fundamental equivalence requires $\nabla^\mu G_{\mu\nu} \propto \nabla^\mu T_{\mu\nu} = 0$, any additional geometric terms introduced to the field equations must also possess a vanishing divergence to avoid violating the conservation laws.

Since the derivative of the metric tensor is zero ($\nabla^\mu g_{\mu\nu} = 0$), you can legally add a term proportional to the metric tensor—specifically $\Lambda g_{\mu\nu}$, where $\Lambda$ is a uniform scalar constant—to the left-hand side. The divergence of this new term is exactly zero, preserving the mathematical consistency of the system.

This yields the most general, complete form of the Einstein Field Equations:

$$R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

Where:

$R_{\mu\nu}$ is the Ricci curvature tensor.
$R$ is the Ricci scalar curvature.
$g_{\mu\nu}$ is the metric tensor.
$\Lambda$ is the cosmological constant.
$G$ is the Newtonian constant of gravitation.
$c$ is the speed of light in vacuum.
$T_{\mu\nu}$ is the stress-energy-momentum tensor.

The Trace-Reversed Form of the EFE

$$R_{\mu\nu} = \frac{8\pi G}{c^4} \left( T_{\mu\nu} - \frac{1}{2}T g_{\mu\nu} \right) + \Lambda g_{\mu\nu}$$

which is very similar to the above standard form.

To obtain the trace-reversed form of the Einstein Field Equations, we begin with the standard formulation including the cosmological constant $\Lambda$:

$$R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

The objective is to isolate the Ricci curvature tensor $R_{\mu\nu}$ entirely on the left side. To achieve this, we must determine the value of the Ricci scalar $R$ in terms of the stress-energy tensor. We do this by taking the trace of the entire equation.

Mathematically, taking the trace means contracting the equation with the inverse metric tensor $g^{\mu\nu}$:

$$g^{\mu\nu} \left( R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} + \Lambda g_{\mu\nu} \right) = g^{\mu\nu} \left( \frac{8\pi G}{c^4} T_{\mu\nu} \right)$$

Applying the contraction to each term yields the corresponding scalars:

$g^{\mu\nu} R_{\mu\nu} = R$ (The Ricci scalar)
$g^{\mu\nu} g_{\mu\nu} = 4$ (The trace of the metric tensor in 4-dimensional spacetime)
$g^{\mu\nu} T_{\mu\nu} = T$ (The trace of the stress-energy tensor, or the Laue scalar)

Substituting these trace values back into the equation:

$$R - \frac{1}{2}R(4) + \Lambda(4) = \frac{8\pi G}{c^4} T$$

$$R - 2R + 4\Lambda = \frac{8\pi G}{c^4} T$$

$$-R + 4\Lambda = \frac{8\pi G}{c^4} T$$

Solving for the Ricci scalar $R$:

$$R = 4\Lambda - \frac{8\pi G}{c^4} T$$

Substitution and Final Form

Now, substitute this expression for $R$ back into the original Einstein Field Equations:

$$R_{\mu\nu} - \frac{1}{2} \left( 4\Lambda - \frac{8\pi G}{c^4} T \right) g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

Distribute the $-\frac{1}{2} g_{\mu\nu}$:

$$R_{\mu\nu} - 2\Lambda g_{\mu\nu} + \frac{1}{2} \frac{8\pi G}{c^4} T g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

Combine the $\Lambda$ terms ($-2\Lambda + \Lambda = -\Lambda$) and isolate $R_{\mu\nu}$:

$$R_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu} - \frac{1}{2} \frac{8\pi G}{c^4} T g_{\mu\nu} + \Lambda g_{\mu\nu}$$

Factoring out the coupling constant yields the Trace-Reversed Einstein Field Equations with the cosmological constant: