Einstein’s Equations

To redefine the theory of gravity, we want the metric to be determined by the distribution of matter and energy ⇒ Einstein’s Equations

In Newtonian gravity, $\vec{F}=-\frac{GMm}{r^2}{\vec{e}}_r$

Since this is a conservative force, $\vec{F}=-m\vec{\nabla }\varphi \text{with }\varphi =-\frac{GM}{r}$ We can think of this as a solution of Poisson’s Equation with a delta function source, ${\nabla }^2\varphi =4\pi GM{\delta }^3(\vec{x})$ In general, for a mass density ρ,

$$\nabla^2\phi=4\pi G\rho$$

In General Relativity, the role of this potential is played by the metric.

Consider the vacuum case, the Poisson’s Equation becomes Laplace’s Equation. Then, the Ricci tensor,

$R_{\mu \nu }=0$

Now, what about sources? In Newtonian theory, the source was a mass density → energy density in non-relativistic theory. In special relativity, the energy of a particle gets generalised to its four-momentum. That is,

$p^{\mu }={\int }_V{\rho }^{\mu }$

Thus, the relativistic generalisation of energy density is theStress-Energy tensor, a (2,0) tensor field which maps a covector to a vector as,

$T:n\to \rho {\rho }^{\mu }=T^{\mu \nu }{n}_{\nu }$

We assume it to be symmetric. Let us now understand the components of the tensor:

• Consider a surface t=0, so the normal has the first component as 1, rest 0. Hence,

${\rho }^{\mu }=T^{\mu 0};T^{00}-\text{energy density }\rho ;T^{i0}-\text{spatial momentum density }{\rho }^i$

The purely spatial components are the stress tensor.

Conservation of Energy: In a relativistic theory,

${\partial }_t\rho +{\partial }_i{\rho }^i=0$

which in integrated form translates to - the change in energy in a region is equal to the momentum flux through its boundaries. This is equivalent to saying that the change inStress-Energy tensor is zero. In a curved spacetime,

${\nabla }_{\mu }{T}^{\mu \nu }=0$

Physically, theChristoffel-symbols terms in this generalisation express the exchange of energy between matter fields and gravitational field.

Examples:

• Cosmological constant: Suppose

$T^{\mu \nu }=\Lambda g^{\mu \nu }$

This is automatically conserved since

${\nabla }_{\mu }{g}_{\rho \sigma }=0$

In QFT, vacuum fluctuations make such a constant contribution where the length scale is naturally the UV cutoff scale.

Observationally, there seems to be such a contribution where the length scale is the size of the observable universe - Cosmological Constant Problem.

• Electromagnetic field:

$T^{\mu \nu }=\frac{1}{4\pi }(F^{\mu \lambda }{F}_{\lambda }^{\nu }-\frac{1}{4}{g}^{\mu \nu }{F}^{\lambda \sigma }{F}_{\lambda \sigma })$

• Perfect fluid:

The stress tensor is a generalisation of energy density. So, it is a natural source for Einstein’s equations. A natural guess is $R_{\mu \nu }=\kappa T_{\mu \nu }$, but the LHS is not conserved. Hence, we replace the Ricci tensor by the Einstein tensor to obtain

$G_{\mu \nu }=R_{\mu \nu }-\frac{1}{2}{g}_{\mu \nu }R=8\pi G{T}_{\mu \nu }$

Remarks:

• LHS is conserved by theBianchi identity ⇒ stress-energy tensor is also conserved.
• Second-order non-linear partial differential equations for the metric.
• The constant is chosen to reproduce Newtonian theory for weak sources, nearly flat metric.