What are the Einstein field equations

[Greg Bernhardt]

Definition/Summary

The Einstein Field Equations are a set of ten differential equations which express the general theory of relativity mathematically: they relate the geometry (the curvature) of spacetime to the energy/matter content of spacetime.

These ten differential equations may be written as a single second-order (two-index) symmetric tensor equation, relating the Ricci curvature tensor $R_{\mu\nu}$ to the stress-energy tensor $T_{\mu\nu}$.

Equations

Short version (using Einstein tensor $G_{\mu\nu}$):

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

Using standard cosmological units with $G\ =\ c\ =\ 1$:

$$G_{\mu\nu}\ =\ 8\pi\,T_{\mu\nu}$$

Long version (using Ricci curvature tensor $R_{\mu\nu}$ and scalar curvature $R\ =\ Tr(R_{\mu\nu})$):

$$R_{\mu\nu}\ -\ \frac{1}{2}\,R\,g_{\mu\nu}\ =\ 8\pi\,T_{\mu\nu}$$ or $$T_{\mu\nu}-\ \frac{1}{2}\,T\,g_{\mu\nu}\ =\ \frac{1}{8\pi}\,R_{\mu\nu}$$

"Symmetric" decomposition, into scalar part:

$$R\ =\ -\,8\pi\,T$$

and traceless symmetric tensor part:

$$R_{\mu\nu}\ -\ \frac{1}{4}\,R\,g_{\mu\nu}\ =\ 8\pi\left(T_{\mu\nu}\ -\ \frac{1}{4}\,T\,g_{\mu\nu}\right)$$

Extended explanation

Cosmological units:

Cosmology is one of the few areas in which practitioners prefer not to use SI units.

Cosmological units are defined so that $G\ =\ c\ =\ 1$

Structure of the EFE:

A second-order (two-index) tensor equation is the simplest possible equation which could describe the relationship between curvature and matter/energy.

The only two-index tensor describing matter and energy is the symmetric stress-energy tensor, $T_{\mu\nu}$.

The only two-index tensors describing the structure of space are the symmetric Ricci curvature tensor $R_{\mu\nu}$ and the symmetric metric tensor $g_{\mu\nu}$.

Also available, as scalar multipliers, are the traces $R\ =\ Tr(R_{\mu\nu})$ and $T\ =\ Tr(T_{\mu\nu})$

The EFE is the only combination of these which, in the weak-field limit, gives the inverse-square law of Newtonian gravity.

A very small multiple of $g_{\mu\nu}$ may also be inserted into the EFE without noticeably affecting the weak-field limit: that multiple is the cosmological constant, $\Lambda$, whose value is estimated at less than $10^{-35}\,s^{-2}$

Trace and traceless:

A symmetric tensor has one scalar invariant: the trace.

By comparison, an anti-symmetric tensor has two scalar invariants, usually written in the form $E^2 - B^2$ and $\boldsymbol{E}\cdot\boldsymbol{B}$

A symmetric tensor equation can be split into two parts, a scalar trace equation, and a symmetric traceless tensor equation.

For the EFE, these show that (except for the factor $8\pi$):

trace of Ricci curvature equals minus trace of stress-energy,
but traceless Ricci curvature equals traceless stress-energy:​

$$Tr(R_{\mu\nu})\ =\ R\ =\ -\,8\pi\,T\ =\ -\,8\pi\,Tr(T_{\mu\nu})$$

$$Notr(R_{\mu\nu})\ =\ R_{\mu\nu}\ -\ \frac{1}{4}\,R\,g_{\mu\nu}\ =\ 8\pi\left(T_{\mu\nu}\ -\ \frac{1}{4}\,T\,g_{\mu\nu}\right)\ =\ 8\pi\,Notr(T_{\mu\nu})$$

The notation "Notr" is not a standard notation.

$Tr(A_{\mu\nu})$ is defined as $g^{\mu\nu}A_{\mu\nu}$

Note that (in a four-dimensional space) $Tr(g_{\mu\nu})\ =\ \frac{1}{4}$ and so $Tr\left(A_{\mu\nu}\ -\ \frac{1}{4}\,g_{\mu\nu}\,Tr(A_{\mu\nu})\right)\ =\ (1\ -\ \frac{1}{4}\,g_{\mu\nu})Tr(A_{\mu\nu})\ =\ 0$ and so it would be more convenient if trace were defined to be one-quarter of its standard definition.

The reason for the factor $8\pi$ is ultimately that [will someone please complete this paragraph? ]

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[Aaron Paulo]

The Einstein Field Equations are a set of ten differential equations which express the general theory of relativity mathematically. These equations relate the curvature of spacetime to the energy/matter content of spacetime, and can be written in two ways. The short version is expressed as a single symmetric tensor equation, with G_{\mu\nu} equaling 8\pi G/c^4 multiplied by T_{\mu\nu}. The long version is expressed as two equations: one scalar equation, R = -8\piT, and one traceless symmetric tensor equation, R_{\mu\nu} - (1/4)Rg_{\mu\nu} = 8\pi(T_{\mu\nu} - (1/4)Tg_{\mu\nu}). Cosmological units are used in cosmology, where G = c = 1. The trace of a symmetric tensor is a scalar invariant, and by splitting the equation into scalar and traceless parts, we can see that the trace of the Ricci curvature equals minus the trace of the stress-energy, while the traceless Ricci curvature equals the traceless stress-energy. The factor 8\pi is ultimately related to the weak-field limit giving the inverse-square law of Newtonian gravity.