# Einstein's constant

Einstein's constant or Einstein's gravitational constant, denoted κ (kappa), is the coupling constant appearing in the Einstein field equation which can be written:

$G^{\alpha \gamma} = \kappa \, T^{\alpha \gamma}~$

where Gαγ is the Einstein tensor and Tαγ is the stress–energy tensor.

This equation relates to the curvature of space and time, telling that stress-energy is what causes the disturbance of spacetime, thus gravitation. Einstein used Newton's law of universal gravitation in his field equations, and the constant of κ is found to have a value of:[1]

$\kappa \, = \, - { 8 \, \pi \, G \over c^2 }~$

N.B.: Writing Einstein's constant depends on how the stress–energy tensor is defined, so the Einstein field equations are always invariant (see details in the section "About the two possible writings" further).

## Calculation

In the following, the value of Einstein's constant will be calculated. To do so, at the beginning a field equation where the cosmological constant Λ is equal to zero is taken, with a steady state hypothesis. Then we use the Newtonian approximation with hypothesis of a weak field and low velocities with respect to the speed of light.

The Newton law will arise and its corollary Poisson's equation.

In this approximation, Poisson's equation appears as the approached form of the field equation (or the field equation appears as a generalization of Poisson's equation). The identification gives the expression of Einstein's constant related to quantities G and c.

### The Einstein field equations in non-empty space

We have to obtain a suitable tensor to describe the geometry of space in the presence of an energy field. Einstein proposed this equation in 1917, written as:

$G^{\alpha\gamma} + \Lambda \mathrm{g}^{\alpha\gamma} = (\mathrm{const}) T^{\alpha\gamma}~$

(const) is what will become Einstein's constant. We will take the cosmological constant Λ equal to zero (one of the requirements of the properties of the gravitational equations is that they reduce to the free-space field equations when the density of energy in space Tαγ is zero, therefore that the cosmological constant λ appearing in this equation is zero) so the field equation becomes:

$G^{\alpha\gamma} = \left( R^{\alpha\gamma} - \frac{1}{2} \mathrm{g}^{\alpha\gamma} \, R \right) = \kappa \, T^{\alpha\gamma}~$

where Rαγ s the Ricci tensor, gαγ is the metric tensor, R the scalar curvature and κ is Einstein's constant we will calculate in the next section..

This equation can be written in another form, contracting indexes:

${R^\alpha}_\alpha - \frac{1}{2} \, {\mathrm{g}^\alpha}_\alpha \, R = {\kappa \, T^\alpha}_\alpha~$

Thus:

$R = -{\kappa \, T^\alpha}_\alpha = -\kappa \, T~$

where T is the scalar Tαα which we shall refer to as the Laue scalar.

Using this result we can write the field equation as:

 $R^{\alpha\gamma} = \kappa \left(T^{\alpha\gamma} - \frac{1}{2} \, \mathrm{g}^{\alpha\gamma} \, T \right)$

### Classical limit of the gravitational equations

It will be shown that the field equations are a generalization of Poisson's classical field equation. The reduction to the classical limit, besides being a validity check on the field equations, gives as a by-product the value of the constant κ.

$|i$ and $|i|j$ respectively indicate $\tfrac{\partial}{\partial x^i}$ and $\tfrac{\partial^2}{\partial x^i \partial x^j}$. Thus, $|i|i$ means $\tfrac{\partial^2}{(\partial {x^i})^2}$

Consider a field of matter with low proper density ρ, moving at low velocity v. The stress–energy tensor can be written:

$T_{{\mu}v}= \rho \begin{pmatrix} 1 & v_x / c & v_y / c & v_z / c\\ v_x / c & v^2_x / c^2 & v_x v_y / c^2 & v_x v_z / c^2\\ v_y / c & v_y v_x / c^2 & v^2_y / c^2 & v_y v_z / c^2\\ v_z / c & v_z v_x / c^2 & v_z v_y / c^2 & v^2_z / c^2 \end{pmatrix}$

If the terms of order $(\tfrac{v}{c})^2$ and $\rho (\tfrac{v}{c})$ are neglected, it becomes:

$T_{\mu\nu}= \begin{pmatrix} \rho_0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$

One assumes the flow to be stationary and therefore expects the metric to be time-independent. We use the coordinates of special relativity ct, x, y, z that we write as x0, x1, x2, and x3. The first coordinate is time, and the three others are the space coordinates.

Applying a perturbation method, consider a metric appearing through a two-term summation. The first is the Lorentz metric, ημν which is that of the Minkowski space, locally flat. Formulating gives:

$\displaystyle \mathrm{d}s^2 = (\mathrm{d}x^0)^2 - (\mathrm{d}x^1)^2 - (\mathrm{d}x^2)^2 - (\mathrm{d}x^3)^2$

The second term corresponds to the small perturbation (due to the presence of a gravitating body) and is also time-independent:

$\displaystyle \varepsilon\gamma_{\mu\nu}$

Thus we write the metric:

$\displaystyle g_{\mu\nu} = \eta_{\mu\nu} + \varepsilon\gamma_{\mu\nu}$

Clarifying the length element:

$\displaystyle \mathrm{d}s^2 = (\mathrm{d}x^0)^2 - (\mathrm{d}x^1)^2 - (\mathrm{d}x^2)^2 - (\mathrm{d}x^3)^2 + \varepsilon\gamma_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}$

If we neglect terms of order $\displaystyle \varepsilon\rho_0$, the Laue scalar $T^{\mu}_{\mu}$ is:

$T^{\mu}_{\mu} = \operatorname{Tr} \begin{pmatrix} \rho_0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0& 0\\ 0 & 0 & 0 & 0 \end{pmatrix} = \rho_0$

And the right side of the field equations is to first order in all the small quantities $\rho_0$, $\tfrac{v}{c}$ and $\varepsilon\gamma_{\mu\nu}$ is written:

\begin{align}C\left(T_{\mu\nu}-\frac{1}{2} g_{\mu\nu}T \right)&\simeq C\left(T_{\mu\nu}-\frac{1}{2} g_{\mu\nu} T \right) \\ & \simeq C \left[ \begin{pmatrix} \rho_0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} - \frac{1}{2}\begin{pmatrix} \rho_0 & 0 & 0 & 0\\ 0 & -\rho_0 & 0 & 0\\ 0 & 0 & -\rho_0 & 0\\ 0 & 0 & 0 & -\rho_0 \end{pmatrix} \right] \\ & \simeq \frac{C\rho_0}{2} \delta_{\mu\nu} \end{align}

Neglecting second-order terms in $\varepsilon\gamma_{\mu\nu}$ gives the following approximate form for the contracted Riemann tensor:

$R_{\mu\nu} \cong \frac{1}{2}\left[\ln(-g)\right]_{|\mu|\nu|}-[\mu\nu,\beta]_{|\beta}$

Thus the approximate field equations may be expressed as:

$\frac{1}{2}\left[\ln(-g)\right]_{|\mu|\nu|}-[\mu\nu,\beta]_{|\beta} = \frac{\kappa \, \rho_0}{2} \, \delta_{\mu\nu}$

At first let us consider the case μ = ν = 0. As the metric is time-independent, the first term of the equation above is zero. What remains is:

$[00,\beta]_{|\beta} = \left(g^{\alpha\beta}\left[00,\alpha\right]\right)_{|\beta} = -\frac{\kappa \, \rho_0}{2}\qquad\qquad (*)$

The Christoffel symbol of the first kind is defined by:

$\left[00,\alpha\right] = \frac{1}{2}\left(g_{0\alpha|0} + g_{\alpha 0|0} - g_{00|\alpha}\right)$

Since the Lorentz metric is constant in space and time, this simplifies to:

$\left[00,\alpha\right] = -\frac{\varepsilon}{2} \gamma_{00|\alpha}$

Moreover $\gamma_{\mu\nu}$ is time-independent, so [00,0] is zero. Neglecting second-order terms in the perturbation term $\varepsilon\gamma_{\mu\nu}$, we get:

$g^{\beta\alpha} \left[00,\alpha\right] = \frac{\varepsilon}{2} \gamma_{00|\beta}$

which is zero for β = 0 (which then corresponds to the derivative with respect to time). Substituting inside (*) we obtain the following approximate field equation for $\gamma_{00}$:

$\varepsilon\sum_{\beta = 0}^{3} \gamma_{00|\beta|\beta} = -\kappa \, \rho_0$

or, by virtue of time independence:

$\varepsilon\sum_{\beta = 1}^{3} \gamma_{00|\beta|\beta} = -\kappa \, \rho_0$

This notation is just a writing convention. The equation can be written:

$\sum_{\beta = 0}^{3} \gamma_{00|\beta|\beta} = \sum_{i=1}^3 \frac{\partial {}^2 \gamma_{00}}{\partial {x_{\beta}}^2} = \frac{\partial {}^2 \gamma_{00}}{\partial {x_1}^2} + \frac{\partial {}^2 \gamma_{00}}{\partial {x_2}^2} + \frac{\partial {}^2 \gamma_{00}}{\partial {x_3}^2} = -\kappa \, \rho_0$

which can be identified to Poisson's equation if we write:

$-\frac{\varepsilon\gamma_{00}}{\kappa} = \frac{\varphi}{4\pi G}$

Therefore it is established that the classical theory (Poisson's equation) is the limiting case (weak field, low velocities with respect to the speed of light) of a relativistic theory where the metric is time-independent.

To be complete, gravity has to be demonstrated as a metric phenomenon. In the following, without detailing all calculation, the simplistic description of the complete calculation is given. Again, at first start from a perturbed Lorentz metric:

$\displaystyle g_{\mu\nu} = \eta_{\mu\nu} + \varepsilon\gamma_{\mu\nu}$

$\displaystyle \mathrm{d}s^2 = (\mathrm{d}x^0)^2 - (\mathrm{d}x^1)^2 - (\mathrm{d}x^2)^2 - (\mathrm{d}x^3)^2 + \varepsilon\gamma_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu}$

Suppose the velocity v to be low with respect to the speed of light c, with a small parameter $\beta = \tfrac{v}{c}$. We have:

$\displaystyle x^0 = ct$

We can write:

\begin{align} \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 &= c^2 - v^2 + \varepsilon\gamma_{\mu\nu} \frac{\mathrm{d}x^{\mu}}{\mathrm{d}t} \frac{\mathrm{d}x^{\nu}}{\mathrm{d}t}\\ &= c^2 \left(1 - \beta^2 + \varepsilon\gamma_{\mu\nu} \frac{\mathrm{d}x^{\mu}}{\mathrm{d}x^0} \frac{\mathrm{d}x^{\nu}}{\mathrm{d}x^0}\right) \end{align}

Limiting to the first degree in β and ε gives:

$\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 \cong c^2(1 + \varepsilon\gamma_{00})$

Then one writes, as a classical calculation, the differential equation system giving the geodesics. Christoffel symbols are calculated. The geodesic equation becomes:

$\frac{\mathrm{d}^2 x^{\alpha}}{\mathrm{d}t^2} + [00,\alpha] c^2 = 0 \qquad (**)$

The approximate form of the Christoffel symbol is:

$[00,i] = \frac{1}{2}\varepsilon\gamma_{00|i}$

Introducing this result into the geodesic equation (**) gives:

$\displaystyle \frac{\mathrm{d}^2 x^i}{\mathrm{d}t^2} = -\frac{c^2}{2} \varepsilon \gamma_{00|i}$

This is a vector equation. Since the metric is time-independent, only space variables are concerned. Therefore the second member of the equation is a gradient.

Coding the position-vector by the letter X and the gradient by the vector ∇, one can write:

$\displaystyle \frac{\mathrm{d}^2 X}{\mathrm{d}t^2} = \frac{c^2}{2} \varepsilon \gamma_{00}$

This is no more than Newton's law of universal gravitation in classical theory, deriving from the gravitational potential φ if one makes the identification:

$\varphi = -\frac{c^2}{2} \varepsilon \nabla \gamma_{00}$

Conversely, if we set a gravitational potential φ, the movement of a particle will follow a space-time geodesic if the first term of the metric tensor is like:

$g_{00} = 1 + \frac{2\varphi}{c^2}$

That step is important. Newton's law appears as a particular aspect of the general relativity with the double approximation:

• weak gravitational field
• low velocity with respect to the speed of light

With the calculation above, we have made the following statements:

• A metric g, solution of the Einstein field equation (with a cosmological constant Λ equal to zero).
• This metric would be a weak perturbation in relation to a Lorentz metric η (relativistic and flat space).
• The perturbation term would not depend on time. Since the Lorentz metric does not depend on time either, that metric g is also time-independent.
• The expansion into a series gives a linearization of the Einstein field equations.
• This linearized form is found to identify to Poisson's equation because a field is a curvature, linking the perturbation term to the metric and to the gravitational potential thanks to the relation:
$\varphi = \frac{c^2}{2} \varepsilon \gamma_{00}$

And this rewards the value of the constant κ, called "Einstein's constant" (which is not the cosmological constant Λ or the speed of light c):

$\kappa \, = \, - { 8 \, \pi \, G \over c^2 }~$

One can then write the Einstein field equation:

$G^{\alpha\gamma} + \Lambda g^{\alpha\gamma} = -\frac{8\pi G}{c^2} T^{\alpha\gamma}~$

## About the two possible writings

We have seen, neglecting the terms of order $(\tfrac{v}{c})^2$ and $\rho (\tfrac{v}{c})$, that the Laue scalar could be written:

$T^{\mu}_{\mu} = \operatorname{Tr} \begin{pmatrix} \rho_0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0& 0\\ 0 & 0 & 0 & 0 \end{pmatrix} = \rho_0$

which gives the corresponding Einstein's constant:

$\kappa \, = \, - { 8 \, \pi \, G \over c^2 }~$

But another valid choice for writing the form of the stress–energy tensor is:

$T_{{\mu}v}= \rho \begin{pmatrix} c^2 & v_x c & v_y c & v_z c\\ v_x c & v^2_x & v_x v_y & v_x v_z\\ v_y c & v_y v_x & v^2_y & v_y v_z\\ v_z c & v_z v_x & v_z v_y & v^2_z \end{pmatrix}~$

Neglecting the same term orders, the corresponding Laue scalar is:

$T^\mu_\mu=\operatorname{Tr} \begin{pmatrix} \rho_0 c^2 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}=\rho_0 c^2~$

which owns an additional term c2, so the corresponding Einstein's constant in the field equations is then:

$\kappa \, = \, - { 8 \, \pi \, G \over c^4 }~$

This is just a question of similar choices, since for each chosen writing the Einstein field equations are the same.

$G \over c^2~$