Einstein’s General Theory of Relativity

Derivations, Applications and Reflections – by Albert Prins

Part VII – Questions and Discussion

7 Answers to Questions

7.1 Derivation of the Schwarzschild Formula in terms of proper time \( \tau \)

Question:
What I find difficult to accept in the general theory of relativity is the derivation with respect to “\(ds\)”. The line element is nothing more than the speed of light multiplied by the locally measured time difference \(dt_{0}\) \((ds = c\,dt_{0})\). I can understand \(dt/ds\) (difference in clock times), but what does \(dx/ds\) mean?

Answer:
The confusion arises from the interpretation of \(ds\). In general relativity we have: \[ ds = c\,d\tau, \] where \( \tau \) is the proper time: the time measured by a clock that is at rest with respect to the object being measured—in other words, the time on a clock that “moves along” with the object itself.

The quantity \(dt\), on the other hand, is the coordinate time in a universal (or external) reference frame, for example the center of a gravitational field (such as the center of the Earth). This time \(t\) is therefore not a directly measured time, but a computational parameter that can be derived from \(d\tau\) via the metric.

The relation between the two is: \[ d\tau = \sigma\,\gamma\,dt, \] where:

Derivation from the Schwarzschild metric

We consider the time interval based on the general form of the metric tensor product in a static, spherically symmetric field:

\[ c^{2} d\tau^{2} = A\,c^{2} dt^{2} - B\,dx^{2} - D\,dy^{2} - E\,dz^{2}, \] where \(A, B, D, E\) are the components of the metric tensor, depending on the position in space (for example on \(r\)).

Divide both sides by \(c^{2} d\tau^{2}\): \[ 1 = A\left(\frac{dt}{d\tau}\right)^{2} - \frac{B}{c^{2}}\left(\frac{dx}{d\tau}\right)^{2} - \frac{D}{c^{2}}\left(\frac{dy}{d\tau}\right)^{2} - \frac{E}{c^{2}}\left(\frac{dz}{d\tau}\right)^{2}. \]

We rewrite the spatial derivatives using the chain rule: \[ \frac{dx}{d\tau} = \frac{dx}{dt}\,\frac{dt}{d\tau}, \qquad \text{etc.} \]

With this, the equation becomes: \[ 1 = A\left(\frac{dt}{d\tau}\right)^{2} - \frac{B}{c^{2}}\left(\frac{dx}{dt}\right)^{2} \left(\frac{dt}{d\tau}\right)^{2} - \frac{D}{c^{2}}\left(\frac{dy}{dt}\right)^{2} \left(\frac{dt}{d\tau}\right)^{2} - \frac{E}{c^{2}}\left(\frac{dz}{dt}\right)^{2} \left(\frac{dt}{d\tau}\right)^{2}. \]

Here, \(x, y, z\) divided by \(t\) in their own frame turn out to be velocities in that frame.

\[ 1 = A\left(\frac{dt}{d\tau}\right)^{2} \left[ 1 - \frac{B}{A c^{2}}\left(\frac{dx}{dt}\right)^{2} - \frac{D}{A c^{2}}\left(\frac{dy}{dt}\right)^{2} - \frac{E}{A c^{2}}\left(\frac{dz}{dt}\right)^{2} \right]. \]

Now define the speed: \[ v^{2} = \frac{B}{A}\left(\frac{dx}{dt}\right)^{2} + \frac{D}{A}\left(\frac{dy}{dt}\right)^{2} + \frac{E}{A}\left(\frac{dz}{dt}\right)^{2}. \]

Substitute \(v\): \[ 1 = A\left(\frac{dt}{d\tau}\right)^{2} \left(1 - \frac{v^{2}}{c^{2}}\right) = A\,\gamma^{2}\left(\frac{dt}{d\tau}\right)^{2}. \]

From this it follows that: \[ \left(\frac{dt}{d\tau}\right)^{2} = \frac{\gamma^{2}}{A} \quad\Longrightarrow\quad d\tau^{2} = A\,\gamma^{2}\,dt^{2} = \sigma^{2}\gamma^{2} dt^{2}. \]

Or: \[ d\tau = \sigma\,\gamma\,dt. \]

This is the relation between the time of the measuring clock and the time at the origin of the universal frame.

Conclusion

This derivation shows the relation between the proper time \(\tau\) (as measured by a moving clock in its own rest frame) and the coordinate time \(t\) (as defined in the global gravitational field). The role of \(dx/ds\) also becomes clear: it describes the rate of spatial change per unit proper time—thus the projection of the four‑velocity onto the spatial coordinates.

This relation is fundamental in general relativity and forms the basis for analyzing time dilation in gravitational fields, such as in the Hafele–Keating experiment and other applications of the Schwarzschild metric.

7.2 Explanation of Einstein’s transformation formula

In general relativity it is fundamental that physical laws remain invariant under coordinate transformations. The relation between old and new coordinate systems is expressed mathematically using a transformation formula based on partial derivatives.

1. Coordinate systems

The formula represents the covariant transformation between two coordinate systems. The old system is denoted by \(x^{\beta}\), with coordinate axes \((x^{0}, x^{1}, x^{2}, x^{3})\). The new system is denoted by \(x'^{\alpha}\), with \((x'^{0}, x'^{1}, x'^{2}, x'^{3})\).

2. Transformation formula

The differential of the new coordinates \(dx'^{\alpha}\) is expressed in terms of the differential of the old coordinates \(dx^{\beta}\) as follows: \[ dx'^{\alpha} = \frac{\partial x^{\beta}}{\partial x'^{\alpha}}\,dx^{\beta}. \]

This formula is written in Einstein notation, which means that there is an implicit sum over \(\beta\).

Written out explicitly, this means: \[ dx'^{\alpha} = \frac{\partial x^{0}}{\partial x'^{\alpha}}\,dx^{0} + \frac{\partial x^{1}}{\partial x'^{\alpha}}\,dx^{1} + \frac{\partial x^{2}}{\partial x'^{\alpha}}\,dx^{2} + \frac{\partial x^{3}}{\partial x'^{\alpha}}\,dx^{3}. \]

For each value of \(\alpha = 0,1,2,3\) this gives a separate equation that expresses each of the new coordinate differentials \((dx'^{0}, dx'^{1}, dx'^{2}, dx'^{3})\) in terms of the old coordinates.

3. Tensor form

Altogether we obtain:

\[ \begin{aligned} dx'^{0} &= \frac{\partial x^{0}}{\partial x'^{0}}\,dx^{0} + \frac{\partial x^{1}}{\partial x'^{0}}\,dx^{1} + \frac{\partial x^{2}}{\partial x'^{0}}\,dx^{2} + \frac{\partial x^{3}}{\partial x'^{0}}\,dx^{3}, \\[6pt] dx'^{1} &= \frac{\partial x^{0}}{\partial x'^{1}}\,dx^{0} + \frac{\partial x^{1}}{\partial x'^{1}}\,dx^{1} + \frac{\partial x^{2}}{\partial x'^{1}}\,dx^{2} + \frac{\partial x^{3}}{\partial x'^{1}}\,dx^{3}, \\[6pt] dx'^{2} &= \frac{\partial x^{0}}{\partial x'^{2}}\,dx^{0} + \frac{\partial x^{1}}{\partial x'^{2}}\,dx^{1} + \frac{\partial x^{2}}{\partial x'^{2}}\,dx^{2} + \frac{\partial x^{3}}{\partial x'^{2}}\,dx^{3}, \\[6pt] dx'^{3} &= \frac{\partial x^{0}}{\partial x'^{3}}\,dx^{0} + \frac{\partial x^{1}}{\partial x'^{3}}\,dx^{1} + \frac{\partial x^{2}}{\partial x'^{3}}\,dx^{2} + \frac{\partial x^{3}}{\partial x'^{3}}\,dx^{3}. \end{aligned} \]

This can also be written in matrix or tensor form:

\[ \begin{pmatrix} dx'^{0} \\ dx'^{1} \\ dx'^{2} \\ dx'^{3} \end{pmatrix} = \begin{pmatrix} \dfrac{\partial x^{0}}{\partial x'^{0}} & \dfrac{\partial x^{1}}{\partial x'^{0}} & \dfrac{\partial x^{2}}{\partial x'^{0}} & \dfrac{\partial x^{3}}{\partial x'^{0}} \\[6pt] \dfrac{\partial x^{0}}{\partial x'^{1}} & \dfrac{\partial x^{1}}{\partial x'^{1}} & \dfrac{\partial x^{2}}{\partial x'^{1}} & \dfrac{\partial x^{3}}{\partial x'^{1}} \\[6pt] \dfrac{\partial x^{0}}{\partial x'^{2}} & \dfrac{\partial x^{1}}{\partial x'^{2}} & \dfrac{\partial x^{2}}{\partial x'^{2}} & \dfrac{\partial x^{3}}{\partial x'^{2}} \\[6pt] \dfrac{\partial x^{0}}{\partial x'^{3}} & \dfrac{\partial x^{1}}{\partial x'^{3}} & \dfrac{\partial x^{2}}{\partial x'^{3}} & \dfrac{\partial x^{3}}{\partial x'^{3}} \end{pmatrix} \begin{pmatrix} dx^{0} \\ dx^{1} \\ dx^{2} \\ dx^{3} \end{pmatrix}. \]

\[ \begin{pmatrix} dx'^{0} \\ dx'^{1} \\ dx'^{2} \\ dx'^{3} \end{pmatrix} = \begin{pmatrix} \dfrac{\partial x^{0}}{\partial x'^{0}} & \dfrac{\partial x^{1}}{\partial x'^{0}} & \dfrac{\partial x^{2}}{\partial x'^{0}} & \dfrac{\partial x^{3}}{\partial x'^{0}} \\[6pt] \dfrac{\partial x^{0}}{\partial x'^{1}} & \dfrac{\partial x^{1}}{\partial x'^{1}} & \dfrac{\partial x^{2}}{\partial x'^{1}} & \dfrac{\partial x^{3}}{\partial x'^{1}} \\[6pt] \dfrac{\partial x^{0}}{\partial x'^{2}} & \dfrac{\partial x^{1}}{\partial x'^{2}} & \dfrac{\partial x^{2}}{\partial x'^{2}} & \dfrac{\partial x^{3}}{\partial x'^{2}} \\[6pt] \dfrac{\partial x^{0}}{\partial x'^{3}} & \dfrac{\partial x^{1}}{\partial x'^{3}} & \dfrac{\partial x^{2}}{\partial x'^{3}} & \dfrac{\partial x^{3}}{\partial x'^{3}} \end{pmatrix} \begin{pmatrix} dx^{0} \\ dx^{1} \\ dx^{2} \\ dx^{3} \end{pmatrix}. \]

This matrix represents the Jacobian of the coordinate transformation and describes how vector components transform between two systems.

4. Interpretation

This formula shows how a vector (or differential) in one system can be expressed in the other system. The key points are:

5. Example: Transformation within the Schwarzschild metric

A practical application is the transformation from spherical coordinates \((t, r, \theta, \phi)\) to Cartesian coordinates \((t, x, y, z)\). The spatial coordinates are transformed via:

\[ x = r\sin\theta\cos\phi, \qquad y = r\sin\theta\sin\phi, \qquad z = r\cos\theta. \]

The corresponding differential transformations for \(dx, dy, dz\) can then be derived using the chain rule, exactly as formalized above:

\[ dx = \frac{\partial x}{\partial r}\,dr + \frac{\partial x}{\partial \theta}\,d\theta + \frac{\partial x}{\partial \phi}\,d\phi, \] \[ dy = \frac{\partial y}{\partial r}\,dr + \frac{\partial y}{\partial \theta}\,d\theta + \frac{\partial y}{\partial \phi}\,d\phi, \] \[ dz = \frac{\partial z}{\partial r}\,dr + \frac{\partial z}{\partial \theta}\,d\theta. \]

These transformations form the basis for rewriting the Schwarzschild metric from spherical to Cartesian coordinates, as Schwarzschild did in his original paper.

7.3 Answer to questions concerning Schwarzschild

Question 1:
Where does the general relativity formula with the Ricci tensor come from, the one that only became widely used after 1916?

Answer:
The full field equations of general relativity, including the Ricci tensor, were part of Einstein’s theory from the beginning. The simplified version with the condition \(g = -1\) was later used to make the equations mathematically simpler, but this restriction reduces the number of possible solutions.

In much of the literature the tensor \(G_{\mu\nu}\) is called the Einstein tensor. Einstein himself presented this tensor as: \[ G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R. \]

Here:

The Ricci scalar is given by: \[ R = g^{\mu\nu} R_{\mu\nu} = g^{00} R_{00} + g^{11} R_{11} + g^{22} R_{22} + g^{33} R_{33}. \]

Contracting the Einstein field equations with \(g^{\mu\nu}\) yields: \[ g^{\mu\nu} G_{\mu\nu} = g^{\mu\nu} R_{\mu\nu} - \frac{1}{2} g^{\mu\nu} g_{\mu\nu} R = R - \frac{1}{2}(4)R = -R. \]

The full Einstein field equations read: \[ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = \frac{8\pi G}{c^{4}}\, T_{\mu\nu}, \] where:

Outside a massive sphere there is no matter or energy. In that case: \[ T_{\mu\nu} = 0, \] and the field equations reduce to: \[ G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 0. \]

We know that: \[ G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} g^{\mu\nu} R_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}(4) R_{\mu\nu} = - R_{\mu\nu}. \]

From which it follows that: \[ G_{\mu\nu} = 0 \quad\text{only if}\quad R_{\mu\nu} = 0 \quad\text{and hence also}\quad R = 0. \]

Background: from Riemann to Ricci

Einstein built on the work of Riemann, who had already developed a mathematical description of curved surfaces. The Riemann tensor: \[ R_{\mu\beta\rho\nu} \] is a rank‑four tensor and difficult to visualize. Since the mass‑energy‑momentum tensor \(T_{\mu\nu}\) has only two indices, the Riemann tensor must be reduced from four to two indices.

Using the metric tensor, the covariant Riemann tensor can be converted into a partially contravariant form: \[ R^{\beta}{}_{\mu\rho\nu} = g^{\beta\alpha} R_{\mu\alpha\rho\nu}. \]

This is needed to perform the desired contraction. By setting \(\beta = \rho\), the contraction can be carried out, yielding the Ricci tensor: \[ R^{\beta}{}_{\mu\beta\nu} = R_{\mu\nu}. \]

Thus, the Ricci tensor is the trace of the Riemann tensor.

The role of Christoffel symbols

The Ricci tensor can also be expressed in terms of the Christoffel symbols: \[ R_{\mu\nu} = R^{\rho}{}_{\mu\rho\nu} = \Gamma^{\rho}_{\mu\nu,\rho} - \Gamma^{\rho}_{\mu\rho,\nu} + \Gamma^{\rho}_{\lambda\rho}\Gamma^{\lambda}_{\nu\mu} - \Gamma^{\rho}_{\nu\lambda}\Gamma^{\lambda}_{\mu\rho}. \]

where the Christoffel symbol itself is: \[ \Gamma^{\rho}_{\mu\nu} = \frac{1}{2} g^{\rho\alpha} \left( \frac{\partial g_{\nu\alpha}}{\partial x^{\mu}} + \frac{\partial g_{\mu\alpha}}{\partial x^{\nu}} - \frac{\partial g_{\mu\nu}}{\partial x^{\alpha}} \right). \]

The derivative of the Christoffel symbol is then: \[ \Gamma^{\rho}_{\mu\nu,\gamma} = \frac{\partial \Gamma^{\rho}_{\mu\nu}}{\partial x^{\gamma}} = - g^{\rho\alpha} \frac{\partial g_{\rho\alpha}}{\partial x^{\gamma}} \Gamma^{\rho}_{\mu\nu} + \frac{1}{2} g^{\rho\alpha} \left( \frac{\partial^{2} g_{\nu\alpha}}{\partial x^{\mu}\partial x^{\gamma}} + \frac{\partial^{2} g_{\mu\alpha}}{\partial x^{\nu}\partial x^{\gamma}} - \frac{\partial^{2} g_{\mu\nu}}{\partial x^{\alpha}\partial x^{\gamma}} \right). \]

When computing the Ricci components \(R_{00}, R_{11}, R_{22}, R_{33}\) using the full Einstein field equations, they all turn out to be zero, which is correct. But when we perform the calculation with the restricted form of the field equations \((g = -1)\), the result is not correct. Thus, the Schwarzschild solution does satisfy the general field equations but not the restricted ones. This is consistent because for the Schwarzschild solution: \[ g \neq -1. \]

On the Schwarzschild solution and the restriction \(g = -1\)

Schwarzschild uses the familiar polar form. The determinant of the metric tensor (here the product of the coefficients) is not \(-1\). This polar form satisfies the Einstein field equations, but not the restricted version of these equations, because in the latter case \(g = -1\) is required.

Schwarzschild derived a transformation based on adapted polar coordinates, choosing the transformation such that \(g = -1\) is obtained. In that case the equation also satisfies the restricted Einstein field equations.

Conclusion

Although Schwarzschild tried to satisfy Einstein’s wish to have the metric trace \(g = -1\), in my view the only relevant requirement is that the Einstein field equations—with \(T_{\mu\nu} = 0\), and hence \[ R_{00} = R_{11} = R_{22} = R_{33} = 0, \] —are fulfilled, regardless of whether \(g = -1\) or \(g \neq -1\). Thus, the requirement \(g = -1\) is an unnecessary restriction.

Question 2:

The consequence of the difference in formulas is significant. In your document I count nine Christoffel symbols, whereas Karl Schwarzschild found ten. In your case, \(\Gamma^{2}_{22}\) appears to be missing.

This is because your definition of the metric tensor \(g\) differs from that of Schwarzschild: for Schwarzschild we have \(g_{22}\) and \(g_{33} = -1\), whereas you include the coordinate \(r\) (for example \(g_{22} = -r^{2}\)).

Droste (1917), Eddington (1921), MWT (1975) and OAS (2007) also imposed \(g = -1\) for the Schwarzschild solution, so that: \[ g_{22} = g_{33} = -1. \]

This raises the question: do you think that \(g = -1\) is required for the Schwarzschild solution?

Answer:

In Schwarzschild’s original derivation, one starts from the Cartesian coordinate system \((x, y, z)\). In that form, the metric has the following components: \[ g_{00} = \sigma^{2}, \qquad g_{11} = -\frac{1}{r^{4}\sigma^{2}}, \qquad g_{22} = -\frac{r^{2}}{\sin^{2}\theta}, \qquad g_{33} = -r^{2}\sin^{2}\theta. \]

In this case, ten (or fourteen) relevant Christoffel symbols are generated. You can also see in my overview of formulas that I derived expressions for both the spherical and the Cartesian \(x, y, z\) form. In the \(x, y, z\) form, \(\Gamma^{2}_{22}\) indeed exists.

For the spherical form, however, the situation is different; there the elements of the metric tensor are:

\[ g_{\mu\nu} = \begin{pmatrix} \sigma^{2} & 0 & 0 & 0 \\ 0 & -\dfrac{1}{\sigma^{2}} & 0 & 0 \\ 0 & 0 & -r^{2} & 0 \\ 0 & 0 & 0 & -r^{2}\sin^{2}\theta \end{pmatrix}. \]

\[ g_{00} = \sigma^{2}, \qquad g_{11} = -\frac{1}{\sigma^{2}}, \qquad g_{22} = -r^{2}, \qquad g_{33} = -r^{2}\sin^{2}\theta. \]

This is exactly the same form as in Schwarzschild’s work. In these spherical coordinates, \(g_{22}\) and \(g_{33}\) depend explicitly on \(r\) and \(\theta\), and therefore cannot be constant, as in \(g_{22} = g_{33} = -1\).

If these values were constant, the partial derivatives \(\partial g_{22}/\partial r\), \(\partial g_{22}/\partial \theta\), \(\partial g_{33}/\partial r\), \(\partial g_{33}/\partial \theta\) would all be zero. As a result, many Christoffel symbols—including crucial ones such as \(\Gamma^{1}_{22}\) and \(\Gamma^{2}_{22}\)—would also vanish.

This also applies to Schwarzschild! The elements \(g_{22}\) and \(g_{33}\) cannot be \(-1\), because in that case \(\partial g_{22}/\partial r\), \(\partial g_{22}/\partial \theta\), \(\partial g_{33}/\partial r\), \(\partial g_{33}/\partial \theta\) would be zero and the number of Christoffel symbols would be limited to only: \[ \Gamma^{0}_{01},\ \Gamma^{0}_{10},\ \Gamma^{1}_{00},\ \Gamma^{1}_{11}. \]

Regarding \(\Gamma^{2}_{22}\)

For spherical coordinates this component is indeed zero, because \(g_{22}\) does not depend on \(\theta\) and its derivative is therefore zero: \[ \Gamma^{2}_{22} = \frac{1}{2} g^{22} \frac{\partial g_{22}}{\partial \theta} = 0. \]

It is important that, when evaluating components at a specific value such as \(\theta = \pi/2\) (i.e. \(90^\circ\)), substitution is only done at the end of the calculations.

For example: \[ \Gamma^{2}_{33} = \frac{1}{2} g^{22} \left(-\frac{\partial g_{33}}{\partial \theta}\right) = -\cos\theta\sin\theta = 0 \quad \text{at } \theta = \frac{\pi}{2}. \]

This value becomes zero when \(\theta = \pi/2\), but for the Ricci component one also needs the derivative of this Christoffel symbol: \[ \frac{\partial}{\partial \theta}\Gamma^{3}_{32} = -\cos^{2}\theta + \sin^{2}\theta = 1 \quad \text{at } \theta = \frac{\pi}{2}. \]

And this is not zero, which is crucial for computing, for example, the Ricci tensor component \(R_{22}\).

Regarding the condition \(\det(g) = -1\)

Why Einstein introduced this restriction is not entirely clear, but it often simplifies the algebra and enforces symmetry. In my view, however, it leads to an unnecessary limitation.

It also depends on the type of coordinate system chosen.

For example, the metric tensor in \((t, x, y, z)\) indeed yields: \[ \det(g) = \sigma^{2} \cdot \left(-\frac{1}{r^{4}\sigma^{2}}\right) \cdot \left(-\frac{r^{2}}{\sin^{2}\theta}\right) \cdot (-r^{2}\sin^{2}\theta) = -1. \]

But in spherical coordinates it becomes: \[ \det(g) = \sigma^{2} \cdot \left(-\frac{1}{\sigma^{2}}\right) \cdot (-r^{2}) \cdot (-r^{2}\sin^{2}\theta) = -r^{4}\sin^{2}\theta. \]

And thus here: \[ \det(g) \neq -1. \] Yet this metric perfectly satisfies the Einstein field equations in vacuum \((T_{\mu\nu} = 0)\), which implies: \[ R_{\mu\nu} = 0 \quad\text{and hence also}\quad R = 0. \]

Conclusion:
The requirement \(\det(g) = -1\) is a coordinate‑dependent convention that may offer mathematical convenience, but is not physically necessary for the correctness of the Schwarzschild solution. What truly matters is that the field equations are satisfied. Schwarzschild’s choice to use a transformation that enforces \(\det(g) = -1\) was mainly intended to meet Einstein’s preference, but is superfluous from a physical point of view.

Question 3:

The field equations in your document on pages 2 and 3, based on the Ricci tensor, differ significantly from those that we (and Karl Schwarzschild) used in Appendix E, based on the \(G\)‑tensor. You also mentioned the \(G\)‑tensor in your document on page 9. My question is: shouldn’t the results be the same?

Answer:

By the \(G\)‑tensor, as you call it in your question, you mean Einstein’s restricted field equations. As I have shown theoretically in my discussion, Schwarzschild satisfies the general field equations but not the restricted \(G\)‑equations. This is because Einstein introduced the additional requirement \(\det(g) = -1\) to obtain simpler formulas, but this led to an unnecessary restriction of possible solutions— including the Schwarzschild solution expressed in spherical coordinates.

Yet this solution is an excellent tool for computing phenomena in vacuum in a relatively simple way.

Question 4:

I still have some difficulty understanding the Schwarzschild solution and the Einstein field equations. Could you elaborate on this?

Answer:

It seems we are once again entering a discussion we have had before. Let me state clearly at the outset: it is not my intention to defend the Schwarzschild or Einstein solution against your approach, nor to criticize your suggestion to modify the Schwarzschild metric.

My aim is to reach a complete understanding. As long as I do not fully grasp the Schwarzschild solution, I will keep searching for insight. Only when I recognize and understand a fundamental error will I consider revising the solution.

Let us therefore first examine the Schwarzschild solution in detail before delving into the Einstein field equations. I do not claim to have the final answer, but I would like to outline how I currently understand the structure.

Einstein’s starting point

Einstein sought a description of gravity in which gravity is no longer a force, as in Newton’s theory, but a consequence of the curvature of space‑time.

He wanted to find a coordinate system in which no forces are felt, so that a freely falling particle moves without acceleration—in a sense “voluntarily”, without an external cause.

In classical mechanics, an object moves with constant velocity along a straight line if no force acts on it. Einstein translated this into relativity: an object without external force moves along a geodesic in curved space‑time. These geodesics are, in a sense, the “straight lines” of curved space‑time.

Einstein therefore looked for a mathematical formulation that is valid in any coordinate system, curved or not, and correctly describes the gravitational field. This led to the field equations: \[ R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R = 0 \] in the vacuum case (outside matter), where:

This equation is covariant: it holds in any coordinate system.

Coordinates, metric and geometry

Although the field equations are coordinate‑independent, the components of the tensors involved do depend on the choice of coordinates. The Ricci tensor and the scalar \(R\) are expressed in terms of the Christoffel symbols, which themselves are derived from the metric \(g_{\mu\nu}\).

The metric describes how the interval \(ds^{2}\) between two infinitesimally close points is computed. In the simplest case (for example flat space in Cartesian coordinates) this is: \[ ds^{2} = c^{2} dt^{2} - dx^{2} - dy^{2} - dz^{2}. \]

But in curved space‑time, \(ds^{2}\) depends on position and on the metric components. The metric encodes the geometric structure of space, including possible cross terms (such as \(dx\,dy\)) if the coordinates are not orthogonal.

By way of comparison:

By analogy, Einstein views space‑time as composed of infinitely many infinitesimally small patches, in which the geometry can locally be treated as flat (via the equivalence principle). In these small regions we still use a coordinate system, but the metric components vary from point to point and encode the curvature.

Schwarzschild’s approach

Karl Schwarzschild found an exact solution of the Einstein equations in vacuum around a spherically symmetric mass. He chose a coordinate system that incorporates as much symmetry as possible:

This yields the Schwarzschild metric: \[ ds^{2} = \left(1 - \frac{2GM}{c^{2} r}\right)c^{2} dt^{2} - \left(1 - \frac{2GM}{c^{2} r}\right)^{-1} dr^{2} - r^{2} d\theta^{2} - r^{2}\sin^{2}\theta\, d\phi^{2}. \]

This formula describes the squared line element \(ds^{2}\) as a function of four coordinates: \(t, r, \theta, \phi\). The coefficients (metric components) depend on \(r\) (and implicitly on \(\theta\) via \(\sin^{2}\theta\)), but not on \(t\) or \(\phi\).

This reflects the physical assumptions: the solution is static (time‑independent) and spherically symmetric.

Each location in space has its own metric components, and thus its own geometric structure. By integrating \(ds\) along a path, we obtain the total distance or duration of the trajectory in this curved space‑time.

Summary

7.4 Detailed derivation of Einstein’s equation (57) from equation (53)

Question:
I am reading Einstein’s original GR paper, which I have attached as a PDF to this e‑mail. (Einstein, Relativity: The Special and General Theory, 1916 (this revised edition: 1924)) (Einstein, The Collected Papers of Albert Einstein, 1997)

In section 18, at the bottom of page 186 of the article (bottom left on page 22 of the PDF), there is an equation that I am trying to derive using the method Einstein proposes in the paper (multiplication …

of equation (53) by the derivative of the metric tensor and using the methods in section 15). Could you derive this equation in the specific way Einstein indicates, based solely on the preceding material in Einstein’s paper? Can you show me the detailed steps you took to obtain that equation according to Einstein’s method?

Answer:

Note: the equation numbers refer to Einstein’s original work on General Relativity.

Derivation of equation (57) from equation (53)

We start from equation (53) in Einstein’s paper: \[ \frac{\partial \Gamma^{\alpha}_{\mu\nu}}{\partial x^{\alpha}} + \Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}_{\nu\alpha} = -\kappa\left( T_{\mu\nu} - \frac{1}{2}g_{\mu\nu} T \right), \] where:

Step 1: Multiply by \(\dfrac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\)

We multiply both sides of equation (53) by \(\dfrac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\), i.e.: \[ \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \left( \frac{\partial \Gamma^{\alpha}_{\mu\nu}}{\partial x^{\alpha}} + \Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}_{\nu\alpha} \right) = \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \left[ -\kappa\left( T_{\mu\nu} - \frac{1}{2}g_{\mu\nu} T \right) \right] \]

This gives: \[ \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma^{\alpha}_{\mu\nu}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}_{\nu\alpha} = -\kappa \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} T_{\mu\nu} + \frac{\kappa}{2} g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} T \]

Step 2: Use \(\dfrac{\partial g^{\mu\nu}}{\partial x^{\sigma}} g_{\mu\nu} = 0\)

From equation (29) in Einstein’s paper we have: \[ \frac{1}{\sqrt{-g }}\frac{\partial(\sqrt{-g })}{\partial x^{\sigma}} = -\frac{1}{2} g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}. \]

Since \(\sqrt{-g } = 1\), its derivative is zero, so: \[ 0 = -\frac{1}{2} g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}, \] and therefore: \[ g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} = 0. \]

This means that the contraction of \(\dfrac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\) with \(g_{\mu\nu}\) is zero, which Einstein uses in the subsequent steps to eliminate terms and eventually arrive at equation (57).

From equation (29) it follows that: \[ -\frac{1}{2} g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} = 0. \]

Substituting, we obtain: \[ -\kappa\,\left( \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} - \frac{1}{2} g_{\mu\nu} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T\right) = -\kappa\,\left( \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} - 0 \right) = -\kappa\, \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu}. \]

The equation then becomes: \[ \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma^{\alpha}{}_{\mu\nu}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma^{\alpha}{}_{\mu\beta}\Gamma^{\beta}{}_{\nu\alpha} + \kappa\, \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0. \]

The next step yields: \[ \frac{1}{2\kappa}\, \left( \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma^{\alpha}{}_{\mu\nu}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma^{\alpha}{}_{\mu\beta}\Gamma^{\beta}{}_{\nu\alpha} \right) + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0 \tag{1} \]

Substitute this into: \[ \frac{1}{2\kappa}\, \frac{\partial}{\partial x^{\alpha}} \left(- 2\kappa\, t_{\sigma}{}^{\alpha}\right) + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0 \tag{1a} \]

This leads to: \[ -\frac{\partial t_{\sigma}{}^{\alpha}}{\partial x^{\alpha}} + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0 \tag{2} \]

Now use Einstein’s equation (56): \[ \frac{\partial \left(t_{\mu}^{\sigma}+T_{\mu}^{\sigma}\right)}{\partial x^{\sigma}}=0 \]

Thus: \[ \frac{\partial t^{\sigma}_{\mu}}{\partial x^{\sigma}} = - \frac{\partial T^{\sigma}_{\mu}}{\partial x^{\sigma}}. \]

Replace \(\sigma\) by \(\alpha\), and \(\mu\) by \(\sigma\): \[ \frac{\partial t_{\sigma}^{\alpha}}{\partial x^{\alpha}} = - \frac{\partial T_{\sigma}^{\alpha}}{\partial x^{\alpha}}. \]

Equation (2) then becomes: \[ \frac{\partial T_{\sigma}^{\alpha}}{\partial x^{\alpha}} + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0. \]

And this is precisely Einstein’s equation: \[ \boxed{ \frac{\partial T_{\sigma}^{\alpha}}{\partial x^{\alpha}} + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0 } \tag{57} \]

This is exactly Einstein’s equation (57), derived according to his own method from equation (53), using only the material from sections 15–18 of his paper.

Derivation of \( \frac{\partial}{\partial x^{\alpha}} \left(- 2\kappa\, t_{\sigma}^{\alpha}\right) \) of equation (1a)

We want to prove that: \[ \frac{\partial}{\partial x^{\alpha}} \left(- 2\kappa\, t_{\sigma}^{\alpha}\right) = \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma^{\alpha}_{\mu\nu}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}_{\nu\alpha}. \]

We use Einstein’s equation (48): \[ \frac{\partial H}{\partial g^{\mu\nu}} = -\Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}{}_{\nu\alpha}, \qquad \frac{\partial H}{\partial g_{\sigma}^{\mu\nu}} = \Gamma^{\sigma}_{\mu\nu}. \]

And Einstein’s equation (47b): \[ \frac{\partial}{\partial x^{\alpha}} \left( \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial H}{\partial g^{\mu\nu}} = 0 \quad\Rightarrow\quad \frac{\partial}{\partial x^{\alpha}} \left( \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) = \frac{\partial H}{\partial g^{\mu\nu}}. \]

The combination \[ \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma^{\alpha}_{\mu\nu}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}_{\nu\alpha} \] can be rewritten as: \[ \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial}{\partial x^{\alpha}} \left( \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial H}{\partial g^{\mu\nu}}. \]

We can now differentiate: \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial g_{\sigma}^{\mu\nu}}{\partial x^{\alpha}} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} - \frac{\partial H}{\partial g^{\mu\nu}} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}. \]

Here we have: \[ \frac{\partial g_{\sigma}^{\mu\nu}}{\partial x^{\alpha}} = \frac{\partial^{2} g^{\mu\nu}}{\partial x^{\alpha}\partial x^{\sigma}} = \frac{\partial g_{\alpha}^{\mu\nu}}{\partial x^{\sigma}}. \]

Insert this into the expression: \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial g_{\alpha}^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} - \frac{\partial H}{\partial g^{\mu\nu}} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}. \]

As stated in Einstein’s text below equation (47a), the quantity \(H\) is treated as a function of \(g^{\mu\nu}\) and \[ g_{\sigma}^{\mu\nu} = \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}, \] so that: \[ \frac{\partial H}{\partial x^{\sigma}} = \frac{\partial H}{\partial g_{,\alpha}^{\mu\nu}} \frac{\partial g_{,\alpha}^{\mu\nu}}{\partial x^{\sigma}} + \frac{\partial H}{\partial g^{\mu\nu}} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}. \]

Insert this into the previous expression: \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial H}{\partial x^{\sigma}} = 0. \]

This is exactly the form Einstein uses to obtain the highlighted step: \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial H}{\partial x^{\sigma}}. \]

Or equivalently: \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial}{\partial x^{\sigma}}(H), \] \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial}{\partial x^{\alpha}} \left( \delta_{\sigma}^{\alpha} H \right), \] \[ \frac{\partial}{\partial x^{\alpha}} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} - \delta_{\sigma}^{\alpha} H\right) \]

According to Einstein’s equation (49): \[ -2\kappa\, t_{\sigma}^{\alpha} = g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} - \delta_{\sigma}^{\alpha} H. \]

Insert this into equation (1): \[ \frac{1}{2\kappa}\, \left( \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma^{\alpha}{}_{\mu\nu}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma^{\alpha}_{\mu\beta}\Gamma^{\beta}{}_{\nu\alpha} \right) + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0 \] becomes: \[ \frac{1}{2\kappa}\, \left\{ \frac{\partial}{\partial x^{\alpha}} \left( -2\kappa\, t_{\sigma}^{\alpha} \right) \right\} + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}}\, T_{\mu\nu} = 0. \]

\[ q.e.d. \]

7.5 Question About an Equation in Einstein’s Original Work (English Edition)

Question:
I again attach the PDF of Einstein’s article for reference (Einstein, Relativity: The Special and General Theory, 1916 – revised edition 1924; also in The Collected Papers of Albert Einstein, 1997). At the bottom of page 191 of the English version, three terms appear separated by equality signs. I do not understand why the first term equals the second. Einstein refers to equation (60), but that does not clarify it for me. Can you explain why these two terms are equal?

Answer:
We first examine equation (60) in Einstein’s original German article.

On page 812 of the German original, there appears to be an error in equation (60): \[ \frac{\partial F_{\rho\sigma}}{\partial x^{\tau}} + \frac{\partial F_{\sigma\tau}}{\partial x^{\rho}} + \frac{\partial F_{\tau\rho}}{\partial x^{\rho}} = 0 \] This is likely incorrect and should read: \[ \frac{\partial F_{\rho\sigma}}{\partial x^{\tau}} + \frac{\partial F_{\sigma\tau}}{\partial x^{\rho}} + \frac{\partial F_{\tau\rho}}{\partial x^{\sigma}} = 0 \tag{60} \] In the English translation (page 189), this has already been corrected.

On page 191 (English edition), we find the following equation: \[ F^{\mu\nu}\,\frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = -\frac{1}{2} F^{\mu\nu}\,\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} = -\frac{1}{2} g^{\mu\alpha} g^{\nu\beta} \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}}. \tag{1} \] We focus on the first equality: \[ F^{\mu\nu}\,\frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = -\frac{1}{2} F^{\mu\nu}\,\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}}. \]

According to equation (60): \[ \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} + \frac{\partial F_{\nu\sigma}}{\partial x^{\mu}} + \frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = 0, \] so that: \[ \frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = -\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} -\frac{\partial F_{\nu\sigma}}{\partial x^{\mu}}. \]

Substituting this into equation (1): \[ F^{\mu\nu}\,\frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = F^{\mu\nu} \left( - \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} - \frac{\partial F_{\nu\sigma}}{\partial x^{\mu}} \right) = -F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} -F^{\mu\nu}\frac{\partial F_{\nu\sigma}}{\partial x^{\mu}}. \]

Now split each term into two equal parts: \[ = -\frac{1}{2}F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} -\frac{1}{2}F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} -\frac{1}{2}F^{\mu\nu}\frac{\partial F_{\nu}{}_{\sigma}}{\partial x^{\mu}} -\frac{1}{2}F^{\mu\nu}\frac{\partial F_{\nu}{}_{\sigma}}{\partial x^{\mu}}. \]

Rearranging dummy indices: \[ = -\frac{1}{2}F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} -\frac{1}{2} \left( F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} +F^{\mu\nu}\frac{\partial F_{\nu}{}_{\sigma}}{\partial x^{\mu}} +F^{\nu\mu}\frac{\partial F_{\mu}{}_{\sigma}}{\partial x^{\nu}} \right) \] Since \(F^{\nu\mu} = -F^{\mu\nu}\), the last term becomes: \[ F^{\nu\mu}\frac{\partial F_{\mu}{}_{\sigma}}{\partial x^{\nu}} = -F^{\mu\nu}\frac{\partial F_{\mu}{}_{\sigma}}{\partial x^{\nu}}. \]

Thus: \[ F^{\mu\nu}\frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = -\frac{1}{2}F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} -\frac{1}{2}\left(F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} +F^{\mu\nu}\frac{\partial F_{\nu}{}_{\sigma}}{\partial x^{\mu}} -F^{\mu\nu}\frac{\partial F_{\mu}{}_{\sigma}}{\partial x^{\nu}} \right) \]

By swapping the indices of \( \frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} \) and changing the sign:

\[ = -\tfrac{1}{2} F^{\mu\nu} \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} - \tfrac{1}{2} \left( F^{\mu\nu} \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} + F^{\mu\nu} \frac{\partial F_{\nu\sigma}}{\partial x^{\mu}} + F^{\mu\nu} \frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} \right) \]

\[ = -\tfrac{1}{2} F^{\mu\nu} \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} - \tfrac{1}{2} F^{\mu\nu} \left( \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} + \frac{\partial F_{\nu\sigma}}{\partial x^{\mu}} + \frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} \right) \]

The expression inside the parentheses is exactly equation (60), which is equal to zero.

Therefore: \[ F^{\mu\nu} \frac{\partial F_{\sigma\mu}}{\partial x^{\nu}} = -\tfrac{1}{2} F^{\mu\nu} \frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} \] \[q.e.d.\]

7.6 Question About Einstein’s Equation (69)

Question:
In equation (69) of Einstein’s book, we find: \[ k = \frac{8\pi K}{c^{2}} = 1.87\times 10^{-27} \quad (E69) \] Why does this number not match the value we use today? And why is there still a \(c^{2}\) in the denominator if Einstein previously stated that one may take \(c = 1\)?

Answer:
Einstein used the CGS unit system (centimeter–gram–second), whereas today we typically use the SI system (meter–kilogram–second). This causes numerical differences in constants such as \(K\) (the gravitational constant), depending on the unit system.

Step 1: Interpretation of Einstein’s notation

In Einstein’s equation: \[ k = \frac{8\pi K}{c^{2}}, \] we have:

Substituting: \[ k = \frac{8\pi \cdot 6.67\times 10^{-8}} { (3.00\times 10^{10})^{2}} \approx 1.87\times 10^{-27}. \] This matches Einstein’s value exactly within the CGS system.

Step 2: Conversion to modern units (SI)

In modern literature, the Einstein field equations are written using: \[ k = \frac{8\pi G}{c^{4}} \approx 2.07 \times 10^{-43}, \] with:

Substituting: \[ k = \frac{8\pi \cdot 6.67\times 10^{-11}} { (3.00\times 10^{8})^{4}} \approx 2.07\times 10^{-43}\ \mathrm{m^{-1}\,kg^{-1}\,s^{2}}. \]

Step 3: Why is there still a \(c^{2}\) in Einstein’s equation?

Although Einstein uses the convention \(c = 1\) (natural units) in other parts of the book, he keeps \(c\) explicit here. This is most likely because he wanted to provide a numerical estimate in concrete physical units, and therefore temporarily refrained from using the convention \(c = 1\).

Conclusion