Einstein’s General Theory of Relativity

Derivations, Applications and Considerations – by Albert Prins

Part VII – Questions and Discussion

7 Answers to Questions

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

Question:
What I find difficult to accept in general relativity is the derivation toward “\(ds\)”. 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:

\begin{equation} ds = c\,d\tau, \end{equation}
where \( \tau \) is the proper time: the time measured by a clock that is at rest relative 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 directly measured time, but a computational parameter that can be derived from \(d\tau\) via the metric.

The relation between both is:

\begin{equation} d\tau = \frac{\sigma}{\gamma}dt, \end{equation}
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:

\begin{equation} c^{2} d\tau^{2} = A\,c^{2} dt^{2} - B\,dx^{2} - D\,dy^{2} - E\,dz^{2}, \end{equation}
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 \(c^{2} d\tau^{2}\):

\begin{equation} 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}. \end{equation}

We rewrite the spatial derivatives using the chain rule:

\begin{equation} \frac{dx}{d\tau} = \frac{dx}{dt}\cdot\frac{dt}{d\tau}, \qquad \text{etc.} \end{equation}

This gives:

\begin{equation} 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}. \end{equation}

Here \(x, y, z\) divided by \(t\) in their own frame represent velocities in that frame.

\begin{equation} 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]. \end{equation}

Now define the velocity:

\begin{equation} 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}. \end{equation}

Substitute \(v\):

\begin{equation} 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}. \end{equation}

This yields:

\begin{equation} \left(\frac{dt}{d\tau}\right)^{2} = \frac{\gamma^{2}}{A} \quad\Longrightarrow\quad d\tau^{2} =\frac{ A}{\gamma^{2}}\,dt^{2} = \frac{\sigma^{2}}{\gamma^{2}} dt^{2}. \end{equation}

Or:

\begin{equation} d\tau =\frac{ \sigma}{\gamma}\,dt. \end{equation}

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

Interpretation:

This derivation establishes the relation between proper time \(\tau\) (as measured by a moving clock in its own rest frame) and 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 — that is, 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 mathematically expressed 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 \( x_{\alpha'} \), 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:

\begin{equation} dx_{\alpha'} = \frac{\partial x_{\beta}}{\partial x_{\alpha'}} \, dx_{\beta} \end{equation}

This formula is written using Einstein notation, meaning that there is an implicit summation over \( \beta \).

This therefore means:

\begin{equation} 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 \end{equation}

For each value of \( \alpha \) (0 to 3), this yields a separate equation expressing each of the new coordinate differentials \( dx_0', dx_1', dx_2', dx_3' \) in terms of the old coordinates.

3. Tensor form

In total we obtain:

\begin{equation} 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 \end{equation}
\begin{equation} 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 \end{equation}
\begin{equation} 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 \end{equation}
\begin{equation} 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{equation}

This can also be represented as a tensor (matrix form):

\begin{equation} \begin{pmatrix} dx_0' \\ dx_1' \\ dx_2' \\ dx_3' \end{pmatrix} = \begin{pmatrix} \frac{\partial x_0}{\partial x_0'} & \frac{\partial x_1}{\partial x_0'} & \frac{\partial x_2}{\partial x_0'} & \frac{\partial x_3}{\partial x_0'} \\ \frac{\partial x_0}{\partial x_1'} & \frac{\partial x_1}{\partial x_1'} & \frac{\partial x_2}{\partial x_1'} & \frac{\partial x_3}{\partial x_1'} \\ \frac{\partial x_0}{\partial x_2'} & \frac{\partial x_1}{\partial x_2'} & \frac{\partial x_2}{\partial x_2'} & \frac{\partial x_3}{\partial x_2'} \\ \frac{\partial x_0}{\partial x_3'} & \frac{\partial x_1}{\partial x_3'} & \frac{\partial x_2}{\partial x_3'} & \frac{\partial x_3}{\partial x_3'} \end{pmatrix} \cdot \begin{pmatrix} dx_0 \\ dx_1 \\ dx_2 \\ dx_3 \end{pmatrix} \end{equation}

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 another system. Important points:

5. Example: Transformation within the Schwarzschild metric

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

\begin{equation} x=r\sin\theta\cos\varphi \quad x=r\sin\theta\sin\varphi \quad x=r\cos\theta \end{equation}

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

7.3 Answer to questions concerning Schwarzschild

Question 1:

Where does the general relativity formula with the Ricci tensor come from, which 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:

\begin{equation} G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R \end{equation}

Here:

The Ricci scalar is given by:

\begin{equation} R=g^{\mu\nu}R_{\mu\nu}=g^{00}R_{00}+g^{11}R_{11}+g_{22}R^{22}+g^{33}R_{33} \end{equation}

Contraction of the Einstein field equations with \( g^{\mu\nu} \) yields:

\begin{equation} 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}4R=-R \end{equation}

The full Einstein field equations are:

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

Where:

Outside a massive sphere there is no matter or energy. In that case \( T_{\mu\nu}=0 \), and the field equations reduce to:

\begin{equation} G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=0 \end{equation}

We know that:

\begin{equation} 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}4R_{\mu\nu}=-R_{\mu\nu} \end{equation}

From which it follows:

\begin{equation} G_{\mu\nu}=0 \text{ only if } R_{\mu\nu}=0 \text{ and thus also } R=0 \end{equation}

Background: from Riemann to Ricci

Einstein built upon the work of Riemann, who had already developed a mathematical description of curved surfaces. The Riemann tensor:

\begin{equation} R_{\mu\beta\rho\nu} \end{equation}

This 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:

\begin{equation} R^{\beta}_{\mu\rho\nu}=g^{\beta\beta}R_{\mu\beta\rho\nu} \end{equation}

This is required to perform the desired contraction. By setting \( \beta=\rho \), the contraction yields the Ricci tensor \( R_{\mu\nu} \).

\begin{equation} R^{\beta}_{\mu\beta\nu}=R_{\mu\nu} \end{equation}

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:

\begin{equation} R_{\mu\nu}=R_{\mu\rho\nu}^{\rho}=\Gamma_{\mu\nu,\rho}^{\rho}-\Gamma_{\rho\mu,\nu}^{\rho}+ \Gamma_{\rho\lambda}^{\rho}\Gamma_{\nu\mu}^{\lambda}-\Gamma_{\nu\lambda}^{\rho} \Gamma_{\rho\mu}^{\lambda} \end{equation}

where the Christoffel symbol itself is:

\begin{equation} \Gamma_{\mu\nu}^{\rho}=\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) \end{equation}

The derivative of the Christoffel symbol is then:

\begin{equation} \Gamma_{\mu\nu,\gamma}^{\rho}=\frac{\partial \Gamma_{\mu\nu}^{\rho}}{\partial x^{\gamma}}= - g^{\rho\alpha}\cdot \frac{\partial g_{\rho\alpha}}{\partial x_{\gamma}}\cdot \Gamma_{\mu\nu}^{\rho} +\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) \end{equation}

When calculating the Ricci components \( R_{00},R_{11},R_{22},R_{33} \) using the full Einstein field equations, they are all zero, which is correct. However, when using the restricted form of the field equations \( (g=-1) \), the result is not correct. Thus, the Schwarzschild equation satisfies the general field equations but not the restricted version. This is consistent because for the Schwarzschild solution \( g\neq -1 \).

On the Schwarzschild solution and the restriction g=−1

Schwarzschild uses the well-known polar equation. 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, since the latter requires \( g=-1 \). Schwarzschild derived a transformation based on modified polar coordinates, choosing the transformation such that \( g=-1 \) is achieved. In that case, the equation also satisfies the restricted Einstein field equations.

Conclusion

Although Schwarzschild attempted to meet Einstein’s requirement of having metric trace \( g=-1 \), in my view the only relevant requirement is that the Einstein field equations, with \( T_{\mu\nu}=0 \), and thus \( R_{00}=R_{11}=R_{22}=R_{33}=0 \), are satisfied, regardless of whether \( g=-1 \) or \( g\neq -1 \). Therefore, the condition \( g=-1 \) 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, the 222 \( (\Gamma_{22}^{2}) \) appears to be absent. This is because your definition of the metric tensor g differs from that of Schwarzschild; \( g_{22} \) and \( g_{33} \) are -1 for Schwarzschild, whereas you include the coordinate r (for example \( g_{22}=-r^2 \)). Droste (1917), Eddington (1921), MWT (1975) and OAS (2007) also adhered to \( g=-1 \) for the Schwarzschild solution, such that: \( g_{22}=g_{33}=-1 \). This raises the question for me: do you think that \( g=-1 \) is required for the Schwarzschild solution?

Answer:

In Schwarzschild’s original derivation, the starting point was the Cartesian coordinate system (x, y, z). The metric in that form has the following components:

\begin{equation} g_{00}=\sigma^2 \quad g_{11}=-\frac{1}{r^4\sigma^2} \quad g_{22}=-r^2\sin^2\theta \quad g_{33}=-r^2\sin^2\theta \end{equation}

In this case, ten (or fourteen) relevant Christoffel symbols are generated. You can also see in my overview of formulas that I have derived expressions for both the spherical and the Cartesian x, y, z forms. In the x, y, z form, 222 \( (\Gamma_{22}^{2}) \) does exist.

However, for the spherical form this is different; there the components of the metric tensor are:

\begin{equation} g_{00}=\sigma^2 \quad g_{11}=-\frac{1}{\sigma^2} \quad g_{22}=-r^2 \quad g_{33}=-r^2\sin^2\theta \end{equation}

This is exactly the same form as in Schwarzschild. In these spherical coordinates, \( g_{22} \) and \( g_{33} \) explicitly depend on \( r \) and \( \theta \), and therefore cannot be constant as in \( g_{22}=g_{33}=-1 \). If these values were constant, the partial derivatives \( \frac{\partial g_{22}}{\partial r}, \frac{\partial g_{22}}{\partial \theta}, \frac{\partial g_{33}}{\partial r}, \frac{\partial g_{33}}{\partial \theta} \) would all be zero. As a result, many Christoffel symbols, including crucial ones such as \( \Gamma_{221} \) and \( \Gamma_{222} \), would also vanish.

This also applies to Schwarzschild! The elements \( g_{22} \) and \( g_{33} \) cannot be -1, because in that case \( \frac{\partial g_{22}}{\partial r}, \frac{\partial g_{22}}{\partial \theta}, \frac{\partial g_{33}}{\partial r}, \frac{\partial g_{33}}{\partial \theta} \) would be zero and the number of Christoffel symbols would be limited to 001 \( (\Gamma^{0}_{01})\), 010 \( (\Gamma^{0}_{10}) \), 100 \(( \Gamma^{1}_{00}) \) and 111 \( (\Gamma_{11}^{1} )\).

Regarding \( \Gamma_{22}^{2} \):

For spherical coordinates, this component is indeed zero, because \( g_{22} \) does not depend on \( \theta \), and the derivative is therefore zero:

\begin{equation} \Gamma_{22}^{2}=\frac{1}{2}g^{22}\frac{\partial g_{22}}{\partial \theta}=0 \end{equation}

It is important, however, that when evaluating components at a specific value such as \( \theta=\frac{\pi}{2} \) (i.e. 90°), substitution may only be performed at the end of the calculations.

For example:

\begin{equation} \Gamma_{33}^{2}=\frac{1}{2}g^{22}\left(-\frac{\partial g_{33}}{\partial \theta}\right) =-\cos\theta\sin\theta=0 \text{ when } \theta=90^\circ \end{equation}

This value becomes zero when \( \theta=\frac{\pi}{2} \), but for the Ricci component we also need the derivative of this Christoffel symbol with respect to \( \theta \), which is:

\begin{equation} \frac{\partial \Gamma_{33}^{2}}{\partial \theta}=-\cos^2\theta+\sin^2\theta=1 \text{ when } \theta=90^\circ \end{equation}

And this is therefore 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 simplifies the algebra in many cases and provides symmetry. However, in my view it leads to an unnecessary restriction. It also depends on the type of coordinate system chosen. For example, the metric tensor elements in t, x, y, z coordinates do indeed yield \( \det(g) = -1 \):

\begin{equation} \sigma^2 \cdot \left(-\frac{1}{r^4\sigma^2}\right) \cdot\left( -\frac{r^2}{\sin^2\theta} \right) \cdot\left( -r^2\sin^2\theta\right) = -1 \end{equation}

But in spherical coordinates:

\begin{equation} \sigma^2 \cdot -\frac{1}{\sigma^2} \cdot \left(-r^2\right) \cdot \left(-r^2\sin^2\theta\right) = - r^4\sin^2\theta \end{equation}

And thus here \( \det(g)\neq -1 \). Nevertheless, this metric perfectly satisfies the Einstein field equations in vacuum \( (T_{\mu\nu}=0) \), which implies that \( R_{\mu\nu}=0 \) and therefore also \( R=0 \).

Conclusion: The requirement \( \det(g)=-1 \) is a coordinate-dependent convention that can provide 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 primarily intended to meet Einstein’s preferences, but is physically superfluous.

Question 3:

The field equations in your document on pages 2 and 3, based on the Ricci tensor, differ significantly from those which 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 refer to in your question, you mean the restricted field equations of Einstein. As I have shown theoretically, Schwarzschild satisfies the general field equations but not the restricted field equations G. This is because Einstein introduced the additional requirement \( \det(g)=-1 \) in order to obtain simpler formulas, but this led to an unnecessary restriction of possible solutions, such as the Schwarzschild solution based on spherical coordinates. Meanwhile, this solution remains an excellent one for calculating physical phenomena in vacuum in a relatively simple way.

Question 4:

I still have some difficulty understanding the Schwarzschild equation and Einstein’s field equations. Could you elaborate further on this?

Answer:

It seems that we are once again entering a discussion that we have had before. Let me first make it clear: it is not my intention to defend the Schwarzschild or Einstein solution against your approach, nor to criticize your proposed modification of the Schwarzschild metric. My aim is to achieve a complete understanding. As long as I do not fully comprehend the Schwarzschild solution, I will continue to seek 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 already know the full answer, but I want to explain how I currently understand the structure.

The starting point of Einstein

Einstein sought a description of gravity in which gravity is no longer a force, as in Newtonian physics, but rather a consequence of the curvature of spacetime. He aimed to find a coordinate system in which no forces are felt, so that a freely falling particle moves without acceleration — in a sense “naturally,” without external cause.

In classical mechanics, an object moves at 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 spacetime. These geodesics are, in a sense, the “straight lines” of curved spacetime.

Einstein therefore sought 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:

\begin{equation} R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=0 \end{equation}

in the vacuum case (outside a mass), where:

This equation is covariant: it is valid in any arbitrary 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 coordinate system. The Ricci tensor and 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 distance \( ds^2 \) between two infinitesimally close points is computed. In the simplest case (for example flat space in Cartesian coordinates), this is:

\begin{equation} ds^2=c^2dt^2-dx^2-dy^2-dz^2 \end{equation}

But in curved spacetime, \( ds^2 \) depends on location and on the metric components. The metric contains information about the geometric structure of space, including possible cross terms (such as \( dx\,dy \)) if the coordinates are not orthogonal.

For comparison:

Analogously, Einstein considered spacetime as composed of infinitely many infinitesimal flat patches, where geometry is locally flat (via the equivalence principle). Within these small regions we still use a coordinate system, but the metric components vary from point to point — encoding 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 with as much symmetry as possible:

This yields the Schwarzschild metric:

\begin{equation} ds^2=\left(1-\frac{2GM}{c^2 r}\right)c^2dt^2-\left(1-\frac{2GM}{c^2 r}\right)^{-1}dr^2-r^2d\theta^2-r^2\sin^2\theta\,d\varphi^2 \end{equation}

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

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

Each point 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 a trajectory in this curved spacetime.

Summary

Interpretation:
The Schwarzschild solution satisfies the full Einstein field equations, regardless of whether \( g=-1 \). The condition \( g=-1 \) only restricts the number of possible solutions, but is not physically necessary.

7.4 Detailed derivation of Einstein Equation (57) from Equation (53)

Question:

I am reading Einstein’s original GR paper. I have attached it as a PDF to this email. (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 (multiplying 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 indicated by Einstein, and based solely on the preceding material in Einstein’s article? Could you show me the detailed steps you took to arrive at 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:

\begin{equation} \frac{\partial \Gamma_{\mu\nu}^{\alpha}}{\partial x^\alpha} +\Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} =-\kappa\left( T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right) \end{equation}

Here:

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

\begin{equation} \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \left( \frac{\partial \Gamma_{\mu\nu}^{\alpha}}{\partial x^\alpha} +\Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} \right) = \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \left( -\kappa \left(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T\right) \right) \end{equation}

This leads to:

\begin{equation} =-\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) \end{equation}

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

From equation (29) in Einstein’s paper:

\begin{equation} \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} \end{equation}

Since \( \sqrt{-g}=-1 \), we have:

\begin{equation} -\frac{1}{2}g_{\mu\nu}\frac{\partial g^{\mu\nu}}{\partial x^\sigma}=0 \end{equation}

Thus:

\begin{equation} = -\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} \end{equation}

The equation becomes:

\begin{equation} \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \frac{\partial \Gamma_{\mu\nu}^{\alpha}}{\partial x^\alpha} + \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} + \kappa \frac{\partial g^{\mu\nu}}{\partial x^\sigma}T_{\mu\nu} =0 \end{equation}

The next step:

\begin{equation} \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 \end{equation}

Substitute this into:

\begin{equation} \frac{1}{2\kappa} \left(\frac{\partial}{\partial x^\alpha}\left(-2\kappa t^\alpha_\sigma \right) \right) + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} T_{\mu\nu} = 0 \end{equation}

(See below for the derivation of the step from equation (50) to (51).)

This leads to:

\begin{equation} - \frac{\partial t_{\sigma}^{\ \alpha}}{\partial x^{\alpha}} + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} T_{\mu\nu} = 0 \end{equation}

Now use Einstein’s equation (56):

\begin{equation} \frac{\partial (t_{\mu}^{\sigma} + T_{\mu}^{\sigma})}{\partial x^{\sigma}} = 0 \end{equation}

Thus:

\begin{equation} \frac{\partial t_{\mu}^{\sigma}}{\partial x^{\sigma}} = - \frac{\partial T_{\mu}^{\sigma}}{\partial x^{\sigma}} \end{equation}

Replace \(\sigma\) by \(\alpha\), and \(\mu\) by \(\sigma\):

\begin{equation} \frac{\partial t_{\sigma}^{\alpha}}{\partial x^{\alpha}} = - \frac{\partial T_{\sigma}^{\alpha}}{\partial x^{\alpha}} \end{equation}

The equation (52) becomes:

\begin{equation} \frac{\partial T_{\sigma}^{\alpha}}{\partial x^{\alpha}} + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} T_{\mu\nu} = 0 \end{equation}

This is exactly Einstein’s equation (57).

Derivation of the step from equation (50) to (51) above:

To prove that:

\begin{equation} \frac{\partial}{\partial x^{\alpha}}\left(-2\kappa t_{\sigma}^{\ \alpha}\right) = \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \frac{\partial \Gamma_{\mu\nu}^{\alpha}}{\partial x^{\alpha}} + \frac{\partial g^{\mu\nu}}{\partial x^{\sigma}} \Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} \end{equation}

We use Einstein equation (48):

\begin{equation} \frac{\partial H}{\partial g^{\mu\nu}} = -\Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} \end{equation}
\begin{equation} \frac{\partial H}{\partial g_{\sigma}^{\mu\nu}} = \Gamma_{\mu\nu}^{\sigma} \end{equation}

Einstein equation (47b):

\begin{equation} \frac{\partial}{\partial x^{\alpha}} \left( \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial H}{\partial g^{\mu\nu}} = 0 \Rightarrow \frac{\partial}{\partial x^{\alpha}} \left( \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) = \frac{\partial H}{\partial g^{\mu\nu}} \end{equation}
The left-hand side of equation (50):

\begin{equation} \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \frac{\partial \Gamma_{\mu\nu}^{\alpha}}{\partial x^\alpha} + \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} \end{equation}

Can be rewritten as:

\begin{equation} \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}} \end{equation}

We can now differentiate:

\begin{equation} \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 g^{\mu\nu}}{\partial x^\sigma} \frac{\partial H}{\partial g^{\mu\nu}} \end{equation}

Here it holds:

\begin{equation} \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} \end{equation}

Substitute this:

\begin{equation} \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} \end{equation}

As stated in Einstein’s document under equation (47a), \( H \) is considered a function of \( g^{\mu\nu} \) and \( g_{\sigma}^{\mu\nu}=\frac{\partial g^{\mu\nu}}{\partial x^\sigma} \), thus:

\begin{equation} \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} \end{equation}

Substitute this:

\begin{equation} \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} \end{equation}

Or:

\begin{equation} \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) \end{equation}
\begin{equation} = \frac{\partial}{\partial x^\alpha} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} \right) - \frac{\partial}{\partial x^\alpha} (\delta_{\sigma}^{\alpha}H) \end{equation}

\begin{equation} = \frac{\partial}{\partial x^\alpha} \left( g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} - \delta_{\sigma}^{\alpha}H \right) \end{equation}

According to Einstein equation (49):

\begin{equation} -2\kappa t_{\sigma}^{\alpha} = g_{\sigma}^{\mu\nu} \frac{\partial H}{\partial g_{\alpha}^{\mu\nu}} - \delta_{\sigma}^{\alpha}H \end{equation}

Substitute this into equation (50):

\begin{equation} \frac{1}{2\kappa} \left( \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \frac{\partial \Gamma_{\mu\nu}^{\alpha}}{\partial x^\alpha} + \frac{\partial g^{\mu\nu}}{\partial x^\sigma} \Gamma_{\mu\beta}^{\alpha}\Gamma_{\nu\alpha}^{\beta} \right) + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^\sigma}T_{\mu\nu} =0 \end{equation}

Becomes:

\begin{equation} \frac{1}{2\kappa}\left[ \frac{\partial}{\partial x^\alpha} (-2\kappa t_{\sigma}^{\alpha})\right] + \frac{1}{2} \frac{\partial g^{\mu\nu}}{\partial x^\sigma}T_{\mu\nu} =0 \end{equation}

q.e.d.

7.5 Question about an equation in Einstein’s original work (English version)

Question:

I am attaching again the PDF file 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, there are three terms separated by equal signs. I do not understand why the first term is equal to the second. Einstein refers to equation (60), but that does not help 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):

\begin{equation} \frac{\partial F_{\varrho\sigma}}{\partial x^\tau} + \frac{\partial F_{\sigma\tau}}{\partial x^\varrho} + \frac{\partial F_{\tau\varrho}}{\partial x^\varrho} =0 \end{equation}

It should likely be:

\begin{equation} \frac{\partial F_{\varrho\sigma}}{\partial x^\tau} + \frac{\partial F_{\sigma\tau}}{\partial x^\varrho} + \frac{\partial F_{\tau\varrho}}{\partial x^\sigma} =0 \end{equation}

In the English translation this is corrected.

On page 191 we find:

\begin{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} \end{equation}

Why does:

\begin{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}? \end{equation}

According to (60):

\begin{equation} \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 \end{equation}

Thus:

\begin{equation} \frac{\partial F_{\sigma\mu}}{\partial x^\nu} = -\frac{\partial F_{\mu\nu}}{\partial x^\sigma} - \frac{\partial F_{\nu\sigma}}{\partial x^\mu} \end{equation}

Substitute:

\begin{equation} 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_{\mu\nu}}{\partial x^\sigma} \end{equation}

Now we split each term into two equal parts:

\begin{equation} = -\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}} \end{equation}
\begin{equation} = -\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_{\nu\sigma}}{\partial x^{\mu}}\right) \end{equation}

Now we rearrange the indices (dummy indices may be swapped):

\begin{equation} = -\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) \end{equation}

Since \(F^{\nu\mu} = -F^{\mu\nu}\), the last term changes sign:

\begin{equation} = -\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) \end{equation}

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

\begin{equation} = -\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_{\sigma\mu}}{\partial x^{\nu}}\right) \end{equation}
\begin{equation} = -\frac{1}{2} F^{\mu\nu}\frac{\partial F_{\mu\nu}}{\partial x^{\sigma}} - \frac{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) \end{equation}

The expression in parentheses is exactly equation (60), which equals zero. Thus:

\begin{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}} \end{equation}

q.e.d.

7.6 Question about Einstein’s equation (69)

Question:
In equation (69) from Einstein’s book it states:

\begin{equation} k = \frac{8\pi K}{c^2} = 1.87 \cdot 10^{-27} \end{equation}
Why does this value not match the one we use today? And why is there still a \( c^2 \) in the denominator if Einstein previously stated that \( c = 1 \)?

Answer:
Einstein worked in this book with CGS units (centimeter–gram–second), whereas today we usually use SI units (meter–kilogram–second). This causes differences in the numerical values of physical constants such as \( K \) (the gravitational constant), depending on the system of units.

Step 1: Interpretation of Einstein’s notation

If we substitute these values:

\begin{equation} k = \frac{8\pi \cdot 6.67 \cdot 10^{-8}}{(3.00 \cdot 10^{10})^2} \approx 1.87 \cdot 10^{-27} \end{equation}
This is exactly the value Einstein gives. So his calculation is correct within the CGS system.

Step 2: Conversion to modern units (SI)

In modern literature, we use for the Einstein field equations:

\begin{equation} k = \frac{8\pi G}{c^4} \approx 2.07 \cdot 10^{-43} \end{equation}
with:

Substituting:

\begin{equation} k = \frac{8\pi \cdot 6.67 \cdot 10^{-11}}{(3.00 \cdot 10^{8})^4} \approx 2.07 \cdot 10^{-43} \, \text{m}^{-1} \text{kg}^{-1} \text{s}^2 \end{equation}

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

Although Einstein uses the convention \( c = 1 \) (natural units) elsewhere in the book, he keeps \( c \) explicit here. This is likely because at this point he wanted to provide a numerical estimate corresponding to physical units, and therefore temporarily did not apply the convention \( c = 1 \).

Conclusion: