Schwarzschild’s metric and the Landau–Lifshitz formulation of general relativity

Abstract

Fromholz, Poisson and Will (Am J Phys 82:295–300, 2013) examine Schwarzschild’s original 1916 derivation of the now famous metric associated with his name, along with a derivation using the Landau–Lifshitz formulation of the Einstein field equations of general relativity. These authors exploit the Landau–Lifshitz formulation to derive a nonlinear second-order ordinary differential equation which they were unable to fully solve, the general solution of which, may or may not, produce further metrics which are distinct from the Schwarzschild line element. In this paper, we examine in some detail the particular Landau–Lifshitz formulation given in Fromholz, Poisson and Will (Am J Phys 82:295–300, 2013), and we obtain both special and general solutions of their nonlinear second-order ordinary differential equation. Clearly, knowledge of the general solution gives rise to the prospect of determining new solutions which may be other than the Schwarzschild line element in disguise. However, the present writer questions the validity of the analysis of Fromholz, Poisson and Will (Am J Phys 82:295–300, 2013) and two specific objections are raised: one related to the harmonic gauge condition and the other related to the determinants connecting the line element metric to the gothic metric.

Appendices

Appendices

A Most general Schwarzschild metric with $A(r) = B(r)$

In this appendix, we determine the most general Schwarzschild metric of the form $(\mathrm{{d}}s)^2 = A(r)(\mathrm{{d}}r)^2 + B(r)r^2(\mathrm{{d}}\sigma )^2 - C(r)(\mathrm{{d}}t)^2$ subject to the requirement $A(r) = B(r)$. If $A(r) = B(r)$, then from (1.4)$_{1}$ we have immediately that $C(r) = 1 - 2\,m/rB^{1/2}$, and on substitution into (1.4)$_{2}$, we may deduce the nonlinear first-order ordinary differential equation

$$\begin{aligned} \frac{(rB')^2}{4} + rBB' + \frac{2mB^{3/2}}{r} = 0, \end{aligned}$$

which on using $F(r) = B(r)^{1/2}$ may be solved as a quadratic equation to deduce $rF' = -F \pm (F^2 - 2mF/r)^{1/2}$, and the subsequent substitution $F(r) = G(r)/r$ produces $rG' = \pm (G^2 - 2\,mG)^{1/2}$. The resulting integral may be evaluated using the successive substitutions $u(r) = G(r)^{1/2}$ and $u(r) = (2\,m)^{1/2}\cosh \theta$ to obtain $\theta = \pm (\log r)/2 + \theta _0$, where $\theta _0$ denotes the constant of integration. On re-tracing these steps, we may eventually deduce that

$$\begin{aligned} A(r) = B(r) = \left( \frac{m}{2}\right) ^2\left( e^{\theta _0} + \frac{e^{-\theta _0}}{r}\right) ^4,\quad C(r) = 1- 4\left( r^{1/2}e^{\theta _0} + \frac{e^{-\theta _0}}{r^{1/2}}\right) ^{-2}, \end{aligned}$$

as the most general Schwarzschild metric of the form $(\mathrm{{d}}s)^2 = A(r)(\mathrm{{d}}r)^2 + B(r)r^2(\mathrm{{d}}\sigma )^2 - C(r)(\mathrm{{d}}t)^2$ with $A(r) = B(r)$, and noting that C(r) may be simplified to give $C(r) = \tanh ^{2}(\theta _0 + (\log r)/2)$, and $\theta _0$ denotes an arbitrary constant.

B Derivation of $\beta ' = 2(\alpha - \beta )/r$ using Cartesian-like coordinates

In Cartesian-like coordinates $(x^{1}, x^{2}, x^{3}, x^{4}) = (x, y, z, t)$, [8] propose the following gothic inverse metric ${\mathfrak {g}}^{ij}$

$$\begin{aligned}&{\mathfrak {g}}^{ij} = \alpha (r)\delta ^{ij} + (\beta (r) - \alpha (r))\frac{x^{i}x^{j}}{r^2} \quad \quad (i, j = 1, 2, 3), \\&\quad \quad {\mathfrak {g}}^{44} = - N(r), \quad \quad {\mathfrak {g}}^{4j} = 0, \quad \quad (j= 1, 2, 3), \end{aligned}$$

where $\alpha (r)$, $\beta (r)$ and N(r) denote functions of $r = (x^2 + y^2 + z^2)^{1/2}$ only, and we emphasise that since, in this appendix, the indices refer to Cartesian-like coordinates, the level of the index is immaterial. Now from these expressions, and the so-called harmonic gauge condition

$$\begin{aligned} \frac{\partial {\mathfrak {g}}^{ij}}{\partial x^{j}} = \frac{\partial {\mathfrak {g}}^{i1}}{\partial x^{1}} + \frac{\partial {\mathfrak {g}}^{i2}}{\partial x^{2}} + \frac{\partial {\mathfrak {g}}^{i3}}{\partial x^{3}} + \frac{\partial {\mathfrak {g}}^{i4}}{\partial x^{4}} = 0, \end{aligned}$$

(B.1)

for $i = 1$ we obtain

$$\begin{aligned} \frac{\partial }{\partial x}\left( \alpha + \frac{(\beta - \alpha )x^2}{r^2}\right) + \frac{\partial }{\partial y}\left( \frac{(\beta - \alpha )xy}{r^2}\right) + \frac{\partial }{\partial z}\left( \frac{(\beta - \alpha )xz}{r^2}\right) = 0, \end{aligned}$$

and on simplifying this equation, we might deduce

$$\begin{aligned} \frac{x}{r}\left( \beta ' + \frac{2(\beta - \alpha )}{r}\right) = 0. \end{aligned}$$

Similarly, for $i = 2$ and $i = 3$ we might obtain the corresponding conditions

$$\begin{aligned} \frac{y}{r}\left( \beta ' + \frac{2(\beta - \alpha )}{r}\right) = 0, \quad \quad \quad \frac{z}{r}\left( \beta ' + \frac{2(\beta - \alpha )}{r}\right) = 0, \end{aligned}$$

and from which the condition $\beta ' = 2(\alpha - \beta )/r$ is apparent.

C Gothic inverse metric ${\mathfrak {g}}^{ij}$ in spherical polar coordinates

In this appendix, we determine the components of the gothic inverse metric ${\mathfrak {g}}^{ij}$, as given by (B.1), in terms of spherical polar coordinates $(x^{1}, x^{2}, x^{3}, x^{4}) = (r, \theta , \phi , t)$ where from (1.3), $(r, \theta , \phi )$ are given by

$$\begin{aligned} r = (x^2 + y^2 + z^2)^{1/2}, \quad \quad \theta = \tan ^{-1}\left( \frac{(x^2 + y^2)^{1/2}}{z}\right) , \quad \quad \phi = \tan ^{-1}\left( \frac{y}{x}\right) . \end{aligned}$$

Formally, under a coordinate transformation, say from $x^{i}$ to $x^{*i}$, a contravariant tensor of rank two, transforms according to the rules (see, for example, [19], page 8 or [6], page 435)

$$\begin{aligned} {\mathfrak {g}}^{*ij} = \frac{\partial x^{*i}}{\partial x^{m}}\frac{\partial x^{*j}}{\partial x^{n}} {\mathfrak {g}}^{mn}, \end{aligned}$$

so that, for example, with $i = j = 1$, we have

$$\begin{aligned} {\mathfrak {g}}^{*11} =&\, \frac{\partial r}{\partial x^{m}}\frac{\partial r}{\partial x^{n}} {\mathfrak {g}}^{mn} = \frac{x}{r}\frac{\partial r}{\partial x^{n}} {\mathfrak {g}}^{1n} + \frac{y}{r}\frac{\partial r}{\partial x^{n}} {\mathfrak {g}}^{2n} + \frac{z}{r}\frac{\partial r}{\partial x^{n}} {\mathfrak {g}}^{3n}, \\ =&\, \left( \frac{x}{r}\right) ^2{\mathfrak {g}}^{11} + \left( \frac{y}{r}\right) ^2{\mathfrak {g}}^{22} + \left( \frac{z}{r}\right) ^2{\mathfrak {g}}^{33} + 2\frac{xy}{r^2}{\mathfrak {g}}^{12} + 2\frac{xz}{r^2}{\mathfrak {g}}^{13} + 2\frac{yz}{r^2}{\mathfrak {g}}^{23}, \\ =&\, \alpha + \frac{(\beta - \alpha )}{r^4}(x^4 + y^4 + z^4 + 2x^2y^2 + 2 x^2z^2 + 2y^2z^2), \\ =&\, \beta . \end{aligned}$$

Similarly, on using the partial derivatives

$$\begin{aligned} \frac{\partial \theta }{\partial x} =&\, \frac{xz}{r^2(x^2 + y^2)^{1/2}}, \quad \quad \frac{\partial \theta }{\partial y} = \frac{yz}{r^2(x^2 + y^2)^{1/2}}, \quad \quad \frac{\partial \theta }{\partial z} = -\frac{(x^2 + y^2)^{1/2}}{r^2}, \\ \frac{\partial \phi }{\partial x} =&\, -\frac{y}{(x^2 + y^2)}, \quad \quad \frac{\partial \phi }{\partial y} = \frac{x}{(x^2 + y^2)}, \quad \quad \frac{\partial \phi }{\partial z} = 0, \end{aligned}$$

along with the formulae

$$\begin{aligned} {\mathfrak {g}}^{*22} =&\, \frac{\partial \theta }{\partial x^{m}}\frac{\partial \theta }{\partial x^{n}} {\mathfrak {g}}^{mn}, \\ =&\, \left( \frac{\partial \theta }{\partial x}\right) ^2{\mathfrak {g}}^{11} + \left( \frac{\partial \theta }{\partial y}\right) ^2{\mathfrak {g}}^{22} + \left( \frac{\partial \theta }{\partial z}\right) ^2{\mathfrak {g}}^{33} \\&+ 2\frac{\partial \theta }{\partial x}\frac{\partial \theta }{\partial y} {\mathfrak {g}}^{12} + 2\frac{\partial \theta }{\partial x}\frac{\partial \theta }{\partial z} {\mathfrak {g}}^{13} + 2\frac{\partial \theta }{\partial y}\frac{\partial \theta }{\partial z} {\mathfrak {g}}^{23}, \\ {\mathfrak {g}}^{*33} =&\, \frac{\partial \phi }{\partial x^{m}}\frac{\partial \phi }{\partial x^{n}} {\mathfrak {g}}^{mn}, \\ =&\, \left( \frac{\partial \phi }{\partial x}\right) ^2{\mathfrak {g}}^{11} + \left( \frac{\partial \phi }{\partial y}\right) ^2{\mathfrak {g}}^{22} + \left( \frac{\partial \phi }{\partial z}\right) ^2{\mathfrak {g}}^{33} \\&+ 2\frac{\partial \phi }{\partial x}\frac{\partial \phi }{\partial y} {\mathfrak {g}}^{12} + 2\frac{\partial \phi }{\partial x}\frac{\partial \phi }{\partial z} {\mathfrak {g}}^{13} + 2\frac{\partial \phi }{\partial y}\frac{\partial \phi }{\partial z} {\mathfrak {g}}^{23}, \end{aligned}$$

we may eventually deduce that ${\mathfrak {g}}^{22} = \alpha /r^2$ and that ${\mathfrak {g}}^{*33} = \alpha /(r\sin \theta )^2$, while for ${\mathfrak {g}}^{*44}$, we have simply

$$\begin{aligned} \quad \quad {\mathfrak {g}}^{*44} = \frac{\partial x^4}{\partial x^{m}}\frac{\partial x^4}{\partial x^{n}} {\mathfrak {g}}^{mn} = {\mathfrak {g}}^{44} = - N(r), \end{aligned}$$

and all off diagonal components of ${\mathfrak {g}}^{ij}$ are zero. Thus, altogether, in spherical polar coordinates $(r, \theta , \phi , t)$, the only nonzero components of the gothic inverse metric ${\mathfrak {g}}^{ij}$ are given by

$$\begin{aligned} {\mathfrak {g}}^{ij} = \begin{pmatrix} \beta (r) &{} 0 &{} 0 &{} 0 \\ 0 &{} {\alpha (r)}/{r^2} &{} 0 &{} 0 \\ 0 &{} 0 &{} {\alpha (r)}/{(r\sin \theta )^2} &{} 0 \\ 0 &{} 0 &{} 0 &{} - N(r) \end{pmatrix}, \end{aligned}$$

(C.1)

and therefore the only nonzero components of the gothic metric tensor ${\mathfrak {g}}_{ij}$ in spherical polar coordinates are given by

$$\begin{aligned} {\mathfrak {g}}_{ij} = \begin{pmatrix} 1/\beta (r) &{} 0 &{} 0 &{} 0 \\ 0 &{} {r^2}/{\alpha (r)} &{} 0 &{} 0 \\ 0 &{} 0 &{} {(r\sin \theta )^2}/{\alpha (r)} &{} 0 \\ 0 &{} 0 &{} 0 &{} -1/N(r) \end{pmatrix}. \end{aligned}$$

D Verifying the first integral using V and $\rho$

In this appendix, for the sake of completeness, we present the details to verify the first integral (4.3) using the V and $\rho$ variables which are defined by $V(r) = W(r)r^{-1/2}$ and $\rho = \log r$. For the first approach, using $U = dV/d\rho + V/2$, equations (4.1) and the first integral (4.3) become, respectively

$$\begin{aligned} \frac{\mathrm{{d}}U}{\mathrm{{d}}\rho } = \left( \frac{3}{2} + \frac{C^{*}}{V^2}\right) U, \quad \quad 2UV = \left( U + \frac{C^{*}}{V}\right) ^2 - k^2e^\rho , \end{aligned}$$

so that on differentiating the latter equation with respect to $\rho$, we have

$$\begin{aligned} 2V\frac{\mathrm{{d}}U}{\mathrm{{d}}\rho } + 2U\frac{\mathrm{{d}}V}{\mathrm{{d}}\rho } = 2\left( U + \frac{C^{*}}{V}\right) \left( \frac{\mathrm{{d}}U}{\mathrm{{d}}\rho } - \frac{C^{*}}{V^2}\frac{\mathrm{{d}}V}{\mathrm{{d}}\rho }\right) - k^2e^\rho , \end{aligned}$$

which may be re-arranged to give

$$\begin{aligned} 2\left\{ V - \left( U + \frac{C^{*}}{V}\right) \right\} \frac{\mathrm{{d}}U}{\mathrm{{d}}\rho } =&\, - 2\left\{ U + \left( U + \frac{C^{*}}{V}\right) \frac{C^{*}}{V^2}\right\} \frac{\mathrm{{d}}V}{\mathrm{{d}}\rho } - k^2e^\rho , \\ =&\, - 2\left\{ U + \left( U + \frac{C^{*}}{V}\right) \frac{C^{*}}{V^2}\right\} \left( U - \frac{V}{2}\right) + 2UV - \left( U + \frac{C^{*}}{V}\right) ^2, \\ =&\, 2\left\{ V - \left( U + \frac{C^{*}}{V}\right) \right\} \left( \frac{3}{2} + \frac{C^{*}}{V^2}\right) U, \end{aligned}$$

where the final line is obtained by expanding out the previous line and re-factorising, and clearly the required result follows.