# Number Theory/Singularities

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A mathematical singularity, is a point (in a equation) at which it is not defined or its behaviour is either not understood or uncontrollable. This can be due to many reasons for example divergencies, not being differentiable or having a (definition/logic) gap. Since physical systems are described by mathematical equations this has great implications as unphysical results can occur as for example divergencies cause in a system of finite quantitative property to tend to infnite or a gap causes the used coordinate system to have a point, from which a physical system cannot be described anymore.

Physical singularities, such as $\pm \infty$ , do not exist in the real world. Usually they are a sign, that current knowledge is insufficient to describe what happens at extreme densities, temperatures, forces, $\dots$ .

## Example

The best example for discussion is the Schwarzschild metric/solution, which appears to have singularities at $r = 0$ and $r = \frac{2GM}{c^2} = r_s$ .

The singularity at $r = r_s$ is an illusion; it is a coordinate singularity, which arises from a bad choice of coordinates or coordinate conditions. As soon as the coordinate system is changed (for example to Kruskal$-$ Szekeres coordinates), then the metric becomes regular at $r = r_s$ and can extend the exterior to values of $r < r_s$ . Thus, one can expand the external metric to the inner $-$ . This is a direct effect of the solution process. To get the Schwarzschild metric the Einstein field equation have to be solved, but they are solved outside of the Schwarzschild radius and in the vaccum. Hence, for $r = r_s$ we have a bad choice of coordinates.

The case $r = 0$ seems different, since the curvature becomes infinite; thus, indicating the presence of a physical singularity. To see, that for $r = 0$ (and other cases) a singularity appears that might be a generic feature of the theory of general relativity, but which does not hold physically, we use the line element for the Schwarzschild metric that has the form:

$\begin{array}{llll} (ds)^2 & = & & c^2(d\tau)^2~+~\displaystyle\frac{2c^2}{\sqrt{1 - \displaystyle\frac{r_s}{R}}} d\tau dT~+~\displaystyle\frac{c^2}{1 - \displaystyle\frac{r_s}{R}}\bigg(1 - \displaystyle\frac{r_s}{r(\tau)}\bigg)(dT)^2~- & r(\tau)^2 ((d\theta)^2 + \sin^2(\theta)(d\phi)^2) \\ \end{array}$ whereby the Schwarzschild radius is $r_s = \frac{2GM}{c^2}$ and the coordinates are given by:

$\left( \begin{matrix} \tau & T & \theta & \phi \end{matrix} \right)^T = \left( \begin{matrix} \tau\\ T\\ \theta\\ \phi \end{matrix} \right)$ where $\tau$ is the proper time of a falling particle, $T$ is the proper time of an observer on the surface of the sphere, $\theta$ is the colatitude (angle from north) and $\phi$ is the longitude. For the orbit of a test particle we get:

$\begin{array}{llll} 0 \leq (ds)^2 & = & & c^2 \Bigg(d\tau + \displaystyle\frac{dT}{\sqrt{1 - \displaystyle\frac{r_s}{R}}} \Bigg)^2~-~\displaystyle\frac{c^2}{1 - \displaystyle\frac{r_s}{R}} \displaystyle\frac{r_s}{r(\tau)}(dT)^2~- & r(\tau)^2 ((d\theta)^2 + \sin^2(\theta)(d\phi)^2) \\ \end{array}$ With $\lim R \to 0$ the sphere disappears (not only for an observer on the sphere but for all!) after $U = 2\pi R$ , but after $E = hf$ $\Rightarrow$ $\lambda = \frac{v_p}{f} > 0$ the wavelength of the particle is greater than zero. Thus, the particle seems to fly out of space. This is of course not true and show a broader problem, namely that the singularity is not physically. The metric holds though for all: $r(\tau) > 0$ .