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If we regard a simple scenario in which a particle is subjected to a constant force, it will take a hyperbolic path on a Minkowski diagram. If we assumed equally spaced particles started accelerating at at same instant, then there will be many hyberbolic lines of same type but shifted horizontally. These hyperbolic lines will act as our lines for the curved space grid, and the reflection of these lines on the line $y = x$ will act as our curved time grid because we can imagine a moving light passing by all the particles to correspond to a tick. After we have constructed this curvilinear coordinate system, we can find the Christoffel symbols by finding the rate of change of the space and time tangent vectors with respect to the space and time axes and expressing them as a linear combination of the space and time axes tangent vector. After we have calculated all the Christoffel symbols, can we use those to calculate the metric tensor and obtain the time-time and radius-radius components of Schwarzschild metric?

Assume the hyperbolic path of a particle starting at x = 0 is given by $\sqrt{(x+a)^2 - a^2}$ where $a = \frac{m'c^2}{F}$

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    $\begingroup$ You've been given a good answer to your question, but I think you may enjoy Greg Egan's article on the Rindler Horizon: "The purpose of this web page, then, is to analyse in detail (using only special relativity) some interesting thought experiments that can be carried out by a constantly accelerating observer, who sees a “Rindler horizon” in spacetime that is very similar to the event horizon of a black hole. This is certainly not a perfect substitute for a full, general-relativistic analysis [...]" $\endgroup$ – PM 2Ring Jan 8 '20 at 3:00
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Can Schwarzschild Metric be derived from curvilinear coordinates in special relativity?

The answer is no, the Schwarzschild metric has a non-vanishing Riemann curvature tensor whereas the Minkowski metric has a vanishing Riemann tensor. Since the Riemann curvature tensor is a tensor, if its components vanish in one coordinate system they vanish in all coordinate systems.

How to see this? A quick way is from the tensor transformation law: $$R^{\mu'}_{\phantom{\mu'}\nu'\alpha'\beta'}=\dfrac{\partial x^{\mu'}}{\partial x^\mu}\dfrac{\partial x^\nu}{\partial x^{\nu'}}\dfrac{\partial x^\alpha}{\partial x^{\alpha'}}\dfrac{\partial x^\beta}{\partial x^{\beta'}}R^\mu_{\phantom{\mu}\nu\alpha\beta}.$$

Notice that if $R^{\mu}_{\phantom{\mu}\nu\alpha\beta}$ vanishes in some coordinate system, by the above, the components will vanish in any other. So you can't define a coordinate system in Minkowski spacetime which will give you the Schwarzschild spacetime.

I should remark that one may argue that the near-horizon region of the Schwarzschild geometry may be shown to be isometric to a portion of the Minkowski spacetime known as Rindler wedge. This is closely related to your reasoning. But this is an approximation in which one is expanding the metric to first order around the horizon. This is argued e.g., in Susskind's "An Introduction to Black Holes, Information and the String Theory Revolution: The Holographic Universe" which I recommend you give a look.

Apart from that, I would like to notice that the two spacetimes are even topologically distinct. The Schwarzchild geometry has topology $$\mathbb{R}\times (0,2M)\cup (2M,+\infty)\times S^2,$$ whereas Minkowsi spacetime has topology $\mathbb{R}^4$.

One may argue that it is possible to extend the Schwarzschild spacetime in order to add the "points at $r = 2M$" since one may show that the metric is not really singular there and since these points are reached by observers at finite proper time. Still, $r = 0$ is a true geometric singularity because a curvature invariant diverges there. This makes it impossible to add in this point making the two solutions truly topologically distinct. For details on this matter, specially concerning the Kruskzal-Szekeres maximal extension, see the Phys.SE thread "Are Minkowski and Schwarzschild spacetimes diffeomorphic?"

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  • $\begingroup$ why does r=0 singularity matters for topology? You can construct homeomorphism of Schwarschild spacetime to $\mathbb{R}^4$ through use of Kruskal-Szekeres coordinates and thus the two spaces should be topologically equal. The singularity not being point of the manifold actually helps in constructing the map (I am not sure about the result if the singularity would be included, since then you need to map singularity to infinity, which makes the map fail to be bijection) $\endgroup$ – Umaxo Jan 8 '20 at 5:22
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    $\begingroup$ It matters because the singularity makes you have a whole line excluded. For more details on this matter please check physics.stackexchange.com/questions/432035/… and the great answer by @ValterMoretti. $\endgroup$ – Gold Jan 8 '20 at 13:47

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