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On this question, from long ago, the Planck length is calculated as the length at which the Reduced Compton Wavelength is equal to the Schwarzschild Radius. However, in the calculation, the scalar "2" in the formula for the Schwarzschild radius is omitted in subsequent calculations. If I follow the calculations to their conclusion, with the 2 included, I get

when the Schwarzschild Radius is equal to the Planck length, the Reduced Compton Wavelength is equal to double the Planck length.

I must be doing something wrong, but I am not sufficiently far along in my studies to know where the problem is. Can anyone help me understand where the 2 went, please?

Qmechanic
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Earl Whitney
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1 Answers1

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A more detailed analysis has to consider the fact, that radial lenghts are not metrically invariants and have no inner radial end inside the horizon to decribe a standing wave.

The Schwarzschild radial coordinate $r$ is defined by the metric surface $4 \pi r^2 $ of a sphere, as is evident by the metric term $r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2$ as in flat space. So the Planck lenght is tied somehow to angular momentum spectrum on a metric sphere, e.g. by Bohr's length of a single wave $e^{i\phi}$ along the equator on the horizon.

roland F
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