4 ms·

Singularity has nothing to do with any distance, it is a divergence at some point. The singularity in a black hole at r = 0 is also much less than a Planck len

Singularity has nothing to do with any distance, it is a divergence at some point. The singularity in a black hole at r = 0 is also much less than a Planck length from the center. And a point mass, I would assume, will also always lead to a singularity. That the Schwarzschild radius is less than the Planck length might mean something or not, all the Planck quantities are just scales derived from fundamental constants. At those scales different effects from our current theories would become comparable if our theories would hold at those scales, but there is currently not really any known significance to them.

nyrikki 3y ago
It may help to add that the Plank length the smallest measurement of length with any meaning. It is about the limitations of describing positions, not an implication that spacetime is discrete. We don't know if spacetime is discrete, but we also have no evidence to suggest it is right now.
- danbruc 3y ago
  That's not actually true, or at least it is not currently known. It is very well possible that distances less than a Planck length have meaning, or maybe distances lose all meaning a couple of orders of magnitude above the Planck length. We just don't know the physical laws at that scales. If you consider at what energy a photon (E = hc/λ) will be able to resolve, i.e. have a wavelength equal to, the Schwarzschild radius of that energy (r = 2GM/c² with E = mc²), then you get the Planck length up to a factor of root two [1], implying that measuring a Planck distance requires a photon so energetic that it turns that region into a black hole. If, very big if, the known laws still apply at those scales. [1] To get the Planck length, the wavelength must be doubled, i.e. be equal to the diameter instead of the Schwarzschild radius. r = 2GM/c² → m = rc²/2G → E = rc⁴/2G (with E = mc²) hc/λ = rc⁴/2G → hc/2r = rc⁴/2G (with λ = 2r) → r² = hG/c³ → lₚ = r = √hG/c³

nyrikki 3y ago

It may help to add that the Plank length the smallest measurement of length with any meaning. It is about the limitations of describing positions, not an implication that spacetime is discrete. We don't know if spacetime is discrete, but we also have no evidence to suggest it is right now.

danbruc 3y ago
That's not actually true, or at least it is not currently known. It is very well possible that distances less than a Planck length have meaning, or maybe distances lose all meaning a couple of orders of magnitude above the Planck length. We just don't know the physical laws at that scales. If you consider at what energy a photon (E = hc/λ) will be able to resolve, i.e. have a wavelength equal to, the Schwarzschild radius of that energy (r = 2GM/c² with E = mc²), then you get the Planck length up to a factor of root two [1], implying that measuring a Planck distance requires a photon so energetic that it turns that region into a black hole. If, very big if, the known laws still apply at those scales. [1] To get the Planck length, the wavelength must be doubled, i.e. be equal to the diameter instead of the Schwarzschild radius. r = 2GM/c² → m = rc²/2G → E = rc⁴/2G (with E = mc²) hc/λ = rc⁴/2G → hc/2r = rc⁴/2G (with λ = 2r) → r² = hG/c³ → lₚ = r = √hG/c³