3 ms·

You capture a quantum black hole, slap a charge on it, and wiggle that: The Hole Man https://wikipedia.org/wiki/The_Hole_Man https://wikipedia.org/wiki/The_Hole

You capture a quantum black hole, slap a charge on it, and wiggle that: The Hole Man https://wikipedia.org/wiki/The_Hole_Man https://wikipedia.org/wiki/The_Hole_Man

pavel_lishin 8y ago
Wouldn't that evaporate rather quickly? (Sure, Niven didn't know that _then_... but science has a habit of catching up to some stories.)
- waterhouse 8y ago
  According to this... https://en.wikipedia.org/wiki/Hawking_radiation#A_crude_analytic_estimate https://en.wikipedia.org/wiki/Hawking_radiation#A_crude_anal... "[F]or instance, a 1-second-life black hole has a mass of 2.28×10^5 kg", and the time taken seems proportional to the cube of the mass. So a 2.28×10^6 kg black hole would take 1000 seconds, and so on. How massive was the black hole in the story?
  - pavel_lishin 8y ago
    The story mentions "ten-to-the-seventeenth grams in mass and ten-to-the-minus-eleven centimeters across", but I'm not sure if that's the black hole in question, or a hypothetical one. They definitely mention it being smaller than an atom, though.
    - waterhouse 8y ago
      Interesting. That's 10^14 kg, which, according to the above, would be between 10^24 and 10^27 seconds to evaporate, which is at least in the quadrillions of years. As for the radius... Looks like the Schwarzchild radius is proportional to the mass, and for the moon (7x10^22 kg) it's 0.11x10^-3 meters, so this 10^14 kg mass should be maybe 1.5x10^-13 meters, which is 1.5 x 10^-11 cm. Nice. It looks like Niven did his homework. Incidentally, Wiki says that Hawking argued for black hole evaporation in 1974, and Niven's story won an award in 1975, which I assume might indicate it was published around then. Seems it'd be a close call whether Niven heard about it before publishing his story.
    - pdonis 8y ago
      The problem for a hole this size isn't the evaporation time; that's way longer than the age of the universe. The problem is the Hawking radiation pressure. Some quick formulas (M is mass of the hole in kg): Evaporation time T = 8.4 x 10^-17 M^3 seconds Hawking radiation power W = 3.6 x 10^32 / M^2 Watts Hawking radiation pressure at 10^-10 meters (roughly one atom) distance P = 9.6 x 10^42 / M^2 Pa (Note that the last formula assumes that the hole's horizon radius is smaller than 10^-10 meters, which it is for Niven's hole.) For M = 10^14 kg, we get: T = 8.4 x 10^25 seconds (which is more than 10^18 years) W = 3.6 x 10^4 Watts (36 kW, comparable to your car's engine traveling on the highway with a family and luggage, but pretty darn bright for something that's basically just a light source) P = 9.6 x 10^14 Pa (almost 10^10 atmospheres!) So this hole won't accrete matter, because anything that gets close to it gets violently pushed away by its Hawking radiation. It will tunnel its way through Mars and out the other side, and then back again, executing simple harmonic motion, indefinitely.

You capture a quantum black hole, slap a charge on it, and wiggle that: The Hole Man https://wikipedia.org/wiki/The_Hole_Man https://wikipedia.org/wiki/The_Hole_Man

pavel_lishin 8y ago

Wouldn't that evaporate rather quickly? (Sure, Niven didn't know that _then_... but science has a habit of catching up to some stories.)

waterhouse 8y ago
According to this... https://en.wikipedia.org/wiki/Hawking_radiation#A_crude_analytic_estimate https://en.wikipedia.org/wiki/Hawking_radiation#A_crude_anal... "[F]or instance, a 1-second-life black hole has a mass of 2.28×10^5 kg", and the time taken seems proportional to the cube of the mass. So a 2.28×10^6 kg black hole would take 1000 seconds, and so on. How massive was the black hole in the story?
- pavel_lishin 8y ago
  The story mentions "ten-to-the-seventeenth grams in mass and ten-to-the-minus-eleven centimeters across", but I'm not sure if that's the black hole in question, or a hypothetical one. They definitely mention it being smaller than an atom, though.
  - waterhouse 8y ago
    Interesting. That's 10^14 kg, which, according to the above, would be between 10^24 and 10^27 seconds to evaporate, which is at least in the quadrillions of years. As for the radius... Looks like the Schwarzchild radius is proportional to the mass, and for the moon (7x10^22 kg) it's 0.11x10^-3 meters, so this 10^14 kg mass should be maybe 1.5x10^-13 meters, which is 1.5 x 10^-11 cm. Nice. It looks like Niven did his homework. Incidentally, Wiki says that Hawking argued for black hole evaporation in 1974, and Niven's story won an award in 1975, which I assume might indicate it was published around then. Seems it'd be a close call whether Niven heard about it before publishing his story.
  - pdonis 8y ago
    The problem for a hole this size isn't the evaporation time; that's way longer than the age of the universe. The problem is the Hawking radiation pressure. Some quick formulas (M is mass of the hole in kg): Evaporation time T = 8.4 x 10^-17 M^3 seconds Hawking radiation power W = 3.6 x 10^32 / M^2 Watts Hawking radiation pressure at 10^-10 meters (roughly one atom) distance P = 9.6 x 10^42 / M^2 Pa (Note that the last formula assumes that the hole's horizon radius is smaller than 10^-10 meters, which it is for Niven's hole.) For M = 10^14 kg, we get: T = 8.4 x 10^25 seconds (which is more than 10^18 years) W = 3.6 x 10^4 Watts (36 kW, comparable to your car's engine traveling on the highway with a family and luggage, but pretty darn bright for something that's basically just a light source) P = 9.6 x 10^14 Pa (almost 10^10 atmospheres!) So this hole won't accrete matter, because anything that gets close to it gets violently pushed away by its Hawking radiation. It will tunnel its way through Mars and out the other side, and then back again, executing simple harmonic motion, indefinitely.