The timedilatationdilation formula is: T' = T/ sqrt(1 - v²/c²), where T' is the time measured by someone on Earth and T the proper time. In this example, the proper time is D/c, where D is the distance Sun-Earth. So: D/v = (D/c)/ sqrt(1 - v²/c²) c/v = 1/ sqrt(1 - v²/c²) sqrt(1 - v²/c²) = v/c 1 - v²/c² = v²/c² 2v² = c² v = c/sqrt(2) This holds for all distances.
That starts well but goes horribly wrong after the first equation. The second equation should be this: (D/v) = T' / sqrt(1 - v²/c²) Here, D is the Earth-Sun distance in the reference frame of the Earth, v your speed in this same reference frame, and (D/v) is how much time you take to cover that distance in the same reference frame. D is given (Earth-Sun distance), T' is given (8 minutes 20 seconds), v is the desired unknown.
8:20 is the proper time that the traveller must measure. Warp choose this value because it's the time that light takes to travel from the Sun until the Earth, as measured by someone on Earth. Therefore, T = 8:20 = D/c
but then proceeds to point out that a possible dismissal for this could be the existence of irrational numbers.
Unless the 'exterior universe' the computer exists in has its own shortest measurable time, even shorter than tP, it would be impossible to manipulate this many bits in one Planck time.
Not withstanding the assumption that time in the exterior universe functions the same as time does in our universe, this assumes that the simulation necessarily runs in real time (from our point of view). There's no real reason to assume that needs to be the case. (...) Even if it would run at real time, the assumption that it is 'impossible' seems based on physical constraints that do not necessarily apply in the exterior universe. Really, with the exterior universe's properties being as much undefined as they are, practically anything would be possible.
It's probably too late now, but this looks like a simple exercise using the chain rule. If u = 1/r then u' = -1/r^2 * r'. You can then use u' = du/dtheta * theta' and the 2nd equation to write theta' in terms of L and r. Finding r'' should be pretty similar and part b should follow from just substituting what you have into the original equation.
Since \lambda lies in p's past, the Turing machine can signal (a solution) to p at any stage of this never-ending task. Meanwhile, the observer takes a quick trip (finite proper time) through spacetime to p, to pick up the solution.
Conservation of angular momentum has the consequence that if you shrink a rotating object (all other parameters remaining the same), it will start rotating proportionately faster (so that angular momentum is preserved). A (rotating) star that collapses into a black hole becomes zero-sized. By this logic it would rotate at infinite speed. Yet that doesn't happen. It will still rotate at a finite speed. Is the reason that the singularity actually becomes a ring rather than a point? (Which makes me think: Does it become a ring because of conservation of angular momentum?)
There is a limit to a black holes density, as you can only pack atoms together so tightly that there is no spaces between them.
However, the center of a black hole cannot have an infinite pull. Otherwise the entire cosmos would have been sucked in by now.
Tecnically speaking, the ring singularity appears on stationary solutions; that is, the ring singularly exists from -infinity to +infinity, amd there is no dynamic solution showing the collapse of a rotating star into a rotating (Kerr) black hole. However, numerical simulations do exist, and they show collapse going towards an oblate intermediate form. But things inside the event horizon tend not to be modeled because they tend to crashnthe simulation or cause severe numerical stability problems. Whether the singularity would end up as a ring or as a disc is harder to say. Moreover, it is likely there will be a lot of angular momentum and mass lost through gravitational waves during the collapse, so they can't be assumed constant. Finally, the radius of the ring singularity of a rotating uncharged black hole is actually quite simple — it is equal to its angular momentum divided by (mass times speed of light).
