Einstein & ITishnikYouth
Einstein Einstein
ITishnikYouth ITishnikYouth
Einstein Einstein
Hey, have you ever tried to write a recursive function that models time dilation, like turning \(t' = t \sqrt{1-v^2/c^2}\) into an algorithm? I'm curious how you would structure that in code and what patterns would emerge.
ITishnikYouth ITishnikYouth
Yeah, just treat the factor as a multiplier and call it recursively. Something like this in JavaScript style: function dilate(t, v, steps) { if (steps === 0) return t; const factor = Math.sqrt(1 - (v*v)/(c*c)); return dilate(t * factor, v, steps - 1); } If you unwind the recursion you get t * factor^steps, so the pattern is just a geometric progression. In a continuous sense you’d replace the discrete steps with an integral, but for a quick simulation this works and shows the exponential decay of proper time.
Einstein Einstein
Nice trick—though you’re really folding the proper time into a tower of little “slow‑down” steps. If you let the steps go to infinity you’ll recover the continuous Lorentz factor, but watch out: if you ever let \(v\) exceed \(c\) the square root will go imaginary—like trying to bake a cake with a black‑hole whisk. Maybe try plugging in some real speeds, like 0.99 c, and see how quickly the factor shrinks. Also, if you wanted to avoid recursion you could just compute \(t \times (1 - v^2/c^2)^{steps/2}\) in one line—less stack overflow and more coffee.
ITishnikYouth ITishnikYouth
Yeah, that 0.99 c runs into the factor 0.141, so 10 steps give you 0.141^10—almost zero. In practice I just drop the recursion and write `t*Math.pow(Math.sqrt(1-v*v/c/c),steps)`; it’s the same geometric pattern, just in one line. If you push v>c, the math breaks, so I add a guard: if(v>=c){console.warn("No real time here");} And always remember: in code, as in relativity, the limit is the limit.
Einstein Einstein
Nice guard clause, but remember that even a single step past light speed in a toy model throws the whole thing into a strange, complex number realm—like trying to solve a geometry problem in an imaginary dimension. It’s a good reminder that your code should stay in the physically meaningful domain, just like your equations. Keep it tidy, and maybe add a sanity check for negative speeds too—those would just flip the sign of time, which is another quirky paradox to keep in mind.
ITishnikYouth ITishnikYouth
Sure thing, just slap a quick clamp in: if (v < 0) v = -v; then proceed. That way the Lorentz factor stays the same, and time doesn’t flip. The only real “paradox” you’ll see is a negative t if you let the sign of t change, but that’s just a unit test bug. Keep the domain clean, and the recursion will behave nicely.
Einstein Einstein
Sounds like a neat trick—just remember that the universe doesn’t care if you flip a sign; it only cares if the physics stays consistent. So keep your c positive, guard against absurd speeds, and let your recursion run free—like a photon zipping through a lattice of equations. Happy coding, and may your time dilation stay as smooth as a jazz improvisation.