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Natural Science Forum / Physics / Particle Physics / September 2005the square root of 2
So I have a right triangle, sides 1 unit in length, that makes the hypotenuse sqrt(2) units long, right? So if the sides are 1mm long, how many decimal places can I calculate sqrt(2) to, to be PHYSICALLY meaningful? ditto, if the sides are 1nm long, 1fm long.... cheers What are you measuring? The difference between mathematical representation and physical reality ;) > What are you measuring? > So I have a right triangle, sides 1 unit in length, > that makes the hypotenuse sqrt(2) units long, right? > > So if the sides are 1mm long, how many decimal places can I calculate > sqrt(2) to, to be PHYSICALLY meaningful? Actually, we have no proof that the real number line has any bearing on any physical dimension. For all we know, the rational numbers might be as suitable a representation. Heck, the integers might also be a suitable representation. > ditto, if the sides are 1nm long, 1fm long.... > > cheers Integers may well be better, if you want to count zerps. >> So I have a right triangle, sides 1 unit in length, >> that makes the hypotenuse sqrt(2) units long, right? [quoted text clipped - 10 lines] >> >> cheers > So I have a right triangle, sides 1 unit in length, > that makes the hypotenuse sqrt(2) units long, right? [quoted text clipped - 3 lines] > > ditto, if the sides are 1nm long, 1fm long.... I'm betting that all lengths are made from the Planck length. So that's about 32 decimal places, and several more for the factor conversion. Further "underlengths" are only meaningful as energy going into fase changes making further particles. -Aut |
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