Replying to a comment on:

Why you don't fall through the floor (Sonnet) by ?-Dave_Mysterious-?

Schrödinger’s equation: -(ħ²/2M)grad²φ(x,y,z) + Vφ(x,y,z) = Eφ( z,y,z) For a particle in a box: V = 0 if 0 < x,y,x < a = ∞ otherwise Try a solution of the form φ = Nsin(2πlx/a)sin(2πmy/a)sin(2πnz/a) Where N is the normalisation factor. φ must go to zero at x,y,z = a, therefore l,m,n are integers. So, E = -(1/2M)(2πħ/a)²(l² + m² + n²) Now, pressure P = ∂E/∂V And dV = d³a = d(a³) = 3a²da Therefore P = (1/3a²)∂E/∂a This is quantum pressure.

?-Dave_Mysterious-? 10-Dec-03/4:02 AM
An equalateral triangle still wouldn't fall through the hole. You're theory is therefore wrong, or at best incomplete. It just so happens that the circle can't fall down the hole, but that doesn't explain why it isn't triangle shaped. Is that why you wrote "Tri-again"? Clearly the real reason must be so that the workman is presented with the same hole no matter what angle he approaches it from. e.g. if it was square, a man approaching from 45 degrees would have more chance of falling in than a man approaching from 0 degrees, which would be unfair.




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