- #1
FrostJax
- 1
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Let's say there was a Lattice L of a certain molecule P. The molecules are spherical and are in static equilibrium. L is in a bucket shape, with the R1 and R2 of the two bounding areas circular areas that are along the x,y plane at z(R1) = Z1 and z(R2) = Z2. The density of the structure is known as well as the structure. There is a surface charge distribution amongst the top layer of the lattice all along the bucket. The radii of the top circle bounding the bucket is large enough so that a certain number N of water molecules won't be effected by the charge distribution along sides. Now we drop a number of water molecules on to the lattice such that there is a one to one ratio between the lattice molecules and the water molecules along a line along the z axis at the varying x,y coordinates of the lattice molecules. The molecules have an initial velocity v(x,y,z) = v_0z (v_0x and v_0y are zero). They are dropped from a height such that the forces such as intermolecular forces and the force of the bounding charged rings are negligible. Do the water molecules vibrate as waves, bounce up and down exlusively and in a random fashion, or other distinguishable or "indistinguishable" pattern of travel until they're "halted" by nonconservative forces? Now instead assume there's an insulator layer between the circle that bounds the lattice and the sides of the bucket, such that there is no charge distribution along the sides. Now the radius of the bottom ring that bounds the lattice is equal to the radius of the top ring, thus the side of the bucket exists along a plane that extends from z1 to z2 at a reasonable bucket radius S with phi from 0 to 2*pi (in cylindrical coordinates). We drop that one to one ratio of water molecules, same question as before. Waves, distinguishable/indistinguishable pattern, or bounce up and down exclusively and randomly? A good answer to this question would be one where I finish reading it understanding that I'm thinking of something the wrong way. Another good answer would be the answer, thank you!