hi,everyone

I am a newcomer to this powerful tool. Now I am confused with some basic knowledge about reduced units.

My modelling object is polyethelyene ( CH2)n. so the density is 940 kg/m3. I want the reduced units for density. I have sigma = 3.93e-10, I have CH2 monomer mass 2.3248e-26. So the reduced density should be calculated like (real density* sigma^3 / mass). That will be around 2.45. However, 0.8~0.9 is the most common value for this kind of system. I confused.

So I also try the book by Daan Frenkel and Berend Smit. On the page 37 they also talked about the reduced units. And an example says reduced density 0.7 is 1680 kg/m3 in the real units. The material is Lennard-Jones Argon sigma= 3.405e-10, M= 0.03994 kg/mol, eplison/Kb= 119.7K. However, again, I got 0.997 by my own calculation instead of 0.7

Totally confused about the reduced units. Why my calculations are wrong? Any suggestion wil be appreciated. Thanks in advance

Justin

Re: LJ, I think it is rho* = 1.0. The book has an error in the
table. Later on that page, they talk about 840 kg/m^3 having a rho* of 0.5.

Re: polyethylene - see the previous 10/10/08 mail by Mahajan for
the same Q. There is another factor for polyatomic systems in
computing rho*, but I'm not remembering it off the top of my head.

Steve

Here is the answer for polyethylene (thanks to Mark Stevens).
The C-C bond length is about 1.5 Angs, so you can't have
an LJ monomer with a sigma = 3.9Angs = 1 CH2.

The coarse-grain model is that 1 LJ monomer = about 3 CH2,
which gives

rho* = (940 kg/m^3) (6.023 particle/mole) (1 mole/ 0.042 kg) (3.93e-10 m^3)
rho* = 0.8

Note that this uses 42 g/mole for 3 CH2 rather than 14 g/mole for 1 CH2,
which makes rho* 3x smaller than you had.

Said another way, your value of rho* = 2.45 was correct, and
it implied that you would have to pack your single CH2 monomers
very densely (overlapping) to give a bond length much less
than sigma.

Steve