md2011-03-note_Part_2

# md2011-03-note_Part_2 - Copper @ 300 K For Cu: a = 3.62 b =...

This preview shows pages 1–3. Sign up to view the full content.

Copper @ 300 K I For Cu: I a = 3 . 62 Å I b = a / 2 2 . 56 Å I M = 63 . 55 a.u. ( u = 1 / N A 1 . 66 × 10 - 27 kg, m = uM ) I At room temperature ( T = 300 K), we thus get Δ t 6 b 5 × 20 q m i 3 k B T 6 2 . 56 × 10 - 10 m 100 r 1 . 66 × 10 - 27 kg × 63 . 55 3 × 1 . 38 × 10 - 23 m 2 kg s 2 K × 300 K 7 . 46 fs (6) Notes

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
In practice, Δ t 6 4 fs is needed for stability. I Therefore, even with the rules-of-thumb, one has to test the suitability of a certain Δ t for every new system , new potential and new process . I The most important criterion is the energy conservation ( d E d t = 0 ) . I In standard MD, Δ t is limited by the fastest process expected to happen during the simulation. I For ﬁnite temperatures ( > 100 K), 10 fs is an absolute upper limit for Δ t a 1 s simulation requires O ( 10 14 ) time steps. With one 2 GHz CPU, this would take 14 hours. I However, this is not a realistic scenario. Notes
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 02/14/2012 for the course CSE 6590 taught by Professor Kotakoski during the Winter '12 term at York University.

### Page1 / 5

md2011-03-note_Part_2 - Copper @ 300 K For Cu: a = 3.62 b =...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online