M10 Angular momentum is conserved or not

Manual 10 for MD++
Angular momentum is conserved or not?

Keonwook Kang and Wei Cai

Jul 06 , 2009

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In molecular dynamics simulations, positions and velocities of atoms are updated following Newton's equations of motion. Naturally, we would expect that linear and angular momenta are conserved during the time integration, or

${\begin{array}{rcl}\mathbf {L} &\equiv &\sum _{i}m_{i}\mathbf {v} _{i}=\mathrm {Const.} \\\mathbf {P} &\equiv &\sum _{i}\mathbf {r} _{i}\times m_{i}\mathbf {v} _{i}=\mathrm {Const.} \end{array}},$

where L is linear momentum and P is angular mumentum. Usually, we subtract ceter-of-mass velocity from the velocity of each atom so that the whole system can not drift and the linear momentume becomes zero.

${\begin{array}{rcl}\mathbf {v} _{i}&:=&\mathbf {v} _{i}-\mathbf {v} _{\mathrm {CM} }\\\mathbf {L} &=&\sum _{i}m_{i}\mathbf {v} _{i}-\mathbf {v} _{\mathrm {CM} }\sum _{i}m_{i}=0\end{array}},$

where v_CM is defined as

$\mathbf {v} _{\mathrm {CM} }\equiv {\frac {\sum _{i}m_{i}\mathbf {v} _{i}}{\sum _{i}m_{i}}}$.

Similarly, we can additionaly subtract velocity component contributing rotation so that the whole system can not rotate and the angular momentum becomes zero.

$\mathbf {v} _{i}:=\mathbf {v} _{i}-\mathbf {\Omega } \times \mathbf {r} _{i}$