# Why not solve molecular dynamics trajectories exactly?

The basis of Molecular Dynamics simulation is the integration of newton's equations of motion:

$$\frac{F}{m} = \frac{d^2r}{dt^2}$$

I understand that there are various methods based on Taylor's expansion, to integrate forward in time and calculate the new position. Doing this always leaves an error term in the expansion, however.

To solve for position exactly, I can integrate both sides:

$$r = \frac{Ft^2}{m} + c_1 t + c_2$$

If $r(t=0) = 0$ then $c_2 = 0$.

If it's possible to determine $c_1$ too (?) we have an exact solution, and so I can calculate the next position at $t+dt$ with perfect accuracy, unlike with Taylor expansions.

Can anyone show me where my understanding is wrong? Thank you.

• You are assuming $F$ is constant. – Kirill Nov 27 '15 at 3:08
• If t is small enough, I can treat F as constant and calculate the change in position, r, in a similar fashion to the Taylor-expansion methods, right? Then update F and repeat. – Struggling snowman Nov 27 '15 at 3:13
• Treating F as constant over a short interval is equivalent to a Taylor expansion truncated at first order. – Jeff Hammond Nov 27 '15 at 3:54
• @Kirill @Jeff Thanks for the comments! I really appreciate your help. But truncation at the square term (Ft^2/m) is a 2nd-order truncation, right (not 1st)? – Struggling snowman Nov 27 '15 at 5:25
• The order of a Taylor expansion refers to the derivative, not the power (i.e., a second-order truncation would keep the constant term and the first derivative, and ignore the second derivative and above). And it bears pointing out explicitly: As soon as you truncate, you will not have perfect accuracy. – Christian Clason Nov 27 '15 at 8:47