# Time reversibility of velocity verlet easy

1. Jan 13, 2012

### dikmikkel

1. The problem statement, all variables and given/known data
Prove that the velocity verlet scheme is time reversible.

2. Relevant equations
r(t+dt) = r(t) + v(t)dt + 1/2a dt^2
v(t+dt) = v(t) + 1/2 a(t)dt + 1/2a(t+dt)dt

3. The attempt at a solution
I substitute -dt in at dt and get:
r(t-dt) = r(t) - v(t)dt +1/2a dt^2
v(t-dt) = v(t) -1/2a(t)dt -1/2a(t-dt)dt

the squared factor of time in the acceleration term is annoying me, can it be done easier?

2. Jan 13, 2012

### xaos

what exactly do you see is wrong with the squared factor?

3. Jan 14, 2012

### dikmikkel

Well, the acceleration should(my opinion) lead to move the position back again if it is time reversible(as it is). Can you help?

4. Jan 14, 2012

### xaos

do the calculations for t''=t'+dt, then do them again for t'=t-dt, everything should simplify back to t''=t.

Last edited: Jan 14, 2012
5. Jan 15, 2012

### dikmikkel

When i do that it still does not make sense to me.
Would you do the first half part, maybe with substitutions.

6. Jan 15, 2012

### dikmikkel

Okay, now i found out i think.

r(t+dt) = r(t)+v(t)dt + 1/2a(t)dt^2
v(t+dt) = v(t)+1/2a(t)dt + 1/2a(t+dt)dt
Now we are at r(t+dt) and want to reverse it:
r(t-dt) = r(t+dt) - v(t+dt)dt + 1/2a(t+dt)dt^2 =
r(t)+v(t)dt + 1/2a(t)dt^2 -(v(t)+1/2a(t)dt + 1/2a(t+dt)dt)dt +1/2a(t+dt)dt^2 =
r(t) . Q.e.d
v(t-dt) = v(t+dt) - 1/2a(t+dt)dt-1/2a(t+dt-dt)dt =
v(t)+1/2a(t)dt + 1/2a(t+dt)dt -1/2a(t+dt)dt-1/2a(t)dt =
v(t), q.e.d
Is that right?

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