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Time reversibility of velocity verlet easy

  1. Jan 13, 2012 #1
    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. jcsd
  3. Jan 13, 2012 #2
    what exactly do you see is wrong with the squared factor?
  4. Jan 14, 2012 #3
    Well, the acceleration should(my opinion) lead to move the position back again if it is time reversible(as it is). Can you help?
  5. Jan 14, 2012 #4
    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
  6. Jan 15, 2012 #5
    When i do that it still does not make sense to me.
    Would you do the first half part, maybe with substitutions.
  7. Jan 15, 2012 #6
    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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