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Classical mechanics, initial conditions question

  1. Mar 11, 2010 #1

    fluidistic

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    I just started CM (I had 2 classes until now) and the professor said that if you know the position and velocity of say all the particles, then you know how the system will evolve.
    This, I already read and knew. I've probably a common question so feel free to redirect me to any similar thread. Here it comes:
    Suppose you're given the position and velocity of 2 particles. They are making contact and their velocity is 0 m/s. How do you know if a moment later they will start to move away from each other because they had a collision or if they will stay motionless, i.e. will remain stuck together? (Suppose the universe is constituted by these only 2 particles)

    If I knew v(t) it would be a different story, but only knowing v at a certain moment makes me wonder.

    Maybe it's like having a second order differential equation (representing the motion of each particles) that is totally unknown, but you have y'(0) and y(0). But I'm not sure at all on this.

    Could you answer my first question in words? Or maybe mathematically? Thanks for any help.
     
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  3. Mar 11, 2010 #2

    tiny-tim

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    Hi fluidistic! :smile:
    Two particles can't make contact … particles don't work like that.

    Two electrons, say, could get very close before they started to repel each other. But they could only be close because there had been a "collision", ie one had been heading for the other.

    When they repel each other, the repulsion will be predictable from their positions when stationary.
     
  4. Mar 11, 2010 #3

    fluidistic

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    Thanks for the reply!
    Ok I see. But I thought CM was a theory that is a "simplified" way of describing the universe (for example, it does not consider electric and magnetic forces, as far as I know). I thought it would work pretty well assuming particles collide (even though as you said they don't make contact).

    How would you answer this question if Lagrange asked you?
     
  5. Mar 11, 2010 #4

    tiny-tim

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    But if they're "simplified" bodies, stationary and in contact, they'd have no reason to separate, would they? :confused:

    To put it another way, if there's no electric etc forces, how did they get to be stationary, if there was a collision?
     
  6. Mar 11, 2010 #5
    I think the point is that you will know v(t) if you have the dynamical equations for the system. The dynamical equations plus the initial conditions tell you the complete future history. This is what makes the system deterministic. This is true even in chaotic systems, although calculations become problematic due to sensitivity to intial conditions, in that case.

    Anyway if there is a colision occuring at the point of your initial conditions, then your dynamic equations will (or, at least should) include the forces in action (compression, electrostatic, electromagnetic, gravitational, etc.).
     
  7. Mar 11, 2010 #6

    fluidistic

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    Hmm I think I was trying to say something else, my English is maybe too limited, hahaha!
    Yes, if the 2 particles or "perfect" balls are in contact with a 0 velocity, there is no reason they would separate. However if you pitch one ball into another one and take a photo at the moment of the collision (in CM I guess this moment is not even a differential amount of time, it's only 1 moment; it's instantaneous. But I could be wrong in this assumption. I think it would hold if we assume no deformation, hence the word "perfect"), you'd see both balls making contact with each other and you'd see they both have a 0 velocity. Thus you are in the same case as when the balls are initially quiet and making contact. The difference is that in the first case there will be or there is a transfer of linear momentum and if the ball you throw was less heavy than the quiet one, the ball would bounce back.
    My question is, how do you determine whether you are in the first case or in the second case, with only 1 photo taken? Namely, the photo showing the 2 balls making contact and with velocity =0.
    An equivalent problem would be to let fall a ball on the floor and take a photo at the moment it touches the ground. The second case would to put the ball on the ground and take a photo. Both photos would show exactly the same thing to my eyes: a ball touching the ground and with velocity worth 0 m/s. How do you distinguish whether the ball will bounce off or not? I repeat, I've been told that knowing only the position and the velocity is enough to determine how the system will evolve.




    The problem is that at first I only know [tex]\vec r(0)[/tex] and [tex]\vec v (0)[/tex]. I've been told it's enough to determine the equation of motion of the whole system if I know this information for each particle. So I can't start with the dynamical equations of the system, I do not have them.
     
  8. Mar 11, 2010 #7
    You must have misunderstood, or you must be dealing with a special case. A special case would be particles that have no forces on them. Otherwise, if you don't know the forces, how can you know what the particles will do after their initial condition?

    Once you know the system constraints and forces, you can use physics (Newtonian, special relativity, etc as the case may be) to develop dynamical equations. These equations are often put in terms of state space equations which would be first order differential equations for time rate of change of the position and velocity vectors, which are the state variables. This is why initial conditions are needed for position and velocity, because you need to use them as the starting values of the system state variables. The rate equations then tell you how those state variables change in time.
     
  9. Mar 12, 2010 #8

    fluidistic

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    Ok thanks a lot, it makes sense.
     
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