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Noncentral potential and Newton third law

  1. Aug 8, 2006 #1
    Hello,
    I was just searching online and discovered this great forum. Please I need help in sorting out a problem with a non-central potential and Newton third law.

    This is not a physics homework problem. I'm a graduate student working on this equation as a reserch project and there are publications out there using this expression in "Analytical Bond order potential" by Pettiffor et al.
    They call it Yukawa potential.

    The form of the potential is:

    (Sorry for the quakky expressions, I'm not sure if there is an equation editor on this forum, a guide will be appreciated).

    U_ij = (A/r_ij)* exp(-S_ij(r_ij - r_c))

    Where r_ij = sqrt((x_i - X_j)^2 + (y_i - Y_j)^2 + (z_I - z_j)^2)

    or simply distance between atom i and atom j in cartesian coordinate

    However,

    S_ij = 1/2(S_i + S_j)

    Where;
    S_i = k_o + (sum(over k not equal to i) exp(-c*r_ik))

    similarly

    S_j = k_o + (sum(over k not equal to j) exp(-c*r_ jk))

    Where r_ik and r_ jk are distances from atoms i to atom k
    and the distance from atom j to atom k respectively.

    This is the real question.
    Suppose we have three atoms arranged in a triangle connected to each other where non of the distances between the triangle is the same; that is rij, rjk and rik are not the same.

    We expect for three body system with no external forces in a closed system to have a conserved force. The forces should sum up to zero; That is

    F_i = - (F_j + F_k).


    I'm not sure if I'm doing something wrong here.:confused:

    My problem is that the above energy expression is not giving a conserved forces that sum up to zero, especially when the environment of atoms i and j are different, That is, S_i is not equal l to S_j (when r_ij , r_ik and r_ jk are not all equal) and therefore F_i /= - F_ J so Newton third law of action and reaction are eqaul and opposite is violated.

    Despite the fact that this force depend only on position, the system does not yield to conservation law.

    The three atom in a triangle is the most trivial case, but the real problem actually involves many atoms surronding atoms i and j with different densities (S_i and S_ j ) in such a way that the force on atom i due to atom j is affected by other atoms k1_i, k2_i, k3_i, ... kn_i surrounding atom i.

    While similarly the force on atom j due to atom i is affected by other atoms
    k1_j, k2_j .....kn_j surrounding atom j.

    I need to use this potential to derive the forces to perform molecular dynamics simulation and it's been alot of headeache moving forward with this problem as the system will not conserve energy ( Give zero total force) in a close system.

    Any help will be appreciated. Sorry for my long post, I'm just trying to make things clear. Thanks.
     
  2. jcsd
  3. Aug 8, 2006 #2
    Wao! 21 view and no reply?

    Am I in the wrong section?
     
  4. Aug 8, 2006 #3

    Physics Monkey

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    Hi Yukawa,

    Welcome to Physics Forums!

    First, the forum supports the use of LateX to make your equations look pretty. Just use the [ tex ] tag around an expression as in [tex] \textbf{F} = m \textbf{a} [/tex] (click to see the code).

    Second, since the potential depends only on the coordinate differences between various atoms, the total momentum of the system must be conserved. This momentum is just the conserved quantity associated with the translation invariance of the Lagrangian (Noether's Thm). Unfortunately, it's not immediately apparent to me what is going wrong, but you should find that the forces do cancel. I imagine it's just an algebra error somewhere.

    Third, that is one crazy potential!
     
    Last edited: Aug 8, 2006
  5. Aug 8, 2006 #4
    Hey thanks for the response.

    Is it the Linear momentum that should be conserved in this case?

    For the 3-body system example that I gave, the forces only cancels when r_ij and r_ik are the same or when all the three atoms are not connected in a triangle.

    When all the three atoms are connected in a triangle, then the total force on the system is no longer zero. I have used the center of mass for the whole system to constrain the velocities to give zero total linear momentum with no success.

    I just don't know what is going wrong here. The fundermental problem here is that when the environment of i and j are different, then the magnitude of the force on i is not equal to the force on j, but I do expect the total system force to sum up to zero which is also not happening here.

    Any help as to what I'm doing wrong will be appreciated.
     
  6. Aug 9, 2006 #5

    Physics Monkey

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    Hi again yukawa,

    Yes, it is the linear momentum of the system which is conserved.

    I can give you a simple proof that the sum of the forces is zero. The total potential energy of the system is [tex] U = \frac{1}{2} \sum_{i \neq j} U_{i j} [/tex] (the 1/2 takes care of double counting). Note that this quantity depends only on the relative separation of the atoms because each [tex] U_{i j} [/tex] depends only on the relative separation. The force on the i-th atom is given by [tex] \vec{F}_i = - \nabla_i U [/tex], where [tex] \nabla_i [/tex] means derivative with respect to the coordinates of the i-th atom. This is, of course, nothing but the standard formula for force in terms of potential energy.

    Now, in a system with N atoms, the sum of all the forces is [tex] \sum_{i = 1}^N \vec{F}_i = \sum_{i = 1}^N - \nabla_i U [/tex]. We need to show that this is zero irrespective of the positions of the atoms, etc. To do this, multiply both sides by an arbitrary vector [tex] \vec{a} [/tex]. The left hand side is just the dot product of [tex] \vec{a} [/tex] with the sum of all the forces. The right hand side looks like [tex] - \sum_i \vec{a}\cdot\nabla_i U [/tex], but we know what this term means. If [tex] \vec{a} [/tex] is small, this is nothing but the approximate difference between [tex] U(\vec{r}_i) [/tex] and [tex] U(\vec{r_i} + \vec{a}) [/tex].

    Now we just have to realize that [tex] U [/tex] depends only on coordinate differences, so a shift of every coordinate by the same amount (by [tex] \vec{a} [/tex]) won't change the value of [tex] U [/tex]. Thus the right hand side is zero no matter what [tex] \vec{a} [/tex] is, and this means that left hand side must also be zero no matter what [tex] \vec{a} [/tex]. Now at last we see that [tex] \vec{a} \cdot \sum_i \vec{F}_i [/tex] can only be zero for all [tex] \vec{a} [/tex] if [tex] \sum_i \vec{F}_i = 0 [/tex]. This simple proof shows that the forces must sum to zero for your potential.

    In practical terms: if you've found that the forces don't sum to zero, it means you've made some kind of algebra or differentiation error somewhere. That's about the best I can do unfortunately. I hope it helps some.
     
    Last edited: Aug 9, 2006
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