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Internal Forces in System of Particles

  1. Oct 4, 2014 #1
    I am following along with Goldstien's Classical Mechanics Book and I am on page 11. The text is breaking down the total potential energy of a system into two parts: the external conservative forces and the internal conservative forces. My question pertains to the internal forces.

    Writing the sum of the internal forces as: $$\sum_{i,j} \int_1^2 \vec F_{ji} \cdot d \vec s_i.$$

    He then uses the fact that the interacting force is conservative to rewrite in a potential $$\vec F_{ji} = - \vec \nabla_i V_{ji}.$$

    Now we have: $$\sum_{i,j} \int_1^2 \vec F_{ji} \cdot d \vec s_i=-\sum_{i,j} \int_1^2 \vec \nabla_i V_{ji} \cdot d \vec s_i.$$

    Still a bit shaky on formally why, but I recognize the integral works out to simplify to: $$\sum_{i,j} \int_1^2 \vec F_{ji} \cdot d \vec s_i = -\sum_{i,j} V_{ji} \Big |_1^2.$$

    However, there is at least one mistake here, as I should have gotten a factor of 1/2 out front. Which I understand the need for because you are double counting potential. However, I am failing to see where it comes out of from the math. From my understanding you do want to "double count" the forces to get the correct total work, but double counting the potential get you twice the work you should have. Any insight would be appreciated.
     
  2. jcsd
  3. Oct 5, 2014 #2

    Jano L.

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    Gold Member

    The last equation is not correct. You have probably used the transformation

    $$
    \int_1^2 \nabla_i V_{ji}\cdot d\mathbf r_i ~-> V_{ji}(\mathbf r_i(2), \mathbf r_j(2))- V_{ji}(\mathbf r_i(1), \mathbf r_j(1))~~~(wrong).
    $$
    This transformation is not correct, because ##V_{ji}## is not a function of the integration variable ##\mathbf r_i## only. That the transformation not correct generally is easily seen from this simple case: consider a process in which the particle ##i## does not move and the particle ##j## changes its position. The integral on the left is zero because there is no change in ##\mathbf r_i##, but the right-hand side is non-zero because ##\mathbf r_j## changed. So the two sides of the transformation cannot be equal.

    The theorem that is actually used in Goldstein is
    $$
    \int_1^2 \nabla_{\mathbf r} V(\mathbf r)\cdot d\mathbf r = V (\mathbf r(2))- V(\mathbf r(1)).
    $$
    which is valid when ##V## is a function of one integration variable ##\mathbf r##.


    Goldstein considers ##V_{ij}## to depend on two variables ##\mathbf r_i, \mathbf r_j## so the theorem cannot be directly applied. However, ##V## is considered to depend on the two vectors in such a way that it can be considered a function of difference of these two vectors. Goldstein denotes this difference ##\mathbf r_{ij} = \mathbf r_i - \mathbf r_j## and the new function as ##V_{ij}##.

    He then rewrites the integrals in the expression for work in terms of this new function and new vector variable. Only then is the above theorem used.
     
  4. Oct 5, 2014 #3
    Thank you. I looked over the integral again and see exactly what you are saying and better understand all of that work. I think I now understand. Here is where I see the 1/2 coming into play:

    $$\sum_{i,j} \int_1^2 \vec F_{ji} \cdot d \vec s_i = -\frac{1}{2} \sum_{i,j} \int_1^2 \left( \vec \nabla_i V_{ij} \cdot d \vec s_i + \vec \nabla_j V_{ji} \cdot d \vec s_j \right ).$$

    The second term on the right side is equal and opposite of the first term, but since we are summing over all i and j, then it will actually work out to the same thing. Thus we have introduce a factor of two we need to remove with a 1/2.
     
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