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Symmetric matrices and Newton's third law

  1. May 18, 2012 #1
    So, I was studying coupled oscillations and came across a statement that I couldn't figure out. It was that a particular matrix was symmetrical by Newton's Third Law. I know what Newton's Third Law is, I know what symmetric matrix is.

    But, for example, a matrix like this:

    -2k/m k/m


    k/m -2k/m

    Being multiplied times a vector like <x1,x2> to produce the acceleration vector <x1'',x2''>. It comes from the equations

    x1''=(-2k/m)x1+(k/m)x2
    x2''=(k/m2)x1+(-2k/m)x2

    I'm trying to see how Newton's third law makes the matrix symmetrical. I mean, I can see why each mass's equation of acceleration takes the same form, because the choice of x1 being x1 and x2 being x2 is arbitrary. Can somebody explain how the force-pair law means that this matrix will be symmetrical?
     
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  3. May 19, 2012 #2

    haruspex

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    Please describe the physical set up and the meanings of x1, x2, k, m in the context.
    Btw, you have an m2 at one point. Should that be m, or should all the other m's be qualified as m1, m2?
     
  4. May 19, 2012 #3

    AlephZero

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    It's certainly not "obvious" how the symmetry of the stiffness relates to the third law.

    If the stiffness matrix is not symmetric, the amount of work done moving between two points in space depends on the path you take when you move. In other words, if the stiffness matrix is not symmetric it is possible to do work (or extract work) from the system by moving around a closed path.

    You can demonstrate this with a 2 DOF system (but the details are fairly tedious, so you will have to work those out for yourself!)

    Suppose the stiffness is
    $$\begin{bmatrix} k_{11} & k_{12} \\ k_{21} &k_{22} \end{bmatrix}$$
    and you move from (0, 0) to (1, 1) by two different paths:
    1. Along the straight line from (0, 0) to (1, 0) and then along the straight line to (1, 1).
    2. From (0,0) to (0, 1) and then to (1, 1).

    Using "work = force x distance" you find the work done in one case is ##k_{11}/2 + k_{12} + k_{22}/2## and in the other case is ##k_{11}/2 + k_{21} + k_{22}/2##. If these are the same, then ##k_{12} = k_{21}##.

    A similar argument works for the mass matrix, using kinetic energy.

    In fact there are situations where the stiffness matrix is NOT symmetric, because there is an external source of energy and you CAN make the system do work by moving around a closed path, without creating a perpetual motion machine. But you don't need to worry about that in a first course on dynamics!
     
    Last edited: May 19, 2012
  5. May 19, 2012 #4
    Sorry, I should have said that the system is the following, in order from left to right (arbitrary, I know):

    <wall><spring of constant k><mass m><spring of constant k><mass m><spring of constant k><wall>

    AlephZero, I tried following your example but I couldn't get the result you were getting. How can you use Fd=W when the force isn't constant? I tried using 1/2<x|K|x> but only got 1/2k11+1/2k22 for both. I couldn't get the middle terms.
     
  6. May 19, 2012 #5

    AlephZero

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    Use integration. Work done = ##\int_{x_1}^{x_2} F(x)\,dx##. Since the k's are constant, this is the same as Work done = (average force) x distance.

    Along the line from (0,0) to (1,0), the first force component goes from 0 to ##k_{11}## so the work done is ##[(0 + k_{11})/2]\times (1-0) = k_{11}/2##. The displacement in the second component direction doesn't change, so the second force component does no work.

    Along the line from (1,0) to (1,1) the first force component does no work. The second force component goes from ##k_{21}## to ##k_{21}+k_{22}##, so the work done is ##[(k_{21} + (k_{21} + k_{22}))/2] \times (1-0) = k_{21} + k_{22}/2##.

    Of course the standard ##x^TKx/2## formula for the work done moving from (0,0) to (1,1) gives ##(k_{11} + k_{12} + k_{21} + k_{22})/2##, and because K is symmetric ##(k_{12} + k_{21})/2 = k_{12} = k_{21}##.
     
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