Symmetric matrices and Newton's third law

In summary, The conversation discusses coupled oscillations and the relationship between a symmetrical matrix and Newton's Third Law. The matrix in question is used to represent a 2 degree of freedom (DOF) system, where x1 and x2 are arbitrary choices for displacement and k and m represent stiffness and mass, respectively. The stiffness matrix is shown to be symmetric due to the force-pair law, which states that the amount of work done in moving between two points in space is independent of the path taken. This is demonstrated by calculating the work done in two different paths and showing that they are equal if the stiffness matrix is symmetrical. However, there are cases where the stiffness matrix may not be symmetric, but this is not relevant in
  • #1
TomServo
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9
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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  • #2
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?
 
  • #3
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!
 
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  • #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.
 
  • #5
TomServo said:
How can you use Fd=W when the force isn't constant?

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}##.
 

FAQ: Symmetric matrices and Newton's third law

What is a symmetric matrix and how is it related to Newton's third law?

A symmetric matrix is a square matrix that is equal to its transpose. In other words, if matrix A is symmetric, then A[i,j] = A[j,i] for all i and j. This property is related to Newton's third law, which states that for every action, there is an equal and opposite reaction. In the context of matrices, this means that for every entry in the matrix, there is a corresponding entry in the symmetric position that has the same value.

2. How can symmetric matrices be used in physics and mechanics?

Symmetric matrices are commonly used in physics and mechanics to represent physical systems and equations. For example, in the study of forces and motion, symmetric matrices can be used to represent Newton's second law of motion, which states that force is equal to mass times acceleration. Additionally, symmetric matrices can be used to represent systems of forces and their interactions, as well as to solve problems related to equilibrium and stability.

3. What is the significance of symmetry in Newton's third law?

The concept of symmetry is important in understanding and applying Newton's third law because it helps to explain why forces act in pairs. When two objects interact, the forces that they exert on each other are equal in magnitude and opposite in direction. This symmetry in the forces is a direct consequence of Newton's third law and is essential in predicting and understanding the behavior of physical systems.

4. Can symmetric matrices be used to solve problems related to Newton's third law?

Yes, symmetric matrices can be used to solve a variety of problems related to Newton's third law. For example, they can be used to calculate the net force on an object, determine the acceleration of a system, and analyze the equilibrium of a system. Additionally, symmetric matrices can be used to model and predict the motion of objects under the influence of multiple forces, making them a valuable tool in solving problems involving Newton's third law.

5. Are there any limitations to using symmetric matrices in the context of Newton's third law?

While symmetric matrices are a useful tool in understanding and applying Newton's third law, there are some limitations to their use. For example, symmetric matrices may not be able to accurately model non-rigid bodies or systems with complex interactions between forces. Additionally, they may not account for factors such as friction or air resistance, which can affect the motion of objects in the real world. It is important to consider these limitations and use other techniques and models when necessary to fully understand and analyze physical systems.

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