Saturday, April 19, 2008

[NR 2] Chain of NKBs

We first consider a chain of N linear Kelvin bodies (LKBs):

Assume that all dashpots are equal, but that the spring constants k_n differ for each body n.

Since all LKBs feel the same force F(t), a separate evolution equation can be written for each individual body:

\dot{x_n}=\frac{1}{\gamma}\left[ F(t)-k_n\;x_n \right]

We are interested in the total length X(t) of the chain, which is given by:

X(t)=\sum_{n=1}^N x_n(t)

The total microstate of the system is described by the vector of the N individual lengths x_n(t), but externally only the global variable X(t) is relevant.

In this linear case, we can define a response function S_n(w) for each individual LKB:

x_n(\omega)=\left[  \frac{1}{i\omega\gamma+k_n} \right] \;F(\omega)=S_n(\omega)\;F(\omega)

Remarkably, we can even define a global response function of the LKB chain in frequency space, which directly connects the total chain length X with the external force F:

X(\omega)=\sum_{n=1}^N x_n(\omega)=\left[ \sum_{n=1}^N S_n(\omega)\right] F(\omega)=S(\omega)F(\omega)

This means that in the linear case, we don't have to keep track of the N microvariables if we just want to know the global X<-->F response.

Now we replace the LKBs by the nonlinear Kelvin bodies described in [NR 1]. All we can do is to write down the separate DGLs for each individual NKB:

\dot{x}_n=\frac{1}{\gamma}\left[ F(t)-N_n(x_n) \right]

There is no abbreviation in this nonlinear case to obtain X(t) from F(t): We have to solve the N coupled differential equations simultaneously and at each time sum over the individual lengths.

So this simple example demonstrates:

We cannot in general treat a non-linear system like a black box and write down response equations which depend only on the externally accessible macro-variables. We rather have to use an evolution equation for the full multi-component micro-state, which will boil down to the solution of a coupled set of nonlinear differential equations.

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