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I remember to have read (a long time ago) that if I have a mechanical system $\ddot x=\frac1mF(x)$ with a high friction, then I can instead study the other system

$$\dot x\sim F(x)$$

to get an approximate qualitative understanding, e.g. an approximate spacial path $x(t)$. Is this true? If Yes, how can this be justified? E.g. if I model my friction like this:

$$\ddot x=\frac 1m F(x)-c \dot x,$$

how can I show that for increasing $c$, the system $\dot x\sim F(x)$ gives better and better qualitative results in some sense?


Extra: What are some keywords or approaches on how to study the energy loss in such a high friction system?

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1 Answer 1

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Physically, the reason for neglecting the $\ddot{x}$ term is that when the viscosity is huge, the system relaxes to its equilibrium very quickly. The crudest approximation would be to just assume that $x(t) = x_0$ does not depend on time. Then we must have $F(x_0) =0$. There both terms $\ddot{x}$ and $-c \dot{x}$ are neglected. This approximation can be improved to take the late-time dynamics into account by including only the largest missing term: $-c \dot{x}$.

I show this here with a (relatively) long calculation. I hope that you can follow it. Please complain if your can't. I focus on the last part of the dynamics. At infinite times, the system relaxes to equilibrium

$$ x(t \rightarrow \infty) = x_0 \, ,$$

with $F(x_0)=0$. The large time dynamics is then included by defining

$$ x(t) = x_0 + \delta x \, ,$$

with $\delta x $ small. Then we can write $F(x) \cong F(x_0) + F'(x_0) \delta x$ and recover a linear equation

$$ \ddot{\delta x} = -\frac{k}{m} \delta x - c \dot{\delta x} \, .$$

I define $k = -F'(x_0)$, which is positive because $x_0$ is a stable solution. The above equation is solved by

$$ \delta x(t) = A \text{e}^{- \frac{c}{2}\left(1+\sqrt{1-\frac{4k}{mc^2}}\right)t}+B \text{e}^{- \frac{c}{2}\left(1-\sqrt{1-\frac{4k}{mc^2}}\right)t} \, . \qquad (*)$$

$A$ and $B$ are (here) unimportant integration constants. I assume that $c$ is large enough for the arguments of both square roots to be positive. I now expand the two exponents to order one in $\epsilon = \frac{4k}{mc^2}$, which is small when $c$ is large, and get

$$ \delta x(t) = A \text{e}^{- \left(c-\frac{k}{mc}\right)t}+B \text{e}^{- \frac{k}{mc}t} \, .$$

We see two different relaxation rates, $\tau_1 = 1/(c-\frac{k}{mc}) \cong 1/c + k/(mc^3)$ and $\tau_2 = mc/k$. $\tau_1$ is much smaller than $\tau_2$. Therefore the second terms describes a much slower process than the first. If we are interested in the long-time dynamics, only the second term is important.

We can now insert $\delta x(t)$ back into the equation of motion and compare the terms. Since it is linear I do it for the two exponentials seperately. The first exponential, proportional to $A$ provides,

$$ \left(c-\frac{k}{mc}\right)^2 = -\frac{k}{m} + c \left(c-\frac{k}{mc}\right) \, .$$

We correctly recover a solution to the equation of motion and learn nothing. The second exponential (proportional to $B$) gives

$$ \left(\frac{k}{mc}\right)^2 = -\frac{k}{m} + c \frac{k}{mc} = 0 \, ,$$

which is only consistent when we neglect the term proportional to $\epsilon^2 \sim 1/c^2$. You can now track back to where this term comes from and find that is is precisely the $\ddot{x}$ term that you want to neglect. This tells us that the inertial part of the equation of motion (the $\ddot{x}$ term) is much smaller at large (but not infinite) times.

I realised while looking for keywords that this question is actually already answered here. There are plenty of keywords as well.

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