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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.