I need to expand an equation, of the form $$\dot{r} = \gamma(a,\mu) F_1 + g(\mu,\ell,h,R) F_2$$

in powers of $\epsilon = a/\ell$.

So I thoughts I would non-dimensionalize it first.

I know that $$\gamma(a,\mu) = 1/6\pi a \mu,$$ with $\mu$ a dynamic viscosity, $a$ a length scale. $g$ is a complicated function of $\mu$ and $\ell,h,R$ which are also length scales. $F_{1,2}$ are forces, $\dot{r}$ is a velocity.

I would like to measure lengths in units of $\ell$, and times in units of $\Omega^{-1}$, where $\Omega$ is a frequency.

It is not clear to me how can I get to a position where I can perform an expansion in $\epsilon$.

  • $\begingroup$ It seems somewhat straight forward to write the equation in dimensionless form. What have you tried ? $\endgroup$ – Adam Jan 24 at 16:33
  • $\begingroup$ The problem arises when I have terms only in $a$, such as in $\gamma$. First of all, that does not give me $a/l$. Should I substitute $a=\epsilon \ell$? Also, I would get $1/\epsilon$, which is not ideal if I am to expand in small $\epsilon$. $\endgroup$ – usumdelphini Jan 24 at 19:49
  • $\begingroup$ If you write explicitly the dimensions of every term (in length and time and ...), it will be easy to sort this out. For example, by dimensional analysis, you can write $g(\mu,l,h,R)=l^{\alpha}\Omega^{\beta} g(\mu l^{\gamma} \Omega^{\delta},1,h/l,R/l)$, etc. This should then make the rest easy. $\endgroup$ – Adam Jan 24 at 20:02

You need to measure quantities in units of $\ell,\Omega$, and μ, so, then, viscosity units instead of mass--which has units of $\mu\ell/\Omega$.

The left-hand-side has units of $\ell \Omega$, which must also agree with the units of the right-hand side.

The forces F have units of $\mu \Omega \ell^2$; and $g$ has the same units as $\gamma$, namely $1/\ell \mu$.

As a consequence, in these units, the right-hand side reads, for dimensionless $f\equiv F/ \mu \Omega \ell^2 $, $$\ell \Omega\left (\frac{f_1}{6\pi \epsilon } + g\mu \ell ~f_2 \right ). $$

The parenthesis now consists of dimensionless quantities. As discussed, the dependence of the otherwise complicated function g on $\mu$ is simple, given its over-all dimension, $$ g(\mu,\ell,h,R)=\tilde g (\ell,h,R) /\mu ,$$ where $\tilde g$ still has units of $1/\ell$, which can be trivially scaled out of all of its length variables, as indicated below.

Defining $\tilde h\equiv h/\ell$ and $\tilde R\equiv R/\ell$, you end up with a dimensionless r.h.-side parenthesis, $$ \left (\frac{f_1}{6\pi \epsilon } + \tilde g(1,\tilde h, \tilde R) ~f_2 \right ). $$

You might experiment with simple notional example functions to appreciate the logic of the scaling.

If, indeed, you need to expand in terms of ε and not 1/ε, the form of the equation is unforgiving, and multiplying the equation by ε merely dictates excessive dominance of the of the first term to zeroth order, leading to its vanishing. But this non-dimensionalization is unique, given your desiderata. Perhaps you wish to modify them by, e.g., measuring lengths in units of a, instead.

Edit on alternate length units as per comment : If you wanted, instead, to use a instead of $\ell$ as your unit length, $\epsilon$ would disappear from the first term, and would only live on the second term. The second term might be proportional to $\epsilon$, depending on the explicit form of g. Recall g must have dimensions of 1/μ×length; so, e.g., for a purely hypothetical toy original
$$ g= \frac{ 1}{\mu \ell} ( 1+ (h/R)^2 + h/\ell +...) $$ the corresponding nondimensionalized function multiplying the nondimensionalized second force $F_2$ would be $$ a\mu g =\epsilon (1+(h/R)^2 + \epsilon (h/a)+...) . $$ and then expansion in $\epsilon$ would be regular... But without knowledge of specifics, one really cannot say much...

  • $\begingroup$ Could you please show how measuring lenghts in terms of $a$ would change the leading term in $\epsilon$? Would this be constant instead? $\endgroup$ – usumdelphini Feb 6 at 9:13
  • $\begingroup$ If you used a instead of $\ell$ in your units, $\epsilon$ would disappear from the first term, and the second term would have a dimensionless $g'(1/\epsilon, h/a,R/a)$ instead of the $\tilde g$ above. If the leading behavior of g were 1/$\ell$, the second term would go like $\epsilon$. You might have to give me a toy exemplary g to illustrate.... $\endgroup$ – Cosmas Zachos Feb 6 at 14:45

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