# When is a parameter considered small for perturbation and how does physical units affect that?

In perturbation theory procedures (not specific to any particular topic) we tend to have (or delibrately insert) some small variable $$\epsilon$$ in an equation that is otherwise difficult to solve if that parameter wasn't small. Then we try solving perturbatively (assuming a solution in the form of power series in $$\epsilon$$), typically by finding the leading coefficient when $$\epsilon \rightarrow 0$$, and thereafter calculating the remaining coefficients recursively. But the entire concept hinges on the relative smallness of $$\epsilon$$, and this leaves room for error when this parameter appears in physical problems with some normalisation or when it appears next to physical quantities that have units taken by convention (e.g. SI units). This is known to cause problems and the user should be careful, but I haven't found a clear, unified explanation on how this should be done.

For example, say I have an equation like $$y''-f(x)y'+\epsilon g(x) -(\epsilon+1)w(x)y/x=0$$, where the functions $$y=y(x)$$ or $$g(x)$$ may refer to, say, a magnetic field magnitude (measured in Tesla) or the velocity field (measured in meter/second), etc. How do we tell which values of $$\epsilon$$ will be considered small here? We cannot simply say it is when $$\epsilon<<1$$, because our physical units for these fields may be orders of magnitude smaller than 1 anyway. How do we find the benchmark of "smallness" here? Are there any general rules or conventions on this?

• ε is dimensionless, and the rest of the equation is dimensionally consistent, no? Commented Feb 6, 2020 at 1:48
• Related answer from fluid dynamics that applies here as well: physics.stackexchange.com/questions/256583/… Commented Feb 6, 2020 at 4:21

In purely technical terms, when we have a perturbation problem we introduce a dimensionless parameter $$\lambda$$ and rewrite the problem as unperturbed part $$+\lambda$$perturbation. Then we expand the solution in power series of $$\lambda$$ and equate coefficients. Using your concrete example to demonstrate, we have $$y''-f(x)y'-w(x)y/x=0$$ as the unperturbed equation, to which we know a set of solutions (otherwise it is useless to use as an unperturbed problem) $$y_n(x)$$ and let's focus on one solution $$y_0(x)$$. Now for the perturbation we now expand $$Y(x) = y_0(x) + \sum_{n>0}\lambda^n q_n(x)$$ and we plug this into the perturbed equation $$Y''-f(x)Y'-w(x)Y/x+\lambda\epsilon[g(x)-w(x)Y/x]=0$$ and we equate powers of $$\lambda$$. Then at the end we set $$\lambda=1$$ and we get the solution. Throughout all this process, we didn't assume anything on the perturbation, besides the fact that it can be expanded in a power series (which is not trivial! A lot of functions do not have such an expansion.) We also didn't have any dimensional discussion - $$\lambda$$ is dimensionless. However, as you can see in your example, $$\lambda$$ always came with $$\epsilon$$, so the power series in $$\lambda$$ will also be a power series in $$\epsilon$$ (this is not always the case, where we can in advance identify a parameter to expand in).
Now, we must analyze our solution, and see what we get. The solution might be a complete nonsense, of course, and no one guarantees that our approximation method works. In order for the solution to be valid, we want that the power series will converge. So we look at the results $$q_n(x)$$ and want them to be smaller than the original result itself $$||q_n(x)|| \ll ||y_0(x)||$$ (note that as $$\lambda$$ is dimensionless these two properties are bound to have the same dimensions). But it might be that this is not necessarily the "smallness" we are looking for. It might be that we want our solutions to remain well separated from the other solutions, in some sense $$||q_n|| \ll ||y_0-y_n||$$. Or we might want some observable property to get a small relative correction $$| E[y_0+\sum_{n>0} q_n]-E[y_0] | \ll E[y_0]$$.
If we have a small dimensionful expansion parameter in advance, as in your case with $$\epsilon$$, then it should correspond to some scale in our solution $$y_0$$, and then we can compare them and assume that for the most cases if $$|\epsilon_{y_0}| \gg |\epsilon|$$ then the expansion in power-series of $$\epsilon$$ will represent small corrections to the true solution. We can make it more formal by dividing the equation with $$\epsilon_{y_0}$$ and get a power-series in a dimensionless small number $$\epsilon/\epsilon_{y_0}$$. This, however, depends on the particulars of the problem.
How well this approximation works depends on how accurate we want our result to be. In general if we expand to $$(\epsilon/\epsilon_{y_0})^n$$ then this is the level of accuracy of our result. Sometimes just the first or two leading orders are enough for what we want, sometimes not.