I am trying to understand scaling arguments.
Imagine one has a physical theory described by an equation whereby the first (spatial) derivative of a quantity, say $G$, equals the second (spatial) derivative of the same quantity, multiplied by some negative coefficient $\beta$.
Scaling is not needed to see the equation has a solution of the form: $$G \sim exp(-x/\beta)$$
The exponential makes it easy to define a "relaxation" length, characteristic of the problem.
If I try to use scaling arguments to find this " characteristic length", the first derivative is dimensionally $L/x$, where $x$ is a characteristic spatial length and $L$ is a length related to the change of $G$ over $x$.
But what can we deduce regarding the second derivative?
It will have dimensions of $1/length$, and will be related to the change of the first derivative, but in what way, I am unsure.
If I simply write $$l/x \sim \beta l/x^2$$ I obtain $x \sim \beta$, and I am not so sure what this means at all.
Best Answer
You are right that the only reason to set
$$\frac{d^2G}{d^2x} \sim \frac{l}{x^2} \, ,$$
is that they have the same dimensionality. The change in the second derivative of $G$ may be very different from the change in $G$ as $x$ changes over $l$.
Another way to do this is to go to dimensionless units. If you rescale $x$ with $l$,
$$ \hat{x} = \frac{x}{l} \, ,$$
and write your equation in therms of the rescaled variable, you will find that $\beta$ is rescaled as
$$ \hat{\beta} = \frac{\beta}{l} \, .$$
Then you can choose $l = \beta$ so that $\hat{\beta}=1$ and there is no scale left in your problem. Whatever the solution it will be of the form
$$ f(\hat{x}) = f\left(\frac{x}{\beta}\right) \, .$$
Since $\hat{\beta}=1$, $f(\hat{x})$ contains no free parameter. Everything is in $\beta$.