In analysis, a statement like $f(x) \ll g(x)$ (as $x\to x_0)$, has a very precise meaning: $$ \lim_{x\to x_0}\dfrac{f(x)}{g(x)}=0. $$ I was wondering, when physicists write $L_1 \ll L_2$, for, say, two length scales, what is the precise meaning, i.e., what is the limiting process lurking?

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    $\begingroup$ Without any further qualification, $L_1 \ll L_2$ just means that $L_1/L_2$ is a small number. Depending on who you ask and the application that might mean around $0.1$ or $0.001$. $\endgroup$ – knzhou Nov 21 '17 at 22:25
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    $\begingroup$ The above condition is valid as long as $L_1/L_2$ is sufficiently small. Whatever approximation is being made under this condition typically gets more and more accurate the smaller that ratio is. $\endgroup$ – probably_someone Nov 21 '17 at 22:25

It depends on the context. As mentioned in the comments, if we had two length scales $L_1$ and $L_2$ then we can say $L_1 \ll L_2$ to mean $L_1/L_2$ is 'small' but what 'small' is depends on what we are using the approximation for, and also the scales of other quantities.

More generally, the meaning may be different. In the case of general relativity, when we speak of a perturbation $h_{ab}$, saying it is 'small' is often written as $h_{ab} \ll g_{ab}$ but the literal meaning is not that the numeric values taken as ratios are small, since a coordinate transformation can change those.

Thus, the meaning of $\ll$ changes based on the context. Sometimes there is no rigorous meaning.

  • $\begingroup$ Thanks for the reply! In the context of GR since we're dealing with fields, can't we drop back to the precise definition and say $h_{ab}(x)\ll g_{ab}(x)$ (as $x\to x_0$)? But then again we would probably have an issue with limits on manifolds...Maybe the concept is purposefully non-rigorous. $\endgroup$ – EEEB Nov 21 '17 at 22:42
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    $\begingroup$ @EEEB No for two reasons. 1. We are not interested in how it behaves at a point $x_0$, we are interested in it being small for all points on the manifold. 2. We are not dealing with a simple numerical comparison (i.e. ratio) to establish smallness, it's more complicated. $\endgroup$ – JamalS Nov 21 '17 at 23:07

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