I have encountered $\gg$ in many physics text books where it's used as a relation between constants or functions but in none of the text books I have read is it properly defined anywhere.

If $A \gg B$ where $A$ and $B$ are constants, or $f(x) \gg g(x)$ does this simply mean that $A \geq 10B$ and $f(x) \geq 10g(x)$ for all $x$?

Note: I'm asking this here because mathematicians don't use the much greater than relation. At least not that I know of.

  • 5
    $\begingroup$ I do think this might be a math question, but we'll see what people think. $\endgroup$ – David Z Sep 21 '15 at 18:09
  • $\begingroup$ Ive seen the much greater than symbol used in calculus.. $\endgroup$ – TanMath Sep 21 '15 at 19:31

There is a consistent definition, but it involves a couple of arbitrary thresholds, so I doubt you'd consider it rigorous. The construction $X \gg Y$ means that the ratio $\frac{Y}{X}$ is small enough that subleading terms in the series expansion for $f\bigl(\frac{Y}{X}\bigr) - f(0)$ can be neglected, where $f$ is some relevant function involved in the calculation. This depends on what you mean by "can be neglected", of course; typically that will be determined by the uncertainties of your data or your theoretical parameters.

Technically it depends on $f$, too, but the functions we use in physics normally have power series with coefficients that are close to 1, or within a couple orders of magnitude at least. As long as the coefficients don't grow exponentially (by which I mean $f_n \approx f_0 k^n$ for some $k$), they don't have much of an effect on which terms in the series are negligible.

As an example, consider a typical condition in relativity, $v \ll c$. (Or $c \gg v$ if you prefer.) The relevant function is, of course, the gamma factor: $$\gamma\biggl(\frac{v}{c}\biggr) = \frac{1}{\sqrt{1 - v^2/c^2}} = 1 + \frac{1}{2}\biggl(\frac{v}{c}\biggr)^2 + \frac{3}{8}\biggl(\frac{v}{c}\biggr)^4 + \cdots$$ so we should examine the subleading term of $\gamma\bigl(\frac{v}{c}\bigr) - \gamma(0)$, which is $\frac{3}{8}\bigl(\frac{v}{c}\bigr)^4$, to see if it's negligible relative to the leading term, which is $\frac{1}{2}\bigl(\frac{v}{c}\bigr)^2$. This is equivalent to checking how the ratio of the two terms, $\frac{3}{4}\bigl(\frac{v}{c}\bigr)^2$, compares to $1$.

plot of ratio of subleading terms to leading term

I've included the ratio of all subleading terms to the leading term, as well as the ratio of only the first subleading term to the first one, but for slow speeds they're nearly the same anyway. And for high speeds you clearly have no business claiming $v \ll c$ in the first place.

Here is where the most obvious ambiguity comes in: where do you draw the line? Interestingly enough there can be an actual line: you draw a horizontal line across the plot at whatever level you're comfortable considering "negligible", and where that crosses the curve, that tells you what value of $v$ sufficiently satisfies $v \ll c$. The value depends on your requirements, of course, but you can see that for $v \lesssim 0.1c$, the curves are nearly indistinguishable from the bottom of the plot. So it's very common to consider $v \ll c$ satisfied when $v \lesssim 0.1c$.

  • 1
    $\begingroup$ +1 for "can be neglected" depends on the situation. As an electrical engineer "much greater than" depends on target system accuracy but often is 100 times the value to keep error to small single digits. $\endgroup$ – Phil Sep 22 '15 at 0:24
  • 2
    $\begingroup$ Note that the last paragraph also means that you can often (not always!) make a precise mathematical statement: Given an accuracy $\varepsilon>0$ of whatever you want to calculate, there will be a threshold $c$ such that for all $Y/X<c$ the statement is true with precision at least $\varepsilon$. The only reason not to do this is because it is often very clumsy. $\endgroup$ – Martin Sep 22 '15 at 15:49

It is a symbol and an idea used in mathematics too. But the important part is just that $B$ is 'ignorable' relative to $A$. This depends on the level of precision that is being used experimentally. If you're working to a precision of 1 part in 100, then $B$ should not effect the answer to that level of precision. If you're working to 1 part in a million, then $B$ should not effect the answer to that level.

For example, we could say that general relativity is ignorable for putting a man on the moon, but not for running a GPS system. The more specific takeaway is to learn that there's no such thing as an exact number in physics. It's always $A\pm err$


I think the answer is no. It generally precedes some approximation method with a bounded error, but there are so many approximations methods in physics -- some rigorous, some nonrigorous -- that it's way too presumptuous to give it a rigorous definition.

Generally, it means one of several things:

  1. If $a\ll b$, expanding in powers of $\frac{a}{b}$ is appropriate (and implicitly -- is within the power series radius of convergence!)
  2. If $a\ll b$, then $ka=b$ for some $k>1$ "sufficiently large". "Sufficiently large" can mean totally different things depending on the application.

That's it. I don't think you can say anything more about it.

Expanding in parameters of a small parameter is rigorous and well defined, and usually gives you an upper bound on the error, but this is not always what is done. Let me give you an example where the actual mathematical "real analysis" details would be difficult to work out. This example is an example where you have no clue what "sufficiently large" means until you finish the problem.

Consider the equation $f''(x)+A^2 f(x)+A f'(x)^2=0$. I know it's contrived, but you can really end up with nasty equations like this one. Note also that the units are consistent if $f$ is in meters, $x$ is in meters, and $A$ is in inverse meters. $f$ could be lots of different things, but the reasoning in many approximation methods is as follows: Say $f(x)$ changes significantly (doubles/halves/what-have-you) only on length scales of order $L$. Then you might expect differentiating to have the effect of dividing by $L$. Then the magnitude of the terms should be like $L^{-2} f(x)+A^2 f(x)+ L^{-2} A f(x)^2$. Suppose $L\ll f(x)$ and $f(x)\ll A$. Well, I don't see any method to prove that terms in here can be neglected, so I'm going to assume the third term can be neglected. Then the remaining ODE is just a sinusoid with a certain small magnitude, maybe $f(x)=\frac{k}{A} \sin(A x)$. Let's check if the assumption is valid by plugging everything back into the ODE:

$f''(x)+A^2 f(x)+A f'(x)^2=-A k \sin(x)+A k \sin(x)+A k^2 \sin^2(A x)$

So, the assumption is good if $A k^2$ is small enough, but what is small enough? We can't say "$A k^2\ll 1$" because then the units wouldn't be consistent, but we know it has to be close enough to zero for us to not care about it.

On that matter, what could $L\ll f(x)$ mean? $\frac{L}{f(x)}$ diverges to infinity periodically, so it isn't even less than one! And I had to assume neglecting something would work in the first place. Plus, what if the ODE has some problem where there's feedback from the nonlinear term we neglected? Then the solution blows up and the whole thing is crap!

There are lots of things you could do to fix all these problems, but my point is: Approximations are an art, and in practice with things like this, the process of solving your problem is sloppy and nonrigorous, but with some luck you can go back later and patch up all of the mathematical holes in your argument. If I wanted to get a real mathematical result out of this, I might start with "Assume $f(x)=\frac{k}{A}\sin(A x)$. Then $|f''(x)+A^2 f(x)+A f'(x)^2| \lt \varepsilon$ for all $x$, where [some relationship between $k$ and $\varepsilon$]." All the approximations I did to get to the formula for $f(x)$ were nonrigorous and severely need to be patched up!

It ain't always just series expansions.


I think there is a rigorous definition of "$\ll$" sign which is opposite to what you are asking but equally useful notion. You should read this "$\ll$" as is negligible compared to. For example, $f(x) \ll g(x)$ near $x=x_0$ (in a more general context) iff $ \frac{f(x)}{g(x)}\to 0$ as $x\to x_0$.

  • $\begingroup$ No, this is not how people use it. What would $M_{\text{sun}} \gg M_{\text{moon}}$ mean then? There's no limit, no "point", nothing. $\endgroup$ – yo' Sep 22 '15 at 6:56
  • $\begingroup$ @yo My answer is particular to functions and I don't see why it is not correct. $\endgroup$ – Physics Moron Sep 22 '15 at 7:13
  • $\begingroup$ $M_{sun} \gg M_{moon}$$\leftrightarrow M_{moon}\ll M_{sun}$ which would mean "The mass of the moon is negligible compared to the mass of the sun." You could also word it as "The mass of the sun is practically infinite compared to the mass of the moon." Both of those are perfectly useful, meaningful definitions, but I'm not sure how they qualify as "rigorous". In the example, what is "near" $x=x_0$? $\endgroup$ – MichaelS Sep 23 '15 at 6:57
  • $\begingroup$ @MichaelS I agree. That's why in my comment I said My answer is particular to functions. $\endgroup$ – Physics Moron Sep 23 '15 at 11:11

I've often encountered the $\ll$ sign in contexts like:

Given $A \ll B$, a certain property $P$ holds - not approximately or as a limit, but exactly. This is then the opposite to that $P$ would be true for $\gtrsim$ or $\gtrapprox$ (depending on style). I don't remember specifific cases, but I've seen the latter in studies of dynamic systems, chaos etc.

The way I've thought of $A \ll B \implies P $ is then just that there exist an $X<B$ such that for all $A < X$, the property holds true.