B What is meant by "first order" and "second order"

[Buckethead]

I see comments such as "explains ... to the first order" or "to the second order" quite a bit in physics discussions. Can someone explain in lay terms, what first order and second order refer to?

 

[fresh_42]

Imagine a function $f(x)=\log(x^3+1)$. To compute function values can be a hard task, but there is a representation as an infinite sum, namely $f(x)=x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3}+\ldots$, here for small function values around $x=0$. Of course we cannot sum up infinitely many terms, so we will have to stop somewhere. E.g. if we write $f(x)=x^3-\dfrac{x^6}{2}+ C\cdot x^9$ with some constant $C$ we say that $f(x)$ is approximated by $x^3-\dfrac{x^6}{2}$ up to first order. If we write $f(x)=x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3}+C\cdot x^{11}$ then we speak of an approximation up to second order. The word approximation is often left out and people say $f(x)\approx x^3-\dfrac{x^6}{2}$ up to first order. That means, it is exact up to a linear approximation, a tangent at a point. The further away we get from this point the less suited is the tangent as an approximation and we might want to calculate more than two summands. The $x^3$ term counts as zeroth order.

Have a look at the graphs at the end of the page: http://www.wolframalpha.com/input/?i=Taylor+f(x)=ln(x^3+1)

 

[anorlunda]

There's no mystery. first order effects are the most significant, and second order effects are refinements or tweaks.

Order can refer to order of magnitude. Therefore first order effects give the right answer to within 10%, second order to within 1% and so on.

 

[Buckethead]

Perfect! Thank you both for your answers. That clears up a lot.

Is it just guesswork as to when a solution is sufficient in the number of orders used? For example I saw a discussion where an effect was not seen at first order but was seen at second order. So in general one should solve to as high an order as practical until you get diminishing returns?

I also seen things such as "an exact solution could not be found" and I imagine this just refers to the order at which the equation was solved and does not refer to a failure at being able to solve the equation?

 

[fresh_42]

Is it just guesswork as to when a solution is sufficient in the number of orders used?

Not quite. It is driven e.g. by the accuracy of measurements. It makes no sense to compute $10$ digits if you can only measure $2$. Or it is given by the purpose. Earth's surface is curved, nevertheless we get along well with flat street maps. The error is just too small. But you better hope your pilot doesn't use flat directions on a trans-Atlantic flight.

For example I saw a discussion where an effect was not seen at first order but was seen at second order. So in general one should solve to as high an order as practical until you get diminishing returns?

Ideally yes, but first order approximations are linear approximations, i.e. tangents. Those are far easier to calculate and really often sufficiently close at small distances, cp. the examples above. They at least carry the tendency.

I also seen things such as "an exact solution could not be found" and I imagine this just refers to the order at which the equation was solved and does not refer to a failure at being able to solve the equation?

No, that's a computational remark. E.g. solve $x^5+c_1x^4+c_2x^3+c_3x^2+c_4x+c_5=0$ can usually not be done via formulas, but only by numerical and thus approximation procedures. And there are a lot more problems, where we don't have closed forms for and only numerical approaches. The reason is that natural processes are often far more complicated than could be described by simple equations, so that algorithms will be necessary - and the computer has no number for $\pi$, only an approximation.

The series I mentioned above are exact, even if they are infinitely long. So in those cases we can work with them and make no mistake. But they are often not nice to handle, so that they are cut after some steps. In the example above, the error for $x=0.1$ in the second order approximation is at the tenth digit, and thus irrelevant for most applications.

 

[Buckethead]

Great! Thank you so much for the detailed answer. Much appreciated.

 

[sophiecentaur]

Can someone explain in lay terms, what first order and second order refer to?

You have to remember that many of the mathematical models that are used to describe the Physical World do not involve simple polynomials. There are many models that involve trigonometrical and exponential functions etc etc.. When we talk about second (or other) order effects from these models, we are actually approximating them with a simple polynomial ( a quadratic or higher order).
This is something that people to take for granted but it can be confusing for someone who isn't familiar with the Taylor Expansion and other mathematical tricks. (Also, you need to bear in mind that it cannot always be done this way.)

 

[Buckethead]

You have to remember that many of the mathematical models that are used to describe the Physical World do not involve simple polynomials.

Thank you for that reminder. It is easy to forget that mathematical models of reality are just that and not the actual mechanisms by which reality operates and are therefore always going to be approximations. Excellent!