# Taylor approximation in physics

In a textbook about semicoductor physics, I came across a passage about deriving the carrier concentration at thermal equilibrium in semiconductors I didn't quite grasp:

The recombination $R(T,n,p)$ rate depends on the ionization energy, temperature $T$, and carrier concentrations $n$, $p$. As the recombination rate must be zero if one of the carrier concentrations $n$, $p$ is zero, it follows for the first non-vanishing term of a Taylor expansion: $$R(T,n,p) = r(T)np$$

The Taylor expansion, as I know it from the mathematics class, is defined only for functions of one variable. Besides that, there is always a point involved about which the expansion is carried out. Strictly speaking, $R$ is a function of three variables. And no point is given. I guess this is the kind of inaccuracy practiced among physicists in order not to bloat their argumentation, and generally understood -- among physicists.

So, how does an expansion of a function of three variables work and which point should I assume here?

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For Taylor series in several variables, see e.g. wikipedia en.wikipedia.org/wiki/… – Qmechanic Nov 20 '11 at 13:19
@Qmechanic thanks for the link. What about the point? Any idea? – artistoex Nov 20 '11 at 13:55

The argument is not a mathematical Taylor expansion, there is an implicit physical argument here which is very well known but not given.

The rate of a reaction (like recombination) at low density should be the product of the densities, with a coefficient that is independent of the density. This happens to be a mathematical Taylor expansion in the densities. But zero is a special point for an expansion, because there are natural power-law change of variables, and you can't determine which power is right without knowing the physics. For example, the period of a pendulum goes to zero as the length of the pendulum goess to zero, does this mean that the period must be proportional to L? Of course not. It's the square-root of L.

### The correct argument

When you have two objects which are moving randomly and have to find each other, the probability each A object finding a B object per unit time is proportional to the density of B objects. If you have twice as many B objects per unit volume, it takes half as long to find one, just from independent search statistics. Similarly, the rate is doubled if you double the concentration of A objects. So the leading term is the so-called mass-action term, the product of the concentrations.

At higher concentrations, the A's and B's start to have excluded volume effects--- it is slightly easier for A to find B because the mass-action assumes indepedent motion of the different B's while the real B's only can travel in the volume not excluded by other B's. There are also corrections to the diffusion rate fron interactions.

The corrections are negligible when the gas is dilute compared to the physical cross-section of interaction, and this limit defines the domain of applicability of mass action dynamics. The argument that the molecules are in the mass-action limit is what is used here, and the Taylor expansion phrasing is suboptimal.

But the result is that the physics of recombination is by the first term in a taylor expansion in the two densities, so the calculation that follows is correct, even though the physical argument is wrong. This is typical of textbooks, and you should probably throw this one away.

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Thank you for your comprehensive answer! So the mass action law is a premise to rather than a conclusion of $R(T,n,p)=r(T)np$? That would be really funny, because after the passage cited above, the textbook goes on to derive the equation for the mass action law :-). – artistoex Nov 23 '11 at 14:09
@artistoex: What is there to derive?? The equation r(T)np is the mass action law. I wouldn't call it a premise or a conclusion--- it is just a self-evident equivalent form of this equation. The argument for mass action is statistics, it is not a Taylor series, because there is no a-priori reason for the rate to be analytic in concentration at zero concentration. This book is written by somebody incompetent, and you should throw it away and get a book by somebody else. – Ron Maimon Nov 23 '11 at 21:00
Maybe It's because I lack deeper understanding, but the mass action law according to Wikipedia, $np=n_i^2$, looks a bit different to me, and the textbook actually claims to "derive" that formula. – artistoex Nov 30 '11 at 13:50
Wrong mass action: en.wikipedia.org/wiki/Law_of_mass_action . The mass action law for the case you are considering has a=1 b=1 in the notation of Wikipedia. The exponents are the number of species of the reactant that go into the reaction. The reasoning is that if you have a reaction where two molecules go to something, if you double the concentration of either molecule, it takes half as long for them to find each other in a solution, so that if you double both, you multiply the rate by 4. For molecules that self-react, double the concentration of this molecule, the rate quadruples. – Ron Maimon Nov 30 '11 at 18:37

The temperature $T$ plays a passive spectator role, so let us leave it out of the notation.

Then the Taylor expansion of $R(n,p)$ at the point $(n,p)=(0,0)$ reads

$$R(n,p)~=~ R(0,0)+ R_n(0,0) n + R_p(0,0) p$$ $$+ \frac{1}{2}R_{nn}(0,0) n^2 + \frac{1}{2}R_{pp}(0,0) p^2+ R_{np}(0,0) np$$ $$+ {\rm third~order~and~higher~terms},$$

where the sub-indices mean partial differentiation wrt. the indices. For instance,

$$R_n(n,p)~:=~ \frac{\partial R(n,p)}{\partial n}, \qquad R_{np}(n,p)~:=~ \frac{\partial^2 R(n,p)}{\partial n\partial p}, \qquad{\rm etc}.$$

The textbook then claims that all of the zeroth-order, first-order and second-order terms must be zero except the $R_{np}(0,0) np$ term, since the process requires both $n$ and $p$ to be present.

Now identify $R_{np}(0,0)$ with $r$.

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I don't follow the zero argument except for the first term. But thank you very much, anyway! – artistoex Nov 20 '11 at 16:38
@artistoex: I have updated the answer. – Qmechanic Nov 20 '11 at 17:02
This answer was prepared with the only intention to faithfully reproduce the Taylor series argument as it was given in the snippet of the textbook. I agree with Ron Maimon that the textbook argument is not optimal. – Qmechanic Nov 21 '11 at 14:04