A Necessary and Sufficient Condition in the Model of Kondepudi and Nelson for the Breaking of Chiral Symmetry

Abstract

The setting is a system containing achiral reactants which form a chiral compound. In 1983, Kondepudi and Nelson proposed a model for the breaking of chiral symmetry. The present article reduces the conditions for bifurcation to a single condition which is shown to be both necessary and sufficient. A number of other papers on this topic also propose models for the breaking of chiral symmetry. These are shown to be essentially special cases of the model of Kondepudi and Nelson, with the same necessary and sufficient condition. The central question of this line of research is: in a racemic mixture of a chiral compound, could an excess of one enantiomer over the other develop on its own? Our answer is yes, if and only if a certain simple condition is satisfied. This answer should prove useful in further research, both theoretical and experimental, into the origin of life.

1. Introduction

One of the difficulties for the theory of chemical evolution has been the homochirality of certain organic molecules. Specifically, it has not been evident how, in an initially racemic solution, one enantiomer could come to exist at a significantly higher concentration than the other.

A mathematical model for the breaking of chiral symmetry was proposed in 1953 by F. C. Frank [1]. This model involves chiral enantiomers, each of which is catalyst for its own production and an anti-catalyst for the production of the other. Solving the resulting system of differential equations shows bifurcation to the excess of one enantiomer or the other. The total concentration of the favored enantiomer increases without bound in either case, and that of the disfavored enantiomer approaches zero.

The Frank model did not allow for the generation of the chiral enantiomers from achiral reactants without the catalytic assistance of the the enantiomers themselves. The model cannot, therefore, account for the very origin of the chiral enantiomers. Neither was the possibility of reverse reactions considered.

In 1983, D. K. Kondepudi and G. W. Nelson proposed a model, now ubiquitous in the literature, whereby symmetry might be broken [2]. The model consists of a hypothetical system of compounds and reactions, each with a certain rate. The Kondepudi–Nelson model removes the deficiencies of the Frank model: both catalyzed and uncatalyzed production of the chiral enantiomers is included, and the reverse reactions are all included. The system is assumed to be locally in equilibrium; that is, the molar reaction rates are essentially constant and independent, and the fluctuation in the rate of change of the concentration of any reactant is due to statistically independent contributions of the reactions involving that reactant.

The doctoral research of R. Bourdon García is an example of how this model is still of current interest. The model is used for the amino acid homochirality generation in prebiotic atmosphere conditions, a relevant issue in the study of the origins of life. The research is based on the production of amino acids via Strecker synthesis and it is adjusted to the Kondepudi–Nelson autocatalytic model ([3] 5.3.2). However, Bourdon García does not obtain the simple necessary and sufficient condition for symmetry-breaking shown in this article. In Section 4, we apply our result to the system considered by Bourdon García.

We present the model, provide proof of some mathematical assertions not justified rigorously by Kondepudi and Nelson, and reduce the conditions for symmetry-breaking to one necessary and sufficient condition.

Nearly 150 articles have cited this work of Kondepudi and Nelson, but an examination of these articles shows that symmetry-breaking conditions have not been determined in general. Research of this sort has been done before on other variants of the Frank model (for example, by Gutman, [4]), but not on the Kondepudi–Nelson model itself. In particular, Hochberg and Ribó have analyzed a model [5] which is seen in Appendix C of this article to be a special case of the Kondepudi–Nelson model, and have also analyzed another, different model [6].

In the appendices we examine three other models for the breaking of chiral symmetry. They are examples of the kind of results we have found here for the Kondepudi–Nelson model (symmetry-breaking conditions, stability, instability, etc.). We compare them with our results for the primary model under consideration—that of Kondepudi and Nelson. They are:

• The model of Ball and Brindley (Appendix A).
• The model of Jafarpour, Biancalani, Goldenfeld et al. (Appendix B).
• The model of Stich, Ribó, Blackmond, and Hochberg (Appendix C).

All three models are shown to function essentially as special cases of the Kondepudi–Nelson model. The results found for these three models are largely the same as what we have found here; that is, our treatment of the Kondepudi–Nelson model subsumes the results for these three models. These models are considered to illustrate by example that a number of symmetry-breaking models considered in the literature amount to special cases of the Kondepudi–Nelson model. However, no comparable analysis has previously been done for the Kondepudi–Nelson model itself.

2. Materials and Methods

The model of Kondepudi and Nelson is investigated using the theory of dynamical systems and differential equations.

3. Results

3.1. The Model of Kondepudi and Nelson

Here ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ are achiral reactants, ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$ are chiral enantiomers (right- and left-handed, respectively) and $\mathbf{Z}$ is an inactive product. The concentrations of ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ are kept constant, and $\mathbf{Z}$ is removed from the system. The reactions are:

$$\left\{\begin{array}{ccc}{\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}\stackrel{k_1}{\underset{k_{-1}}{\rightleftarrows }}{\mathbf{X}}_{-}& & {\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}\stackrel{k_1}{\underset{k_{-1}}{\rightleftarrows }}{\mathbf{X}}_{+}\\ {\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}+{\mathbf{X}}_{-}\stackrel{k_2}{\underset{k_{-2}}{\rightleftarrows }}2{\mathbf{X}}_{-}& & {\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}+{\mathbf{X}}_{+}\stackrel{k_2}{\underset{k_{-2}}{\rightleftarrows }}2{\mathbf{X}}_{+}\\ & {\mathbf{X}}_{-}+{\mathbf{X}}_{+}\stackrel{k_3}{⟶}\mathbf{Z}\end{array}\right.$$

We shall assume for the time being that all reaction rates are non-zero:

$$k_1,k_{-1},k_2,k_{-2},k_3\ne 0$$

This matter will be reconsidered in Appendix A on the Ball–Brindley model, in Appendix B on the Jafarpour model, and in Appendix C on the Stich model.

3.1.1. Intuitive Description

It may be of some use to describe the system intuitively. One may imagine a cavern filled with water. The walls on one side are made of ${\mathbf{A}}_1$, and on the other side made of ${\mathbf{A}}_2$. The substances ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ dissolve in the water and react with each other to form a product $\mathbf{X}$, which decomposes back into ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$. The product $\mathbf{X}$ is chiral, and each molecule is either right-handed (${\mathbf{X}}_{+}$) or left-handed (${\mathbf{X}}_{-}$). The formation and decomposition of ${\mathbf{X}}_{+}$ are both accelerated by existing ${\mathbf{X}}_{+}$ in the water. The same is true of ${\mathbf{X}}_{-}$. Additionally, ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$ react with each other to form a precipitate $\mathbf{Z}$, which settles to the bottom of the cavern. The total mass of the system is kept constant by the dissolving of more ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ from the cavern walls.

3.1.2. Differential Equations

Denoting by $\left[\mathbf{R}\right]$ the concentration of reactant $\mathbf{R}$, the resulting differential equations are:

$$\left\{\begin{array}{c}\frac{d\left[{\mathbf{X}}_{+}\right]}{dt}=k_1\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]-k_{-1}\left[{\mathbf{X}}_{+}\right]+k_2\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]\left[{\mathbf{X}}_{+}\right]-k_{-2}{\left[{\mathbf{X}}_{+}\right]}^2-k_3\left[{\mathbf{X}}_{+}\right]\left[{\mathbf{X}}_{-}\right]\\ \\ \frac{d\left[{\mathbf{X}}_{-}\right]}{dt}=k_1\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]-k_{-1}\left[{\mathbf{X}}_{-}\right]+k_2\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]\left[{\mathbf{X}}_{-}\right]-k_{-2}{\left[{\mathbf{X}}_{-}\right]}^2-k_3\left[{\mathbf{X}}_{+}\right]\left[{\mathbf{X}}_{-}\right]\end{array}\right.$$

3.2. Mathematical Results of the Model of Kondepudi and Nelson

In this subsection, we prove the mathematical results of the model. These results include those proved by Kondepudi and Nelson, those results stated or assumed but not proved, and one significant result not found by them. Even our discussion of results already proved by Kondepudi and Nelson should be of some value here, as the original article [2] has a number of typographical errors that make the argument difficult to follow.

3.2.1. Typographical Errors

The consistent notation we use in this article to describe multiple models is different from that used in the original article of Kondepudi and Nelson [2], but for completeness, we list the typographical errors we have identified in the original article.

• In the paragraph containing Equation A.12, the inequality $s^2-4K_2^2{K}_{-2}^2\ge 0$ should be $s^2-4K_2^2{K}_{-1}^2\ge 0$.
• In the next paragraph, the inequalities $\left(AB\right)>{\left(AB\right)}_{\mathrm{c}}$ and $\left(AB\right)<{\left(AB\right)}_{\mathrm{c}}$ should be switched.
• Later in the same paragraph, after Equation A.16, the equation $K_2\left(AB\right)=2K_2{\beta }_{\mathrm{A}}+K_{-1}$ should instead be written $K_2\left(AB\right)=2K_{-2}{\beta }_{\mathrm{A}}+K_{-1}$.
• The same error occurs in Equation A.17 , where $2K_2{\beta }_{\mathrm{A}}$ should instead be $2K_{-2}{\beta }_{\mathrm{A}}$, and
• The sentence containing that equation should reference A.8 instead of A.9.
• The first term in Equation B.13 should be $F_{11}\alpha \delta \lambda$ instead of $F_{11}\alpha \lambda$.
• In the sentence preceding the one containing Equation B.16,
  “letting $\nu \to 0$" should instead be “letting $\eta \to 0$".

3.2.2. A Change of Variables

Let

$$\alpha =\frac{\left[{\mathbf{X}}_{+}\right]-\left[{\mathbf{X}}_{-}\right]}{2},\beta =\frac{\left[{\mathbf{X}}_{+}\right]+\left[{\mathbf{X}}_{-}\right]}{2},\lambda =\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right].$$

Our system of differential equations becomes the following:

$$\left\{\begin{array}{c}\frac{d\alpha }{dt}=\left(-k_{-1}+k_2\lambda \right)\alpha -\left(2k_{-2}\right)\alpha \beta \hfill \\ \\ \frac{d\beta }{dt}=\left(k_1\lambda \right)+\left(-k_{-1}+k_2\lambda \right)\beta +\left(-k_{-2}+k_3\right){\alpha }^2+\left(-k_{-2}-k_3\right){\beta }^2\hfill \end{array}\right.$$

The constraints imposed by physical reality are

$$\alpha ,\beta ,\lambda \in \mathbb{R},\mathrm{with}\lambda \ge 0\mathrm{and}\beta \ge \left|\alpha \right|$$

3.2.3. Asymmetric Steady States

First, we search for asymmetric steady states. For such a state, $\alpha \ne 0$, and of course, $\frac{d\alpha }{dt}=\frac{d\beta }{dt}=0$, so after dividing out $\alpha$ from the first equation, our system becomes the following:

$$\left\{\begin{array}{c}0=\left(-k_{-1}+k_2\lambda \right)-\left(2k_{-2}\right)\beta \hfill \\ \\ 0=\left(k_1\lambda \right)+\left(-k_{-1}+k_2\lambda \right)\beta +\left(-k_{-2}+k_3\right){\alpha }^2+\left(-k_{-2}-k_3\right){\beta }^2\hfill \end{array}\right.$$

From the first equation we obtain

$${\beta }_{\mathrm{a}}=\frac{k_2\lambda -k_{-1}}{2k_{-2}},$$

and its equivalent form

$$k_2\lambda =2k_{-2}{\beta }_{\mathrm{a}}+k_{-1},$$

which when substituted into the second equation produces

$$\begin{array}{cc}\hfill 0=& \left(k_1\lambda \right)+\left(-k_{-1}+2k_{-2}{\beta }_{\mathrm{a}}+k_{-1}\right){\beta }_{\mathrm{a}}+\left(-k_{-2}+k_3\right){\alpha }^2+\left(-k_{-2}-k_3\right){\beta }_{\mathrm{a}}^2\hfill \\ \hfill =& \left(k_1\lambda \right)+\left(k_3-k_{-2}\right){\alpha }^2-\left(k_3-k_{-2}\right){\beta }_{\mathrm{a}}^2,\hfill \end{array}$$

which implies

$$\alpha =\pm \sqrt{{\beta }_{\mathrm{a}}^2-\frac{k_1\lambda }{k_3-k_{-2}}}$$

(unless $k_3=k_{-2}$ and $\lambda =0$, which from the expression for ${\beta }_{\mathrm{a}}$ above would force ${\beta }_{\mathrm{a}}=\frac{k_2\lambda -k_{-1}}{2k_{-2}}=\frac{-k_{-1}}{2k_{-2}}<0,$ a physical impossibility). Denote the positive value by ${\alpha }_{\mathrm{a}}$. Recall that $\beta \ge \left|\alpha \right|$ must be true for the solution to be physically possible. Therefore we must have

$$\frac{k_1\lambda }{k_3-k_{-2}}\ge 0,$$

which is equivalent to the denominator $k_3-k_{-2}$ being positive; that is, to

$$\begin{array}{|c|}\hline k_3>k_{-2}\\ \hline\end{array}.$$

(1)

This is a necessary condition for the existence of asymmetric steady states.

However, for which values of $\lambda$ are our solutions $(\alpha ,\beta )=(\pm {\alpha }_{\mathrm{a}},{\beta }_{\mathrm{a}})$, where

$$\begin{array}{|c|}\hline {\beta }_{\mathrm{a}}=\frac{k_2\lambda -k_{-1}}{2k_{-2}},{\alpha }_{\mathrm{a}}=\sqrt{{\beta }_{\mathrm{a}}^2-\frac{k_1\lambda }{k_3-k_{-2}}}\\ \hline\end{array},$$

(2)

possible? To answer this question we observe that

$${\alpha }_{\mathrm{a}}=\sqrt{{\beta }_{\mathrm{a}}^2-\frac{k_1\lambda }{k_3-k_{-2}}}$$

is real and non-zero if and only if

$${\beta }_{\mathrm{a}}>\sqrt{\frac{k_1\lambda }{k_3-k_{-2}}}.$$

Since

$${\beta }_{\mathrm{a}}=\frac{k_2\lambda -k_{-1}}{2k_{-2}},$$

the last condition is equivalent to

$$\frac{k_2\lambda -k_{-1}}{2k_{-2}}>\sqrt{\frac{k_1\lambda }{k_3-k_{-2}}},$$

which is equivalent to

$$\frac{{\left(k_2\lambda -k_{-1}\right)}^2}{4k_{-2}^2}>\frac{k_1\lambda }{k_3-k_{-2}}$$

if $k_2\lambda -k_{-1}>0$, and impossible otherwise. Assuming $k_2\lambda -k_{-1}>0$ and $k_3>k_{-2}$, we restate the last condition as

$$k_2^2{\lambda }^2-s\lambda +k_{-1}^2>0,$$

(3)

where

$$\begin{array}{|c|}\hline s=2k_2{k}_{-1}+\frac{4k_{-2}^2{k}_1}{k_3-k_{-2}}\\ \hline\end{array}>2k_2{k}_{-1},$$

(4)

the inequality being implied by Condition (1). The roots of the quadratic polynomial in $\lambda$ on the left-hand side of Inequality (3) are

$${\lambda }_0=\frac{s-\sqrt{s^2-4k_2^2{k}_{-1}^2}}{2k_2^2}\mathrm{and}\begin{array}{|c|}\hline {\lambda }_{\mathrm{c}}=\frac{s+\sqrt{s^2-4k_2^2{k}_{-1}^2}}{2k_2^2}\\ \hline\end{array},$$

(5)

which are both real and positive. (The reason for our names for these two values will be apparent shortly.) Since $k_2^2{\lambda }^2-s\lambda +k_{-1}^2\to \infty$ as $\lambda \to \pm \infty$, then Inequality (3) holds for $\lambda <{\lambda }_0$ and for $\lambda >{\lambda }_{\mathrm{c}}$.

Kondepudi and Nelson ([2]) state that $s^2-k_2^2{k}_{-1}^2\ge 0$ (which is necessary for ${\lambda }_{\mathrm{c}}\in \mathbb{R}$ (second of Equation (5))) “gives another condition [besides $k_3>k_{-2}$] on the kinetic constants” (Page 489). They fail to notice that $s^2-k_2^2{k}_{-1}^2\ge 0$ is a consequence of $k_3>k_{-2}$, as shown above (Inequality (4)).

For $\lambda <{\lambda }_0$, we have

$$2k_2\lambda <\frac{s}{k_2}-\sqrt{{\left(\frac{s}{k_2}\right)}^2-{\left(2k_{-1}\right)}^2},$$

which by the Triangle Inequality on a right triangle is $<2k_{-1}$, so that

$$2k_2\lambda <2k_{-1},$$

which contradicts $k_2\lambda -k_{-1}>0.$

On the other hand, for $\lambda >{\lambda }_{\mathrm{c}}$, we have

$$2k_2\lambda >\frac{s}{k_2}+\sqrt{{\left(\frac{s}{k_2}\right)}^2-{\left(2k_{-1}\right)}^2},$$

which implies the condition $k_2\lambda -k_{-1}>0.$

We now see (assuming Condition (1)) that the condition $\lambda >{\lambda }_{\mathrm{c}}$ is both necessary and sufficient for the existence of the asymmetric steady states (which must be as in Equation (2)), for it subsumes the condition $k_2\lambda -k_{-1}>0.$ We are thus justified in calling ${\lambda }_{\mathrm{c}}$ the “critical value” of $\lambda$.

We summarize our results on asymmetric steady states as follows. Condition (1) is a necessary and sufficient condition on the reaction rates for asymmetric steady states to be possible. Those asymmetric steady states will appear when $\lambda >{\lambda }_{\mathrm{c}}$ as in Definition (5), where s is as in Definition (4). The steady states themselves will be given by Equation (2).

3.2.4. Symmetric Steady States

Second, we search for symmetric steady states. For such a state, $\alpha =0$, and of course, $\frac{d\alpha }{dt}=\frac{d\beta }{dt}=0$, so the first differential equation of our system is satisfied and the second becomes the following:

$$0=\left(k_1\lambda \right)+\left(k_2\lambda -k_{-1}\right)\beta -\left(k_{-2}+k_3\right){\beta }^2$$

The solution to this equation is

$$\beta =\frac{2k_{-2}{\beta }_{\mathrm{a}}\pm \sqrt{{\left(2k_{-2}{\beta }_{\mathrm{a}}\right)}^2+4(k_{-2}+k_3)k_1\lambda }}{2\left(k_{-2}+k_3\right)},$$

where, as before, ${\beta }_{\mathrm{a}}=\frac{k_2\lambda -k_{-1}}{2k_{-2}}$. (Note that ${\beta }_{\mathrm{a}}$ is well-defined whether or not the asymmetric steady states actually exist.) Only the root where “±” is “+” is possible, for if “−” were chosen, $\beta$ would be $<0$. This solution

$$\begin{array}{|c|}\hline {\alpha }_{\mathrm{s}}=0,{\beta }_{\mathrm{s}}=\frac{2k_{-2}{\beta }_{\mathrm{a}}+\sqrt{{\left(2k_{-2}{\beta }_{\mathrm{a}}\right)}^2+4(k_{-2}+k_3)k_1\lambda }}{2\left(k_{-2}+k_3\right)}\\ \hline\end{array}$$

(6)

is physically possible regardless of the reaction rates $k_i>0$ and any value of $\lambda >0$ (since ${\beta }_{\mathrm{s}}\ge 0=\left|{\alpha }_{\mathrm{s}}\right|$).

3.2.5. Possible Points of Bifurcation

Third, we search for possible points of bifurcation. These are values of $\lambda$ where the symmetric and asymmetric steady states coincide. For such a state,

$${\alpha }_{\mathrm{a}}={\alpha }_{\mathrm{s}}=0\mathrm{and}{\beta }_{\mathrm{a}}={\beta }_{\mathrm{s}}.$$

Replacing ${\beta }_{\mathrm{s}}$ by ${\beta }_{\mathrm{a}}$ in the formula for ${\beta }_{\mathrm{s}}$ above (Equation (6)), we obtain

$$\begin{array}{cc}& 2\left(k_{-2}+k_3\right){\beta }_{\mathrm{a}}=2k_{-2}{\beta }_{\mathrm{a}}+\sqrt{{\left(2k_{-2}{\beta }_{\mathrm{a}}\right)}^2+4(k_{-2}+k_3)k_1\lambda },\hfill \\ \hfill \iff & 2k_3{\beta }_{\mathrm{a}}=\sqrt{{\left(2k_{-2}{\beta }_{\mathrm{a}}\right)}^2+4(k_3+k_{-2})k_1\lambda },\hfill \\ \hfill \iff & \left(k_3^2-k_{-2}^2\right){\beta }_{\mathrm{a}}^2=(k_3+k_{-2})k_1\lambda ,\hfill \\ \hfill \iff & {\beta }_{\mathrm{a}}^2=\frac{k_1\lambda }{k_3-k_{-2}},\hfill \end{array}$$

which is exactly the description of the ${\lambda }_{\mathrm{c}}$ where we found the asymmetric solutions began to exist (for $\lambda >{\lambda }_{\mathrm{c}}$). Note that Condition (1) is a necessary and sufficient condition for this point of possible bifurcation (where symmetric and asymmetric steady states conincide) to exist.

We conclude the following, assuming Condition (1): For $0<\lambda <{\lambda }_{\mathrm{c}}$, there is a single possible steady state, $({\alpha }_{\mathrm{s}},{\beta }_{\mathrm{s}})$. For $\lambda >{\lambda }_{\mathrm{c}}$, there are three possible steady states, $({\alpha }_{\mathrm{s}},{\beta }_{\mathrm{s}})$ and the two $({\alpha }_{\mathrm{a}},{\beta }_{\mathrm{a}})$. For $\lambda ={\lambda }_{\mathrm{c}}$, the three possible steady states coincide. Here there is one possible point of bifurcation.

Let us see how ${\beta }_{\mathrm{a}}$ compares to ${\beta }_{\mathrm{s}}$ for various values of $\lambda$. We have

$$\frac{{\beta }_{\mathrm{s}}}{{\beta }_{\mathrm{a}}}=1$$

at the critical point $\lambda ={\lambda }_{\mathrm{c}}$. Furthermore, from the second of Equation (6) we obtain

$$\begin{array}{cc}\hfill \frac{{\beta }_{\mathrm{s}}}{{\beta }_{\mathrm{a}}}=& \frac{2k_{-2}+\sqrt{{\left(2k_{-2}\right)}^2+4(k_{-2}+k_3)k_1/\left({\beta }_{\mathrm{a}}\left(\frac{{\beta }_{\mathrm{a}}}{\lambda }\right)\right)}}{2\left(k_{-2}+k_3\right)}\hfill \\ \hfill =& \frac{2k_{-2}+\sqrt{{\left(2k_{-2}\right)}^2+4(k_{-2}+k_3)k_1/\left(\frac{k_2\lambda -k_{-1}}{2k_{-2}}\left(\frac{k_2}{2k_{-2}}-\frac{k_{-1}}{2k_{-2}\lambda }\right)\right)}}{2\left(k_{-2}+k_3\right)}.\hfill \end{array}$$

From this last expression we can see that as $\lambda$ increases, $\frac{{\beta }_{\mathrm{s}}}{{\beta }_{\mathrm{a}}}$ decreases, meaning

$${\beta }_{\mathrm{a}}>{\beta }_{\mathrm{s}}\mathrm{for}\lambda >{\lambda }_{\mathrm{c}}.$$

(7)

This last fact (7) is stated by Kondepudi and Nelson [2], and illustrated by a sample graph of ${\beta }_{\mathrm{s}}$ and ${\beta }_{\mathrm{a}}$ vs. $\lambda$ ([2], Figure 1). However, the values of the various reaction rates $k_i$ used for the graph are not stated, and no proof of the claim is given.

3.2.6. Stability

Fourth, we investigate the stability of the system at the possible steady states. This will tell us, as $\lambda$ increases past the critical point ${\lambda }_{\mathrm{c}}$, what might happen to the system. The system, which for $\lambda <{\lambda }_{\mathrm{c}}$ had only one possible steady state (the symmetric one, $({\alpha }_{\mathrm{s}},{\beta }_{\mathrm{s}})$) has now two possible asymmetric steady states $(\pm {\alpha }_{\mathrm{a}},{\beta }_{\mathrm{a}})$, and one possible symmetric steady state $({\alpha }_{\mathrm{s}},{\beta }_{\mathrm{s}})$.

A steady state is stable if the real parts of the eigenvalues of the Jacobian of the differential operator are all negative, and unstable if any are positive. The Jacobian of the differential operator is

$$\begin{array}{cc}\hfill J=& \left(\frac{\partial (\dot{\alpha },\dot{\beta })}{\partial (\alpha ,\beta )}\right)=\left(\begin{array}{cc}\frac{\partial \dot{\alpha }}{\partial \alpha }& \frac{\partial \dot{\alpha }}{\partial \beta }\\ \frac{\partial \dot{\beta }}{\partial \alpha }& \frac{\partial \dot{\beta }}{\partial \beta }\end{array}\right)\hfill \\ \hfill =& \left(\begin{array}{cc}-k_{-1}+k_2\lambda -2k_{-2}\beta & -2k_{-2}\alpha \\ 2(k_3-k_{-2})\alpha & -k_{-1}+k_2\lambda -2(k_{-2}+k_3)\beta \end{array}\right)\hfill \end{array}$$

Since

$$-k_{-1}+k_2\lambda =2k_{-2}{\beta }_{\mathrm{a}}$$

from the first of Equation (2), the Jacobian is

$$J=\left(\begin{array}{cc}2k_{-2}({\beta }_{\mathrm{a}}-\beta )& -2k_{-2}\alpha \\ 2(k_3-k_{-2})\alpha & 2k_{-2}{\beta }_{\mathrm{a}}-2(k_{-2}+k_3)\beta \end{array}\right).$$

3.2.7. Stability of Symmetric Steady States

At the symmetric steady state, $\alpha =0$ and $\beta ={\beta }_{\mathrm{s}}$, and since

$$2k_{-2}{\beta }_{\mathrm{a}}-2\left(k_{-2}+k_3\right){\beta }_{\mathrm{s}}=-\sqrt{{\left(2k_{-2}{\beta }_{\mathrm{a}}\right)}^2+4(k_{-2}+k_3)k_1\lambda }$$

from the second of Equation (6), the Jacobian becomes

$$J=\left(\begin{array}{cc}2k_{-2}({\beta }_{\mathrm{a}}-{\beta }_{\mathrm{s}})& 0\\ 0& -\sqrt{{\left(2k_{-2}{\beta }_{\mathrm{a}}\right)}^2+4(k_{-2}+k_3)k_1\lambda }\end{array}\right),$$

whose eigenvalues are the diagonal entries. The $(2,2)$-entry is always negative, so that the stability of the symmetric steady state depends on the $(1,1)$-entry. However, we have seen in (7) that ${\beta }_{\mathrm{a}}-{\beta }_{\mathrm{s}}$ is $<0$ for $\lambda <{\lambda }_{\mathrm{c}}$ and is $>0$ for $\lambda >{\lambda }_{\mathrm{c}}$. Thus, the symmetric steady state is stable for $\lambda <{\lambda }_{\mathrm{c}}$ and unstable for $\lambda >{\lambda }_{\mathrm{c}}$.

3.2.8. Stability of Asymmetric Steady States

At the asymmetric steady states (which exist if and only if $\lambda >{\lambda }_{\mathrm{c}}$), $\alpha =\pm {\alpha }_{\mathrm{a}}$ and $\beta ={\beta }_{\mathrm{a}},$ so that the Jacobian becomes

$$J=\left(\begin{array}{cc}0& \mp 2k_{-2}{\alpha }_{\mathrm{a}}\\ \pm 2(k_3-k_{-2}){\alpha }_{\mathrm{a}}& -2k_3{\beta }_{\mathrm{a}}\end{array}\right).$$

The characteristic polynomial of this matrix is

$$\begin{array}{cc}\hfill J=& |rI-J|\hfill \\ \hfill =& \left|\begin{array}{cc}r& \pm 2k_{-2}{\alpha }_{\mathrm{a}}\\ \mp 2(k_3-k_{-2}){\alpha }_{\mathrm{a}}& r+2k_3{\beta }_{\mathrm{a}}\end{array}\right|\hfill \\ \hfill =& r^2+2k_3{\beta }_{\mathrm{a}}r+4k_{-2}(k_3-k_{-2}){\alpha }_{\mathrm{a}}^2,\hfill \end{array}$$

which by the second of Equation (2) is

$$=r^2+2k_3{\beta }_{\mathrm{a}}r+4k_{-2}((k_3-k_{-2}){\beta }_{\mathrm{a}}^2-k_1\lambda ).$$

The roots of this polynomial are

$$r=-k_3{\beta }_{\mathrm{a}}\pm \sqrt{k_3^2{\beta }_{\mathrm{a}}^2-4k_{-2}((k_3-k_{-2}){\beta }_{\mathrm{a}}^2-k_1\lambda )}.$$

By the second of Equation (2),

$${\beta }_{\mathrm{a}}^2>\frac{k_1\lambda }{k_3-k_{-2}},$$

which by Condition (1) implies

$$(k_3-k_{-2}){\beta }_{\mathrm{a}}^2>k_1\lambda .$$

Therefore,

$$\begin{array}{cc}& 4k_{-2}((k_3-k_{-2}){\beta }_{\mathrm{a}}^2-k_1\lambda )>0,\hfill \\ \hfill \Rightarrow & k_3^2{\beta }_{\mathrm{a}}^2-4k_{-2}((k_3-k_{-2}){\beta }_{\mathrm{a}}^2-k_1\lambda )<k_3^2{\beta }_{\mathrm{a}}^2,\hfill \\ \hfill \Rightarrow & \sqrt{k_3^2{\beta }_{\mathrm{a}}^2-4k_{-2}((k_3-k_{-2}){\beta }_{\mathrm{a}}^2-k_1\lambda )}\mathrm{is}\mathrm{either}\mathrm{real}\mathrm{and}<k_3{\beta }_{\mathrm{a}},\mathrm{or}\mathrm{purely}\mathrm{imaginary},\hfill \\ \hfill \Rightarrow & r=\frac{-k_3{\beta }_{\mathrm{a}}\pm \sqrt{k_3^2{\beta }_{\mathrm{a}}^2-4k_{-2}((k_3-k_{-2}){\beta }_{\mathrm{a}}^2-k_1\lambda )}}{2}\mathrm{both}\mathrm{have}\mathrm{negative}\mathrm{real}\mathrm{part}.\hfill \end{array}$$

This means that the asymmetric steady states are stable if they exist.

The stability of asymmetric steady states is assumed by Kondepudi and Nelson [2] but not shown.

3.2.9. Conclusion: One Necessary and Sufficient Condition

Let us summarize our findings so far.

• The system has a single symmetric steady state $(\alpha ,\beta )=(0,{\beta }_{\mathrm{s}})$ for any value of $\lambda$.
• If Condition (1) ($k_3>k_{-2}$) does not hold, there are no asymmetric steady states.
• If Condition (1) ($k_3>k_{-2}$) does hold, then there exists ${\lambda }_{\mathrm{c}}$ such that:

  -

  Asymmetric steady states exist exactly when $\lambda >{\lambda }_{\mathrm{c}}$.

  -

  If $\lambda >{\lambda }_{\mathrm{c}}$, there are exactly two asymmetric steady states $(\pm {\alpha }_{\mathrm{a}},{\beta }_{\mathrm{a}})$.

  -

  Asymmetric steady states$(\pm {\alpha }_{\mathrm{a}},{\beta }_{\mathrm{a}})$ are stable whenever they exist.

  -

  The symmetric steady state is stable for $\lambda <{\lambda }_{\mathrm{c}}$ and unstable for $\lambda >{\lambda }_{\mathrm{c}}$.

  -

  All three steady states $({\alpha }_{\mathrm{s}},{\beta }_{\mathrm{s}}),(\pm {\alpha }_{\mathrm{a}},{\beta }_{\mathrm{a}})$ coincide for $\lambda ={\lambda }_{\mathrm{c}}$.

In short, Condition (1)—$\begin{array}{|c|}\hline k_3>k_{-2}\\ \hline\end{array}$—is both necessary and sufficient for bifurcation.

3.2.10. Intuitive Understanding of Conclusion

It may be helpful to comment on why our conclusion—our one necessary and sufficient condition—is consistent with inuition. The rate $k_{-2}$ is of the catalyzed decomposition of the chiral enantiomers ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$. The reactions with this rate reduce the concentrations of ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$, but since they are catalyzed by the respective enantiomers themselves, the enantiomer with the higher concentration will be the one whose concentration is more affected. Thus, it is believable that a greater value of $k_{-2}$ would tend to reduce enantiomeric excess. On the other hand, the elimination reaction with rate $k_3$ eliminates equal numbers of molecules of ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$, which will augment any enantiomeric excess. To see this, imagine that there are 3 molecules of ${\mathbf{X}}_{+}$ and 2 molecules of ${\mathbf{X}}_{-}$. If one molecule of each combine to form the inactive product $\mathbf{Z}$, there are now 2 remaining molecules of ${\mathbf{X}}_{+}$ and 1 molecule of ${\mathbf{X}}_{-}$. The ratio of right-handed to left-handed molecules has increased from 3:2 to 2:1.

So we can see that a greater value of $k_{-2}$ works against enantiomeric excess, while a greater value of $k_3$ works for enantiomeric excess. Of course, these are by no means the only relevant characteristics of the Kondepudi–Nelson system, and they certainly do not make obvious the condition $k_3>k_{-2}$ for bifurcation. Yet at least this condition is seen to be consistent with intuition.

4. Application

R. Bourdon García has used the model of Kondepudi and Nelson for the amino acid homochirality generation in prebiotic atmosphere conditions, a relevant issue in the study of the origins of life. The research is based on the production of amino acids via Strecker synthesis and it is adjusted to the Kondepudi–Nelson autocatalytic model ([3] 5.3.2). However, Bourdon García does not obtain the simple necessary and sufficient condition for symmetry-breaking shown in this article. Our result here can be applied to the reactions considered by Bourdon García, to obtain a criterion for symmetry-breaking in the production of aminonitrile ${\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}$ (Reactions 5-10, 5-11, 5-12, 5-13). Bourdon García’s reaction rates $k_{-\mathrm{a}}=k_6=k_8$ and $k_{\mathrm{f}}=k_9=k_{10}$ are our $k_{-2}$ and $k_3$, respectively. Our criterion $k_3>k_{-2}$ predicts that symmetry will be broken if and only if $k_{\mathrm{f}}>d_{-\mathrm{a}}$; that is, the equal reaction rates ($k_{\mathrm{f}}$) of

$${\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}\right)}_{-}+2{\mathrm{H}}_2\mathrm{O}\stackrel{k_9}{⟶}{\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{COOH}\right)}_{-}+{\mathrm{NH}}_3$$

and

$${\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}\right)}_{+}+2{\mathrm{H}}_2\mathrm{O}\stackrel{k_{10}}{⟶}{\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{COOH}\right)}_{+}+{\mathrm{NH}}_3,$$

are larger than the equal reaction rates ($k_{-\mathrm{a}}$) of

$${\mathrm{CH}}_3\mathrm{CHNH}+\mathrm{HCN}+{\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}\right)}_{-}\underset{k_6}{⟵}2{\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}\right)}_{-}$$

and

$${\mathrm{CH}}_3\mathrm{CHNH}+\mathrm{HCN}+{\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}\right)}_{+}\underset{k_8}{⟵}2{\left({\mathrm{CH}}_3\mathrm{CH}\left({\mathrm{NH}}_2\right)\mathrm{CN}\right)}_{+}$$

where the subscripts “−” and “+” represent left- and right-handed enantiomers, respectively.

5. Conclusions

The model of Kondepudi and Nelson ([2]) describes the reactions necessary for the breaking of chiral symmetry in a chemical system. We have examined this work carefully, correcting typographical errors, justifying all unproved assertions, and finding a necessary and sufficient condition not observed by the authors.

Recent work in models of the breaking of chiral symmetry ([5,7,8,9,10]) amounts to special cases of the Kondepudi–Nelson model.

In summary, the condition (1) is both necessary and sufficient for the breaking of chiral symmetry; that is,

$$\begin{array}{|c|}\hline k_3>k_{-2}\\ \hline\end{array},$$

where $k_{-2}$ and $k_3$ are the respective rates of the reverse autocatalytic and elimination reactions in the Kondepudi–Nelson system:

$$\left\{\begin{array}{ccc}{\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}\stackrel{k_1}{\underset{k_{-1}}{\rightleftarrows }}{\mathbf{X}}_{-}& & {\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}\stackrel{k_1}{\underset{k_{-1}}{\rightleftarrows }}{\mathbf{X}}_{+}\\ {\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}+{\mathbf{X}}_{-}\stackrel{k_2}{\underset{k_{-2}}{\rightleftarrows }}2{\mathbf{X}}_{-}& & {\mathbf{A}}_{\mathbf{1}}+{\mathbf{A}}_{\mathbf{2}}+{\mathbf{X}}_{+}\stackrel{k_2}{\underset{k_{-2}}{\rightleftarrows }}2{\mathbf{X}}_{+}\\ & {\mathbf{X}}_{-}+{\mathbf{X}}_{+}\stackrel{k_3}{⟶}\mathbf{Z}\end{array}\right.$$

Appendix A.1. Comparison of the Models of Ball–Brindley and Kondepudi–Nelson

In the Ball–Brindley model, the two enantiomers ${\mathbf{Y}}_{+}$ and ${\mathbf{Y}}_{-}$ of hydrogen peroxide deactivate the respective enantiomers ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$ of the chiral molecule, and catalyze each other’s deactivation reaction. Since ${\mathbf{Y}}_{+}$ and ${\mathbf{Y}}_{-}$ are presumably in fixed and equal concentration, say $\left[{\mathbf{Y}}_{+}\right]=\left[{\mathbf{Y}}_{-}\right]=y$, the rates of elimination of both ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$ from the bottom two reactions are $k_3\left[{\mathbf{X}}_{+}\right]\left[{\mathbf{X}}_{-}\right]y$. This compares with a rate of $k_3\left[{\mathbf{X}}_{+}\right]\left[{\mathbf{X}}_{-}\right]$ of elimination of both ${\mathbf{X}}_{+}$ and ${\mathbf{X}}_{-}$ in the original Kondepudi–Nelson system (by the reaction ${\mathbf{X}}_{+}+{\mathbf{X}}_{-}\stackrel{k_3}{⟶}\mathbf{Z}$). Thus, $k_{3}y$ here plays the role of the original $k_3$, and the two reactions here together perform exactly the same function as the deactivation reaction in the Kondepudi–Nelson model.

Furthermore, the achiral reagent $\mathbf{A}$ of the Ball–Brindley model performs exactly the same function as the achiral reactants ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ of the Kondepudi–Nelson model.

Thus, the Ball–Brindley model is seen to be essentially a special case of the Kondepudi–Nelson model. Accordingly, the Kondepudi–Nelson differential equations reduce to the Ball–Brindley differential equations upon replacing “$\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]$” by “$\left[\mathbf{A}\right]$”, replacing “$k_3$” by “$k_{3}y$”, and setting $k_1=k_{-1}=0$.

Appendix A.2. Mathematical Results of the Model of Ball and Brindley

In our discussion of mathematical results of the Kondpudi–Nelson model, we made the assumption that all reactions considered had non-zero reaction rates—that

$$k_1,k_{-1},k_2,k_{-2},k_3\ne 0.$$

However, as observed above, the Ball–Brindley model is essentially the special case of the Kondepudi–Nelson model where $k_1=k_{-1}=0$. Let us revisit the mathematical results obtained in Section 3.2 and see what changes if $k_1=k_{-1}=0$. We shall continue to assume $k_2,k_{-2},k_3\ne 0$.

Our analysis of the Ball–Brindley model is a straightforward modification of that applied to the more general Kondepudi–Nelson model. Accordingly, results are stated but details omitted.

Appendix A.3. Conclusions: One Necessary and Sufficient Condition

• The system has a single symmetric steady state $(\alpha ,\beta )=(0,{\beta }_{\mathrm{s}})$ for any value of $\lambda$.
• If $k_3=k_{-2}$, all states $(\alpha ,\beta )=(\alpha ,{\beta }_{\mathrm{s}})$ are steady states, none of which are stable.
• If $k_3\ne k_{-2}$, there are two asymmetric steady states $(\alpha ,\beta )=(\pm {\beta }_{\mathrm{a}},{\beta }_{\mathrm{a}})$ for any $\lambda >0$.

  -

  If $k_3>k_{-2}$, then

  • Asymmetric steady states are stable.

  • The symmetric steady state is unstable.

  -

  If $k_3<k_{-2}$, then

  • Asymmetric steady states are unstable.

  • The symmetric steady state is stable.

  In short, Condition (1)—$\begin{array}{|c|}\hline k_3>k_{-2}\\ \hline\end{array}$—is both necessary and sufficient for bifurcation. In this case bifurcation occurs at $\lambda =0$.
  Ball and Brindley observe neither the necessity nor the sufficiency of Condition (1). They illustrate the bifurcation at $\lambda =0$ by displaying several different possible bifurcation diagrams ([7], Figure 2). The mathematics of this bifurcation is considered only in general terms, and the authors seem not to notice that negative values of $\lambda$, which are depicted in each of the different diagrams, are physically meaningless in this situation.

Appendix B.1. Comparison of the Models of Jafarpour, et al. and Kondepudi–Nelson

In the Jafarpour model, the achiral molecule A performs exactly the same function as the two achiral reactants ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ in the Kondepudi–Nelson model.

Thus, the Jafarpour model is seen to be essentially a special case of the Kondepudi–Nelson model. Accordingly, the Kondepudi–Nelson differential equations reduce to differential equations governing the Jafarpour model upon replacing “$\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]$” by “$\left[\mathbf{A}\right]$” and setting $k_{-2}=k_3=0$.

The authors examine the role of random noise in moving the system to an asymmetric state, and drop the requirement that the state itself must be stable. The authors make an unsupported simplifying assumption as well. In the end, they succeed only in showing that under their spurious assumption, the system will tend to drift to one asymmetric state or the other. Then, the authors offer a specious verbal explanation for why it should stay there. Examinations of the original differential equations show that only a symmetric state can be stable.

Appendix B.2. Mathematical Results of the Model of Jafarpour, et al.

The Jafarpour model is a special case of the Kondepudi–Nelson model with

$$k_{-2}=k_3=0.$$

Our analysis of the Jafarpour model proceeds as with the more general Kondepudi–Nelson model. Our findings agree with those of the authors, until we come upon the following.

Appendix B.2.1. A Simplifying Assumption

The authors make the unsupported simplifying assumption that $\beta$ (defined as before) stays constant, saying only that “simulations show" this.

Appendix B.2.2. Random Noise

The authors examine the role of random noise in moving the system to an asymmetric state, and drop the requirement that the asymmetric state must be stable.

Appendix B.3. The Authors’ Conclusion

In the end, the authors succeed only in showing that under their spurious assumption, the system will tend to drift to one asymmetric state or the other. Then, the authors offer a specious verbal explanation for why it should stay there. Our examination above of the original differential equations, by contrast, shows that there is exactly one equilibrium state, which is symmetric and stable. The authors of the next paper considered agree that stochastic considerations like those of Jafarpour, et al. are significant only for small volumes and small numbers of molecules [5].

Appendix C.1. Comparison of the Models of Stich et al. and Kondepudi–Nelson

The achiral reagent $\mathbf{A}$ of the Stich model performs exactly the same function as the achiral reactants ${\mathbf{A}}_1$ and ${\mathbf{A}}_2$ of the Kondepudi–Nelson model.

Thus, the Stich model is seen to be essentially a specialization of the Kondepudi–Nelson model. Accordingly, the Kondepudi–Nelson differential equations reduce to differential equations governing the Stich model upon replacing “$\left[{\mathbf{A}}_1\right]\left[{\mathbf{A}}_2\right]$” by “$\left[\mathbf{A}\right]$” and setting $k_3=0$.

Our analysis of the Stich model is a straightforward modification of that applied to the more general Kondepudi–Nelson model. Accordingly, results are stated but details omitted.

Appendix C.2. Mathematical Results of the Model of Stich, Ribó, Blackmond, and Hochberg

The system has a single steady state $(\alpha ,\beta )=(0,{\beta }_{\mathrm{s}})$, which is symmetric and stable.

Appendix C.3. Conclusions

Our conclusions concerning the Stich model agree with those of the authors of [5], but whereas the authors reach their conclusions by choosing arbitrary and convenient reaction rates and displaying graphs of the resulting systems, we have verified the conclusions here with rigorous mathematics.