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Suppose $a$ and $m$ are real variables and they satisfy the following two coupled Langevin equations: $$ \dot{a}=F_a(a,m)+\eta_a(t);\quad\dot{m}=F_m(a,m)+\eta_m(t); $$ where $\eta_a$ and $\eta_m$ are white noises with strength $\Delta_m$ and $\Delta_a$. According to this, the detailed balance condition is $$\Delta_m\partial_mF_a=\Delta_a\partial_aF_m,$$ but I have no idea how this is derived, supposing it is right.

Edit: in response to the request for more references.

I am well aware that if $i,j$ are two arbitrary discrete states of a system, the detailed balance condition will be $$P(i)W(i\rightarrow j)=P(j)W(j\rightarrow i),$$ where $W(i\rightarrow j)$ is the transition probability from state $i$ to $j$. The continuous version can be derived in the same spirit. Once we know the probability of finding the system at a particular state and transition probabilities, the detailed balance condition should be fairly easy to derive. However, it does not seem trivial to derive the detailed balance condition from a general Langevin equation, and this is what I ask for.

Edit 2: the two noise terms $\eta_a$ and $\eta_m$ are uncorrelated.

The equation comes from here.

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  • $\begingroup$ Comment to the question (v1): Consider adding more references in order to receive useful and focused answers. $\endgroup$
    – Qmechanic
    Dec 17, 2014 at 15:13
  • $\begingroup$ @Qmechanic, I stated what I know and what I don't know. As for the equation, I cannot be more specific because the paper cited did not impose any other restrictions on the equation. I don't have much experience working with Langevin equation, so I don't even know what kind of information is vital. Hope this helps. $\endgroup$
    – wdg
    Dec 17, 2014 at 15:49

2 Answers 2

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Definitions

Define $W(2|1)$ as the transition probability per unit time from $1$ to $2$. This gets us the Master equation

$$\partial_t p(a, m) = \int\!\!\!\!\int\! \left(W(a,m|a',m')p(a',m') - W(a',m'|a,m)p(a,m)\right)\mathrm{d}a'\mathrm{d}m'$$

Define further, for fun, $\mathbb{W}(a,m|a',m') = W(a,m|a',m') - \delta(a-a')\delta(m-m')\int\!\!\int W(a'',m''|a,m)\mathrm{d}a''\mathrm{d}m''$. With direct substitution we have $\partial_t p = \int\!\!\int\mathbb{W}(a,m|a'm')p(a',m')\mathrm{d}a'\mathrm{d}m'$, an operation we denote by

$$\partial_t p = \mathbb{W}p$$

Now the continuous detailed balance condition is: $$W(a,m|a',m')p^\text{eq}(a',m') = W(a',m'|a,m)p^\text{eq}(a,m)$$

where $p^\text{eq}(a,m)$ is the equilibrium probability distribution, defined by the equation $\mathbb{W}p^\text{eq} = 0$. The detailed balance can be written in the form

$$\frac{W(a,m|a',m')}{p^\text{eq}(a,m)} = \frac{W(a',m'|a,m)}{p^\text{eq}(a',m')}$$

We may now multiply this by some function $f(a,m)$ and integrate over $a$ and $m$. If we require the relation to hold for all $f$, we have the equivalent form

$$\int\!\!\!\!\int\frac{W(a,m|a',m')}{p^\text{eq}(a,m)}f(a,m)\mathrm{d}a\mathrm{d}m = \int\!\!\!\!\int\frac{W(a',m'|a,m)}{p^\text{eq}(a',m')}f(a,m)\mathrm{d}a\mathrm{d}m, \quad \forall f$$

For symmetry's sake, let's do the same with another function $g(a', m')$:

$$\int\!\!\!\!\int\!\!\int\!\!\!\!\int\frac{W(a,m|a',m')}{p^\text{eq}(a,m)}f(a,m)g(a',m')\mathrm{d}a\mathrm{d}m\mathrm{d}a'\mathrm{d}m' = \int\!\!\!\!\int\!\!\int\!\!\!\!\int\frac{W(a',m'|a,m)}{p^\text{eq}(a',m')}f(a,m)g(a',m')\mathrm{d}a\mathrm{d}m\mathrm{d}a'\mathrm{d}m', \quad \forall f, g$$

Rearranging and throwing $\mathbb{W}$ into the mix, we have

$$\int\!\!\!\!\int\frac{f(a,m)\ \int\!\!\int \mathbb{W}(a,m|a',m')g(a',m')\mathrm{d}a'\mathrm{d}m'}{p^\text{eq}(a,m)}\mathrm{d}a\mathrm{d}m = \int\!\!\!\!\int\frac{\int\!\!\int \mathbb{W}(a',m'|a,m)f(a,m)\mathrm{d}a\mathrm{d}m\ g(a',m')}{p^\text{eq}(a',m')}\mathrm{d}a'\mathrm{d}m', \quad \forall f, g$$

Now using the definitions from above, we conclude that the detailed balance is satisfied if and only if

$$(\mathbb{W}g, f) = (g, \mathbb{W}f), \quad \forall f,g$$

where the inner product is defined as

$$(g, f) = \int\!\!\!\!\int\frac{g(a,m)f(a,m)}{p^\text{eq}(a,m)}\mathrm{d}a\mathrm{d}m$$

Derivation

Phew. Finally, time to do the hard part. The Fokker-Planck equation corresponding to your pair of stochastic differential equations is (assuming there is no correlation between the noise sources):

$$\partial_t p = -\partial_a(F_a p) -\partial_m(F_m p) + \frac{1}{2}\sigma_a^2\partial_a^2p + \frac{1}{2}\sigma_m^2\partial_m^2p$$

This also conveniently defines the operator $\mathbb{W}$.

The strategy now is to take $(\mathbb{W}f, g)$ and integrate it by parts (only once), then do the same with $(f, \mathbb{W}g)$. Finally we're going to equate the expressions. Let's do these steps one at a time:

$$(\mathbb{W}f, g) = \int\!\!\!\!\int \left((F_a f - \frac{\sigma_a^2}{2}\partial_a f) \partial_a\!\!\left(\frac{g}{p^\text{eq}}\right) + (F_m f - \frac{\sigma_m^2}{2}\partial_m f) \partial_m\!\!\left(\frac{g}{p^\text{eq}}\right)\right)\mathrm{d}a\mathrm{d}m$$

Writing out the derivative, we have inside the integral

$$\frac{F_af\partial_ag}{p^\text{eq}} - \frac{F_afg\partial_a p^\text{eq}}{(p^\text{eq})^2} - \frac{\sigma_a^2\partial_a f\partial_ag}{2p^\text{eq}} + \frac{\sigma_a^2\partial_a f g\partial_a p^\text{eq}}{2(p^\text{eq})^2} + \frac{F_mf\partial_mg}{p^\text{eq}} - \frac{F_mfg\partial_m p^\text{eq}}{(p^\text{eq})^2} - \frac{\sigma_m^2\partial_m f\partial_mg}{2p^\text{eq}} + \frac{\sigma_m^2\partial_m f g\partial_m p^\text{eq}}{2(p^\text{eq})^2} $$

Subtracting from this the once-integrated-by-parts $(f, \mathbb{W}g)$, all the symmetric terms drop and we are left with (after multiplying by $p^\text{eq}$):

$$\left(F_a - \frac{\sigma_a^2\partial_a p^\text{eq}}{2p^\text{eq}}\right)(f\partial_ag - g\partial_af) + \left(F_m - \frac{\sigma_m^2\partial_m p^\text{eq}}{2p^\text{eq}}\right) (f\partial_mg - g\partial_mf) = 0$$

Note that on the left we have stuff that just depends on $a$ and on the right things that depend on $m$. This is to say that each should be zero independent of the other. After straightforward manipulation, we arrive at the wanted result

$$\sigma_m^2\partial_m F_a = \sigma_a^2\partial_a F_m$$

I've never seen this relation before, nor did I find a reference for it. I'm a bit surprised it looks this simple.

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  • $\begingroup$ Nicely done! I still have one question with regard to the second-to-last equation: the term on the LHS of the plus side is a function of both $m$ and $a$, since $F_a$ is a function of both $m$ and $a$. How come that term just depends on $a$? $\endgroup$
    – wdg
    Dec 18, 2014 at 9:24
  • $\begingroup$ Perhaps we can argue that since $f$ and $g$ are both arbitrary functions of $a$ and $m$, $(f\partial_ag - g\partial_af)\neq(f\partial_mg - g\partial_mf)\neq0$ in general. Therefore, the two factors in the second-to-last equation have to be zero independent of the other. $\endgroup$
    – wdg
    Dec 18, 2014 at 9:38
  • $\begingroup$ @wdg I suppose I should've been more careful with my language (but then again, this is the physics.SE, not the maths one). Indeed what I suggest is very similar to your argument: The equations have to be valid for all $f$, $g$ and $f\partial_a g-g\partial_a f$ is independent of $f\partial_m g-g\partial_m f$. The latter assertion could be reasoned from noting that the first part of the equation loses information on $m$ after derivation and the other on $a$, but this is not a very rigorous proof. $\endgroup$
    – alarge
    Dec 18, 2014 at 10:04
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What worries me, if I add a constant term $F_a = F_a + F$. I clearly drive the system out of equilibrium and detailed balance should be broken. However such a constant term does not influence the equation: $\sigma^2_m \partial_m F_a =\sigma^2_a \partial_a F_m $

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    $\begingroup$ Welcome to Physics.SE! Please note that answers are for answering the question, not adding additional questions. This could be posted as a comment on the question or new questions can be asked using the "Ask Question" link at the top $\endgroup$
    – Jim
    Mar 31, 2015 at 18:08
  • $\begingroup$ sorry, I cannot comment, I don't have enough points $\endgroup$
    – flo
    Mar 31, 2015 at 18:21
  • $\begingroup$ Yes, by design, users with less than 50 reputation cannot post comments. A natural assumption from this point would then be that it is also not acceptable to try to circumvent our system by posting a comment as an answer $\endgroup$
    – Jim
    Mar 31, 2015 at 18:29
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    $\begingroup$ sorry about that, I should have rather posed a new question then? However, my question is a follow up question. It also wouldn't make sens to start a new question. $\endgroup$
    – flo
    Mar 31, 2015 at 19:29
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    $\begingroup$ You can always include a reference to this question in a new question $\endgroup$
    – Jim
    Mar 31, 2015 at 19:37

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