Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 27 insertions, 27 deletions

diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md
index 9745959..b89d5da 100644
--- a/source/know/concept/detailed-balance/index.md
+++ b/source/know/concept/detailed-balance/index.md
@@ -24,9 +24,9 @@ since all net transition rates are zero.
 We will focus on the case where both time and the state space are continuous.
 Given some initial conditions,
 assume that a component's trajectory can be described
-as an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
-with a time-independent drift $f$ and intensity $g$,
-and with a probability density $\phi(t, x)$ governed by the
+as an [Itō diffusion](/know/concept/ito-calculus/) $$X_t$$
+with a time-independent drift $$f$$ and intensity $$g$$,
+and with a probability density $$\phi(t, x)$$ governed by the
 [forward Kolmogorov equation](/know/concept/kolmogorov-equations/)
 (in 3D):
 
@@ -37,7 +37,7 @@ $$\begin{aligned}
 
 We start by demanding **stationarity**,
 which is a weaker condition than detailed balance.
-We want the probability $P$ of being in an arbitrary state volume $V$
+We want the probability $$P$$ of being in an arbitrary state volume $$V$$
 to be constant in time:
 
 $$\begin{aligned}
@@ -56,11 +56,11 @@ $$\begin{aligned}
     = - \oint_{\partial V} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
 \end{aligned}$$
 
-In other words, the "flow" of probability *into* the volume $V$
-is equal to the flow *out of* $V$.
+In other words, the "flow" of probability *into* the volume $$V$$
+is equal to the flow *out of* $$V$$.
 If such a probability density exists,
-it is called a **stationary distribution** $\phi(t, x) = \pi(x)$.
-Because $V$ was arbitrary, $\pi$ can be found by solving:
+it is called a **stationary distribution** $$\phi(t, x) = \pi(x)$$.
+Because $$V$$ was arbitrary, $$\pi$$ can be found by solving:
 
 $$\begin{aligned}
     0
@@ -70,7 +70,7 @@ $$\begin{aligned}
 Therefore, stationarity means that the state transition rates are constant.
 To get detailed balance, however, we demand that
 the transition rates are zero everywhere:
-the probability flux through an arbitrary surface $S$ must vanish
+the probability flux through an arbitrary surface $$S$$ must vanish
 (compare to closed surface integral above):
 
 $$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
     = - \int_{S} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
 \end{aligned}$$
 
-And since $S$ is arbitrary, this is only satisfied if the flux is trivially zero
+And since $$S$$ is arbitrary, this is only satisfied if the flux is trivially zero
 (the above justification can easily be repeated in 1D, 2D, 4D, etc.):
 
 $$\begin{aligned}
@@ -93,7 +93,7 @@ but fortunately often satisfied in practice.
 The fact that a system in detailed balance appears "frozen"
 implies it is **time-reversible**,
 meaning its statistics are the same for both directions of time.
-Formally, given two arbitrary functions $h(x)$ and $k(x)$,
+Formally, given two arbitrary functions $$h(x)$$ and $$k(x)$$,
 we have the property:
 
 $$\begin{aligned}
@@ -109,9 +109,9 @@ $$\begin{aligned}
 <div class="hidden" markdown="1">
 <label for="proof-reversibility">Proof.</label>
 Consider the following weighted inner product,
-whose weight function is a stationary distribution $\pi$
+whose weight function is a stationary distribution $$\pi$$
 satisfying detailed balance,
-where $\hat{L}$ is the Kolmogorov operator:
+where $$\hat{L}$$ is the Kolmogorov operator:
 
 $$\begin{aligned}
     \inprod{\hat{L} h}{k}_\pi
@@ -128,8 +128,8 @@ $$\begin{aligned}
     = -\nabla \cdot (\vb{u} \pi k - D k \nabla \pi - D \pi \nabla k)
 \end{aligned}$$
 
-Since $\pi$ is stationary by definition,
-we know that $\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$,
+Since $$\pi$$ is stationary by definition,
+we know that $$\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$$,
 meaning:
 
 $$\begin{aligned}
@@ -138,7 +138,7 @@ $$\begin{aligned}
     = \nabla \pi \cdot (D \nabla k) + \pi \nabla \cdot (D \nabla k)
 \end{aligned}$$
 
-Detailed balance demands that $\vb{u} \pi = D \nabla \pi$,
+Detailed balance demands that $$\vb{u} \pi = D \nabla \pi$$,
 leading to the following:
 
 $$\begin{aligned}
@@ -150,9 +150,9 @@ $$\begin{aligned}
     = \pi \hat{L}\{k\}
 \end{aligned}$$
 
-Where we recognized the definition of $\hat{L}$
+Where we recognized the definition of $$\hat{L}$$
 from the backward Kolmogorov equation.
-Now that we have established that $\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$,
+Now that we have established that $$\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$$,
 we return to the inner product:
 
 $$\begin{aligned}
@@ -170,7 +170,7 @@ $$\begin{aligned}
 
 Now, consider the time evolution of the
 [conditional expectation](/know/concept/conditional-expectation/)
-$\mathbf{E}\big[ k(X_t) | X_0 \big]$:
+$$\mathbf{E}\big[ k(X_t) | X_0 \big]$$:
 
 $$\begin{aligned}
     \pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big]
@@ -184,10 +184,10 @@ $$\begin{aligned}
 
 Where we used the forward Kolmogorov equation
 and the definition of an adjoint operator.
-Therefore, since the expectation $\mathbf{E}$
-does not explicitly depend on $t$ (only implicitly via $X_t$),
+Therefore, since the expectation $$\mathbf{E}$$
+does not explicitly depend on $$t$$ (only implicitly via $$X_t$$),
 we can naively move the differentiation inside
-(only valid within $\mathbf{E}$):
+(only valid within $$\mathbf{E}$$):
 
 $$\begin{aligned}
     \pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big]
@@ -195,9 +195,9 @@ $$\begin{aligned}
     = \mathbf{E}\bigg[ \hat{L}\{k(X_0)\} \bigg| X_0 \bigg]
 \end{aligned}$$
 
-A differential equation of the form $\ipdv{k}{t} = \hat{L}\{k(t, x)\}$
-for a time-independent operator $\hat{L}$
-has a general solution $k(t, x) = \exp(t \hat{L})\{k(0,x)\}$,
+A differential equation of the form $$\ipdv{k}{t} = \hat{L}\{k(t, x)\}$$
+for a time-independent operator $$\hat{L}$$
+has a general solution $$k(t, x) = \exp(t \hat{L})\{k(0,x)\}$$,
 therefore:
 
 $$\begin{aligned}
@@ -220,8 +220,8 @@ $$\begin{aligned}
     = \mathbf{E}\big[ h(X_t) \: k(X_0) \big]
 \end{aligned}$$
 
-Where the integral gave the expectation value at $X_0$,
-since $\pi$ does not change in time.
+Where the integral gave the expectation value at $$X_0$$,
+since $$\pi$$ does not change in time.
 </div>
 </div>