# Is there a physical or mathematical symbol for “happens when”?

How would you denote symbolically, "Equilibrium happens given that ..."?

• I believe you are looking for the implication $\Rightarrow$ or $\Leftarrow$. – ACuriousMind Feb 8 '15 at 19:48
• @ACuriousMind that should be an answer – David Z Feb 8 '15 at 19:49

The statement "$A$ happens given that $B$" is equivalent to "If $B$, then $A$", which is symbolically represented as an implication

$$B \Rightarrow A$$

or, if you want to preserve the order of $A$ and $B$ in the original statement

$$A \Leftarrow B$$

If you want to make statements over (discrete) time, linear temporal logic may be worth looking at. For instance,

$\qquad\displaystyle \Box (A \implies B)$

means that whenever $A$ holds, $B$ has to hold at the same time.

$\qquad\displaystyle \Box (A \implies \Diamond B)$

means that whenever $A$ holds, $B$ will hold at some point in the future. Or yet another option,

$\qquad\displaystyle \Box (A \implies \Diamond \Box B)$

means that whenever $A$ holds, $B$ will hold from some point in the future on, forever.

So if you want to express than some condition triggers the equilibrium which then occurs eventually and is stable (i.e. never stops), the last formula expresses this more clearly then a pure propositional formula.

For completeness I would like to pose an alternative answer to those already given. When doing calculations in physics, in particular in particle physics where particles decay into other particles (but only when certain conditions are fulfilled, see below), one often comes across the (Heaviside) step function (which is also widely used in engineering) defined by some as

$$\tag{1}\theta(x-a)=\begin{cases} 0, & x < a, \\ 1, & x \ge a, \end{cases}.$$

This tells you that whatever $\theta$ multiplies (often inside integrals) is zero unless your integration variable $x$ is larger than some parameter $a.$ Examples of this in e.g. particle physics is when your deriving formulas for a one to two particle decay rate. Then you will get one of these $\theta$'s in the form of

$$\tag{2}\theta(M-m_1-m_2),$$ where $M$ is the mass of the decaying particle and $m_1$ and $m_2$ are the masses of the two (different in this case) final state particles. In this case $(2)$ arises because the decay is not kinematically allowed, i.e. impossible to happen unless $M\geq m_1+m_2$.

For your specific example of equilibrium: suppose equilibrium happens when some certain time $T_{equil}$ has passed and never before. Then $$\tag{3}\theta(t-T_{equil}),$$ where $t$ is the parameter that counts time, would be the answer to your question: "equilibrium happens given that" $t\geq T_{equil}$.

## protected by Qmechanic♦Feb 9 '15 at 9:10

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).