# Can two distinct events occur at precisely the same moment in time?

I am writing a simulation and am having difficulty resolving the order in which two distinct forces occur. The simulation will give different results if the forces are applied to the state in different orders. However, computationally I cannot make them occur at the same time.

I believe I am missing some understanding about whether the world works in continuous time or discrete. I suspect this has something to do with the speed of propagation of information, and relativity, but I'm not sure.

If two distinct events can occur at the same time, how does one write a discrete time simulator without prioritising forces?!

Some more background:

Suppose the state is represented by $\\S_i \in \Omega$ at time $\\i$, where $\\i \in \mathbb{N}$. Suppose there is a state transition function $\\f: \Omega \to \Omega$ such that

$$\ S_{i+1} = f(S_i)$$

$\\f$ is modelled by the combination of two forces, G and H. G and H can each be modelled by two transition functions $\\g: \Omega \to \Omega$ and $\\h: \Omega \to \Omega$.

As far as I can determine, there is no way to compute the resultant force of two arbitrary forces (G and H). What I would like to do is define $\\f$ to be the composition of $\\g$ and $\\h$, the question is whether that should be $\\g \circ h$ or $\\h \circ g$.

• Define "event". The only physics usage of that word is in relativity, where it simply denotes a spacetime point, and doesn't "occur". Also, why is your question tagged with GR/SR tags? – ACuriousMind Dec 27 '14 at 13:41
• @ACuriousMind, what about events in colliders? :D – Constandinos Damalas Dec 27 '14 at 13:43
• But, what's your problem with the forces? If they act simultaneously on your system, don't they have a resultant? What does that have to do with the world time being discrete or continuous? For the computer you use some time-step, and if the time-step is enough small, probably the result is good. Can't you tell us what is the problem for which you write the simulator? Maybe we can help with some advice. – Sofia Dec 27 '14 at 13:45
• @PhotonicBoom: Ah, these crude experimentalists. I admittedly forgot that usage, but it also doesn't really fit here. – ACuriousMind Dec 27 '14 at 13:46
• @PhotonicBoom : it seems to me that Cammil has just computer problems. – Sofia Dec 27 '14 at 13:49

Take the operation that gives you $g$ from $G$ and apply it to $(G+H)$.