# Dynamics: why do physicists include derivatives like $\dot{\theta}$ in the state space for a system like a pendulum?

I come from statistics, so my experience with physics is spotty, especially on some simple stuff. I have been working on some applications related to control theory lately, and was looking at some simple pendulum or double pendulum systems to understand control theory methods. For a reference, I believe this is how the Strogatz book on Nonlinear Dynamics and Chaos defines the pendulum.

One question was, why do physicist include the derivatives of say $$\dot{\theta}$$ in the state space of these models. From what I remember, we include in the state vector only those items needed to completely characterize the system. That is a somewhat ambiguous definition though, since it does not give any rules for sufficiency, etc. So if I just had a state vector with the $$x, y$$, then I could tell exactly where the pendulum was at any given time--with the caveat for period behavior. So what makes including the derivatives necessary?

If I was just using the dynamics definition of the pendulum, then I would define my vector by $$\dot{x}, \dot{y}$$, because the dynamics system could be represented that way, or it could be represented in equivalent polar coordinates. But in this case, the state vector only has derivatives and not the actual integrated solution for $$x(t), y(t)$$.

I can understand from a control theory perspective that the control or actuator may be applied to a particular derivative of the system--so a controller may affect the acceleration of a motor on pendulum, or may affect the dampening force on the pendulum, etc.

So the above gives some plausible reasons, but I was hoping someone might be able to give me a more certain answer--rather than my guesses. I imagine that the answer has to do with "identification" in the sense that we need to have a bijection between the states of the system and the state vector--meaning that we cannot have the same state vector represent more than one state of the system. Of course, then there is nothing to stop us from adding additional elements to the state vector, as long as the system retains that bijection property.

Any help is certainly appreciated.

• The pendulum obeys a second order differential equation, so we need both the initial position and velocity to specify solutions.
– d_b
Commented Aug 10, 2022 at 7:09
• @d_b Oh right. So you need both the initial position and velocity to solve the problem--meaning integrating the solution. Yes, that makes sense. The answer is even more clearn when I convert the second order equation into a first order system. So then the answer to the original quesiton is that the state vector needs to include only those elements needed to solve the system of ODEs, i.e., the dynamics? Commented Aug 10, 2022 at 7:14
• $(x,y)$ tells you the state of the system right now. $({\dot x},{\dot y})$ tells you the state of the system in the next instance. You need to know both to predict the behavior of the system in time. Commented Aug 10, 2022 at 8:02

From what I remember, we include in the state vector only those items needed to completely characterize the system ... So what makes including the derivatives necessary?

The derivatives are necessary to fully characterize the state of the system. Consider the pendulum with $$\theta =0$$ and $$\dot \theta =0$$ versus the pendulum with $$\theta =0$$ and $$\dot \theta = v_0 \ne 0$$.

Those have the same $$\theta$$ but are clearly two different states. The first state leads to a system that does not evolve with time, the pendulum just hangs there in place. The second state leads to a system that swings back and forth periodically over time.

So the state of the system is not characterized by $$\theta$$ alone. Both $$\theta$$ and $$\dot \theta$$ at some moment are required in order to fully characterize its state. This is related to the fact that the equations of motion are a second order differential equation in $$\theta$$ so you have two undetermined coefficients required to define the particular solution

• yes I get what you mean now. I was not thinking of the different initial conditions and how that mean that at $\theta=0$, that I can have different velocities; and then those velocities generate different dynamics, etc. That makes sense now. Thanks for providing the intution behind the mathematical definition of the pendulum problem. Commented Aug 10, 2022 at 18:54

the dynamic behavior of a pendulum is describe by second order differential equation

$$\ddot\theta=f(\theta~,\dot\theta)$$

with the state vector $$~\mathbf y$$

\begin{align*} \mathbf{y}=\begin{bmatrix} \theta \\ \dot{\theta} \\ \end{bmatrix} \end{align*}

you obtain :

\begin{align*} \mathbf{\dot{y}}=\begin{bmatrix} \dot \theta \\ \ddot{\theta} \\ \end{bmatrix}=\begin{bmatrix} \dot{\theta} \\ f(\theta~,\dot{\theta}) \\ \end{bmatrix}=\mathbf{g}(\mathbf{y}) \end{align*}

• Yes, thank you @Eli, I appreciate the response. This makes a lot more sense now. Commented Aug 10, 2022 at 18:55