Is the state of a system represented by a point $\textbf{q}=(q_1,q_2,q_3...q_n)^T$ in configuration space?

Question

I was reading the lecture notes titled: 'An introduction to Lagrangian and Hamiltonian mechanics'.

In these notes, he writes at one place:

We consider mechanical systems that are holonomic and conservative (or for which the applied forces have a generalized potential). For such a system we can construct a Lagrangian $L(q, \dot q,t)$, where $q = (q_1, . . . , q_n) ^T$, which is the difference of the total kinetic $T$ and potential $V$ energies. These mechanical systems evolve according to the n Lagrange equations. These are each second-order ordinary differential equations and so the system is determined for all time once $2n$ initial conditions $(q(t_0), \dot q(t_0))$ are specified (or $n$ conditions at two different times). The state of the system is represented by a point $\textbf{q} = (q_1, . . . , q_n)^T$ in configuration space.

The last line confused me since as far as I know and even as Wikipedia says, a point in the phase space gives the state of the system while the above gives only the configuration. So, is the above line wrong?

Also, it seems that when $q$ is known $p=\frac{\partial L}{\partial\dot q} $ can be calculated. So, shouldn't specifying $q$ itself specify the state completely since that is uniquely related to conjugate momenta $p$?

I just want to know whether the state of a system can be represented by a point in configuration space as that line indicates? 2. Okay Yes I see your point. But then let's say in the case of a 1-D Harmonic Oscillator when I know $q=x=ASin(w*t)$, do I not know the complete specification of the state of the system? — Lost, Jun 26 '21 at 15:07
Related: What is the definition of how to count degrees of freedom? — Qmechanic, Jun 26 '21 at 15:21
You might be interested in reading (the end of) this answer of mine, since your comment indicates what you're really confused about is the distinction between points $(q,\dot{q})$ and a solution to the equations of motions $q(t)$. — ACuriousMind, Jun 26 '21 at 15:30
Thank you for the link. I read it but unfortunately, I am yet not familiar with the symplectic abstraction so I could not properly understand your very informative answer. Maybe could you clarify why two points are absolutely necessary for giving the whole information of the state? I mean it's clear that we need two initial condns of position and velocity to find the eqns of motion but its not very clear to me that why when we are given $(q,\dot q )$ we have totally specified the state? Is it simply because only then can we find the eqns of motion from E-L eqns? — Lost, Jun 26 '21 at 15:46
In other words, what does it mean to specify $(q,\dot q)$ for a system? Can you give an example of a toy system and show how $(q,\dot q)$ are both separately necessary? — Lost, Jun 26 '21 at 15:54

score 2 · Accepted Answer · answered Jun 26 '21 at 16:22

Usually, by the "state" of a system at an instant of time we mean all information necessary to determine uniquely a solution to the equations of motion that then determines all future (and past) states of the system.

For the Lagrangian formalism, such states are tuples of numbers $(q,\dot{q})$, i.e. the generalized coordinates and velocities at an instant of time, for the Hamiltonian formalism, such states are tuples of numbers $(q,p)$, i.e the generalized coordinates and momenta at an instant of time. Note that a function $q(t)$ determines tuples $(q(t),\dot{q}(t))$ and $(q(t),p(t))$ at all instants of time, since the functions $\dot{q}(t)$ and $p(t)$ can be determined from $q(t)$ either by differentiation or use of $p(t) = \partial_{\dot{q}} L(q(t),\dot{q}(t),t)$. See also this answer of mine for more discussion on how the abstract (and time-independent!) coordinates $q,\dot{q},p$ relate to functions $q(t)$

That is, by solving the equations of motion (and assuming they indeed have unique solutions for fully-specified initial conditions, unlike e.g. Norton's dome), the state of a mechanical system at all instants of time can in principle be specified by any of the three things above: A tuple of numbers $(q,\dot{q})$ given at an instant of time $t_0$, a tuple of numbers $(q,p)$ or a function of time $q(t)$.

That your text claims the "state" of a system is given just by a tuple of numbers $q$ means it is using "state" to mean something different, but there is no telling what exactly. The configuration $q$ at an instant of time is of course all you would need to say what one would see on a (idealized) photograph of the system at that instant, but photographs do not capture velocity - the configuration at an instant of time alone is not enough to tell how the system will evolve in time (see e.g. this question for an explicit example).

When u say tuple of numbers does it mean something like this (2,5)? — Lost, Jun 26 '21 at 16:38
@Lost ...yes, that's what a tuple of numbers is (of course there should also be correct units attached to them when we want to actually compare our computations to experiments) — ACuriousMind, Jun 26 '21 at 16:40
So, in our 1-D Harmonic Oscillator case if somebody tells me $(2,5)$ as the $(x,v)$ how would I know the trajectory (past or future)? I would also in addition need $q(t)$ ryt? I am just confirming since you mentioned any of the three $q(t)$ or a tuple of numbers $(q,p)$ which indicates I think that while knowing the tuple I don't need to know $q(t)$ — Lost, Jun 26 '21 at 16:43
@Lost Note the "by solving the equations of motion" that paragraph starts with. You know the equation of motion of the HO. Solve it for the initial condition $(x,v)$ to get $x(t)$ describing the HO at all times. — ACuriousMind, Jun 26 '21 at 16:47

score -1 · Answer 2 · answered Jun 26 '21 at 15:12

-1

Configuration space is basically analogous to the 3-dimensional space we know and love. On the other hand, phase space is a mathematical construct to represent the state of a given system by specifying its position and momentum, classically at least.

The need for phase space arises from the fact that configuration space fails to account for the velocities of the particles.

So, yes, the last statement has an error. However, it might have been used lightly but not to define a rigorous physical concept.

However, in the question you have made a mistake in equating $p=\frac{\delta L}{\delta q} $, in fact $p=\frac{\delta L}{\delta \dot{q}}$

We need to know both the position, $q$ and velocity, $\dot{q}$ to be able to determine the trajectory of a particle.

answered Jun 26 '21 at 15:12

Anagh Venneti

95

Thanks for pointing out the typo. A state of the system is specified by $(q,p)$ which is a point in phase space. Now, if I know $q$, I can just differentiate it to get $\dot q$ and then get $p$. So I just needed to know $q$ right ? – Lost Jun 26 '21 at 15:25
You would have to know the time dependence of $q$ to differentiate it with respect to time, which is not generally known. This is the reason Lagrangian formulation treats it as a distinct variable. – Anagh Venneti Jun 26 '21 at 15:27
@lost what i mean is that ultimately, the question of mechanics is to solve for $q$, so your question becomes a circular one. Once we know $q$, we would have solved the problem – Anagh Venneti Jun 26 '21 at 16:00
In other words, what does it mean to specify (q,q˙) for a system? Can you give an example of a toy system and show how (q,q˙) are both separately necessary? – Lost Jun 26 '21 at 16:18
Say a ball... I specify that its position is at some 4 m from a wall. This would not be enough to determine its future position. We would need the changes in its position with respect to another variable we call time to determine its future trajectory. This change with respect to time is velocity. So either to determine the future trajectory, we need a table of positions and the respective times or positions and respective velocities. – Anagh Venneti Jun 26 '21 at 16:47

Is the state of a system represented by a point $\textbf{q}=(q_1,q_2,q_3...q_n)^T$ in configuration space?

2 Answers2