Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm watching some Physics lectures on the internet by Leonard Susskind:
http://www.youtube.com/watch?v=pyX8kQ-JzHI&feature=BFa&list=PL189C0DCE90CB6D81&lf=plpp_video

In this lecture, and also in Wikipedia and other places phase space is described as the space of all the states we need to know to determine the configuration of the system infinitely into the future.

But I don't understand how is position and velocity enough to determine this, what about forces?

Let's say we have a particle and we want to know where it will be in 10 seconds, we obviously need it's starting position, it's starting velocity and also all the forces acting on it, or it's acceleration.

Where is my misunderstanding?

share|improve this question
1  
Possible duplicate of Why are only derivatives to the first order relevant? and links therein. –  Qmechanic Jan 10 '12 at 19:40
3  
Forces are functions of positions and velocities, so in each point of the phase space, one may calculate the forces that act on any object. In this sense, the phase space contains both forces and accelerations. –  Luboš Motl Jan 10 '12 at 20:42
add comment

2 Answers

up vote 3 down vote accepted

If you are looking for a detailed answer as to why in general Lagrangians depend only on first derivatives, then you should read the answer in this question, as Qmechanic rightfully said.

However, I suspect that you are asking something else:

Given that the equations of motion depend only on first derivatives (like Newton's law), why aren't second derivatives necessary in order to completely describe the state of the system?

The answer (if this is indeed your question) is that the accelerations can be computed from the first derivatives and the coordinates.

To make things concrete, if your system is described by a vector $\vec X(t)$, and the equations of motions are $$\partial^2_t\vec X=f(\vec X,\partial_t \vec X)\ ,$$ then knowing $\vec X$ and $\partial_t \vec X$ at $t=0$ completely determines the state of the system at later (or earlier) times*. The function $f$ describes a flow in phase-space. This means intuitively that if you place yourself somewhere in phase-space (i.e. - you know $\vec X$ and $\partial_t \vec X$), $f$ tells you where to flow to, and the future (and history) of your system is completely determined.

*I assume for simplicity that $f$ is nice. There may arise singularities of all kind of sorts, and you can read about it in any ODE textbook, and also here and here.

share|improve this answer
add comment

If you know the positions of all the particles in a system, what else do you need to know to calculate their positions a time $\delta t$ later?

The answer is just their velocities, giving $\delta \vec r = \vec v\delta t$. The accelerations contribute $\frac 1 2 \vec a \delta t^2$ which is second order in $\delta t$ and can be ignored.

So the positions and velocities carry enough information to characterise a unique state of a system and how it evolves in phase space.

share|improve this answer
    
This is wrong - it's only good for an infinitesimal $\delta t$ later (i.e. to first order in $\delta t$). If you want to know the trajectory you need to know the accelerations. But these can be calculated from the position and velocity. –  yohBS Jan 11 '12 at 18:44
    
It's not wrong: by integrating those infinitesimals, you get the actual result. You may be thinking of a Taylor series...but that avoids integrating. Either way works: either integrate the first order infinitesimals or use an infinte series (the Taylor series of the one-parameter group of time evolution). –  joseph f. johnson Jan 20 '12 at 7:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.