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Generalised Coordinates

  1. Oct 9, 2004 #1

    cepheid

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    I have the following statement in my course notes:

    "For a system of N particles moving in 3 spatial dimensions, we can determine the entire motion as a function of time provided we know:

    [tex] \begin{array}{cc} \mathbf{Q} = (q_1, q_2, ... , q_{3N}}) \\ \mathbf{\dot{Q}} = ({\dot{q}_1, \dot{q}_2, ... , \dot{q}_{3N}}) \end{array} [/tex]

    at some given time t

    "One does not require Cartesian coordinates here, provided the 3N coordinates [itex] \{q_j\} [/itex] completely define the system position i.e. they span the 3N degrees of freedom. Likewise for the velocity coordinates [itex] \{\dot{q_j}\} [/itex]."

    The notes go on to point out that the fact we need only positions and velocities in specifying the state of the system is not a mathematical truth but a fact about nature discovered by Newton, and should be clear from his 2nd law, but much clearer in the Lagrangian and Hamiltonian formalism that we are about to launch into.

    Fair enough. I'm just trying to understand why this fact about nature is so "clear." For instance, if I don't know anything about Lagrangians and want to solve the system using Newton's Second Law, the statement in bold above asserts that I can do so, provided I have the 6N required boundary conditions. Here's the question: doesn't this bold statement implicitly assume that I already know the force laws governing all the forces that act on the system? If not, then how could I possible solve for any equation(s) of motion? What good are 6N boundary conditions if I can't even formulate 3N second-order differential equations to apply them to? For each differential equation, I'd be able to go no further than:

    [tex] m\frac{d^{2}q_j}{dt^2} = \ \ ??? [/tex]

    :confused:
     
    Last edited: Oct 9, 2004
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  3. Oct 9, 2004 #2

    Fredrik

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    Newton's second law looks almost exactly like a formula that appears in a theorem that you can find in any textbook on differential equations. I don't remember the exact form of this theorem, and I'm feeling too lazy to look it up right now, but it goes something like this:

    Suppose that

    [tex]x(t_0)=x_0[/tex]

    and

    [tex]x'(t_0)=v_0[/tex]

    (where the quantities on the right-hand sides are just any real numbers). If F is a "nice" function, then the differential equation

    [tex]x''(t)=F(t,x(t),x'(t))[/tex]

    has exactly one solution.

    So when your course notes say that Newton "discovered" this fact about nature, they really mean that he discovered that position as a function of time always seems to obey a differential equation of that form. It is a mathematical truth that a differential equation of that form has exactly one solution for each initial condition, but it's a physical (experimental) fact that the function that represents a particle's position as a function of time satisfies a differential equation of this simple form.

    I don't know why any of this would be more clear in the Lagrangian formalism. I don't think it is.

    Yes, you would have to know the force to calculate how the generalized coordinates change with time, but you don't have to know the force to know that there's exactly one solution. You only have to know that the force (in any realistic situation) is a nice enough function to satisfy the requirements of the theorem that guarantees the existence of a unique solution.

    By the way, what you called a boundary condition should be called an initial condition. A boundary condition would be a specification of Q at two different times. That would also be sufficient to ensure the existence of a unique solution.
     
    Last edited: Oct 9, 2004
  4. Oct 9, 2004 #3

    cepheid

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    Thanks for the clarification about the theorem involving second-order d.e.'s. I'm not sure, because I haven't read on, but could it be that the "truth" we are trying to explain is clearer in the Lagrangian formalism because it (seems to) bypass this notion of "force" (which Newton never defines), instead explaining it in terms of the more fundamental principle of least action?

    As for the comment about initial conditions...HUH? Where in my post do I say that the 6N numbers represent position and velocity coordinates at an *initial* time t = t0? I don't. In fact all the notes state is that they must be given for some random time t. So how could they be called initial conditions, and what else would you call them? Is there some rule that a boundary condition must specify the *same* quantity at two different times? Because my prof used the term in a much more generic sense, saying that as long as you have 6N numbers (or however many you need), you'll be okay. He called those numbers the boundary conditions. Any thoughts on the correct terminology? Anyone?

    Incidentally, the specific example of a boundary condition that you gave (Q at two different times) is the one he later used in the illustration of the prinicple of least action and the derivation of Lagrange's equations.
     
  5. Oct 10, 2004 #4

    Fredrik

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    I suppose the principle of least action can be thought of as more fundamental than Newton's second law, since it can be used to find other interesting equations, but I don't think that makes this subject any clearer. I think what they probably have in mind is that in the Hamiltonian formulation, the state of a physical system is represented by a pair (Q,P), where Q is the Q you defined, and P is the conjugate momenta to the generalized coordinates. The set of all possible (Q,P) is called the phase space, and the equations of motion (two of them, in the Hamiltonian formulation) generate a "flow" in phase space.

    Of course, you could think of the set of pairs (x,v) or (x,p) as a phase space of Newtonian mechanics, and the force F as the function that defines the flow on that phase space, but one usually doesn't. I think the reason is that the Hamiltonian formalism suggests a way to add structure to the phase space. The phase space is not a vector space, or even a metric space, but it's not just a set of points. Instead of a metric, you can define something called a symplectic form. (I never learned this stuff perfectly, and I've forgotten a lot of what I did learn, so I won't try to explain this further. You'll probably know more than me about this in a few weeks anyway).

    Should the condition be called an "initial condition", a "boundary condition", or something else? I'm actually not 100% sure about what the standard terminology is, but to me it seems strange (even wrong) to use the term "boundary condition" when we're talking about a point in the interior of the domain of the function x. I agree of course that it also sounds strange to call it an "initial condition", for the reason you stated, but at least to me that sounds a lot less strange than calling it a "boundary condition". ("Initial" sounds strange, "boundary" sounds wrong). Perhaps someone whose ODE book isn't under a pile of junk in a room where the light bulb is out :smile: can enlighten us about what the standard terminology is.
     
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