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Equations of Motion

  1. Jun 13, 2013 #1
    In Landau's Mechanics it states "If all co-ordinates and velocities are simultaneously specified, it is know from experience that the state of the system is completely determined and that its subsequent motion can, in principle, be calculated. Mathematically, this means that, if all the co-ordinates [itex]q[/itex] and velocities [itex]\dot{q}[/itex] are given at some instant, the accelerations [itex]\ddot{q}[/itex] at that instant are uniquely defined."

    My question is why is this so. I understand that from knowing the co-ordinates of a mechancial system the future evolution of a system is not uniquely determined. But how does the additional knowledge of the velocities uniquely determine the acceleration of the system and hence its future mechanical state?
     
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  3. Jun 13, 2013 #2

    SteamKing

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    Say you have a spring-mass system. What equation determines the position of the mass after it is deflected? If only an initial position is specified, does this equation have a unique solution? If an initial velocity is also specified, what happens to the solution of the equation then? Re-read the passage from Landau after performing this mental exercise.
     
  4. Jun 13, 2013 #3
    SteamKing: I have thought about this and come to my own conclusion and am interested in testing it along with hearing other opinions. I didn't come here for a lecture and know that you need two initial conditions to solve the differential equation. Why this is so is my question.
     
    Last edited: Jun 13, 2013
  5. Jun 13, 2013 #4

    WannabeNewton

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    SteamKing thoroughly answered your question. The equations of motion are a set of second order ordinary differential equations ##m\ddot{q}^{i} = F^{i}(q^1,...,q^n,\dot{q}^1,...,\dot{q}^n)##. If an arbitrary initial data set ##\{q_0^{1},...,q_0^{n},\dot{q}_0^{1},...,\dot{q}_{0}^{n}\}## is specified at initial time ##t_0## then one of the most standard theorems of ODEs tells us that there exists a neighborhood of ##t_0## on which there exist unique trajectories ##q^1,...,q^n## solving the above differential equation and satisfying the initial conditions. This then uniquely determines the states of the system within that neighborhood of the initial time.
     
  6. Jun 13, 2013 #5
    I don't want this to degenerate polemics so I'm going to write my thoughts. I would appreciate any comments.

    Since force is proportional to acceleration it seems strange that acceleration would depend on acceleration. That would be like the force exerted on something is proportional to the force. This seems somewhat self-referential which is not, a priori, a disqualification since we can define recursive function but this is my intuition. This leaves only position and velocity. If we say all forces are either contact or non-contact (i.e. field) forces then we can say contact forces obviously depend on where something since it must be determined if two things are in contact. Field forces like Newton's gravity only depend on position while the Lorentz force depends also on velocity so here we need to acknowledge some forces depend on position and velocity. If all conceivable forces in classical mechanics either depend on position and/or velocity it seems their knowledge is a sufficient (thought not necessary) condition for determining the forces acting on a particle and hence the acceleration. For example, if the gravitational force only depends on the relative position of two particles why do we need knowledge of the particles velocity to predict its future position. Obviously because something at the same place with different speeds will be in different places after some times interval if acted upon by the same force. So one thing my explanation seems not to explain is the necessity of knowing both initial position and velocity even for forces that do not depend on velocity such as universal gravitation.

    What I know is wrong is to simply appeal to the differential equation and say for N particles you need 2N initial conditions. That is not an explanation because that is exactly what is in question.
     
    Last edited by a moderator: Jun 13, 2013
  7. Jun 13, 2013 #6

    WannabeNewton

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    It is not in question because the unique determination of the states from the well-posted initial value formulation of Newton's 2nd law is a basic result of ordinary differential equations. The proof of this can be found in any decent textbook on the subject of ODEs.
     
  8. Jun 13, 2013 #7
    It is in question because that is my question. I'm not looking for some cookie cutter solution you can find in an ODE text. I've studied all that and know the difference between an explanation based on insight and merely giving some theorem. Besides, we are talking about physics here, not math.
     
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  9. Jun 13, 2013 #8
    This is a response to #6. Think of the simple problem of throwing a ball vertically up into the air. You need to know the initial velocity in order to predict the height of the ball at any subsequent time. For different values of the initial velocity, the ball's vertical position vs time will be different. So you have N = 1, and you need two initial conditions: initial vertical location and initial vertical velocity.

    [Mentor's note: Post #6 is now post #5.]
     
    Last edited by a moderator: Jun 13, 2013
  10. Jun 13, 2013 #9
    Chestermiller, thanks your post is getting close to the questions since you are looking at things physically. But I made your point in post #6. What I am looking for is a general proof and not specific examples. [Mentor's note: Post #6 is now post #5.]

    The attitude I detect here is one "pouring the water into the glass." I have BS in physics and MS in math from a very good school since credentials seem to matter more here than ability to think. Now that we've gotten that out of the way maybe some people here can show some humility and acknowledge my question is not so easily answerable. I will reiterate here:

    I understand that just knowing where something is does not uniquely determine where it will be if acted upon by a certain force. The velocity at the time of the force acting on it also needs to be considered for obvious reasons that have been pointed out here. My question is how does knowing something's position and velocity uniquely determine the force acting on it?
     
    Last edited by a moderator: Jun 13, 2013
  11. Jun 13, 2013 #10

    jbriggs444

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    So your question is why all known force laws are functions of position and velocity? And why no known force laws have a term for acceleration (except to the extent that acceleration eventually alters velocity and position)?

    I would suggest that

    a. That's just the way it is and
    b. It's an idealization since an accelerating ball may be squashed flatter than a non-accelerating ball and may, as a result, experience greater air resistance.
     
  12. Jun 13, 2013 #11

    mfb

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    It is an experimental result.

    You can invent a framework where the third derivative of the position depends on position, velocities and accelerations, where you have to know all three to get full knowledge about the system. According to all measurements done so far, we do not live in such a universe-
     
  13. Jun 13, 2013 #12

    micromass

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    The problem is that we don't know what you would accept as an answer. If you ask "why" about something in mathematics, then citing the theorem and proving it is the correct way to go. But somehow, this doesn't satisfy you.

    So, what kind of answer would please you??

    See this video on "why" questions:

    https://www.youtube.com/watch?v=wMFPe-DwULM
     
  14. Jun 13, 2013 #13

    WannabeNewton

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    You can certainly have Lagrangians which depend on second time derivatives of position. You can have Lagrangians which depend on arbitrary orders of time derivatives. In field theory you can have Lagrangian densities which depend on arbitrary orders of partial derivatives of the field variable(s). There will be analogous Euler-Lagrange equations for such cases. The issue of why we choose to stop at first order is not very clear cut. See here:http://physics.stackexchange.com/questions/4102/why-are-only-derivatives-to-the-first-order-relevant. The point is that including higher than first order time derivatives will lead to internal inconsistencies in the otherwise well-posed initial value formulation of Newton's 2nd law.

    Also, don't start acting defensive and belligerent just because you couldn't elucidate your question properly, it is sophomoric and only makes you look bad.
     
  15. Jun 13, 2013 #14
    Well, there are two sides to the equation. The left side is the mass times the acceleration. But the right side of the equation is the net force, imposed by either body forces or contact forces, or by a combination of both. Empirical experience has shown that body forces acting are a function of position and velocity. Contact forces are more complicated, but are not encountered in collisions between atoms and molecules, because the interactions are fully captured by body forces.
     
  16. Jun 13, 2013 #15

    Fredrik

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    Note that he says that it's known from experience, not that there's a theory-independent mathematical proof.

    It doesn't. I mean, it certainly does when the equation of motion is a differential equation of the type that satisfies the assumptions of the theorem that guarantees a unique solution for each initial condition on the position and velocity. But you can certainly define a theory of physics with equations of motion of a type that doesn't satisfy those assumptions.

    It seems that what you want to know is why all classical equations of motion that have been found to be useful, are differential equations of the type that satisfies the assumptions of that theorem. I think it's possible to make an argument based on quantum field theory, but I don't understand it well enough to really explain it. The idea is that terms with higher order derivatives in quantum field theory Lagrangians are non-renormalizable, and non-renormalizable terms can be shown to have a negligible effect on interactions at low energies (where classical theories are less wrong). If we include higher order derivative terms in the QED Lagrangian for example, Maxwell's equations are replaced by something that includes higher order derivatives, and that could ensure that the new equations are not of the type that has a unique solution for each initial condition on the field and its time derivative. It would probably do something similar to Coulomb's law as well.

    There's no such thing. I think the best one can do is to make a non-rigorous argument based on quantum field theory.

    I don't know how you got that idea, but it's very wrong.
     
  17. Jun 13, 2013 #16

    WannabeNewton

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    Fredrik, I definitely think it is related to the physical theory being well-posed, as you say. Take a look at chapter 11 of Wald for example. He shows that Maxwell's equations and Einstein's equations have a well posed initial value formulation using two theorems regarding certain types of second order PDEs on curved manifolds (theorem 10.1.2 and 10.1.3). I don't know if there are similar theorems for higher order PDEs of analogous types but his comments about how even 2nd order systems of such nature are hard to analyze don't give much hope. Without a well posed initial value formulation, things like continuity of maps taking initial value sets to solution sets will not be guaranteed and, more importantly, causality won't be guaranteed.
     
  18. Jun 13, 2013 #17
    jbriggs444: OK, now we are getting somewhere. But "just the way it is" really doesn't satisfy me although ultimately you will arrive at this answer. I think we haven't reached that point though. There are two statements I can not reconcile. (1) I see clearly from the 2nd order ODE that you need two constants and they turn out to be position and velocity. (2) Say I know the position and velocity of some particle. What does that knowledge have anything to do with knowing the force (and hence the acceleration) acting on the particle.

    mfb: Maybe so but we should at least look for a principle.

    micromass: I was reading about Feynman back in the early 90's so please don't lecture me. Furthermore, I don't care what RF thinks about why questions. I think for myself, much as RF did. There is one video where he first has some critical words for non-scientists then says he was wrong. Going on to say he respects anyone who takes questioning as far as it can go in their discipline is someone he respects. So I think he would be on my side here.

    WannabeNewton: This discussion is focused on classical mechanics so forces dependent on higher derivatives from field theory, etc. are not relevant. From Landau's Mechanics: "The fact the Lagrangian contains only [itex]q[/itex] and [itex]\dot{q}[/itex], but not hider derivatives [itex]\ddot{q}[/itex], [itex]\dddot{q}[/itex], etc., expresses the result already mentioned, that the mechanical state of the system is completely defined when the co-ordinates and velocities are given."

    Before you reply please see my two statements in first paragraph.
     
  19. Jun 13, 2013 #18

    WannabeNewton

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    Sure ngawang but I also mentioned how you can technically have Lagrangians for classical systems which depend on higher order time derivatives than order one. But let me try to see if I understand your question: you are asking why physically, given Newton's 2nd law for a single particle, the two initial conditions needed to get a unique solution to the 2nd law are cited by Landau as being one of initial position and one of initial velocity as opposed to some other combination of physical quantities?
     
  20. Jun 13, 2013 #19
    "There's no such thing. I think the best one can do is to make a non-rigorous argument based on quantum field theory." People please. This has nothing to do with QFT. We are talking about the equations of motion of classical mechanics.

    Second and for the last time I am not "looking for an answer" in the sense most here think. I came here to debate my view and since I've asked a question (it seems) no one here as ever considered woudn't it be a sign of humility to think about it. I may not know as much technical physics as most here but I am sure I can think much more clearly and so we both have something to learn from one another.
     
  21. Jun 13, 2013 #20
    wannabenewton: Let's be clear. I am not here to "ask a question" in the sense that I am seeking someone to tell me what the answer is. I want to challenge others' ideas of why this is so: "In the context of a single particle in classical mechancis, how, physically, does knowing the position and velocity of a particle give one the force acting on it." Whoever understands this will gain real insight into classical mechanics.

    My solution is that since all forces in classical mechanics, in terms of kinematical quantities, depend only on position and velocity no other kinematical quantities are necessary to specify the force acting on a particle.
     
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