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I Is there any theoretical basis for laws being 2nd order

  1. Nov 26, 2018 #1
    Hi, I’m just wondering about this:

    Are there any theoretical reasons why physical laws take the form of 2nd order (in time) differential equations?

    Or is it just observed to be that way?

    Are there ANY laws (even in a limited context) which are 3rd (or higher) order in time?
     
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  3. Nov 26, 2018 #2

    Vanadium 50

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    If an object has an acceleration growing linearly with time, its position will be cubic in time, e.f. d^3x/dt^3 = a constant.
     
  4. Nov 26, 2018 #3
    Ok......... I know things can travel higher order paths... that’s not what I meant.

    In classical mechanics (at least nonrelativistic) the differential equations of motion describing any closed system always come out second order in time, do they not? Of course the most general solutions are not quadratics...

    I don’t confidently know much physics, so I am asking: is this the case for all theories? And it just an observed phenomenon or are there underlying explanations?

    For instance maybe there is some esoteric mathematical way to show 3rd order equations of motion are non physical?
     
  5. Nov 26, 2018 #4
    To maybe put it another way: why is full knowledge of a system of particles the positions/velocities (in classical theory)?
    Why does nature not give freedom in initial acceleration as well as position/velocity? Why can’t the gradient of potentials determine the 3rd time derivative of positions?
    (Is it “just the way it is” or is it related to deeper principles?)
     
  6. Nov 27, 2018 #5

    Mister T

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    That occurs by assumption. It's assumed that the force is proportional to the second time derivative of position. Therefore, when you look at the equations that relate force and position, they will be second order equations.
     
  7. Nov 27, 2018 #6

    anorlunda

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    Newton's Laws of motion can also be derived from the principle of least action (PLA) and the Euler Lagrange equations. That makes the PLA the underlying assumption.
     
  8. Nov 27, 2018 #7

    ZapperZ

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    But this all depends on the starting point here.

    When we use the force equation, one starts solving the differential equation of motion via the established definition of what force is, which is proportional to the second time derivative of displacement. We do this because we think we can account for all the forces on a system. For example, for a simple 1D mass-spring system, we get

    [tex]
    m \ddot{x} = -kx
    [/tex]

    That is the starting point. You find the equation of motion by solving the above differential equation.

    However, this is not the only starting point, as any physics student knows. We also have the Lagrangian/Hamiltonian formulation of it, whereby the starting point here is NOT in terms of forces, but rather the KE and PE of the system. In the same case as the mass-spring system, the Lagrangian can be written as

    [tex] L = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} kx^2 [/tex]

    Look, ma! No second time derivative!

    Now of course, to solve for the equation of motion, you need to apply the Lagrangian to the Euler-Lagrange equation, and voila, you get the same differential equation as you did for Newton's law of forces, which is very comforting. But if we did not teach you "forces", but instead started you off with finding the Lagrangian of a system, then you'd never think that the second time derivative is always there, or even significant enough to be asked in a physics forum. All you see is that the 2nd time derivative is merely something you passed by along the way in solving the problem, not even worth giving a second look.

    Zz.
     
  9. Nov 28, 2018 #8
    Thanks, good discussion, but I’m not satisfied. (Probably won’t be.)
    Yes it’s an assumption but we assume it because it works. It’s not arbitrary. We don’t take F = m(d3r/dt3) because we couldn’t make it work.
    What I am asking is if there is insight as to why we need not let acceleration be a “state variable” (like position and velocity) for the models to work.
    I feel this just pushes the question one step back.
    In the principle of least action we assume the lagrangian can be expressed as a function of (x, dx/dt, t) which (maybe I’m misinterpreting this?) corresponds to the assumption that the state (at a particular time) can be fully described by positions and velocities.
    So then the question becomes, why not have an action principle which minimizes the integral of L(x, dx/dt, d^2x/dt^2, t)?
    “Because then we can’t integrate-by-parts to get the Euler-Lagrange equation” is not the point.
    (Or is it the point? What happens if you search for critical functions of this slightly more general action? I don’t know.)

    Anyway my preferred way to rephrase the question is:
    Why can’t we have a theory where acceleration is a “state variable” like velocity and position?
     
  10. Nov 28, 2018 #9

    Delta2

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    We could define "artificial " quantities for which the laws become 3rd order or more, for example we can define ##X(t)=\int x(t)dt## where ##x(t)## the displacement, and then Newton's second law becomes 3rd order with respect to ##X(t)##, and it could become 4th order with respect to ##\int X(t) dt## and so on...

    BUT it is very interesting fact that almost all laws of the universe take the form of at most 2nd order with respect to the "default" quantities we have defined.
    For example Maxwell's equations are first order with respect to electric and magnetic field, and become 2nd order in the potential formulation. Navier- Stokes equations are also at most second order with respect to velocity field of the fluid. Schrodinger equation or Klein - Gordon equation are at most second order w.r.t to the wave function...

    Can someone give a law that is 3rd order or more with respect to the "default" quantities defined by the theory?
     
  11. Nov 28, 2018 #10

    Mister T

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    Originally it did seem to work quite well. We now know, of course, that it doesn't.

    Acceleration is just another name for second time derivative of position.I don't understand your question.
     
  12. Nov 28, 2018 #11

    ZapperZ

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    You seem to have lost track of WHY we solve for these things. We often solve for them because we need to know their dynamics, i.e. how do various information about the system changes over time.

    So in basic kinematics, we want to know the location of an object with respect to time, which is the "equation of motion". Once you have the equation of motion, everything else can be derived more easily from there. It is like this is the "wavefunction" of the system, where you can find everything else that you need out of it. Just telling me your acceleration doesn't tell me how long it will take for you to get here from Los Angeles. It may be a good starting point to solve thing, but it isn't the END point that is useful enough.

    Zz.
     
  13. Nov 28, 2018 #12

    anorlunda

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    Higher order derivatives in the Lagrangian would violate locality.
    https://en.wikipedia.org/wiki/Principle_of_locality

    I don't understand what you mean by that. If position is a state, aren't its derivatives and integrals also states?

    You're ignoring the most compelling answer
    If we just make up equations that do not agree with observations, that would be pointless.
     
  14. Nov 28, 2018 #13

    anorlunda

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  15. Nov 28, 2018 #14

    Vanadium 50

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    This thread suffers from having a rather wooly question that people are struggling to interpret and understand.

    As far as I can tell, the answer to the question "Why do we usually see equations of motion that are only second-order in time derivatives?" is "Because we usually deal with problems with constant acceleration". I don't think there is any more or less than that to it. There certainly are problems that don't have constant acceleration, and they have higher derivatives, as does electromagnetism in non-linear media.
     
  16. Nov 28, 2018 #15
    That’s an interesting statement! But I don’t see how it follows from allowing accelerations in the argument of the Lagrangian?
    Anyway, is non-locality even the true issue? Isn’t the entire classical theory (with universal time, i.e. nonrelativistic) non-local anyway? (Non-local in the sense that the motion of a particle instantaneously changes the related potentials (i.e. it’s force on other particles) everywhere.)
    Right... my question is about why acceleration isn’t a free variable in the dynamics? If we know the initial positions and velocities of a system then classical theory can (in principle) calculate the subsequent motion. Why does nature not give freedom in initial acceleration as well?
    With all due respect you are severely misunderstanding. It has NOTHING to do with constant acceleration. I know you know that second order equations lead to more than constant acceleration... just look at any simple example, Hooke’s law for a spring, Newton’s law for gravity... not sure why I need to belabor this point...
    When I say “state variable“ I specifically mean:
    a variable that needs to be specified in order for subsequent motion to determined.
    So in classical theory position/velocity are the state variables, and acceleration is not.
    As far as I can tell, this aspect of the theory has no deeper connections. It seems to just be an assumption which works. That’s fine and all, I’m just asking if there’s logic behind why acceleration (and higher) need not be a “state variable.”

    I really don’t think it is that bizarre of a question but maybe I’m just too inarticulate.

    Just to (not) get my point across has been very draining /:
     
  17. Nov 28, 2018 #16

    ZapperZ

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    But why should nature give such freedom? She doesn't give freedom for us to do whatever we want in numerous situations. Why must "c" be a constant?

    In all of this, you never once indicated WHY you would want such a thing. I can understand if there is a need to explore this to make the problem more tractable, or give another new insight that never was available before. But so far, all I can see is that you want this because... well... just because! There was never any rational reason why this is any more useful than what we already have.

    Isn't this a total waste of time?

    Zz.
     
  18. Nov 28, 2018 #17
    Well as I loosely understand, you can follow arguments that have nothing to do with light and find that the transform has to be lorentz with some “universal speed limit.” So then it’s a question why light should match that. Then it’s maybe a question of massless-ness. Such questions are typically related, and I don’t know everything!
    With all due respect, I think I made it clear that I would be content with an answer of “it just is.”
    Further, I only know classical theory. So a main part of my question was how far this assumption (or analagous) goes.
    Maybe I should just ask quantitative questions here...
     
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