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Differentiability in nature (how many levels typically occur? )

  1. May 24, 2009 #1
    Here's a question I've thought about on several occasions:

    How many levels of derivatives (rates of change) typically occur for objects in nature?

    For instance, a car has a position, velocity (1st derivative), and acceleration (2nd derivative), but it can also be said to have a rate of change of acceleration (3rd derivative), 4th derivative, and so on. For example if you push down on your accelerator faster and faster then you are creating a 3rd derivative of the cars motion.

    Is there a threshold at which further derivatives stop being common in the real world? I would imagine so considering that it would likely require more energy with each level of differentiation applicable.

    As far as practice is concerned is there a point at which higher derivatives either stop being useful or stop occurring?

    In the meantime, this reminds me of a related thought: If an object has a constant non-zero 3rd derivative will an person or device inside or attached to that object always be able to feel the force of increasing acceleration? Am I right in saying that the person/device would perceive a gravitation-like effect of increasing magnitude?

    My current thoughts are perhaps: A person can "get used to" (stop "feeling") the force of acceleration (2nd deriv.) because a person's body is familiar with constant acceleration (e.g. because gravity is always there). However in the case of the 3rd derivative a person would always perceive a "tugging" force on their body and eventually be squashed if a (positive) 3rd derivative remains in effect long enough. Is this the right way to think about it?

    Anyway, that was a tangent from the topic. The main question remains how many levels of differentiability typically occur in natural forces, and also perhaps what are the practical consequences.
     
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  3. May 24, 2009 #2

    diazona

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    Differentiation is really just a mathematical operation, it's not something that physically takes any energy. We tend to assume that the functions we invent to describe the natural world are continuous and infinitely differentiable (although sometimes we approximate them with functions that are not). But it's pretty rare to use anything higher than a second derivative in a meaningful equation.

    And if a person is experiencing a changing acceleration, I guess they would just feel themselves getting heavier and heavier (or lighter and lighter) over time. You could do an experiment about this, actually: go ride a roller coaster and pay special attention to what you feel as you go over the curved parts of the track. (By the way, while you can get used to a constant acceleration, you don't stop feeling it.)
     
  4. May 24, 2009 #3

    Pengwuino

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    You can easily imagine a 4th derivative being necessary. A simple model of a rocket can require a 4th time derivative of position. The mass of a rocket is changing and considering a constant thrust, you actually have a force that is changing over time but is a constantly changing acceleration. This is called the "jerk". I forget what the 4th derivative is called but consider that acceleration change changing! That might be a bit artificial as I can't really think of something in a simple model that changes the force behind the acceleration.

    Beyond that I can't really imagine what could depend on a 5th derivative but that doesn't mean there isn't something thats 5th derivative dependent.

    As for the energy, as already asked, why do you think it requires more energy to have more derivatives? As for the question, you will simply feel the force changing. If gravity was constantly increasing, you would constantly feel a pull downwards and you'd have a constant force increasing to keep you standing.
     
  5. May 25, 2009 #4

    atyy

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    How about sinusoidal motion?
     
  6. May 25, 2009 #5

    ZapperZ

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    This is a very puzzling question.

    As atyy has stated, what if a system is described by a sinusoidal function (example: harmonic oscillator)? In fact, in many systems, one could easily end up with an infinite series of sinusoidal function as the solution!

    Zz.
     
  7. May 25, 2009 #6

    vanesch

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    Funny question :smile:

    In a way, the answer to that question is always going to be arbitrary to some point, as one can change any set of differential equations into a set of coupled first order equations, by introducing new variables.

    But if I have to give an answer, I'd say, two. The reason seems to be that the highest order of differentiation that occurs in fundamental physical equations, is two. That doesn't mean that higher-order practical equations do not occur (on the contrary: the equation for the center line of a beam under load is 4th order for instance), but this is because one has reduced a set of more fundamental equations to one more practical equation.

    The Lagrangian formulation of most physics is second-order. Einstein's equation of GR is second-order (in the curvature tensor). I've never seen anything beyond second order in any fundamental formulation of known physics.
    Of course, up to a point this is arbitrary, and maybe it is just that from the moment people would need a third order, they are going to introduce a new physical concept which allows the overall set of equations to be second order.
     
  8. May 25, 2009 #7
    Yes, it was an odd question I admit. It probably sounded fairly naive in several respects but I felt like posting it anyway as a thought.

    I was well aware that differentiability is (mathematically) easily infinite (such as trig. functions for example). I was thinking more along the lines of forces and such. Infinite differentiability of a physical force seems to perhaps imply that the force can contain infinite information with respect to change, but a physical object is finite as far as molecules are concerned. Thus it's existence would almost seem to imply that a finite object holds infinite information. Or perhaps that's true, perhaps the object can hold infinite information because space is a continuum. But the continuum being there doesn't completely answer the question of forces being transmitted in collisions. Can a single molecule "store" change greater than the force and acceleration levels? You can certainly make a single particle have 3rd derivative change by hitting it with several particles in succession, but is that the same really to say that it has a "stored" 3rd derivative "force" associated with it or is it just a discrete sum of of 2nd derivative forces. It's certainly true that higher derivatives can easily be applicable to various complex systems, but the question is do they exist as a fundamental force at the simplest level.

    Are there any planets out there that emit gravitation at the 3rd derivative or higher levels? Think about it, if there's no tendency or limit of differentiation occurring in nature then why aren't there any masses in space that emit a gravity field having an effect higher than the second derivative (acceleration).

    Anyway, sorry if the question was a bit stupid, I was being a bit random at the time and I'm new to this. There's just something about infinite differentiability that seems a bit odd to me as far as information density in nature is concerned.
     
  9. May 25, 2009 #8

    Pengwuino

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    How do you formulate your boundary/initial conditions for a 4th order equation physically? I remember doing a problem in my upper division E/M class and i forget the exact example but it was something fairly simple. It dealt with the current in a wire or something. The problem is what I eventually had was a 2nd order equation in terms of the current. I couldn't figure out the problem and I asked my professor and he said I needed to do the problem in terms of the charge and not the current. I asked him why couldn't I do it the way I was doing (aka why was the problem screwing up?) and he couldn't really figure it out at the time but he said it probably had something to do with the fact that when I took the third differentiation with respect to the current, I lost some information that made the problem unsolvable.

    Now that I look back, I assume it is because I didn't have the 2nd order boundary condition (or first order since i was doing it with respect to current) or initial condition. Now that seems fine since we know the forces at the 2nd order. However, you bring up the 4th order. What I assume you would need is the 3rd order boundary/initial conditions but physically, what are they? How would we find the third order conditions physically?
     
  10. May 26, 2009 #9

    vanesch

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    For the beam, that's simple: you give, for instance, position AND tangent direction at both sides of the beam.

    See the wiki entry on the Euler-Bernoulli equation for instance:
    http://en.wikipedia.org/wiki/Euler-Bernoulli_beam_equation
     
  11. May 26, 2009 #10
    I remember working on an Euler-Bernoulli beam problem in my Numerical Analysis class last semester. I'm not quite sure how the equation was derived though, the course focused more on computation concerns like methods of solving and errors and so forth (as one would expect from a Numerical Analysis class). I believe it though, as any arbitrary derivative can be useful for analysis.

    On a different subject: Is there any gravity emitting body that emits gravity that pulls with a force stronger than the 2nd derivative.

    Has it actually been proven that there's not any body with a gravity field of higher than 2nd order, or is that an assumption? Does it have anything to do with relativity or the number of spacial dimensions we have (i.e. 3)?

    If a body is pulling in more and more objects you could argue that it's gravity is increasing thus it has a 3rd derivative and analytically this can be true (and useful for computation/simulation). However, at any given instant the body would have a 2nd derivative gravity field (it just happens to be increasing).

    I guess the question then is: If a body is left on its own is space is it possible for it to have a 3rd derivative gravity behavior? I.e. is gravity fundamentally 2nd order or can it be higher (or perhaps even lower order)?

    The arbitrary nature of the recent "dark matter theory" (which seems like it might just perhaps assign gravity on a whim in order to make things work) makes one suspect that perhaps physics has actually not yet gained a really good understanding of gravity. Or perhaps its just computation error, or other indirect errors propagating in the analysis process. Then again I know very little physics so who am I to say?

    Just some thoughts.
     
  12. May 26, 2009 #11

    Andy Resnick

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    I've long been intrigued by the ubiquity of 2nd order differential equations in Physics. A recent article in Science, by Schmidt and Lipson

    http://www.sciencemag.org/cgi/content/abstract/324/5923/81

    mentions the following (paraphrased)-

    If you measure the positions of an object, you can deduce the structure of phase space. If you also measure the velocity, you can deduce the Lagrangian and Hamiltonian. If you also measure the acceleration, you can deduce the force laws.

    To me, that implies something very fundamental about second-order equations. To be sure, there are plenty of third and fourth order equations that are useful; but there seems to be something special about second-order equations.
     
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