- #1

Ksitov

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I'm asking why there is no higher derivative than two in physics ? I never encountered a third (time or space) derivative in physics.

Have you some litterature about this?

Thank you.

Regards.

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- #1

Ksitov

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I'm asking why there is no higher derivative than two in physics ? I never encountered a third (time or space) derivative in physics.

Have you some litterature about this?

Thank you.

Regards.

- #2

Bandersnatch

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Google for jerk, jounce, crackle and pop.I'm asking why there is no higher derivative than two in physics ?

- #3

Ksitov

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Why EDP are always in term of second derivatives ? (diffusive - convective - wave equations)

Even in General Relativity, Riemann curvative tensor is "the derivative" of the Christoffel symbols which is "the derivative" of the metric tensor => order 2 in term of spatial derivative.

- #4

A.T.

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You can use these quantities in equations too.I knew these quantities but I'm speaking about equations.

- #5

sophiecentaur

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You are asking the forbidden "why" question, for which there is never a proper answer. All we can say is that 'I'm asking why there is no higher derivative than two in physics ?

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- #7

cosmik debris

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Cheers

- #8

olivermsun

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Look for "wave equation in stiff string," "Korteweg-de Vries (KdV) equation," and similar.

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- #9

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- #10

Gigaz

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- #11

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This is not true. Differentiability is always a local property, no matter how often it is possible. "Relatively far away" is definitely a misinformation.But higher derivatives ultimately describe the behaviour of a function relatively far away from the current point.

- #12

Asymptotic

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I don't recall seeing snap (jounce), crackle, and pop in servo drive manuals and parameter lists, but jerk is available on all modern servos, and (although often called "s-curve") in many induction motor VFDs, too. Jerk increases overall motion profile move time, but is worth the trade-off in applications where high positioning accuracy is required, and also helps prevent machinery from tearing itself apart where high inertia loads are involved.You are asking the forbidden "why" question, for which there is never a proper answer. All we can say is that 'we find that'the vast majority of phenomena can be characterised in equations requiring just a second derivative. Variables such as Jerk etc. are part of Science but not needed for most purposes so you don't come across them too often.

- #13

DrDu

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- #14

sophiecentaur

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Every car journey exhibits a third order time derivative as your accelerator pedal takes time from one setting to another.

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- #15

Gigaz

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This is not true. Differentiability is always a local property, no matter how often it is possible. "Relatively far away" is definitely a misinformation.

Yes, in math. But have you done numerical differention, perhaps even for solving partial differential equations? For higher derivatives at x0, you need to evaluate the function at points further away from x0. And ultimately, the infinitesimally small doesn't really exist in physics.

- #16

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You are implicitly suggesting that there are two different concepts of differentiability in physics and in mathematics. This is also wrong! To cover one misinformation by the next one is nonsense.Yes, in math. But have you done numerical differention, perhaps even for solving partial differential equations? For higher derivatives at x0, you need to evaluate the function at points further away from x0. And ultimately, the infinitesimally small doesn't really exist in physics.

You can have a perfect smooth function at a point which is completely wild and not even continuous "further away". To say you need points farther apart is a contradiction to the concept of a local phenomena and very misleading. E.g. in a often used Taylor expansion in physics, all derivatives are evaluated at the same point. It should not be told that "other points are needed", because it is simply wrong. And you do not need "infinitesimal small" or "dx". One only needs a small neighborhood of a point to have an approach, which normally does exist even in physics. This is already needed for the first derivative.

- #17

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- #18

Gigaz

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You are implicitly suggesting that there are two different concepts of differentiability in physics and in mathematics. This is also wrong! To cover one misinformation by the next one is nonsense.

You can have a perfect smooth function at a point which is completely wild and not even continuous "further away". To say you need points farther apart is a contradiction to the concept of a local phenomena and very misleading. E.g. in a often used Taylor expansion in physics, all derivatives are evaluated at the same point. It should not be told that "other points are needed", because it is simply wrong. And you do not need "infinitesimal small" or "dx". One only needs a small neighborhood of a point to have an approach, which normally does exist even in physics. This is already needed for the first derivative.

Yes, and if you want to have the second derivative, you'll have to use a larger neighborhood than what you took for the first derivative. Higher orders even more so. And as long as the taylor series converges, the derivatives will yield more information about the larger neighborhood of the point.

Let's say you're solving Navier-Stokes or pressure waves. Obviously, the derivatives are not "real derivatives" in a mathematical sense. Length scales below the size of an atom are unreasonable, then the equation doesn't apply anymore. The derivatives only occur in those equations because atoms are in fact pretty small.

- #19

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I assume that the answer has something to do with energy. Energy underlies nevery concept in physics-- whether you are dealing with classical mechanics, quantum mechanics or relativity. Work (a concept extremely closely related to energy) is very easy to express in terms of force, a quantity associated with acceleration, the second derivative of position (W=F⋅d for a constant force or ∫F⋅dx for a varying force). Now, I could express this in terms of yank (Y, mass times jerk, the third derivative of position) as W=F

- #20

jbriggs444

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There is no minimum size for the required mathematical neighborhood. Any neighborhood of non-zero extent will do and can allow derivatives of all orders to be obtained (if said derivatives exist at all).Yes, and if you want to have the second derivative, you'll have to use a larger neighborhood than what you took for the first derivative. Higher orders even more so. And as long as the taylor series converges, the derivatives will yield more information about the larger neighborhood of the point.

If you are interested in estimating a derivative using imperfectly accurate physical measurements then using a larger interval may be of some benefit. Is that the point that you are trying to make?

- #21

rumborak

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This is also wrong! To cover one misinformation by the next one is nonsense.

As a general comment, I really really dislike these types of responses. They are the antithesis to learning, because they are beligerent in nature and don't even bother trying to understand the other side.

@Gigaz makes an interesting, and not entirely untruthful statement that in any approximation of a derivative, no matter how small your delta, the second derivative inevitably considers a neighborhood of 2*delta. You can use the limit of Delta going to zero, but that is evading the point, as the delta is never *actually* zero.

- #22

jbriggs444

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If you are taking a limit, it is improper to talk about what the delta *actually* is. The delta is out-of-scope for any such comment since it is inside a quantifier.You can use the limit of Delta going to zero, but that is evading the point, as the delta is never *actually* zero.

- #23

rumborak

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- #24

jbriggs444

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You are mistaken. In the traditional epsilon/delta definition of a a limit, the delta and the epsilon are inside quantifiers. They have no independent existence. It is a scope error to speak about their "actual" value.

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- #25

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As a general comment, I really really dislike these types of responses. They are the antithesis to learning, because they are beligerent in nature and don't even bother trying to understand the other side.

But higher derivatives ultimately describe the behaviour of a function relatively far away from the current point.

To tell students, that the degree of differentiability takes you further away from the point in question, is a very bad way to look at it and ...To say you need points farther apart is a contradiction to the concept of a local phenomena and very misleading.

Yes, in math.

... makes it even worse. This is not what differentiability means.... implicitly suggesting that there are two different concepts of differentiability in physics and in mathematics.

- #26

olivermsun

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However, it

- #27

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Higher derivatives are not "rare" in physics in the sense that a whole bunch of physical phenomena that one encounters every day depend on behaviors described by higher derivatives. Some examples have already been given in the thread.

Yeah, for instance the streaks ("fingers") that rainwater forms when flowing down a window glass are described by the 4th order equation I mentioned earlier. Usually when there's this kind of higher order behavior somewhere, the differential equation is derived with an assumption that the solution function is a slowly varying function of position - therefore the function and its derivatives are not going to change a lot when moving a distance of 4 finite difference step lengths.

- #28

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Of course, saying "physics would be so much harder if the fundamental laws involved third derivatives" isn't really a satisfying reason why. It's not like God designed the universe so that beginner physicists wouldn't be overwhelmed.

- #29

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If there were higher-order derivatives involved, then you would need a different notion of phase space (where you would specify position, velocity and acceleration of each particle). Such a change would make a huge change to a lot of the formalism of physics (both classical and quantum). I don't know whether it's been studied much.

Actually, it occurs to me that you can formally treat higher-order derivatives as first-order derivatives involving more complex objects.

For example, suppose you have a third-order equation of motion:

[itex]\frac{d^3 x}{dt^3} = F(x, \dot{x}, \ddot{x})[/itex]

Then you make that into a first-order equation for a vector-valued object:

[itex]\frac{d \vec{x}}{dt} = \vec{Q}(\vec{x})[/itex]

where [itex]\vec{x} = \langle x, \dot{x}, \ddot{x} \rangle[/itex]

and [itex]\vec{Q}(\vec{x}) = \langle \dot{x}, \ddot{x}, F(x, \dot{x}, \ddot{x}) \rangle[/itex]

There's probably some sense in which that's cheating, but I'm not sure formally how you would define cheating.

- #30

Chestermiller

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Another example in fluid mechanics is the biharmonic equation for the stream function in creeping flow past a sphere.

- #31

swampwiz

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I'm asking why there is no higher derivative than two in physics ? I never encountered a third (time or space) derivative in physics.

Have you some litterature about this?

Thank you.

Regards.

I would say the short answer is that the all phenomena can be represented as a perturbation upon a differential equation having constant coefficients. The solutions of such a differential equation map out to functions that are the exponential functions (i.e., of the natural logarithm base, e) of the input variable scaled by the values that are the root of the polynomial equation that corresponds to the original differential equation such that the order of the differential is the power of the polynomial term (this is standard material covered by a course in differential equations). Now, for a polynomial with real coefficients, the roots must either be real values or pairs of complex conjugate values, and those pairs when multiplied together yield a quadratic factor; hence whatever is modeled mathematically can be resolved down to a function in which is no more than quadratic in nature, and hence, all that is needed to describe it is the use of at most the 2nd differential.

- #32

Eric Bretschneider

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The Blasius equation relates to boundary layer flow on along a semi-infinite plate

2f''' + f''f = 0

2f''' + f''f = 0

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