# Why no higher derivative in physics?

• I
• Ksitov
Ksitov
Dear all,

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

Thank you.

Regards.

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

Ksitov

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.

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

Gold Member
I'm asking why there is no higher derivative than two in physics ?
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.

• Stephen Tashi, symbolipoint and cnh1995
Mentor
2022 Award
The fun part is, that either the first and second derivative is considered (sometimes third, as in curve sketching) or it is required to have infinitely often differentiable functions. So implicitly the other derivatives are often used, as smooth functions play a big role in physics.

• cnh1995
cosmik debris
Third and higher derivatives of position are used in things like camshafts. In order for the valve to have the maximum area under the valve changes in acceleration have to be carefully managed. As an example when the cam goes past the maximum valve opening the acceleration changes sign.

Cheers

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

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Gold Member
There are equations in the theory of elastic bending of plates where there's a double Laplacian operator ##(\nabla^2 )^2## acting on an unknown function - a fourth order PDE. Another example is the equation of motion of thin liquid films for low-Reynolds number flow.

Gigaz
The question is nevertheless reasonable. Derivatives beyond 2 are really rare in physics. I think ultimately the reason for this is Locality. The physics at (x,t) is determined mostly by events at (x+dx, t-dt). The further away stuff is, the weaker the interaction. But higher derivatives ultimately describe the behaviour of a function relatively far away from the current point.

Mentor
2022 Award
But higher derivatives ultimately describe the behaviour of a function relatively far away from the current point.
This is not true. Differentiability is always a local property, no matter how often it is possible. "Relatively far away" is definitely a misinformation.

Asymptotic
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.
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.

There are plenty of situations, where higher derivatives appear. But maybe the scope of the question can be reduced to why there are usually no higher derivatives than the first in the Lagrangian? This will turn up quite a lot of results on Google.

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

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• Telemachus
Gigaz
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.

Mentor
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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 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.

• Telemachus
Gold Member
I have made simulations of the thin-film flow system that is described by a 4th order PDE, and the high order doesn't cause any significant trouble in the implicit finite differencing - the shape of the liquid surface approaches a paraboloid of revolution as it sets into static equilibrium (just as it physically should do to minimize surface energy).

• Telemachus
Gigaz
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.

First off, as previously mentioned, there are third and higher derivatives in physics. However, you are completely correct that they are rare-- jerk is occasionally used but beyond that I've never really seen a problem using snap (jounce), crackle or pop; perhaps their rare use is why they were named after a cereal slogan. I do not know exactly why they are generally trivial, but I do have a guess (of course, nobody can know exactly why something in physics exists).

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=F0⋅d+Yt⋅d for a constant yank and force not varying wrt d, and in the general case W=∫∫Ydt⋅dx. I find the equation in terms of force much more appealing to use, don't you agree? In fact, you can also express kinematics for a constant jerk, but at that point I think I'd rather use calculus.

Homework Helper
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.
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).

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?

rumborak
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.

• Isaac0427
Homework Helper
You can use the limit of Delta going to zero, but that is evading the point, as the delta is never *actually* zero.
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.

rumborak
It is not improper, since the delta is never zero. Infinitesimals are not zero, and that is the important point in the argument. @Gigaz has a perfectly valid point here.

Homework Helper
It is not improper, since the delta is never zero. Infinitesimals are not zero, and that is the important point in the argument. @Gigaz has a perfectly valid point here.
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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• Stephen Tashi
Mentor
2022 Award
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 say you need points farther apart is a contradiction to the concept of a local phenomena and very misleading.
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 ...
Yes, in math.
... implicitly suggesting that there are two different concepts of differentiability in physics and in mathematics.
... makes it even worse. This is not what differentiability means.

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.

However, it is true that terms involving higher-order derivatives tend to get dropped whenever possible, and hence it's rarer to see them retained (or even acknowledged) in a problem. Hence, the real question is why can we often get away with dropping those terms? Here it may be valid to talk about ratios of terms, expressed in units of deltas, although it's probably getting into the weeds to argue about the effect of molecule sizes on higher-order derivatives in N-S...

• hilbert2
Gold Member
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.

• olivermsun
Staff Emeritus
I think that the original question could be sharpened up a bit. It's not so much that higher derivatives are never used, but that the fundamental equations of physics don't involve them. Newton's equations of motion, Maxwell's equations, Einstein's field equations, Schrodinger's equation. In the case of Newtonian physics, this means that the state of a system, no matter how complicated, can be described by its trajectory in phase space, where phase space specifies the location and momentum of each particle. Since momentum involves velocity, the trajectory in phase space involves time derivatives of momentum, which means second derivatives of position. 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.

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.

Staff Emeritus
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:

$\frac{d^3 x}{dt^3} = F(x, \dot{x}, \ddot{x})$

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

$\frac{d \vec{x}}{dt} = \vec{Q}(\vec{x})$

where $\vec{x} = \langle x, \dot{x}, \ddot{x} \rangle$

and $\vec{Q}(\vec{x}) = \langle \dot{x}, \ddot{x}, F(x, \dot{x}, \ddot{x}) \rangle$

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

Mentor
I have made simulations of the thin-film flow system that is described by a 4th order PDE, and the high order doesn't cause any significant trouble in the implicit finite differencing - the shape of the liquid surface approaches a paraboloid of revolution as it sets into static equilibrium (just as it physically should do to minimize surface energy).
Another example in fluid mechanics is the biharmonic equation for the stream function in creeping flow past a sphere.

swampwiz
Dear all,

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