d3x/dt3

[Gonzolo]

The differential equations that are mostly used in physics are second order, so I am wondering about the third order (or more)? It is clear that in real life, like when driving a car, acceleration changes many times and continuously, but do people ever add d3x/dt3 to Newton's second law? Is it ever necessary?

[Locrian]

Well you don't see it added to Newton's second law because Newton's second law isn't defined with it.

The third derivative certainly has useful meaning, but you really don't see it in fundamental principles because fundamental forces aren't known to change in a measurable way without some other variable changing.

For instance, if you changed the voltage on an anode the position of the electrons moving to it would not be your usual square graph of distance over time. However, that doesen't mean there is a d3x/dt3 in the electromagnetic equations; maxwell's equations didn't change, just the situation.

[jcsd]

It's important in some areas of engineering especially rollercosater design (I saw program on the telly about this) and it's usually called 'jerk' (though it is also called 'jolt' and 'surge'). Infact sometimes even higher order derivatives of postion and time are useful and are also known by names (such as jounce for the the fourth derivative).

edited to add: you may suprised to learn on a rollercoaster in general even the jounce is not constant.

[Galileo]

It's not necessary to consider higher order derivatives.
Galileo (not me) and Newton ofcourse knew that position is relative furthermore
they have shown experimentally that velocity is also relative. You wouldn't
know if you were moving if you are in a train (or in Galileo's case a ship) if
you didn;t look outside. All physical measurements give the same results.
Contrast the case when accelerations are present.
Acceleration is NOT relative, so that should be your object of study.

[Integral]

In engineering I believe that the Beam equation is 3rd order.

Edit:
Humm.... a quick web search shows a 4th order equation.. no time for further research at this time.

[HallsofIvy]

Yes, integral, the beam equation and, more generally, equations in elasticity are 4th order.

[jamesrc]

As far as the names of those derivatives go, for kinematic quantities, as far as I know it goes:

velocity, acceleration, jerk, snap, crackle, pop, (never heard of a term for anything higher)

I've heard of jolt and jounce (I believe as UK alternatives for jerk and snap). And I guess it's so rare in practice to see those derivatives that the names aren't official (at least beyond jerk and definitely beyond snap). Anyway, when I saw the thread I though I'd just point that out as a point of interest.

[Chronos]

Just in case fourth order derivatives are too boring, You can always try your hand at finite element analysis with the Gear algorithm, which uses fifth order derivatives. Personally, it only takes third order derivatives to put me into a coma.

[zefram_c]

A well known third order equation is the Lorentz-Dirac equation; it describes the motion of a classical relativistic point charge interacting with its own fields. Its solutions show problematic results: either runaway acceleration (ie the particle spontaneously accelerates to arbitrary speeds), or non-causal acceleration (the particle accelerates before an external force is turned on). The latter occurs for an interval of time roughly equal to the time light takes to travel the 'classical radius' of the object.

[Gonzolo]

Wow! Thanks! I'm all primed up to learn how to solve them now!

[arildno]

In fluid mechanics, it is occasionally convenient to work with the vorticity equation expressed in terms of the stream function; this is a 4th order differential equation.

[Gonzolo]

Actually, I'm not sure I'll go there anytime soon. But it's nice to know they are not forgotten and actually useful.

[arildno]

I tried to forget it right after my exams..:biggrin:

[dedaNoe]

I think that a phenomenon in physics cannot be described with differential equation.
If you have dx=Vdt then you have actually differentiated x=Vt assuming V=const.
I just don't know any other way to come up with dx=Vdt except by derivating x=Vt assuming V=const.
This means that your describtion of the phenomenon represents special case.
If you have complete differential equation (dx=Vdt+tdV) then you actually have nondifferential one (x=Vt).

[Locrian]

You are integrating incorrectly. Firstly, you aren't including the limits of integration, such as they might be. Secondly, you are assuming in your integration that V is a constant, and then acting suprised when it turns out that way.