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What would happen if instead of $F=m*d^2x/dt^2$, we had $F=m*d^3x/dt^3$ or higher?

Intuitively, I have always seen a justification for ~1/r^2 forces as the "forces beeing divided equally over the area of a sphere of radius r".

But why the 2 in $F=m*d^nx/dt^n$ ?

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The two answers to appear so far are certainly correct, but they may not feel very satisfying. Yes, it's true that the fact that second derivatives are involved can be seen as arising from the fact that you need to specify both $x$ and $v$ to give the initial conditions for a dynamical system, but I'm not sure that that really answers the "why" question. Why don't you have to specify more (or fewer) derivatives in the initial value problem? Of course, this is a common problem with "why" questions: they just lead to another "why" question (as anyone who's talked to a toddler can testify). –  Ted Bunn Feb 2 '11 at 16:43
    
@Ted: Agree. In my opinion the only "why" questions that can be answered without troubles are about mathematical formulations of observed phenomena. –  Kostya Feb 2 '11 at 16:52
    
Besides, F=ma' has inconsistent units. It can never be correct. –  drxzcl Feb 2 '11 at 22:54
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6 Answers

up vote 7 down vote accepted

In statics, you can still have a force without acceleration so $F$ is independent of $a$. $F$ is the cause of the change in the position of an object initially at rest in some frame. To give it physical meaning, you have to define how it's to be measured and one way would be to define 1 unit of F causing one unit of compression in some standard spring.

Now if $F$ causes a body at rest to change its position, then over a time dt the postion has changed by dx. Your job as a physicist is to construct an equation relating F to the change in velocity of the body.

So with all this in mind, what would happen if $F=m*d^3x/dt^3$ ?

It would mean that even though $F$ is the cause behind the change in velocity of a body, there are some changes in the velocity possible where $F = 0$ such as for $a = const$. You would end up with particles accelerating in arbitary directions for $F = 0$.

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Can this be taken one step further and say that if $F=0$ and $a = k$ then energy conservation is violated? Or would we merely observe a different conserved quantity? –  Sklivvz Feb 3 '11 at 9:14
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@Skilvvz, the energy and momentum conservation theorems come from multilplying f by dx, dt. If you do it for $F=m*d^3x/dt^3$, you end up with new "energy" and "momentum" conservation theorems. For "momentum", $fdt = m*da, F1*t2 - F2*t1 = m*a2 - m*a1$. By assuming action = reaction and the forces are all internal, you get $\sigma ma = const$. Same thing for "energy" but more involved which I'll leave to you ;) –  John McVirgo Feb 3 '11 at 23:33
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Because the second derivative is fully determined by the external force, it is an experimental fact. The force can be directly measured with a dynamo-meter. It is not an abstraction. So the acceleration is known as soon as the force is known and vice versa. The trajectory depends also on the initial conditions which are independent from the force but reference frame-dependent.

In CED there is an equation (Lorentz-Abraham one) with a third derivative in time. It has non physical (runaway) solutions.

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by CED you mean Classical Electrodynamics, right? –  user346 Feb 2 '11 at 16:37
    
Right, CED = Classical Electrodynamics. –  Vladimir Kalitvianski Feb 2 '11 at 16:43
    
+1 Ultimately, all laws are laws because they have been proven experimentally. –  Noldorin Feb 2 '11 at 18:28
    
How does a dynamometre measure a force directly? As far as I know it measures a distance when equilibrium is reached, so it measures velocity (in that it ensure it is zero) and distance. –  Sklivvz Feb 3 '11 at 9:10
    
I am not an experimentalist and I am not good at dynamo-meters, sorry. –  Vladimir Kalitvianski Feb 3 '11 at 15:32
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Rephrasing in a Lagrangian manner, this is equivalent to asking about higher-derivative theories. See: Why are only derivatives to the first order relevant?

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There is a deeper reason for $F~=~\frac{d^2x}{dt^2}$ Within the Galilean group it is an invariant with respect to all changes of frame $x’~=~x~+~vt$. The acceleration of a body is not something which can be made to vanish by boosting to another Galilean frame as $$ F’~=~\frac{d^2x’}{dt^2}~=~\frac{d^2x}{dt^2}~+~\frac{d^2vt}{dt^2} $$ where for constant $v$ the second term is clearly zero. The next higher derivative $dF/dt~=~mda/dt$, called a jerk” is also invariant, as are all $d^nx/dt^n$, but the acceleration contained in $T^2_p$ is the lowest element on the jet $T^n_p$ $n~\ge~2$ which is invariant. Further, odd powers of $n$ would not be time reverse invariant under $t~\rightarrow~-t$

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It is because the evolution of mechanical system is fully determined by initial coordinates and speeds. Therefore, your equation must be of second order, otherwise setting initial accelerations and "speeds of increase of accelerations" would be necessary.

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...which we confirm experimentally to the classical approximations. There is of course no reason why one couldn't simulate a fantasy universe in which positions + speeds are not the entire state. –  romkyns Feb 2 '11 at 20:00
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But the fact that the evolution of the system is determined by the initial coordinates and velocity follows from the fact that second order equations are used. So it cannot be given as a reason. –  MBN Feb 2 '11 at 20:29
    
The only "reason" in physics is the consistence with experiment. Question was not about reason, but about justification. –  Kostya Feb 3 '11 at 15:08
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Another reason is that if n did not equal 2, some of the symmetries of the Newton equation would be lost. For example, classically, physics on a microscopic scale is time reversal invariant. We can see this from the Newton equation because if x(t) is sent to x(-t), the 2nd time derivative ensures that the negatives cancel. If n was an odd number, we would not observe this symmetry.

If n was smaller than 2, then Galilean transformations would not leave the equation invariant. If we send x(t) to x(t) + vt, then the second time derivative kills the vt term on the LHS. If n=1, then this would not be possible, and relative velocity would be meaningless (even non-relativistically). If n was some number larger than 2, then transformations that send x(t) to x(t) + b(t^m), where m is less than n, would be a symmetry of the Newton equation. But relative accelerations, jerks, etc. produce observable discrepancies, so that should not be possible either.

Again, this may not really answer "why" in the sense that you are asking, but the fact that the equation matches observations is enough in science to justify it. The Newton equation is basically experimental fact, like Vladimir says.

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