In classical determinism we need to know $2n$ quantities of our system and the equation of motion to predict it's future. In Lagrangian mechanics this is equivalent to knowing $q$ and $\dot q$, the generalised coordinate and the generalised velocity. However if we knew the acceleration $\ddot q$ would this be enough?

My reasoning is that the acceleration is a function of both $q$ and $\dot q$ such that $\ddot q^i=\ddot q^i(q^1,\dots ,q^n;\dot q^1,\dots ,\dot q^n)$, therefore wouldn't we have to know both $q$ and $\dot q$ anyway?

Therefore can classical determinism be encapsulated by knowing the accelerations, therefore implicitly knowing both $q$ and $\dot q$? Does this statement violate the conditions necessary for solving PDEs and ODEs? It seems wrong however that this is only $n$ quantities, but in reality we would be getting $3n$ quantities. The reason I ask is I wan't to wrap classical determinism up in a way that students would understand in one sentence.

  • $\begingroup$ Where does the dependence of $\ddot{q}$ on $(q, \dot{q})$ come from? This is not valid in the general case. $\endgroup$
    – auxsvr
    Jul 12, 2015 at 13:06

2 Answers 2


Observe that, knowing $\ddot{q}$, to get $\dot{q}$ and then $q$ you have to integrate twice. This introduces $2n$ integration constants you have to know to fully describe the system, which is the same amount of freedom you get when solving the Euler-Lagrange equations, where you need initial conditions for $q$ and $\dot{q}$.


I'd like to point out that knowing $2n$ quantities and the equations of motion are not enough to determine the solution. Even at the level of $L=T-U$ for just one particle ($n=1$).

Consider $T=\frac{1}{2}m\left(\frac{dx}{dt}\right)^2,$ and $U=-C\frac {9m}{2}x^{4/3}.$ Then, for your equations of motion, you get $m\frac{d^2x}{dt^2}=\frac{9m}{2}C\frac{4}{3}x^{1/3}.$

For many pairs $(x_0,v_0)$ you get a unique solution. But there are many solutions for the pair $(0,0)$ in particular the solution $x(t)=0$ works and so does the solution $x(t)=C^{3/2}t^3.$ Among many others.

So you need additional assumptions to have $2n$ quantities determine a unique solution. If you attempt to wrap up classical determinism in one sentence by concluding something that is not true, then you will have failed no matter how memorable your sentence.

But let's get to the heart of acceleration. To have something like $ma=F=-dU/dx$ you need to have a $U$ that is differentiable in space, and a position $x(t)$ that is twice differentiable in time. Otherwise it doesn't make sense. And if that is all you require, then you don't have determinism. Full stop.

If you want to make further requirements, such as that your potential has even more spatial derivatives or that your path has even more time derivatives, then you are restricting your allowed potentials/forces or your allowed solutions.

Newton's first law already restricts your solutions. It singles out the solution $x(t)=0$ because it says that if the object is at rest ($v_0=0$) and the force is zero, then the object stays at rest ($x(t)=0$). But most people want to ignore what Newton's law literally says. And then you have to accept a lack of determinism in classical mechanics, or else restrict yourself to avoid considering the potentials/forces where Newton's first law tells you which solution of the many solutions to $F=ma$ is the true solution.

If I were teaching students I'd tell them that Newton's third law tells you to ignore potentials/forces that don't conserve momentum. In other words, Newton's third law excludes hypothetical force laws before you even look at their dynamics.

Then I'd tell them that Newton's second law tells you how to associate dynamical solutions with a particular force law so that you can experimentally test whether a force law agrees with observation and experimentation. In other words, Newton's second law tells you how to make a force law empirical outside of statics.

And finally I'd tell them, that Newton's first law gives you a unique solution because in the instantaneous rest frame of the object, the stays at rest if there is no force in that frame. In other words, all $F=ma$ deviations from uniqueness for scalar potentials locally look like the example above and Newton's first law singles out a unique solution amongst the many that satisfy $F=ma$ so Newton's first law allows us to predict motion knowing the initial conditions, and without Newton's first law we could not do so.

But then uniqueness is a physical principle on par with conservation of momentum.

So we postulate uniqueness by calling some solutions better than others. And there is good reason. We want to say that if something accelerates there should be a cause, a force, responsible. Contrary to uniform motion or remaining at rest, needs no cause (no force) or explaination. If you don't insist on Newton's first law, and only use $F=ma$ then it is entirely posible for something to start to move in any direction and at any moment no matter how long it sat somewhere with no force acting on it and no change in the potential.

So we postulate a whole new principle to exclude that, so that things happen for reasons. No matter how popular it is to pretend that Newton's first law has nothing new to say that isnt' already said by $F=ma,$ it does. It tells us we can have uniqueness.

As for your question about whether knowing the accelerations means you know everything. Even in the case of zero acceleration, you don't know where the particle is, or what its velocity is. So knowing the acceleration simple does not tell you the position or the velocity. And this is actually where frames comes from. If you organize your physics to be about accelerations, which can't tell you where something is or what velocity you have, then you can have frames with any origin and moving with any relative velocioty comapred to each other.

So it is a focus on accelerations from the that leads naturally to frames since the accelerations can't tel lthe osition or the velocity they hold equally well relative to a family of frames. So frames are part fo the second law, not the first law. I would go as far as to say that the existence of frames is part of the second law for the exact reason that you are trying to extract empirically testable results from a force law, so you ned a frame to have that. And the first law still has content, it gives determinism where there was not determinism before. It makes things have reasons for what they do.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.