Suppose you have a smooth hill defined by two parametric equations:
e.g.
x=asinq

y=acosq A particle is resting at the top of the hill, about to roll down it. Is there a way of working out what the position of the particle is after time t.
I hope this is clear, I'll try to elaborate if it isn't.

Thank you
James Thimont

Do you know anything about intrinsic coordinates? If your answer is yes, then you can state the problem in a similar way, and will be easier to tackle by resolving tangential and normal (or energy) components.

I'm not sure how efficient this method is...

Consider the motion of the particle between theta = 0 and theta = x (using the theta in the parametric equation).

The loss in gravitational potential energy is mga(1-cosx) and the gain in kinetic energy is

1/2 m(aw)²


Because the velocity of the particle is aw, where w is dx/dt.

By conservation of energy:

2g(1-cosx) = aw²

The solution of this differential equation is:

integral of 1/sqrt(1-cosx) wrt x equals
sqrt(2g/a)t + C

where t is time and C is an arbitary constant.

Using the substitution 1-cosx = 2sin² t the integral is:

ln tan x/4

so the general solution is:

x = 4 arctan (Ae^sqrt(g/a)t

where A is constant.

So this gives the angle the particle descends to after a certain time.

I think I'd better stop here to check that all that's right. By the way do we have any initial conditions for the problem? By what means does the particle deviate from the centre?

Thanks,

Michael

If the initial conditions are: the particle is released from rest at a point where q = x₀ where x₀ is very small then a good approximation for x, the angle the particle moves through after time t, is:

x = 4 arctan(x₀ e^sqrt(g/a)t /4)

because tan x₀ is approximately x₀ for small x₀ .

Taking the limit as x₀ tends to 0 we get:

x tends to zero for every t.

So if we take a particle on a plane and displace it very slightly then it will take a very long time to reach any fixed point. If you displace it even less then the time will continue to go up. As the displacement tends to zero, the time taken to get to any fixed point will tend to infinity.

Interestingly you could also try this in reverse. If you throw a particle onto a smooth sphere tangentially with exactly the right speed, it is possible for it to take an infinite amount of time to get to the top!

Michael

In fact, a particle resting on top of any hill will always take an infinite time to reach any fixed point (if displaced by an amount that tends to zero) as long as the curve of the hill in the plane in which the displacement takes place is differentiatable at the top and has derivative zero. This can be shown fairly easily by energy considerations. If you're interested I'll post a copy of my reasoning.

Michael

Hi there !

I've just been in a lecture on differential equations, and it's actually quite interesting to note the same kind of behaviour also occurs when we consider a pendulum which is not restricted to moving through small angles. If you give it exactly the correct initial velocity, as the pendulum approaches the top of its trajectory, its velocity decreases such that it gets nearer and nearer the top without ever actually reaching it. The parallels of a circular trajectory and gravitational forces are quite obvious (?!).

Bye 4 now !


Andrew R

Yes, that is quite strange. In fact the energy equations above would be identical for a pendulum expect that the a² would be replaced by the moment of inertia divided by the mass.

Also I suppose the same thing must happen if you try to overbalance a chair on a rough surface but give it the border line velocity.

In fact it is tempting to guess that the same phenomena will occur whenever you push a system towards a high energy equilibrium with exactly the right velocity, as long as the track the centre of mass can move along is differentiatable. But I'm not sure about this yet as I haven't thought about it much.

Michael

It isn't that strange...

Suppose your system is time reversible (ie. it look like it satisfies the same motion equations if a video of it is played backwards) which is a very normal condition for simple physical systems.

If you can hit a particle to an equilibrium point so that it gets there and stops in a finite time then what happens when it is played backwards? Well the particle sits happily on the equilibrium point for ages and then decides to wander off. Clearly this is highly unsatisfactory for many physical models and so the behaviour described just can't happen.

AlexB.

By strange I meant slightly counter-intuitive rather than flawed. Anyway I'm not sure about the last part of the argument. At the top of the hill it is true that the resultant force on the particle is zero. But if it does accelerate from rest here then the initial instantaneous acceleration is zero (as the velocity starts off at a minimum). Therefore F = ma holds and Newton's 1st and 2nd laws are not violated. Of course a bit later the particle will be accelerating but now there will also be a resultant force.

In fact Newton's laws do not uniquely describe what would happen ? they give multi-valued results here. It is impossible to predict exactly when the particle will start deviating ? which is interesting as Newton's laws are normally thought of as being deterministic. With a sphere it does not matter (we have established that the time the particle takes to get to any fixed point is infinite anyway so you could not observe the particle wandering off by itself).

However I can give an example when the scenario described above does occur ? if you have a hill which is formed from two lines crossing, each at 45 degrees to the downward vertical. If the hill was perfectly symmetric and the particle rests at the very top then any time the particle could fall down either side without any external forces acting.

This all reminds me of a thought I had when I met the exponential function for the first time almost exactly one year ago. I thought: is it possible that we could have an exponential function that stays at zero for a while and then rises up and takes positive values? My reasoning was that clearly, staying at zero for the whole time was one possible solution. But suppose we had an increasing curve that touched zero at the origin and had gradient zero here. It is not obvious that dy/dx is not proportional to y. After all at the origin the curve has no value and is not increasing. Slightly further on the curve has a positive value and is increasing. Well before long I could show that exponential functions can't behave in the way I described although there is nothing wrong with the principle.

Exponential functions aren't exactly the same as a particle rolling down a sphere. But one thing that they do share with the distance the particle has travelled is that the rate of change of the value increases with the value itself.

An analogous thing to my exponential function rising from nothing actually does occur if you have a particle balanced on two perpendicular lines. One solution to the resulting differential equation will be y = 0, another will be y = 0 for 0 < x < 10 and beyond 10 y increases.

One thing I'm still interested to know is under exactly what circumstances a system will "snap out" of an unstable equilibrium in a finite time if displaced by an amount tending to zero. For a particle, the answer appears to be all to do with whether the curve of the particle's path is differentiatable - is there a similar condition for other equilibria?

Thanks,

Michael

Firstly - my reply was to show why it is actually intuitive...

What my argument says is that if you can hit a particle to an equilibrium point so that it reaches there with 0 velocity then it must take infinite time. Otherwise playing the video backwards would clearly violate Newton's laws.

Second, you seem to have a little misunderstanding of solution of the differential equations that turn up in the sorts of situations that we are dealing with here. If a system is independent of time (by which I mean the the force on a particle is determined only by its position and velocity but doesn't explicitly depend on time - which all the ones described above are) then if a particle is at a point and is at rest and there is no overall force on it then it will stay at that point forever. So if I start a particle at rest at an equilibrium point then it will remain there forever. There is absolutely no uncertainty. In order for it to start moving it would have to be acted on by a force and as the system is independent of time then this would have had to be there initially.

The same thing holds for your two lines at 45 degrees case. At the top at rest the particle will remain there forever.

You may be confused by the fact that this is a mathematical situation that can not be set up in practice. In practice there are random forces exerted by the atmosphere which would deviate the particle from the equilibrium point and from here it would roll down. Now the atmosphere forces are 'random' and so you can't predict the initial direction of motion.

For your "y=0 for 0 < x < 10 and 10y beyond" --- are you sure that this satisfies the differential equation. It doesn't look differentiable at x=10 and certainly not twice differentiable. Please check this!

I believe that you are also constructing the differential equation for the system by resolving forces in the diagram. How can you do this at the point where the two lines meet? There is no well defined concept of normal and tangent for you to use. Exactly which way does the normal force act?? What about if the diagram is rotated a little??

AlexB.

I've thought of a much simpler example which shows that if the force is determined uniquely by v and x then the particle can still decide to wander off from equilibrium by itself from rest. We are considering 1-D motion with x the position and v the velocity.

Let the force be equal to m xsqrt(v) where m is the mass. This implies that a, the acceleration, satisfies:

a = sqrt(v)

For initial conditions I'll say when t = 0 then x = v = 0.

Now what can the particle do? It could stay in equilibrium forever (x = 0 satisfies the equations). Another solution? x = 1/12t³ also works. Or alternatively the particle could decide to stay at the origin in equilibrium for the first 25 seconds (for example) and then afterwards follow the path x = 1/12(t-25)³ .

It has infinitely many options. Classical physics is unable to answer the question: how long will the particle take before it decides to wander off out of the equilibrium?

The crucial thing is that the particle is never actually accelerating when it is at the equilibrium position. The acceleration comes later.

Here is why I thought that the perpendicular lines will behave similarly. I want to consider an idealised mathematical model and ignore any real-world complications. By the way, it is true that at the top of the hill, it isn't obvious which direction the normal force is in. I just want to assume it is in the opposite direction to the weight in this model by symmetry.

Now consider rolling the particle up the hill with EXACTLY the right velocity to just get to the top. (Only something you could do in this sort of model.) In other words, 1/2mv² = mgh where h is the vertical height of the hill and v is the initial velocity.

So what happens? The particle must get to the top in a finite time or not at all. (As deceleration everywhere before the top is fixed.) But if it did not get to the top then it must have stopped at some point before the top and this contradicts energy considerations.

Therefore after a finite time the particle has reached the top. Here it be at (at least instantaneous) rest. View this backwards ... and the particle will have started accelerating despite being at equilibrium and at rest.

By the way I would be interested if you could give your proof that if dy/dx = y for all x, and y = 0 when x = 0 then y = 0 for all x; I seem to remember mine was very tedious.

Thanks,

Michael

I've not got much time so I'll answer the latter questions for now...

My argument about time reversibility needs that the system is physical... ie. differentiable sufficiently many times at certain points. Also I was implicitly assuming that force was a function of position only (as all the examples you were discussing were of this form - my fault for not making this clear). It was not as put rigorous but just a heuristic to explain the infinite time property.

The problems that occur when normals and tangents do not exist lead to silly things. They aren't real world complications but mathematical complications. Nothing nasty like this ever happens in the real world! It is only people as perverse as mathematicians who consider them!! Again I emphasise that if you are going to draw conclusions about 'classical mechanics' then you should really only consider the type of situations that come about in the realm of classical mechanics.

For your sqrt(v) (which I assume you mean to be sqrt(|v|) otherwise you'll run into large negative square root problems) case you are right that the things you give are solutions (except for the t³ over all time one which becomes |t|^3). However physicists (and anyone using classical mechanics) would rule the nasty ones out by being 'physically unreasonable'. Again this is hard to do as a concrete mathematical axiom but you can sort of view Newton's first law as being it. Basically it is saying that if you are sitting at a force equilibrium then you can only escape if an actual force is applied at the position you are at. Does this make some sense??

I know that you are asking mathematical questions and I'm giving you physical answers. I am actually a mathematician! But when it come to physical theories I'm a very big fan of not being too mathematical about things... as recent developments in the two fields have shown --- you'll lose a lot more than you gain.

If you totally drop the physical analogy and just say that you want the solutions to a = sqrt(|v|) then I'm happy again. And in this case you get a lot because your RHS has nasty singularities in at the origin. The study of differential equations with singularities and problems is a very important branch of mathematics and I'm sure there is someone else here who is more qualified to tell you about them than me.

AlexB.

Hi!

I was interested to hear your view on the independence of maths and physics. Out of interest, what are these "recent developments in the two fields" that support your view?

Anyway, I admit that a = sqrt(v) isn't a great example as I can't think of an (idealised) situation that this could model. And as you say, the force depends on v.

So a far better example is a = abs(sqrt(x)) sgn(x) [where sgn(x) returns 1 for positive x, 0 for 0 and ?1 for negative x]. a = acceleration and x = distance travelled.

(I'm afraid I haven't been able to work out how to get the modulus symbol, so I'll use abs(x) for the meantime.)

With x = v = 0 at t = 0, two solutions immediately emerge: x = 1/144 t⁴ or x = 0. (Of course these are not the only two). What's more this can actually be modelled with a particle rolling down a hill. I haven't worked out the equation of the curve, but the simplest way of doing it would be to resolve forces horizontally, so k/(1+k² ) is equal to sqrt(x) where k is the gradient and x is the horizontal displacement.

If you now release the particle from rest at the top of its hill it will stay still for an indeterminate amount of time. In fact you cannot even derive a probability distribution for when it will move ? you simply do not know anything about when it will start. [This is an interesting idea in itself.] I am fairly certain that it has an evens chance of falling down either side of the hill, given that it will move at all.

Now I'm not suggesting that this phenomenon can ever be observed in the real world ? the main two reasons for this are:

1) It's impossible to have a hill with any exact equation because its measurement will always involve an error. Also as you pointed out there will be atmospheric forces and vibrations in the surface etc along with variations in gravitational field strength. Sadly "nearly ideal" is not good enough to observe the phenomenon.

2) The classical laws of physics are nothing more than approximations anyway. Even if there was no experimental evidence to counter them, you could never be sure that there aren't errors in the 10⁴⁰ th decimal place of its predictions ? and this is enough to spoil the effect.

What I am trying to say though is that in the axiomatic system where the classical laws of physics are true and setups are ideal, things aren't always deterministic. I am not at all sure about why you should rule out any solutions because of Newton?s 1st law. What Newton's first law actually states is that if the resultant force is zero the acceleration is zero, thus making it a special case of the 2nd law. But here there is a rather subtle difference between accelerating and speeding up/changing direction ? in much the same way as there is a difference between a curve increasing and having positive gradient. Anyway, I'd be interested to hear what you think.

Hope you had a good Christmas,

Michael

I'll do the stuff about recent developments at the end (if I have time!)...

[Probability distribution] You are right that if you let things behave in the way you say then you can not derive a probability distribution so you can't really say that there is an even chance of it doing anything as that really doesn't mean anything! There do exist slightly nasty things to do with probability distributions (eg. there is one which exists and is symmetric about x=0 but doesn't have a mean).

[Points 1+2] I agree.

The way you write Newton's 1st law it is a special case (except when m=0) of Newton's 2nd law but I'm trying to say that this may not be the best interpretation of it. There quite a bit of philosophy behind his laws - for example do you think that the second law says that 'Force is equal to mass times rate of change of velocity' or 'Force is equal to the rate of change of momentum'. Because these two things are certainly different if your object is changing its mass (like in a rocket). So perhaps Newton's 1st law can be taken to mean 'Object will not change its speed unless acted upon by a non-zero force' which could allow us to keep this thing at the top of the hill. Or perhaps it can even be taken to be a restriction on the types of force that you are allowed to apply the other laws to. Most physical and mathematical laws come with restrictions attached ('you may only use this if it is continuous / smooth / holomorphic / analytic ...') and so Newton's 1st law might be thought of as this sort of thing.

If you do allow yourself to use any type of force then you do get the solutions you are talking about, I'm just trying to point out that this isn't really in the 'spirit' of Newton's theory and that if it were to be written down formally mathematically then this system would probably rule out these nasty cases.

---

Maths and theoretical physics share some very large bits in common and when ideas have traded places (in either direction) over the years vast leaps in understanding have occured. It sort of happened last for GR and quantum theory when mathematics of manifolds and Hilbert spaces which where invented by mathematicians because they were interesting to study turned out to be vital to physics.

Why should an object that is simple to describe mathematically and interesting mathematically turn out to be fundamental to a description of physics? This is a very big philosophical question: Why should maths be any good whatsoever at describing physics? It wouldn't be too dificult to come up with some physical laws that have no 'nice' approximations and so no-one would ever be able to 'crack' them... but our current laws have some absolutely beautiful approximations. The symmetry, neatness and beauty of the mathematical descriptions is astounding (the more you think about it!). If I ever find a decent book on this I'll let you know but try and have a think about it. Perhaps intelligent technological life can only exist in a mathematically beautiful universe??!!

Currently there is a big trade going on in the reverse direction (physics to mathematics) via string and superstring theories. Many bizarre and diverse mathematical results turn out to be connected in this conjectured physics description. Areas of mathematics that have always been considered utterly different now seem to be basically the same! And very slowly mathematicians are beginning to see why. However, at the beginning there were very large mathematical problems with the physicists theory (it was totally non-rigorous). But it all turned out to be true - lots of things could be predicted if we pretended that certain integrals converged and we could swap them, or if two different infinities would cancel and so on. Some of these can be put on a rigorous footing but this makes them much harder to use and so less predictive.

So if we were to always be prefectly rigorous then we would have lost a lot of beautiful mathematics that has come about recently. There are many current conjecture that everyone knows are true but can't yet work out how to actually prove. But without this 'physics clue' I doubt that many of the conjectures would have ever been discovered.

AlexB.

Many, many thanks for your reply. I was very interested to read about how maths and physics affect one another. I agree that it is a very hard question: why should the physical laws be solvable mathematically? (Which I would prefer rephrased as, why should the maths used to solve physical theories be as easy as it is.) I think that there is an element of: our so called "pure maths" is shaped by the physical universe we live in. Most of the basic operations and axioms would not have been invented if the universe really was that different.

And also I can think of many examples when looking at mathematical problems physically helps. Mainly with second order differential equations (e.g. showing f''(x) = -sin(f) oscillates for f(x)) and finding functions to optimise a property. (Though of course these things are not nearly as abstract or sophisticated as what you were talking about, I guess the same principle holds.)

Anyway, the main reason that I initially pointed out that you do get solutions when the ball rolls down the hill from equilibrium was because I was not sure the time reversibility argument worked. (Where you say if the ball takes a finite time to reach the top of the hill, this means it could also fall from a static equilibrium which is impossible.) I still believe that you do have to work it out for each curve you meet.

Newton's first law taken literally will not always resolve the problem I think. For example you could have the hill in a upward accelerating lift rather than in a g-field and my argument for multi-valued solutions would still hold. Now Newton's first law does not apply directly as there is acceleration! (It does hold if you decide to believe in the equivalence principle but Newton doesn't say anything about this.)

I am not very comfortable with F = rate of change of momentum. After all if you expand this out you get a term in v dm/dt. The fact it includes v makes the force non-relativistic (because you need to assume that the mass that the object gains is moving initially with the same speed as your frame of reference).

Anyway, thank you for your time,

Michael

It is a little more than just physical laws being do-able with reasonably easy maths but this is also a good part of it - it is also why are the physical theories that seem right actually beautiful in an abstract mathematical sense. It is possible to propose nasty differential equations which can still be solved but for some reason nature doesn't do this. All the major advances in physical understanding seem to make the laws simple, symmetric and worth studying from an abstract mathematical point of view.

I'll see what I can find out about the f=ma vs. f=change of momentum problem.

AlexB.

Hi,

I've finally had time to work out an equation of a hill with the property we were discussing. It is the Maclaurin expansion (in abs(sqrt(x))s) of:

y = sqrt(abs(x)-x² ) - arcsin(abs(x))

I say the Maclaurin expansion just to remove the problem with the derivative at the origin.

As we agree, ideal situations like these are impossible. Suppose we now concede this, but stick to believing that Newton's laws govern the universe exactly, can we show that the universe is very unlikely to naturally encounter a scenario when there is a chance of choosing two routes consistent with F = ma etc? One prerequisite would be we need a particle to be in exact equilibrium in any inertial frame of reference (Galilean / freefall etc). I'm not sure how likely this is?

I see what you were saying about probability being meaningless here. The real problem is that there is no symmetry (like when you roll a die you can say that by symmetry each number is equally likely to come up, therefore the chances are 1/6). But you may still be able to say that by symmetry the chances of rolling down either side of the hill (given it rolls at all) is ½ - it just depends on the definition.

With Newton's second law, I think that F = ma always works, provided you use it on an isolated object under an external force. If the object splits up then the total force on the object will still equal its mass x the acceleration of its COM. F = d/dt(mv) also works if you're careful. I just find it horrible to look at as it contains a v as well as a dv/dt.

I am still a little bit confused about the perpendicular lines hill. Suppose we make it an entirely pure mathematical problem:

d² x/dt² = sgn(x)

Given x = dx/dt = 0 when t = 0, then does this mean that x is always 0? This is what the pure maths seems to suggest, but time reversibility says otherwise.

Finally you say that physical theories seem beautiful in an abstract mathematical sense. I think that a good part (but perhaps not all) of this is again because the techniques/definitions used in pure maths are shaped by the universe. There is no reason nature should be governed by differential equations, or that Euclid's axioms should hold. But then it just may have been possible to invent other equivalent concepts that we have never yet dreamed about! These may seem equally beautiful. But I agree that this does not solve the problem entirely.

Many thanks,

Michael

Again - no to the mathematical constructs being governed by physics... I think an example will be best here:

Faraday was the first person to really notice what was happening with electricity and magnetism. These are two really different forces of nature - electricity attracts/repels and magnitism acts perpendicularly to the direction of motion. For ages these were thought to be two different forces (why not they are observably very different). When Maxwell wrote down his EM laws there was however a very large amount of similarity between E and M. They would basically be identical equations except for the fact that we don't see magnetic monopoles. Then with Einstein's special relativity it was discovered exactly why they look so similar: there are not two forces there is only one - the electromagnetic force.

Now there are a few other known fundamental forces (gravity, weak nuclear, strong nuclear) and again they all look totally unlike each other. Gravity only attracts, they all have vastly different strengths, weak nuclear can break certain symmetries (called C,P,T symmetries) strong nuclear doesn't, etc. Yet over the last century theoretical physicists have seen that electromagnetic and weak are infact one force called the electroweak force - and something called symmetry breaking has occured to make them look different to us. I can't remember whether it is conjectured or shown that electroweak and strong are also the same force with a higher energy symmetry breaking. With the advent of supersymmetry physicists are attempting to show that there is only one fundamental force. So even though they look absolutely distinct to us there is infact only 1 force in the universe!

This is what I mean by simplicity - why 1 force and why not more?

If (probably when) they get a working version of supergravity then it is strongly suspected that there will be only one possible theory which will work. Our current theory of gravity doesn't by any means need to be in (3,1)-dimensional spacetime it could easily be in (17,1). These sort of choices annoy physicists and mathematicians - they hope that there theory has as few arbitrary parameters as possible otherwise we are left with the question: 'why does our universe have this choice and not that choice?'. And, recent work on supergravity seems to say that it is a high dimensional theory (~10 but the number of choices is really small) and that they compactify down to appear like 4 dimensional spaces. So the belief is that there will be only one possible theory (no choice of parameters) with only one force. Now that is physical and mathematical simplicity. However its equations will be very difficult to write down or compute with (even for people who had invented different maths!).

Does that make what I'm saying clearer?

AlexB.