Edited by author 01-16-2007 06:48 PM
tyler - Must schmust. Air fills in space and provides a force to the left. There's more of it per square inch through the hole than the molecules hitting the surface, and it's going in faster. If you've got a reason to say it's still not big enough to balance the forces to the right, then give it. If you haven't, then there ain't no must about it.

Still, I think you're right - I think it does move right. Even if the book says it doesn't (as Paul seem to imply before he vanished). But I don't see any good reason yet.

But I'll give you the pressure thing - gas pressure is completely and utterly caused by momentum, I can't deny that - so long as you promise not to use thermodynamic pressure in calculations for wildly anisotropic streams into a vacuum.

Sab - you ain't causing no trouble.

Don't forget, though, that 'pressure' (which is completely and utterly caused by momentum) is the sum of the rate of change of momentum per unit area of gas molecules hitting the surface PLUS those rebounding from the surface. The molecules come in from all angles at the can whether there's a hole or not; but when there's a hole, they don't rebound as much from the right. That's why I thought the can moves right (post 341)

Question is: how much more gas goes through the hole than would have hit the surface if the hole hadn't been made, and how fast does it go through? Paul (who vanished) (post 384) convinced me that both are greater. I done da equations like (post 390). But I don't see why a=b and c=d.

As you say, the chaos inside and around doesn't matter if we look at momentum flowing in and out of the system.

Penguins? At the North Pole? With the polar bears? Sir, you have a cruel streak.
I do hope they are not evacuated penguins, as a polar bears could puncture one in a split-second. Before you could say J.

Double quadruple half whatever it's still proportional. An increase in pressure is an increase in the aggregate momentum of all the molecules and vice versa. We can monitor the change in average momentum of the molecules by measuring the pressure or temperature, which are related by the ideal gas law as you say.
In summary,
Air moves left into can : Can must move right : Air is filling in space left behind moving can (this counts as continued momentum of air to left) : can must continue right until friction wins.

Sorry, I didn't mean to cause any trouble. Forget about the disappearing mass of the puncture. Regarding timescales, for any reasonable size of puncture, I would guess that the can would fill with air before you can say 'Jack Robinson' or even just 'J', because of the immense pressure of the Earth's atmosphere. The fact that the can has not already been crushed means that it must be made of some seriously strong (but according to you, very light!) stuff. [I wrote this message before I read your latest one, s forgive any repetitions/apparent plagiarism].

Note also that the air entering the can will impart the same (leftwards) momentum on the left side of the can as it would have done on the right side had the puncture not been there. The only effect of the puncture is to generate a small delay in momentum transfer as the air has to now travel the distance from the puncture to the left side of the can before hitting it. As the can fills with air, the average delay in momentum transfer will tend to zero. It is during this short period of time that the can will have a net rightwards force acting on it that will cause it to accelerate in that direction. I don't see any way that the force due to the air entering the puncture could be any greater than it would have been had the puncture not been there at all, which means to say that the net rightwards force (neglecting friction) will decrease monotonically to zero - so in the frictionless case, I would expect the can to end up moving rightwards with constant velocity (and of course the surrounding air to have a net negative velocity). [The reason I thought a spherically symmetric system would be appropriate for the thought experiment was because it is easier to visualise the air going round and round with constant (angular) velocity].

Okay, now I have read your post, I would say that there is probably no way that v2 could be negative - the total leftward force on the can (ignoring friction) will always be less than or equal to the rightward force on the can, and friction being what it is, cannot only slow the can down, and certainly cannot make it change direction. (Even though the velocity distribution of the air molecules entering the puncture is different from the bulk, I don't see how this can change/increase the momentum imparted to the can). Even though the air in the can is in a state of chaos, momentum must be conserved, so for our purposes, the chaotic motion does not matter.

And yes, how can one have a frictionless ice rink without polar bears? Penguins too.

Edited by author 01-16-2007 05:01 PM
Ok Sab,

I'm not with you on the centre of gravity shift and dissolving with acid thing - it complicates things further: what's happened to the mass, where's it moved to and how? A puncture can take place without any shift of the centre of gravity, and I propose that we take it as such. Generally if something punctures, like a can, mass doesn't disappear. If you'd like an image, imagine a little pair of vacuum-sealed lift doors opening, very quickly, on a small patch on the side of the can. As they go in opposite directions, there's no shift of centre of gravity, no jolt or torque or anything. But wouldn't it be better just to imagine a hole appearing, without anything else changing, and keep it as simple as possible?

But that aside, I seem to totally agree with everything in your post, including the lack of a conclusion!

The can will certainly accelerate right at the very beginning (before the first molecules have crossed the can), reaching some velocity v1 to the right at time t1 when the first (fastest) molecules strike.

The force will rapidly decrease as the stream of molecules entering the can begins to pile into the wall and (later) into the air contained in the can. Whether or not this force will become negative, causing the can to decelerate until it reaches a standstill by the time the can is full, we don't seem to have a definitive answer to. So all we've got is that there are several forces acting during the filling of the can, and the rightward velocity approaches some value v2 as the can approaches being full, say at t2.

We know the can - and the air - will have stopped at the very end (due to air resistance and the self-damping of turbulence). Say at t3.

(Both t2 and t3 are subject to an arbitrary definition of when an exponential process can be said to 'end' - but we can define them precisely if we need to; and I can explain why we don't need to if you need me to.)

However, what we want to know is whether v2 is (a) negative, (b) positive or (c) zero.

What would be the 'force' applied by a stream of molecules, moving slightly faster than the mean speed in the gas and entering the hole at a rather higher flux (numbers per second per unit area) than they would normally strike the sides, hitting the inside back of the can? It's not described well by the concept of gas pressure, which is a parameter of semi-equilibrium states. Later, as the can starts to fill, what happens when the slowing stream piles into the gas that's already there? What happens is chaotic - turbulence and a big mess. Inside the can is chaos.

Here then (for tyler and anyone else interested in forces) are what I think are the main forces acting during filling:

F1: a constant force externally on the left, pushing right, which is simply the atmospheric pressure over the area of the hole on the opposite side.
F2: an unknown force internally pushing left as the air streams in, which varies with time in an unknown way but settles to a value equal and opposite to F1 when the can is full.
F3: an unknown reduction in the leftward force externally on the left - the pressure acting in the vicinity of the hole - which also contributes in the right hand direction, and decreases to zero when the can is full.
F4: air resistance, acting in the opposite direction to any motion, and much smaller than the above (during filling) unless the hole is very tiny. We can ignore this before t2 (unless everything else cancels perfectly to zero, in which case we can bring it back in.)

These forces act over time. Integrating each one over time will give the impulse of that force; when we know that we can see how the impulses of F2 and F3 (which act to the left) compare with the impulse of F1 (which acts to the right), and that'll tell us whether or not the can is moving at time t2.

It's all a bit difficult.

Alternatively, we can use the fact that impulse equals change in momentum, and look instead at the flow of momentum into and out of the system. It's not mysterious, it's exactly the same thing, and it makes a lot more sense. All we need is some much more elementary properties of the flow of air through the hole and externally in the vicinity of the hole during the filling of the can.

I just don't know how to do it, as I don't know what form a(t), b(t), c(t) and d(t) will take (ref post 390).

So... whaddya think?

Sab - are there any polar bears in your model? I like polar bears.

Ah - I see where you're coming from tyler. Yes, pressure is defined in terms of force per unit area, and force is defined as rate of transfer of momentum, so they are related.

The reason they are not proportional, though, is that if you increase the momentum of molecules, you also increase their speed, which means more of them hit per second. If dt is the mean time between collisions in which a mean momentum of dD is transferred, doubling momentum means we double dD and we half dt, and so, all else being equal, the pressure quadruples.

Hence pressure ends up being proportional to the mean kinetic energy, which in turn is proportional to the temperature, as per the old PV=nRT thing.

Bob, pressure and momentum are directly related:
To avoid confusion I will use D to represent momentum instead of the convetional P, which will represent pressure.
P=pressure
dD=change in momentum
F=force
dt=change in time
A=area

F=dD/dt Force equals the change in momentum over time
P=F/A Pressure equals a force acting over an area

So if F=PA and F=Dd/dt
Then PA=dD/dt
Thus change in momentum directly proportional to pressure.

Sab, I think you touched on an important point. Many have argued that the can moves and then needs to stop to satisfy a momentum balance. However, when the can is in motion, there is a equal diplacement of air flowing around and filling up the space behind it as it moves, thus still satisfying a momentum balance. This is why objects in motion tend to stay in motion (in a frictionless universe) even in fluids. So if you say the can moves, it does not suddenly stop once the air stops flowing into it. There is still air flow around and behind the can as it moves to satisfy the balance. Then only friction will stop it not some mysterious conservation of momentum force.

Hello Sab!

First quick read through your post and I haven't a clue what you're on about at all, but I sort of expected that. Greenwich? I'll have a proper look this evening.

mBG (& Sab) - Re friction / air resistance / etc. please see my response to Huge in post 365 re timescales. The problem as stated has only a can and a presumably infinite extent of air (as no pressure gradients, boundaries or topological oddities are specified), and asks only whether or not the can is moving once it's filled.

We're free to assume there is no gravity if we like (which makes sense considering there's nothing else in the system) - and indeed it's silly not to, as if we don't, the can will blatantly be moving down. So the ice rink is useful as an image, and inasmuch as friction is similar to air resistance anyway - but without losing sight of the problem as posed.

I'm getting a little lost.

When you start considering surface friction and turbulance the problem becomes very complex. Energy is lost (in the form of heat?) from both of these.

There is probably more turbulance inside the can than in the left side air due to the can moving, so this would probably cause the can to tend to move right after filling since the incoming air would lose momentum.

The surface friction would cause the can to move right less while filling, so would tend to cause the can move leftward after filling since the internal air (turbulant though it may be) would travel further and gain more momentum.

So, which is more significant? I think the problem could be adjusted to make either direction correct.

So, what if we only consider a frictionless surface and a small enough hole to avoid turbulance. Is it agreed, in that ideal system, that the can would initially move right and then stop? Or, am I missing something.

Okay Bob, I have thought a bit more about this, and this is my halfpenny's worth on the matter.

Think of the system (initially assumed to be stationary) as being the Earth surrounded by air arranged about it in a spherically symmetric manner, and with the can hovering above the North Pole (say) breaking the spherical symmetry. Suppose that the can is oriented such that its 'right' side (i.e the one to be punctured) is facing due south along the great circle passing through Greenwich. Let's just ignore the downward force on the can due to gravity as it unnecessarily complicates things. Perhaps it is best to think of the surface of the Earth as a frictionless ice rink for this purpose.

Then in the unpunctured case, the centre of mass of the system (assuming an appropriately symmetric can) will be in the centre of the can just above the North Pole. After the (circular, say) puncture has been created at time t=0 (I won't ask how, but let's assume the small circular piece of tin is quickly dissolved by some strong acid), then the centre of mass of the system at t=0 is actually slightly further from Greenwich than in the unpunctured case (i.e. to the left of the centre of the can if the puncture is on its right). Since there are no external forces on the system (ignoring the downward force of gravity), the centre of mass in the final state, whatever that happens to be, must be slightly to the left of the centre of the initial position of the can. Note also that the initial momentum is zero so the net final momentum of the can and the surrounding air must also be zero. Note also that there is quite a bit of potential energy initially, which must eventually be converted to kinetic and internal energy.

Presumably everyone is agreed that the net force on the newly punctured can from the air on its left side will cause it initially to accelerate rightwards (i.e. towards Greenwich). When the can is eventually filled with air, there will be no net force on it, except due to friction from the air outside, which will eventually cause it to slow down to a halt if it is moving. When it stops, the configuration of the air must be such that its net momentum is zero, and that it is distributed in such a way as to ensure that the centre of mass is where it should be. The motion of the air will in general be very complex.

So which way does the can end up going after its initial rightward acceleration? Well, once the air entering the puncture starts hitting the inside of the left wall of the can, the net force causing the can to accelerate rightwards will start to decrease, until eventually there is no net force (If the force on the left side of the can pushing it rightwards is more or less constant - which I think it would be - then I cannot imagine a scenario in which the total leftward force due solely to the air entering the can can be greater than the total rightward force, which might cause a net leftward acceleration). Actually that is not quite correct - there will eventually be a net leftward force due to external frictional forces opposing the rightward motion of the can, which will contribute to eventually slowing the can down to a halt. To summarise, the can will initially accelerate rightwards towards Greenwich, reach some limiting velocity and finally decelerate slowly to zero velocity due to friction.

Conservation of momentum in turn means that the air surrounding the can must have a net motion leftwards (i.e. away from Greenwich). I am no expert on fluid dynamics, but this can possibly be partly understood by the low pressure region created around the puncture as air enters it, sucking more air from right to left, but there will also be a low pressure region in the wake of the rightward moving can, which will also tend to such air towards it - whatever the case I can only imagine that the motion of the air will be very complex complete with vortices etc. The frictional force which reduces the momentum of the can will also serve to reduce the momentum of the surrounding air.

I am happy to be corrected, but that's what comes to mind.

Pressure isn't a great deal to do with average momentum in a gas. If you double the average momentum in a gas, you won't simply double the pressure - you're likely to roughly quadruple it. Pressure is also a quantity that only really has a definition for a gas in a state at least reasonably resembling equilibrium, which is not what we have.

Here the question asks whether or not the can is moving at the end - ie. whether or not there has been a net transfer of momentum to it. Which makes it essentially a momentum question. Furthermore, as the only other thing in the universe is air, it's essentially a question about whether or not the momentum transferred between the can and the air adds up to zero or not.

Just like a can punctured identically in both sides simultaneously has absolutely no reason to move at all, and will not, despite the big changes in energy, pressure, flow of air, etc., there is no reason why the end result of this process need be motion. Unless, that is, you have a valid momentum argument to show it.

In the case of the apple falling, when the string is cut, the forces are well out of balance in one direction so long as it doesn't hit anything. Here there are a number of forces acting in a number of different directions. If we looked at an apple that falls for a distance under gravity and is then also subject to some force opposing gravity (for example an apple landing in some sand) - there may well be a perfectly good reason why the result at the end of the time-frame in question should be a motionless apple.

However, I can't see what that reason is either.

If it really does take statistical mechanics to unravel this, then unless there's a proper statisctical mechanic on hand to give us an outline of what is involved, I'll have to just go away and try and work out how to do it. I've done a fair bit of statistical mechanics - just none in the last, er, xx years is all. And even if I could do that calculation, I may still be left thinking, "yeah, but WHY?" And I'd not know if I'd done something wrong. I'm sure there are folk lurking out there who know and can say why they know...

Bob, since you seem to insist we look at the air (though I believe just looking at the can will solve it). Lets look at ALL of the air's momentum for the time frame of this problem.
The momentum that the air gains rushing into the can slowly turns into the what the can 'sees' as internal pressure. Pressure can be defined as an average of all the little air molecules 'momentum'. The internal pressure(average momentum) for the time frame is always less than the pressure(average momentum) of the air acting on the outside.
So for the time frame involved, the pressure(average momentum)outside is greater than the pressure(average momentum)inside. In order for the momentums to be balanced at the end of time frame, something else (hey, maybe the can!) must be moving.

Bob, your apple on a string analogy is apt. As you say, cutting the string, at t=0 'now the forces are out of balance', which is very much the same as saying, 'now the momentums are out of balance'. In order to balance the momentum, the apple drops, right? Now, lets suppose some time later, we magically turn off gravity, which is analogous to the can being in a state of equal pressure in/out. No more unbalanced acceleration due to forces. Does the apple then stop having its momentum gained from the time span of the forces acting? No, even when gravity is gone, the apple will still be moving.

tyler's just made me realise that my a=b condition (for the impulses balancing and the can not moving in the end) can't be exactly true for every moment during the filling of the can. The fact that no molecules are leaving the can at t=0 would then imply that twice as many are entering as would normally impact on that part of the surface. This can't be true at t=0, as they haven't had time to speed up yet.

This doesn't make a great deal of difference to the overall rate of momentum flow into the can, though. The molecules would be accelerated to their streaming-in speed within a few mean free paths, which is a tiny tiny distance compared to the size of the can (mfp for air is 68nm, according to Wikipedia).

meBigGuy

Yes, then it would move left (assuming Paul's right to say the overall impulses of the can filling should balance, which I'm not entirely convinced about). This is relevant to the question as it's stated, because although there is no solid block preventing the can from moving right, there is some air resistance. This is why I initially said the can would jump right, but end up with a very tiny momentum to the left (post 103), due to this small external resistance to the motion of the jump.

Edited by author 01-15-2007 01:21 PM
I see your point, tyler, but at the moment immediately after the can has punctured, the velocity, momentum and position of every air molecule is identical to what they were at the moment immediately before the can was punctured, when the gas was in a state of equilibrium with the unpunctured can. So it is a valid initial condition for the state of the air. And if we're assuming that it's possible for the mass, postion and velocity of the can to be unaffected by whatever causes the hole to appear (so the question is purely about the physics of a can filling), then it's certainly a valid initial condition for the can too.

A given 'state' of a dynamic system generally means the positions and momenta of the bodies involved. We can introduce a force at a given time without affecting that state at all in that instant. Forces act over time, changing momenta at a given rate. At t=0 in the new system, the force will have had no effect.

An example might be: suspend an apple (or an orange) from the ceiling on a thread. We specify the state of the system exactly, by measuring where it is and observing that it is not moving (it's in equilibrium). Then at t=0 we cut the thread. Now the forces are out of balance, and they act on that same initial state (an apple at height h with mass m and velocity 0 at time 0), changing it over time.