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On 11/13/2018 at 7:57 PM, Strange said:

You may be right although I don't see where wind comes into it.

It doesn't.

5 hours ago, Strange said:

There is no wind inside the room I am in (or on the International Space Station, if you are going to claim it is the wind outside the building) and yet the air exerts pressure in all directions.

Air is a fluid. The constant collisions between molecules distributes the momentum (and therefore pressure) in all directions.

I am a little puzzled by your response above.

Of course, your statement in itself is correct, but I was referring to the pressures acting horizontally from the vertical column of air, as discussed in the treatment given by studiot.  Two or more such adjacent columns would, presumably, have equivalent horizontal pressures, which would cancel out, so making no overall horizontal pressure within the body of the gas. 

I think you may not have understood what I was saying.

On 11/11/2018 at 8:53 PM, swansont said:

I don't see what the problem is. The center of mass doesn't move in the transverse direction. (Momentum, of course, is not conserved in the z direction, as there is a net force)

Very good.  Thank you for contribution.

Thank you  studiot  for your recent contribution.

17 hours ago, mistermack said:

Because of the constant movement of molecules, the pressure at the bottom operates in all directions, but all lateral pressure is balanced by the equal and opposite pressure of the surrounding air.

Thank you for your contribution.

10 hours ago, swansont said:

A change in momentum (in some time interval) is a force. 

Agreed; a change in momentum is, in fact, an impulse, which is force*time.

6 hours ago, NortonH said:

I think that is the key to this problem.

Thank you.

The horizontal pressure problem has gone away.

I realized a few hours ago that there are no such pressures in the atmosphere, of course, provided there is no wind.

Thank you everyone for your help. 

2 hours ago, Strange said:

OK. Let's take it step by step. I am at one and of the table. My cue is like the force of gravity: it is going to push one ball (the cue ball) directly towards the other end of the table (in this analogy, my cue ball only ever points directly along the length of the table - like gravity).

You must not think of your cue as the force of gravity.  The cue imparts momentum to the snooker ball, whereas gravity causes the ball to have weight.

2 hours ago, Strange said:

Now, assume there is another ball between the cue ball and the other end of the table.

Your description of how gravity and momentum works implies that the only possibility is that momentum can only be transferred in straight line from the cue to the opposite edge of table.

No, this is confusing and irrelevant.

What we should be considering is the possible effects on molecules travelling downwards under gravity of the "sideways" travelling molecules in the gas.  Sure, the "sideways" molecules exert equal pressures sideways, but there will be no overall sideways movement unless the whole gas body is in movement, eg as in a wind.  The total sideways momentum  of the total molecules is conserved no matter how many collisions occur.  The initial condition of the downwards molecules has no sideways component to take into account.  

So the total effect on the "sideways" pressure is unchanged by the collisions, including the initial "downwards" molecules.

2 hours ago, Strange said:

But if my cue ball hits the other ball off-centre, then both the cue ball will fly off to the left and right (at 90º to one another because of conservation of momentum) and will hit the sides of the table. That is, they will exert pressure on the sides of the table.

Again, NO!

If you miss- cue and strike the ball off-centre,  or your ball hits a second ball off-centre, then naturally the two balls will travel apart sideways, although continuing to some extent forwards, and the various resulting motions will have momentum conserved in any direction you choose to consider. 

2 hours ago, Strange said:

I'm sure you are genuinely struggling to understand. As my explanations are not working, hopefully someone else can do better. Good luck.

And I am sure you are genuinely trying to help me.  But, as I've said,  some of your posts are confusing and I cannot understand them.  Basically, I think we are in agreement about the "sideways" pressures, but my problem is to understand how these pressures are, presumably, equal to the pressure acting downwards as discussed in "studiot's" excellent post.

So, perhaps someone else can post the missing explanation.

On 11/8/2018 at 8:08 PM, Strange said:

On 11/8/2018 at 7:30 PM, AEBanner said:

Simply because the density of the air decreases with altitude

Why?

The density of the air decreases with altitude because of gravity.

On 11/8/2018 at 8:08 PM, Strange said:

On 11/8/2018 at 7:56 PM, StringJunky said:

All the molecules in the brick are in a rigid formation. He can't see how an unconnected column of air molecules give a combined pressure (weight) on the Earth.

This is very well explained in the excellent post by studiot

On 11/8/2018 at 8:08 PM, Strange said:

On 11/8/2018 at 7:30 PM, AEBanner said:

You seem to be talking about the molecular collision theory here, which has nothing to do with the "air has weight" dilemma I have to explain how the molecules up in the air can "communicate" their weights to the Earth's surface.

The collisions transfer the weight of the air.

Agreed.

As above.  Please refer to the post by studiot

You seem to have missed my post about studiot's excellent treatment dealing with the effects of the distributed mass of the air molecules.  I have gratefully acknowledged this explanation.  Any other problems you may have with my earlier posts on this matter would, I hope, be considered before you post again about my earlier ideas.  

Nevertheless, I should still be pleased to receive suggestions to explain what perhaps I might call the "sideways pressure" issue. 

54 minutes ago, Strange said:

So you are saying that I can't make a snooker ball hit the side cushion, only the cushion at the far end of the table?

I never made any such suggestion.  Please give a reference to any of my posts which gave you this idea, because this needs putting right!  Does it not depend which way you point the cue?

59 minutes ago, Strange said:

So the overall change is zero. Let's say we start with zero momentum to the left and the right. After some number of collisions there will be an equal number of particles heading towards the left and the right. As you say, momentum is a vector quantity so these will cancel out leaving zero net momentum to left and right but a pressure exerted on the left and right sides by the particles heading in each direction.

There, momentum conserved. Everyone is happy.

Sorry, but this is confusing. 

I think we are discussing the effects of collisions with the molecules proceeding vertically downwards under gravity. 

On your assumption, there will be no net CHANGE in momentum as a result of the collisions with the"sideways" molecules, so the "sideways" pressures will be the same as before the collisions.

So I'm not happy yet.  I still should like an explanation for the overall "sideways" pressure, which presumably in practice would be equal to the downwards pressure, even though this is caused by the weight of the air above.

46 minutes ago, Strange said:

56 minutes ago, AEBanner said:

This is fundamental to our discussion, so please are you willing to accept this  Law?

Of course.

I didn't even think it needed saying. I'm not even sure why you would need to ask. 

Thank you very much for your response.  I needed to know your position on this because it is fundamental to the discussion, and as I mentioned, your previous posts had sadly confused me about your thinking on this matter.

34 minutes ago, Strange said:

The momentum will eventually be "distributed" in all directions. So the pressure will be the same in all directions. 

No! This violates the Law of Conservation of Momentum!  

It is true that the collisions can make the snooker balls bounce off and go in all directions, but the total momentum must be conserved.  This means that although some may go right, then some will also go left, and when the various angles and components of momentum are taken into account, the overall change is zero.  The momentum of each ball is mass*velocity, but since velocity is a vector, so is momentum.  If you define the positive direction to be to the right, then the direction to the left is negative, and these signs must be applied in the calculations.

So I think we still need an explanation for the "sideways" pressure.

1 hour ago, Strange said:

angles. The momentum will eventually be "distributed" in all directions. So the pressure will be the same in all directions. 

No!! Please refer to my post of just a few minutes ago.

5 hours ago, swansont said:

Conservation of momentum only tells you for any single collision, there will be equal momenta in the sideways directions (for both x and y). As Strange has implied, you have to consider collisions at all angles. This will quickly make the motion stochastic.

There is more to it than that.

Multiple collisions consist of separate collisions, and momentum conservation applies to each and all. So there is no overall change in momentum.  Please refer to my very recent response to "Strange", where I have gone into it in rather more detail.

3 hours ago, swansont said:

5 hours ago, AEBanner said:

 As for sideways motion, surely Newton's Law of Conservation of Momentum applies here.  I suppose there is another factor I must have missed, and I should be pleased to learn about it.   

Conservation of momentum only tells you for any single collision, there will be equal momenta in the sideways directions (for both x and y). As Strange has implied, you have to consider collisions at all angles. This will quickly make the motion stochastic.

Certainly momentum is conserved for one collision, but so it is conserved for all the collision under consideration.  

So the total momentum after the collisions is equal to the total momentum before the collisions.  But it must be remembered that we are dealing with vectors, so we need to use the vector components in any chosen direction, which necessarily takes the angles into account.

1 hour ago, Strange said:

1 hour ago, AEBanner said:

I have been thinking more about your reply to my post.  As in the quote above, you seem to be implying that because snooker and billiards ARE possible, then that means that Newton's Law of Conservation of Momentum is not true.  

Why would you conclude that? In fact, it is (as noted in another thread) conservation of momentum that defines the angles that the two balls will move: https://mechasco.wordpress.com/2014/05/23/billiard-balls-and-the-90-degree-rule/

I came to that conclusion because of my reading of your earlier comment, which I quote below, together with my post which occasioned your reply.  I have put your comment in bold, hopefully to avoid confusion.

4 hours ago, Strange said:

4 hours ago, AEBanner said:

But this is necessarily in the vertically downward direction because of the gravitational force vector. 

As for sideways motion, surely Newton's Law of Conservation of Momentum applies here.

If there were any truth to that, then games such as snooker or billiards would be impossible.

But snooker and billiards ARE possible.  Therefore, this supports conservation of momentum.

I'm sorry that things got rather confused.  However, now in view of your latest reference to "mechaso", I am lead to believe that you DO, in fact, accept the Law of Conservation of Momentum.   

This is fundamental to our discussion, so please are you willing to accept this  Law?

 I await your reply.

2 hours ago, Strange said:

If there were any truth to that, then games such as snooker or billiards would be impossible.

I have been thinking more about your reply to my post.  As in the quote above, you seem to be implying that because snooker and billiards ARE possible, then that means that Newton's Law of Conservation of Momentum is not true.  

So, I repeat my question.  

Do you not accept that Newton's Law of Conservation of Momentum is true?

24 minutes ago, Strange said:

If there were any truth to that, then games such as snooker or billiards would be impossible.

Do you not accept Newton's Law of Conservation of Momentum?

Given a "perfect" set of snooker balls and table, this law would certainly apply.  But you have to take into account ALL the balls involved in the overall collision process.

23 hours ago, Strange said:

There is a simplification in the descriptions so far, that the forces only act vertically. As the collisions are random and the molecules can be treated as (crudely) spherical. This means that when the molecules collide, they will bounce of in random directions. This distributes the momentum (and therefore force) in all directions. This is what makes gases and liquids a fluid.

Thank you for your post.  

Yes, the mathematical analysis provided by studiot shows clearly the effect of gravity on the air molecules and the consequent air pressure, which I understand.  But this is necessarily in the vertically downward direction because of the gravitational force vector. 

As for sideways motion, surely Newton's Law of Conservation of Momentum applies here.  I suppose there is another factor I must have missed, and I should be pleased to learn about it.   

Thank you again, studiot, for what is clearly a very informative post.  It will take me some time to study it properly.

And no, I didn't come across the book by C J Smith;  no doubt before my time.

Regards.

3 hours ago, studiot said:

Meanwhile I note you have posted in Earth Science not Physics, so is your interest in this really in Earth Science?

I thought that my choice was appropriate for the topic, but maybe I was wrong. 

2 hours ago, studiot said:

Yippee, progress. Another +1 for that.

But please let us have some more feedback for this and any further questions.

You say you understood the mathematical development. This is good as I don't care what  level you are at, I just want to post that that level.

Did you also understand the later post about the difference between fluid statics and fluid dynamics and in particular the bit about shear and normal stresses?

If I talk about random walks for molecules would you have heard of these?

Finally I note that a few posts back you correctly noted that the conditions for momentum to change, resulting in a force, is when a molecule is what you called reflected ie bounces back from a boundary.

Thank you for your friendly and helpful response.  I fully understand the reason for your questions above.  Here is my feedback. 

I am a (very) retired physicist. I received my degree in 1959; and perhaps this partly explains some of my posts.

I have no experience with fluid statics or fluid dynamics. 

I am only slightly aware of the random walks ideas.

On 11/8/2018 at 7:52 PM, studiot said:

The force exerted on the ground by a pile of bricks of standing area 1 square metre is equal to the weight of the bricks.
The pressure is the weight divided by that area.

On 11/8/2018 at 7:52 PM, studiot said:

Are you saying that this does not apply to a column of air standing on the gorund?

Of course, you are correct, it does apply.

16 hours ago, studiot said:

To add to the questions you refuse to answer.

1) What stops the atmosphere evaporating off into space?

2) What is weight?

3) How does the impact of molecule on molecule affect the surface of the Earth?

 

Sorry, I must have missed these questions.  My answers are below.

1) Gravitational force of attraction of the Earth

2)Mass*Gravitational Force

3)When a molecule of air impacts upon the Earth's surface, it is "reflected" off again.  There is a change of momentum in this collision, which applies an impulse to the Earth's surface, that is (force*time). When summed for all the impacting molecules over a specified area, we get the total force on that area, and hence the pressure.

OK, ? 

16 hours ago, studiot said:

Enough hand waving. We need some mathematics.

There is a connection between the kinetic theory and the fluid mechanics.

I have very quickly scribbled out a simple mathematical derivation of this connection in the attachments.

Thank you for an excellent post.

You have solved my misunderstanding for me, and I am truly grateful.  I now see clearly where I went wrong in trying to deal with the mass of the molecules.  Thank you, too, to everybody else for their help, and I am pleased to say I no longer need references to standard work on this matter.

But, I now have a new problem.  I thought that pressure at any point in a gas acts equally in all directions.  The analysis above deals with the downward effects of gravity, so what happens "sideways"?  

 

 

14 hours ago, Strange said:

So is that the problem, you are not bothering to read any explanations, just posting the same replies without even thinking?

Please refer to my very recent reply to studiot

6 minutes ago, studiot said:

Why waste swansont's time when you have refused to acknowledge the standard Physics expalantion I wrote out for you, complete with calculations?

I further state categorically that the atmosphere's pressure on the surface of the Earth can be directly linked to the weight of the air molecules and would be zero if they had no weight.

What a pity you don't want to know.

Thank you for response and for your efforts on my behalf.  But to be fair, I really do want to know; it is simply a case of finding the required time when so many posts need answering.

42 minutes ago, Strange said:

You know how you said that pressure is due to the collisions of the molecules?

They also collide with each other and so are not independent.

And these collisions are how they are in contact with the surface.

It's really not that hard.

[Please note that this response is meant to replace a recent one where I made a mistake with the attribution.]

The collisions between molecules do not transfer mass from one to another, only momentum.

And YES !!!, some collide with the surface and are reflected, and the change in momentum provides an impulse.  When integrated over the surface for all the impacting molecules, this creates the atmospheric pressure.  Its easy.

16 minutes ago, Strange said:

You know how you said that pressure is due to the collisions of the molecules?

They also collide with each other and so are not independent.

And these collisions are how they are in contact with the surface.

It's really not that hard.

The collisions between molecules do not transfer mass from one to another, only momentum.

And YES !!!, some collide with the surface and are reflected, and the change in momentum provides an impulse.  When integrated over the surface for all the impacting molecules, this creates the atmospheric pressure.  Its easy.

3 minutes ago, Strange said:

26 minutes ago, AEBanner said:

I have tried to find published work on the internet for the standard physics explanation for atmospheric pressure.  There is very little, basically Wiki and National Geographic websites. 

Maybe because you made it up?

26 minutes ago, AEBanner said:

So I should really be grateful if you could please provide me with chapter and verse references for the standard physics  explanation.

You are the one claiming that the "standard physics explanation" is that pressure is just caused by weight. If you can't provide a reference to it, maybe it doesn't exist.

Oh, please!

I would never claim to have made up the standard physics explanation for atmospheric pressure.  This was first raised by swansont and I was asking him for help in finding appropriate references.

What I said was in the references I found on the internet, the main explanation they gave was that of the weight.

7 minutes ago, beecee said:

Weight is the force of gravity  acting on objects/molecules etc. The Earth is pulling on all the atmosphere while at the same time, the molecules of the gases in the atmosphere are creating pressure while jiggling. Both play a part, and both are relevant. 

So far, so good.  But how is that pressure exerted on the surface except by molecular collisions?

9 hours ago, swansont said:

That pressure is solely due to weight (collisions playing no part), and exerted only in a downward direction. That seems to be your assertion.

No, that is not my assertion.  I believe that the pressure is caused by molecular collisions.

If the explanation for atmospheric pressure was that the weight of the molecules presses down on the Earth's surface due to gravity, then the resulting pressure could only act vertically downwards, and so there would be no pressure sideways or any other direction, which is clearly not the case.  So it would seem that atmospheric pressure is due to molecular collisions, which is my belief.

6 hours ago, swansont said:

I don't know how to be clearer. The standard physics explanation is that the pressure is equal to the weight of the molecules and the mechanism is molecular collisions. Both of these are part of the explanation. It is not one or the other.

I have tried to find published work on the internet for the standard physics explanation for atmospheric pressure.  There is very little, basically Wiki and National Geographic websites.  The standard explanation seems to be that the cause is the weight of the atmosphere. But no attempt is made to explain how this weight is applied to the Earth's surface.  And that is my problem.

So I should really be grateful if you could please provide me with chapter and verse references for the standard physics  explanation.

As far as I can understand this idea, the weights of the individual air molecules combine somehow, and so the total weight then acts upon the surface.

I simply cannot see this, because the molecules are totally independent, they are "up in the air", and not in contact with the surface.

As an example, consider a man standing on bathroom scales on the Earth's surface.  His weight is applied through his feet to the scales, and the scales show his weight.  Now suppose he jumps straight up into the air.  Clearly, he still has the same weight, but it is not connected to the scales, so they show no weight.  So contact with the surface is required if pressure is to be exerted on the surface.

2 hours ago, Phi for All said:

Or then again, perhaps you're just trying to wind me up.

I wish to apologize to the moderator,  and to swansont , and to everyone else for this remark.

I fully accept that the posts are intended to help me understand the explanation for atmospheric pressure.  But I am still not convinced by some of the explanations offered.    

I shall try to deal with these points in answer to the various individual posts.

16 minutes ago, swansont said:

That pressure is solely due to weight (collisions playing no part), and exerted only in a downward direction. That seems to be your assertion.

You have totally misunderstood my posts on this matter.  My opinion is that atmospheric pressure is due to the collisions of the air molecules with the Earth's surface.  I do not believe that it is caused by the weight of the molecules, and if you read my posts again perhaps then you too will be convinced. 

My first objection to the "air has weight" explanation is that I cannot see a mechanism whereby the weight of air in the column is applied to the surface, since almost all the molecules are not in contact with it.  Each molecule certainly has its own independent weight but it's up there in the air, not touching the surface.  

Another objection is that, as I'm sure you will agree, air pressure acts in all directions, but with the "air has weight" theory the pressure would only be downwards, in line with gravity.

Or then again, perhaps you're just trying to wind me up.