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Cause of Atmospheric Pressure


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2 hours ago, StringJunky said:

That was covered in the discussion I linked: "Note also that this difference in molecular speeds at the top and the bottom manifests itself as a difference in pressure between the top and bottom of the container, as seen by the pressure-depth relationship for a static fluid: ΔP=ρgΔh. – Michael Seifert"

It's indeed wrong, but I couldn't see a refutation in the thread.

Classically (i.e. if heat was not subject to gravity), if you have a tall column of gas in a vertical gravitational field at equilibrium, the temperature of the gas is constant while its density (and pressure) decreases with height.

Otherwise you could run a small perpetual motion machine using the difference in temperature.

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30 minutes ago, Carrock said:

It's indeed wrong, but I couldn't see a refutation in the thread.

Classically (i.e. if heat was not subject to gravity), if you have a tall column of gas in a vertical gravitational field at equilibrium, the temperature of the gas is constant while its density (and pressure) decreases with height.

Otherwise you could run a small perpetual motion machine using the difference in temperature.

OK. Thanks for the explanation.

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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. 

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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.   

Edited by AEBanner
missed out a word
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14 minutes 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.

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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.

Edited by AEBanner
wrote "Motion", should be "Momentum"
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32 minutes ago, AEBanner said:

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.

You seemed to be suggesting that if I hit a snooker ball with the cue ball at any position, it will go straight ahead instead of at an angle 

The same applies to molecules. When they collide they bounce off at "random" directions. Therefore the gravitational force does not just operate vertically. It is distributed evenly in all directions. This is why a gas is a fluid.

(I'm sure studiot can provide the maths if you need it)

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1 hour 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.

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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?

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24 minutes 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/

On the other hand, you appeared to be saying that conservation of momentum means that forces can only be transmitted in a single direction. Which is obviously false, as snooker proves.

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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.

 

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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.

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8 minutes ago, AEBanner said:

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.

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

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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.

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19 minutes ago, AEBanner said:

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

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?

19 minutes ago, AEBanner said:

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. 

OK. 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.

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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.

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36 minutes ago, AEBanner said:

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?

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).

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.

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.

Quote

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

Even if you consider the (very unlikely) case of a molecule falling directly downward under gravity, it will at some point hit another molecule. At this point, both molecules will bounce off at an angle (like the snooker balls) and could hit the sides of the container, exerting sideways pressure on it.

43 minutes ago, AEBanner said:

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.

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

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2 hours ago, AEBanner said:

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.

Pressure doesn't have any sideways asymmetry. The net transverse momentum of a gas sample is zero.

 

3 hours ago, AEBanner said:

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.

And?

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)

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1 hour ago, AEBanner said:

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.

We are not talking about a change in sideways momentum. I am just trying to explain why there IS sideways momentum and therefore pressure.

1 hour ago, AEBanner said:

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.

Because some of the molecules hit the sides.

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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.

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2 hours ago, AEBanner said:

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.

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

Quote

 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.

You have a huge number of molecules undergoing a large number of collisions, effectively randomizing the motion.

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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. 

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5 hours ago, AEBanner said:

I realized a few hours ago that there are no such pressures in the atmosphere,

I think there ARE, but in still air they are perfectly balanced by the surrounding pressure, so there is no NET horizontal pressure.

As for the downward force, if you picture it as a really simplified column of ten molecules, (each of weight M) stacked one on top of another like bricks, then each molecule has a different force at the top than the bottom. The top one has zero at the top, and M at the bottom. The second has M at the top, and 2M at the bottom. The third has 2M at the top, and 3M at the bottom. In each case, the resultant force on the molecule is M upwards, and it's balanced by it's own weight of M downwards. 

At the bottom, there's a force between the last molecule and the Earth of 10 M. And that's the reason for the Atmospheric pressure at sea level. 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.

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