# "Air Pressure" is not "the weight of the air above you"

Doug1943
All the popular definitions of 'air pressure' that I have seen say: "air pressure is the weight of the column of air above you". This seems misleading to me.
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I believe it is much more correct to say that air pressure is the collective force of squillions of air molecules slamming into you. There is a relationship between the number of molecules above the molecules hitting you, each one being pulled towards the Earth's surface and slamming into the one below, but saying that the resulting pressure on you 'is' the 'weight of the air above you' is one of those pat answers that discourage children from thinking -- just a form of words that actually does not explain anything. (The kind of 'science answer' that drove Richard Feynman into a fury, and rightly so.)
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Consider: I stand in a cylinder which is, say, 100 km tall , closed at the bottom, open at the top. There is a column of air above me. Now, remove all of the air, except for one molecule. Is there any 'weight' at all on me? Add a second molecule. And a third... when does the 'weight' start? If it were a column of ball bearings, balanced on my head, then it would be correct to say that the pressure on the top of my head was due to their weight -- each ball bearing's downward force adding to the next one's -- .. .and removing, or adding, a ball bearing would indeed change the total weight of the column (a wee bit).
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But fluids are different. Now reduce the height of the cylinder (of air) from 100 km, to 5 meters. And close the top of the cylinder as well. Make the cylinder very very strong, and 'airtight'. Lift this closed cylinder up a few kilometers above the Earth's surface. The air pressure outside the cylinder goes down. But inside the cylinder, it remains the same.
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Can we now say that the air pressure inside the cylinder is due to the 'weight' of the air above it? (Same experiment: what causes the air pressure inside the International Space Station? Weight?) If I have a column of bricks on top of me -- yes, the pressure on me is due to their weight. If I get inside a strong box, with the bricks on top, I no longer feel their pressure. But the same is not true for air , or any fluid.
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[However, surely the mechanism of pressure must be different for gases, on the one hand, and liquids, on the other. (Let's postpone discussion of pressure in liquids vs gases to a second thread, later.)]
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Dale

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What induces higher or lower atmospheric pressure (the kind described by the weather man/woman each day, and especially when referring to hurricanes)?

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Summary:: The very common statement, "Air pressure = 'the weight of the air above'" is misleading.

Yes. You are essentially and fundamentally incorrect. I will not count the ways. Please read any introductory physics book.

vanhees71, Vanadium 50, davenn and 1 other person
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I guess it's the pressure of the cooling water coming out of the hose that crushes the 55-gallon drum in this demonstration.

vanhees71 and davenn
Summary:: The very common statement, "Air pressure = 'the weight of the air above'" is misleading.

I believe it is much more correct to say that air pressure is the collective force
the pressure is not a force anyway: the force is a vector ;the pressure is not

vanhees71, etotheipi and Doug1943
Doug1943
Yes. You are essentially and fundamentally incorrect. I will not count the ways. Please read any introductory physics book.
I've read lots of physics books. You don't know what you're talking about, unfortunately -- you're one of the types of student that Feynman was talking about: satisfied with some memorized words that will get you the "correct answer" on the exam, and uninterested in actually understanding reality.

davenn, DaveE, weirdoguy and 1 other person
Doug1943
the pressure is not a force anyway: the force is a vector ;the pressure is not
Yes. It's sloppy to talk of pressure as a force when in fact it's the ratio of force to the area to which the force is applied. That settled, what do you think about the argument that saying air pressure is due to the 'weight of the air above' is misleading?

in fact it's the ratio of force to the area
a ratio of a vector to a scalar is still a vector

davenn, DaveE and etotheipi
Doug1943
I guess it's the pressure of the cooling water coming out of the hose that crushes the 55-gallon drum in this demonstration.
You didn't understand the original statement. The argument is not "Air pressure doesn't exist". Rather, the argument is: it is misleading to talk of the 'weight of the air above' as the cause of this pressure, as if we had a pile of bricks above. But taking your example (which I reproduce for my students using soft drink cans with a bit of water in them which I then boil, and plunge the can into cold water), if we reproduced this demonstration inside the International Space Station, it the drum would still be crushed. So .. you're saying that in those circumstances, it's "the weight of the air" above the drum that crushes it?

The notion "pressure" is a subtle enough thing if consider it carefully. It was a prompt to sit reading books before publishing home-made theories

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(Also, let's stick with the examples in the OP. Introducing imploding cans is a red herring (Duhn-duhn -DUHN).)

Consider one of the examples you yourself made:

"I stand in a cylinder which is, say, 100 km tall , closed at the bottom, open at the top. There is a column of air above me."

OK, now take that cylinder out in micro-gravity or even zero gravity. How much pressure will you feel?

Zero.
Because all the air that was in the cylinder will escape out the open end. The only reason why you felt any air pressure when it was on the Earth was because gravity was pulling it all downward. i.e. it is in a very real way, the weight of the air, induced by gravity.

davenn, PeterDonis and kuruman
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You didn't understand the original statement. The argument is not "Air pressure doesn't exist". Rather, the argument is: it is misleading to talk of the 'weight of the air above' as the cause of this pressure, as if we had a pile of bricks above. But taking your example (which I reproduce for my students using soft drink cans with a bit of water in them which I then boil, and plunge the can into cold water), if we reproduced this demonstration inside the International Space Station, it the drum would still be crushed. So .. you're saying that in those circumstances, it's "the weight of the air" above the drum that crushes it?
It is true that my reading of the original statement was hasty. I started my reply but @DaveC426913 got there before me and said it with fewer words. A problem I see with your model is this. Consider a balloon filled with helium and tied to a string. On Earth, the string will be under tension but not in the space station. I can't see how your kinetic theory model can come up with Archimedes's principle.

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I've read lots of physics books.
Perhaps more slowly this time.

davenn
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Summary:: The very common statement, "Air pressure = 'the weight of the air above'" is misleading.

All the popular definitions of 'air pressure' that I have seen say: "air pressure is the weight of the column of air above you". This seems misleading to me.
It is misleading. A correct statement would be: for a fluid in hydrostatic equilibrium the change in pressure from the top to the bottom is equal to the weight of the fluid.

Frabjous
Doug1943

(Also, let's stick with the examples in the OP. Introducing imploding cans is a red herring (Duhn-duhn -DUHN).)

Consider one of the examples you yourself made:

"I stand in a cylinder which is, say, 100 km tall , closed at the bottom, open at the top. There is a column of air above me."

OK, now take that cylinder out in micro-gravity or even zero gravity. How much pressure will you feel?

Zero.
Because all the air that was in the cylinder will escape out the open end. The only reason why you felt any air pressure when it was on the Earth was because gravity was pulling it all downward. i.e. it is in a very real way, the weight of the air, induced by gravity.
Yes. That's why I said, "Now reduce the height of the cylinder (of air) from 100 km, to 5 meters. And close the top of the cylinder as well. Make the cylinder very very strong, and 'airtight'. Lift this closed cylinder up a few kilometers above the Earth's surface. The air pressure outside the cylinder goes down. But inside the cylinder, it remains the same."
Perhaps more slowly this time.
I'm sorry if your reading speed is slow. It's not your fault. No one is responsible for their genes.

davenn
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Yes. That's why I said, "Now reduce the height of the cylinder (of air) from 100 km, to 5 meters. And close the top of the cylinder as well. Make the cylinder very very strong, and 'airtight'. Lift this closed cylinder up a few kilometers above the Earth's surface. The air pressure outside the cylinder goes down. But inside the cylinder, it remains the same."
I don't see how this is germane to the issue. I don't see how it negates the logic in my scenario, which is quite a bit simpler and more directly addresses the issue of air pressure being compared to weight.

Even simpler:
A hundred mile tall column of air on top of your head in a zero gravity environment provides zero pressure.
A hundred mile tall column of air on top of your head in the gravity of Earth provides 14.7 lb. pressure (per square inch).

That 14.7 pounds of pressure (on each square inch) is a direct result of Earth's gravity - i.e. weight.

Doug1943
It is misleading. A correct statement would be: for a fluid in hydrostatic equilibrium the change in pressure from the top to the bottom is equal to the weight of the fluid.
But "air pressure is due to the weight of the air above" is the way it is explained in almost every popular science book, and on the first dozen or so web links you get if you Google the phrase. Your definition would confuse beginners. We want to teach kids how to think ... not to memorize nice phrases.
It's a shame that not a single person here, so far, has even tried to address the question. It's the problem Richard Feynman was addressing. (If you don't know who he was, look him up on Wiki.)

Here's a sample of his approach to science, which is how science should be taught. He's writing about an elementary science book for kids:

“For example, there was a book that started out with four pictures: first there was a wind-up toy; then there was an automobile; then there was a boy riding a bicycle; then there was something else. And underneath each picture, it said "What makes it go?"
I thought, I know what it is: They're going to talk about mechanics, how the springs work inside the toy; about chemistry, how the engine of an automobile works; and biology, about how the muscles work.
...

I turned the page. The answer was, for the wind-up toy, "Energy makes it go." And for the boy on the bicycle, "Energy makes it go." For everything "Energy makes it go."

Now that doesn't mean anything. Suppose it's "Wakalixes." That's the general principle: "Wakalixes makes it go." There is no knowledge coming in. The child doesn't learn anything; it's just a word

What the should have done is to look at the wind-up toy, see that there are springs inside, learn about springs, learn about wheels, and never mind "energy". Later on, when the children know something about how the toy actually works, they can discuss the more general principles of energy.

It is also not even true that "energy makes it go", because if it stops, you could say, "energy makes it stop" just as well. What they're talking about is concentrated energy being transformed into more dilute forms, which is a very subtle aspect of energy. Energy is neither increased nor decreased in these examples; it's just changed from one form to another. And when the things stop, the energy is changed into heat, into general chaos.”

But our crummy educational system turns out generations of rote-learners, for whom physics is just algebra and memorized definitions. Their native curiosity has been drilled out of them. "Wakalixes make it go." "Air pressure is the weight of the air above." Whatever the book says.

davenn
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It's a shame that not a single person here, so far, has even tried to address the question.
Which question? Your OP started with a statement. A statement that has not been granted to be true, as of yet. We can't yet move on to any other discussion that is predicated on that.

Here's a sample of his approach to science, which is how science should be taught. He's writing about an elementary science book for kids:
It is pointless to provide an example that assumes your first claim is true (to-wit: that education is busted), until/unless your first claim is granted as true (you haven't shown that yet).

You submitted this:

"Now reduce the height of the cylinder (of air) from 100 km, to 5 meters. And close the top of the cylinder as well. Make the cylinder very very strong, and 'airtight'. Lift this closed cylinder up a few kilometers above the Earth's surface. The air pressure outside the cylinder goes down. But inside the cylinder, it remains the same."

To which I say: what does this have to do with the original scenario? You've changed the parameters.

To put a fine point on it: The column of air we are speaking of is not closed at the top.

The only thing that is keeping the air in that open-topped column is its weight.

So again: yes, the pressure we feel on the surface of the Earth is the weight of the (open-ended) column of air on top of us that is only able to apply pressure to us because of its weight pressing down on us.

davenn, PeterDonis and berkeman
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How about "the pressure of the air surrounding you must support the weight of the column of air above it". Does that pass your "I'm channeling Feynman" test?
You are correct that RP Feynman was very careful not to dumb things down nor try to explain by giving things a name. And yes most everyone here has read the great man. A few likely even knew him.

davenn
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This all strikes me as a straw man. Most all of physics is taught using analogies and approximations.

Atmospheric pressure being described as the weight of the air above you isn't a bad analogy. But, analogies are always wrong, or we wouldn't use them. There wouldn't be any introductory physics texts, or any thermodynamics, if the answer to all questions was to solve the wave function of the universe and everything in it (that's redundant, isn't it?). So, you're correct, it's not a perfect explanation. But then the whole concept of pressure is meaningless at the most detailed level.

You need to allow physics teachers to lie at least a little bit, or you won't have any physics students.

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@Doug1943 ,
You are correct about the force of the impact of molecules but have failed to connect that in any practical way to the important fact that there is less pressure at higher altitudes. Like all aerodynamics, there are several ways to describe what is going on. Some are more useful than others.

nasu
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Most all of physics is taught using analogies and approximations.
While that is true, this is neither. It is a Gedankenexperiment where one thinks of isolating an imaginary cylinder of gas from your surface to outer space. This facilitates the understanding of the concept of the weight of the column as the source. As such it requires no apology in my physics world.

It's the problem Richard Feynman was addressing. (If you don't know who he was, look him up on Wiki.)
Gedankenexperiments were often used by a physicist named Einstein. (If you don't know who he was, look him up on Wiki.)

phinds, nasu, davenn and 1 other person
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It's a shame that not a single person here, so far, has even tried to address the question.
The question was, and I quote you, “Comments?” Which we did.

Now that doesn't mean anything. Suppose it's "Wakalixes." That's the general principle: "Wakalixes makes it go." There is no knowledge coming in. The child doesn't learn anything; it's just a word
Frankly, this is precisely why I don’t like your approach. Unless you already have a method of calculating or measuring the force of the “squillions” of molecules, then it is a useless factoid. The collisions are Feynman’s wakalixes/energy. A correct explanation that cannot actually be used for further understanding.

Teaching about the weight is good because it is a preexisting concept that they can actually use to make experimental predictions.

The problem isn’t bringing in the concept of the weight of the column of fluid. The problem is saying it gives the pressure instead of the change in pressure and in not specifying that it applies to static fluids only.

Unless they already have the ideas of atomic collisions then it doesn’t help them. How can they connect the idea that pressure is driven by molecular collisions with the observation that pressure is higher at the bottom than at the top? That connection will inevitably come back to weight anyway.

Now reduce the height of the cylinder (of air) from 100 km, to 5 meters. And close the top of the cylinder
By the way, my small correction to the weight concept works perfectly well with this scenario. When two explanations work then take the simpler.

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nasu, davenn, russ_watters and 6 others
OP is wrong, but it is worth explaining why. The Navier-Stokes equation$$\rho \frac{D\mathbf{u}}{Dt} = - \nabla p + \mu \nabla^2 \mathbf{u} + \mathbf{f}$$is the general equation of motion for a fluid. There is a pressure term ##p## - it is a scalar field ##p(\mathbf{x}, t)##. Generally this term is decomposed into two entities,$$p = p_{\mathcal{H}} + p'$$and here ##p_{\mathcal{H}}## is the hydrostatic pressure and is due to the weight of the fluid above. This is in fact the best way to think about ##p_{\mathcal{H}}##. If ##\mathbf{u} = \mathbf{0}## then write ##p = p_{\mathcal{H}}## and from Navier-Stokes you find that ##\nabla p_{\mathcal{H}} = \mathbf{f} = \rho \mathbf{g}## or integrated to ##p_{\mathcal{H}} = \rho \mathbf{g} \cdot \mathbf{x} + p_0##, and ## \rho \mathbf{g} \cdot \mathbf{x}## is up to a constant factor nothing but the weight of fluid supported above.

For fluid in motion i.e. ##\mathbf{u} \neq \mathbf{0}## there are generally pressure gradients either driving the motion or generated by the flow itself; this is encapsulated by the second term ##p'##. Substituting ##p = p_{\mathcal{H}} + p'## into the Navier-Stokes equation yields$$\rho \frac{D\mathbf{u}}{Dt} = -\nabla p' + \mu \nabla^2 \mathbf{u}$$since ##-\nabla p_{\mathcal{H}}## has canceled with the ##\mathbf{f}## term. Thus when solving this equation one may ignore hydrostatic pressure and consider only the deviation ##p'##.

hutchphd, Frabjous and Dale
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the hydrostatic pressure and is due to the weight of the fluid above. This is in fact the best way to think about pH. If u=0 then from Navier-Stokes you find that ∇pH=f=ρg or integrated to pH=ρg⋅x+p0,
Note that although your words say the “pressure and is due to the weight of the fluid above”, your math says that it is the change in pressure that is due to the weight.

nasu and hutchphd
Note that although your words say the “pressure and is due to the weight of the fluid above”, your math says that it is the change in pressure that is due to the weight.
Here I fixed ##p_{\mathcal{H}}(\mathbf{x})## absolutely by writing ##p(\mathbf{0}) = p_0##. But yeah ##\rho \mathbf{g} \cdot \mathbf{x} = p_{\mathcal{H}}(\mathbf{x}) - p_0## is the change in hydrostatic pressure upon translating by ##\mathbf{x}## from ##\mathbf{x}_0##

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Dale
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1) I think we should stop with the insults.
2) Atmospheric pressure is pressure caused by a gravitational field. Atmospheric pressure and pressure are not identical. You can have pressures that are not gravitationally induced.
3) Would the OP agree that if he integrated the density of air above his head times gravitational attraction he would get the correct atmospheric pressure?
4) By sealing the cylinder or using a sealed space station, the bc’s have changed and the conclusions do not necessarily follow.
5) The small number of molecules argument is not applicable, because under this condition, it is not clear that atmospheric pressure can be meaningfully defined. Let’s not get diverted into a discussion of rarified gases.
6) Atmospheric pressure is taught using a continuum description. Yes one could define it using molecules, but that does not invalidate the continuum approach.

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