# I Derivation of the change of air pressure with height

Summary
During the derivation of variation pressure with height in atmosphere why don't we take the weight of the air which is above the air slab which we are using?
If we take a slab of air with cross-sectional area of $A$ and height $dz$ in our atmosphere. Now, what we do is make an argument like this :-

Pressure from below must balance both the weight and Pressure from above to keep the slab at rest. ( I have added an attachment for clarification)

And from this are argument we do something like this :
$P_{above} = P$
$W_{slab} = mg$ , Pressure from weight is $\rho \times A \times dz \times / A = \rho ~dz~g$ ( $\rho$ is the density of air)

Therefore, $P_{below} = P + \rho~dz~g$

$dP = P_{above} - P_{below} = - \rho ~dz~g$

$\frac{dP} {dz} = - \rho ~g$

Now, my question is why we didn't not take into account the weight of air which is above the slab? I mean the air which lies above our sample slab must have weight this weight would have been pushing our slab downwards and therefore pressure from below must also balance this. The mass of air which lies above the slab must have number molecules and gravity would attract them and we should center of mass and assume the gravity to act over there and this force of gravity should cause a pressure on our slab.

I tried to argue that pressure from above is due to weight of air only, but after sometime I thought that this argument is not good as fluid causes pressure due to collision of it's molecules .

So, I want to restate my problem again : Why in our derivation we didn't take into account the pressure caused by the weight of the air that lies above our slab?

Thank you, any help will be much appreciated. I tried to be as clear as I could be.

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#### DrStupid

So, I want to restate my problem again : Why in our derivation we didn't take into account the pressure caused by the weight of the air that lies above our slab?
What do you think what $P_{above}$ is?

What do you think what $P_{above}$ is?
It is the collision of air molecules.

#### DrStupid

It is the collision of air molecules.
That's the microscopic view. From a macroscopic point of view pressure is force per area. In case of $P_{above}$ this force comes from the air above. Do you agree?

#### Dale

Mentor
Now, my question is why we didn't not take into account the weight of air which is above the slab?
The weight of the air above does not act on the slab of air. Why would we include it in the forces on the slab?

When you do a free body diagram you include only forces acting on the body in question. In this case there are three such forces on the slab: the pressure from above the slab, the pressure from below the slab, and the weight of the slab. The weight of the air above does not act on the slab. Including it would be wrong.

Last edited:

That's the microscopic view. From a macroscopic point of view pressure is force per area. In case of $P_{above}$ this force comes from the air above. Do you agree?
Yes.

The weight of the air above does not act on the slab of air. Why would we include it in the forces on the slab?

When you do a free body diagram you include only forces acting on the body in question. In this case there are three such forces on the slab: the pressure from above the slab, the pressure from below the slab, and the weight of the slab. The weight of the air above does not act on the slab. Including it would be wrong.
That's what I want to know. Why including it will be wrong?

#### Dale

Mentor
Why including it will be wrong?
Because it doesn’t act on the slab. Only forces acting on the slab should be included.

Because it doesn’t act on the slab. Only forces acting on the slab should be included.
So on what it acts?

#### Dale

Mentor
So on what it acts?
The weight of the air above acts on the air above. Is that somehow not obvious?

The weight of any object is the gravitational force acting on that object. The thing that it acts on is literally part of the definition of the weight. How can this be unclear?

What do you think weight is?

#### DrStupid

OK, the air exerts a downward force. According to the 3rd law there must be a corresponding counterforce, acting upward on the earth and according to the 2nd law the air would be accelerated if this would be the only force acting on it. But in the static case the air doesn't accelerate. Therefore there must be another force acting on the air. Do you agree? If yes, which force is it?

#### Chestermiller

Mentor
The weight of the air above is captured in the pressure above, $p_{above}$

The weight of the air above is captured in the pressure above, $p_{above}$
Really? thank you I had same thought but I was fearing that I would be wrong.

The weight of the air above acts on the air above. Is that somehow not obvious?

The weight of any object is the gravitational force acting on that object. The thing that it acts on is literally part of the definition of the weight. How can this be unclear?

What do you think weight is?
If I hold a heavy sack on my head won't I feel that something is pushing me down? That feeling of pushing down is because I'm barring the sack from falling due gravity and hence applying a force equal and against the gravity.

OK, the air exerts a downward force. According to the 3rd law there must be a corresponding counterforce, acting upward on the earth and according to the 2nd law the air would be accelerated if this would be the only force acting on it. But in the static case the air doesn't accelerate. Therefore there must be another force acting on the air. Do you agree? If yes, which force is it?
Sir, what I'm saying is: the air which lies above the slab would have it's weight just like our slab has a weight and weight always acts downwards, therefore slab must be bearing the weight of the air above it (just like we bear the weight of something if we hold it over our head).

#### DrStupid

Sir, what I'm saying is: the air which lies above the slab would have it's weight just like our slab has a weight and weight always acts downwards
I just assume that to be the answer to my last question.

therefore slab must be bearing the weight of the air above it (just like we bear the weight of something if we hold it over our head).
"bearing the weight" means exerting an upward force that compensates the weight. In this case it is the upward force I was talking about in #11. Does that tell you something about the origin of $P_{above}$?

#### Mister T

Gold Member
If I hold a heavy sack on my head won't I feel that something is pushing me down?
Sure. It's the sack pushing on your head. It's a contact force, often called the normal force in elementary physics problems. It's not the weight of the sack (which is a force exerted on the sack by Earth).

That feeling of pushing down is because I'm barring the sack from falling due gravity and hence applying a force equal and against the gravity.
You're applying a force on the sack that's equal and opposite to the force the sack exerts on you (Newton's Third Law). The reason the sack doesn't fall is because the force you exert on it is equal to the force Earth exerts on it (Newton's Second Law).

#### Dale

Mentor
If I hold a heavy sack on my head won't I feel that something is pushing me down? That feeling of pushing down is because I'm barring the sack from falling due gravity and hence applying a force equal and against the gravity.
You are misunderstanding the physics. The feeling of something pushing you down is entirely due to the contact force (pressure) on your head. If you reduce the contact force without changing the weight then the feeling reduces. If you reduce the weight without changing the contact force then the feeling is unchanged.

The feeling is not due to the weight, it is due to the pressure. The weight of the sack acts on the sack, not on me. Only the contact force from the sack acts on me, not it’s weight.

I just assume that to be the answer to my last question.
Means $P_{above}$ is just the weight of the air lying above it, okay. I was thinking that it is the pressure caused due to collision of molecules of air with our sample slab. Thank you. You have explained it in a nice way

"bearing the weight" means exerting an upward force that compensates the weight. In this case it is the upward force I was talking about in #11. Does that tell you something about the origin of $P_{above}$?

You are misunderstanding the physics. The feeling of something pushing you down is entirely due to the contact force (pressure) on your head. If you reduce the contact force without changing the weight then the feeling reduces. If you reduce the weight without changing the contact force then the feeling is unchanged.

The feeling is not due to the weight, it is due to the pressure. The weight of the sack acts on the sack, not on me. Only the contact force from the sack acts on me, not it’s weight.
Sir, I think due to my mistake of using the word “weight” as the acting force on something is causing the confusion.
I meant by “weight” the force $mg$ which I must provide to something to keep it from falling. So if I’ holding a sack on my head, I need to apply a force equal to its weight and this force is being applied by itself due Newton’s 3rd Law but I have to stay up for 3rd Law to work. Since both forces are acting ( my force and sack’s weight) on different objects therefore I would feel something heavy or being pushed down feeling.
Let me know if my essence of understanding is wrong over here. I have got a great chance to amend it ( if it’s wrong) from great mentors like you. Thank you.

#### Dale

Mentor
I meant by “weight” the force mg which I must provide to something to keep it from falling.
First, that force is the contact force, not the weight. Second, in many situations the contact force will not be equal to mg.

therefore I would feel something heavy or being pushed down feeling.
Yes, that is the contact force, or in other words the pressure. Any influence that the weight of the upper object might have is transmitted to the lower object through the pressure. Once you have accounted for the contact force (pressure) then you have already accounted for the weight of the upper object.

First, that force is the contact force, not the weight. Second, in many situations the contact force will not be equal to mg.

Yes, that is the contact force, or in other words the pressure. Any influence that the weight of the upper object might have is transmitted to the lower object through the pressure. Once you have accounted for the contact force (pressure) then you have already accounted for the weight of the upper object.
What about the pressure caused due to collision of molecules?

#### Dale

Mentor
What about the pressure caused due to collision of molecules?
Can you be more specific with your question?

Can you be more specific with your question?
There may be two kinds of pressure:-

1. Defined by you as the contact force.

2. Pressure due to collision of molecules with the walls of container that we use in thermodynamics . So the air above the slab must be colliding with our slab and hence may be causing some collisional pressure.

#### vanhees71

Gold Member
I think, you already had the right ansatz in #1. Just use the general equation
$$\vec{\nabla} p=-\rho g.$$
We need an equation of state. A good ansatz is the polytrope
$$p=p_0 (\rho/\rho_0)^n,$$
where $p_0$ and $\rho_0$ are the pressure at height $z=0$. Since the pressure and density change only along $z$, we have
$$\mathrm{d}_z p=-g \rho_0 (p/p_0)^{1/n}$$
with the solution
$$p(z)=p_0 \left (1-\frac{n-1}{n} \frac{\rho_0}{p_0} g z \right)^{n/(n-1)}.$$
For $n=1$ (isothermal equation of state for an ideal gas) we have
$$\mathrm{d}_z p=-g p \rho_0/p_0$$
and thus
$$p(z)=p_0 \exp \left (-\frac{g \rho_0}{p_0} z \right).$$
On the other hand
$$p_0=\rho_0 k T/m,$$
where $k$ is Boltzmann's constant, $T$ the (absolute) temperature, and $m$ the mass of the gas molecules,
$$p(z)=p_0 \exp \left (-\frac{g m}{k T} z \right).$$
This is the barometric formula for an isothermal atmosphere.

"Derivation of the change of air pressure with height"

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