Does Pressure Gradient Exist with Uniform Pressure on All Sides?

[UMath1]

I am having trouble with the notion that pressure is applied to all sides uniformly. It seems to contradict with my understanding of why there is a pressure gradient in the atmosphere. The way I understood it, the reason you experience more pressure at lower altitudes, is because the parcel of air right on top of you experiences more pressure because there is a lot of air on top of it. But if pressure is exerted on all sides, then the parcel of air above you exerts the same pressure on the air above it as it does on you. Wouldn't that mean there is no pressure gradient?

 

[Orodruin]

What you are saying holds when the pressure is constant. The actual statement about pressure acting equally in all directions does not involve a small volume but a small surface. Regardless of the orientation of the surface, the force from the pressure on the surface will be of the same magnitude and antiparallel to the surface normal.

 

[UMath1]

So the pressure applied by the surface of air on top of me is the same both upwards and downwards?

 

[Orodruin]

So the pressure applied by the surface of air on top of me is the same both upwards and downwards?

This statement makes no sense. The top of you is a surface with a well defined normal and therefore there is a well defined pressure force acting on top of you.

 

[UMath1]

What I mean is, if my head was below the xy plane (surface) as depicted in the picture, wouldn't the pressure on the +z and the -z direction be equivalent?

https://upload.wikimedia.org/wikipedia/commons/b/bd/Charged_plane_with_Gauss_surface.svg

 

[Dale]

So the pressure applied by the surface of air on top of me is the same both upwards and downwards?

Consider the pressure force on opposite sides of a thin horizontal piece of paper instead of a human head. Hopefully it is clear that the air pushes on both the top and bottom.

 

[russ_watters]

I am having trouble with the notion that pressure is applied to all sides uniformly. It seems to contradict with my understanding of why there is a pressure gradient in the atmosphere. The way I understood it, the reason you experience more pressure at lower altitudes, is because the parcel of air right on top of you experiences more pressure because there is a lot of air on top of it. But if pressure is exerted on all sides, then the parcel of air above you exerts the same pressure on the air above it as it does on you. Wouldn't that mean there is no pressure gradient?

This does not contradict the idea of a pressure gradient. The statement is that pressure is applied in all directions uniformly, at a point. It should be easy enough to see that pressure at a point in a pressure gradient only has one value even though pressures at two slightly separated points will have different values.

 

[UMath1]

I am still confused. The reason the air below has higher pressure is because it subject to greater pressure from the air above it..but then it applies the same pressure above it. So it wold seem that the air above should be subject to the same pressre.

 

[russ_watters]

I am still confused. The reason the air below has higher pressure is because it subject to greater pressure from the air above it..but then it applies the same pressure above it. So it wold seem that the air above should be subject to the same pressre.
View attachment 91469

That picture shows two different points/surfaces at two different elevations. The principle we are discussing is the pressure at a single point.

Let's go back to your original statement in the OP:

I am having trouble with the notion that pressure is applied to all sides uniformly.

This statement is not a correct statement of the law you are trying to investigate. The correct statement is pressure at a point is the same in all directions.

 

[UMath1]

So let's say we take the point to be the black dot in between the two pressure vectors in my diagram. Then the pressure applied from that point is the same upwards on the blue surface as it is downwards. So then shouldn't the blue dot be subject to the same pressure as the black dot?

 

[russ_watters]

So let's say we take the point to be the black dot in between the two pressure vectors in my diagram. Then the pressure applied from that point is the same upwards on the blue surface as it is downwards. So then shouldn't the blue dot be subject to the same pressure as the black dot?

I guess I'm not clear on what the diagram shows. I see a black dot on a black surface and a blue dot on a blue surface above it. You've drawn two force vectors connected to the black dot, equal in magnitude and opposite in direction. All fine.

The blue and black surfaces are at different elevations, so the pressure at the blue and black dots are different. Why would you think otherwise?

You seem to be saying the black dot is applying a force on the blue dot. How? They aren't touching each other.

 

[UMath1]

I don't understand conceptually why the pressure is different at different altitudes. If we say its because air closer to the surface of the Earth (S1) is subject to the total pressure of the air (Pt) above it, than I have some kind of paradox because then the air at the bottom (S1) exerts a pressure upwards that is equal to the total pressure (Pt) of all the air above it. Then the air right above S1 is also subject to a pressure Pt, and then it should also exert a pressure Pt.

 

[russ_watters]

I don't understand conceptually why the pressure is different at different altitudes. If we say its because air closer to the surface of the Earth (S1) is subject to the total pressure of the air (Pt) above it, than I have some kind of paradox because then the air at the bottom (S1) exerts a pressure upwards that is equal to the total pressure (Pt) of all the air above it. Then the air right above S1 is also subject to a pressure Pt, and then it should also exert a pressure Pt.

The paradox is that you are improperly mixing zero dh and non-zero dh in the same statement.

 

[CrazyNinja]

The paradox is that you are improperly mixing zero dh and non-zero dh in the same statement.

Add the fact that density of air is not a constant to that and you will figure it out.

 

[russ_watters]

Add the fact that density of air is not a constant to that and you will figure it out.

Well, whether density is constant changes the shape of the gradient, but doesn't change the fact that there IS a gradient.

The gradient is caused ultimately by gravity acting on particles with mass that occupy non-zero volume. If the volume (density) is constant or nearly so like in water, the gradient is constant. If they change like in air, it gets steeper with depth.

 

[UMath1]

I am not sure I follow. Where am I mixing the two up?

 

[russ_watters]

I am not sure I follow. Where am I mixing the two up?

Ok:

I don't understand conceptually why the pressure is different at different altitudes.

Clearly, that's a non-zero dH. Different altitude = different height. So you want to know about non-zero dH.

...the air at the bottom (S1) exerts a pressure upwards that is equal to the total pressure (Pt) of all the air above it.

That's true and it's all at point S1. One point = no height difference. You've already violated your own premise/question.

Then the air right above S1 is also subject to a pressure Pt...

The air above S1 is at a different height. So it must have a different pressure. You're back to your original premise/question but using the faulty premise of the previous quote and thus getting the wrong answer.

Something above something else must have a different height!

If you want to zoom way in on the scenario and look at individual packets of air, each has a weight W. If there's 100,000 stacked on top of each other, they exert a force on the surface they are sitting on of 100,000W. But the packet at the bottom only has 99,999 packets of air sitting on it, so the force applied to the top of that backet is only 99,999W. The forces on the top and the bottom of the packet are different! They are at different heights, so the statement that pressure at a point is the same in all directions does not apply there.

The problem, again, is that you are flipping back and forth between non-zero and zero height differences. Possibly based on non-zero and zero particle/air volume sizes.

 

[UMath1]

That makes more sense. What I don't understand though is that if the forces on the top and the bottom of the packet are different, how come the packet doesn't experience a net upward force and accelerate upwards? I think this is what is confusing me.

 

[russ_watters]

That makes more sense. What I don't understand though is that if the forces on the top and the bottom of the packet are different, how come the packet doesn't experience a net upward force and accelerate upwards? I think this is what is confusing me.

Draw a free body diagram of the packet. Make sure you include all three forces. I assure you, they will sum to zero.

 

[UMath1]

Oh..I forgot to include the force of gravity on the packet. Why is though that we calculate the pressure of a packet based solely on the packet on top of it. If we look at the second to last packet on the ground. It experiences a downward pressure of 99,998W/a and an upward pressure of 99,999W/a. So doesn't have a net pressure of 1 W/a?

 

[russ_watters]

Oh..I forgot to include the force of gravity on the packet. Why is though that we calculate the pressure of a packet based solely on the packet on top of it.

[edit: misread]
There is no such thing as "pressure of a packet". There is only pressure at a point or on a surface (when uniform). The packet of air has more than one surface that cover an infinite number of points. There are an infinite number of different pressures being applied to it.

If we look at the second to last packet on the ground. It experiences a downward pressure of 99,998W/a and an upward pressure of 99,999W/a. So doesn't have a net pressure of 1 W/a?

I don't think I've ever heard of a concept called "net pressure" applied in such a case. I might use it when describing a differential pressure between two different places though.