Consider the velocity field of the particles in the air mass in a 2D projection of the X(forward) and Z(up) axes. For each particle, Integrate over area and time, to derive the center of air-mass momentum (p) before and after the passage of the airplane : dp/dt. (On a very calm morning, with no wind or turbulence, the center of air-mass and its momentum is stationary in Z(assume level un-accelerated flight), and equal to the True Airspeed in X pointing in the aft -X direction. Integrate over the area and you will find that the center and momentum of the particle and vector field has changed, with passage of the plane. This center of air mass and center of momentum will move forward(+X) and downward(-Z) relative to its original state. The equal and opposite momentum change with time dp/dt of the airplane is a force. We might label the -X component "drag" and the +Z component "lift"(careful: the airplane coordinate system is different from the stationary airmass). This is a dissipative system, so don't wait too long after the plane passes to record the vector field. We can observe this process in contrails on clear days when the high altitude air is cold and relatively moist. Sadly since we mostly view them from below with a projection along the Z, we miss the downward component of the momentum field. You can see this as a test pilot flying as chase wing-man in formation. Expand this model to 3D to include lateral or Y axis flow and effects! I suggest this "p-dot"(dp/dt) of momentum-change explanation is better, than "pushing" or "pulling" the air downwards, because the later may confuse position and momentum in the view of the reader. This is also the first term(LHS) in the beautiful Euler-LaGrange equation, which would lead to an even more elegant analysis of this question! As a new user, I will need to figure out how to attach the appropriate Figures and Equations to this post...-thanks

Note: The drag equation is really the ideal gas law, except density replaces m/V.

P/rho = R T : <v^2>

Consider the velocity field of the particles in the air mass in a 2D projection of the X(forward) and Z(up) axes. For each particle, Integrate over area and time, to derive the center of air-mass momentum (p) before and after the passage of the airplane : dp/dt. (On a very calm morning, with no wind or turbulence, the center of air-mass and its momentum is stationary in Z(assume level un-accelerated flight), and equal to the True Airspeed in X pointing in the aft -X direction. Integrate over the area and you will find that the center and momentum of the particle and vector field has changed, with passage of the plane. This center of air mass and center of momentum will move forward(+X) and downward(-Z) relative to its original state. The equal and opposite momentum change with time dp/dt of the airplane is a force. We might label the -X component "drag" and the +Z component "lift"(careful: the airplane coordinate system is different from the stationary airmass). This is a dissipative system, so don't wait too long after the plane passes to record the vector field. We can observe this process in contrails on clear days when the high altitude air is cold and relatively moist. Sadly since we mostly view them from below with a projection along the Z, we miss the downward component of the momentum field. You can see this as a test pilot, flying as chase wing-man, in formation (projection in the Y-Z plane from behind or X-Z from the side). Expand this model to 3D to include lateral or Y axis flow and effects! I suggest this "p-dot"(dp/dt) of momentum-change explanation is better, than "pushing" or "pulling" the air downwards, because the later may confuse position and momentum in the view of the reader. This is also the first term(LHS) in the beautiful Euler-LaGrange equation, which would lead to an even more elegant analysis of this question!

As a new user, I will need to figure out how to attach the appropriate Figures and Equations to this post...-thanks

Note: The drag equation is really the ideal gas law, except density replaces m/V.

P/rho = R T : <v^2>