Why does moving air have low pressure?
It's all about conservation of momentum, $F=ma$. Fluid can only change velocity by experiencing a force, and the only force it can feel is a pressure difference. So if there's a velocity difference, there's a pressure difference, and vice-versa (neglecting other things, like gravity). Read this beautiful exposition.
If wikipedia says that, it's wrong. There is no physical requirement for the air to move faster or slower on either side of the wing. The best analogy I've found so far is to think of two layers of cotton batting, or two blankets, of infinite length. The aircraft wing separates them, but there is no need for the blankets to "line up" after the wing has passed.
If you read any half-decent text on aerodynamics, you will find that lift is almost 100% due to the angle of attack. The air below the wing is forced downwards, so momentum conservation forces the wing upwards. Aircraft can and have flown just find upside down.
The shape of the wing, with camber and trailing edge taper, etc., is almost entirely designed to minimize drag, eliminate stall vortices, and all sorts of other messy mathematical problems in turbulent flow. Lift is absolutely NOT due to reduced air density on top of the wing.
There are two main explanations, since I don't know your education level, first I will explain it in layman's terms even though this is not fully correct but you can encounter this explanation in some maybe most of the undergraduate physics textbooks. Then I will explain why this picture is wrong and how to fully understand the physics behind aircrafts.
- Airplane wings are designed in such a way that they are more bulgy at the top of an airplane's wing compared to the bottom, this plays a significant role. Have a look at the illustration:
Air molecules at the top of the airplane's wing should move faster than those at the bottom to meet at the back edge of the wing. Bernoulli's principle written as
$$P_1 +\rho g h_1 +\frac{1}{2} \rho v_{1}^2 = P_2 +\rho g h_2 +\frac{1}{2} \rho v_{2}^2$$
where subscript $2$ refers the bottom and subscript $1$ refers the top of the wing. Since the thickness of the wing thus the difference $h_2-h_1 \approx 0$ one can cancel those terms out and obtain
$$P_1 +\frac{1}{2} \rho v_{1}^2 = P_2 +\frac{1}{2} \rho v_{2}^2 $$
$$P_2-P_1 = \Delta P = \frac{1}{2} \rho (v_{1}^2-v_{2}^2)$$
Then (lifting) force on the aircraft
$$F=\Delta P \cdot A = \frac{1}{2} \rho (v_{1}^2-v_{2}^2)\cdot A$$
This also gives a good intuition why aircrafts malfunction when they reach a certain altitude since $\lim_{\Delta P \to 0} F =0$ so there will be no lift.
- $\textbf{1.}$ This above explanation is not fully correct because even when the length of the airplanes' wings were the same at the top and at the bottom, there would be a lift (consider paper-planes where you don't have bump on top of the wing).
- $\textbf{2.}$ Experiments have shown that the air molecules at the leading edge (from the top and bottom of the of the wing) do not meet at the end of the wing.
To have a complete understanding of the physics behind the working mechanism of aircrafts one should solve the Navier-Stokes equations. Since Navier-Stokes equations are generally too complex to solve, one can use an approximation by solving Euler equations which doesn't take the viscosity into account.
They then further simplify the Euler equations by assuming that compressibility is negligible. Thus in incompressible form, density $\rho$ is constant thus it oversimplifies to
$$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$$
$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-\frac{1}{\rho}\frac{\partial p}{\partial x}$$
$$u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}=-\frac{1}{\rho}\frac{\partial p}{\partial y}$$
But even these oversimplified set of differential equations are difficult to solve therefore one solves them numerically, write a program to simulate the behavior of air molecules at the top and bottom of the wings.