# Bernoulli equation and wind

Bernoulli equation is often used to explain how rooftops are blown off during tornados: the usual explanation is that the high speed of the wind il related to a lower pressure outside the house, compared to the one inside; hence we have a net force lifting the roof.

I was trying how this description agrees with relativity. I tried to consider this simple example: consider two tunnels separated by a wall. In tunnel $$1$$ the air flows with a certain velocity $$\vec{v}$$ in one direction, while in tunnel $$2$$ the air is still. I wold expect a lower pressure in tunnel $$1$$, and hence a net force acting on the wall "from $$2$$ to $$1$$". Considering the reference frame moving with velocity $$\vec{v}$$, I would expect the opposite situation, i.e. a net force acting on the wall "from $$2$$ to $$1$$".

Where is the mistake in this naive reasoning? I've read something about Bernoulli equation in different reference frames, but the key point there was the work done by the surfaces, and I don't see how it could play a role here.

• You need to think of the velocity of the air in each tunnel relative to the tunnel wall. In one frame the wall is still. In the other, not. Commented Jun 12, 2023 at 13:59
• What I don't understand is how the movement of the wall plays a role since is perpendicular to the flow... Commented Jun 12, 2023 at 14:06
• "the usual explanation is that the high speed of the wind il related to a lower pressure outside the house, compared to the one inside; hence we have a net force lifting the roof." Typical misunderstanding of Bernoulli Commented Jun 13, 2023 at 23:17
• Items moving at relativistic speeds are way outside the regime where the Bernoulli relationship is valid. Bernoulli is not even valid when flow speed approaches the speed of sound, let alone the speed of light. I think you may be trying to find a common thread between apples and oranges Commented Jun 14, 2023 at 1:16
• @RC_23 OP is asking just about relative motion, not special relativity Commented Jun 14, 2023 at 2:23

Bernoulli's equation does not state that a fluid in motion is at a lower pressure than a fluid at rest. Rather, it relates pressure changes to changes in velocity. If the upstream and downstream velocities are the same, Bernoulli's equation indicates that the pressures are the same.

I think that this is what is happening in your case, with perfect fluid flowing along a wall parallel to the flow. If you change your reference frame, the velocities changes but they remain the same and the pressures are not modified.

In the case of a pipe whose section changes, Bernoulli's equation is no longer applicable in the frame of reference in which the pipe is in motion.

See the following link: The Bernoulli equation in a moving reference frame

Hope it can help and sorry for my poor english.

This is a subtle little problem that has to do with what Bernoulli's equation actually states; that there exists a relationship between the pressure $$p$$ and the velocity $$\vec{v}$$ such that, in the absence of external work/forces, $$p + \frac{\rho}{2}\vec{v}\cdot\vec{v} = C$$ where $$C$$ is a constant, representing a sort of energy density, and $$\rho$$ is the density.

You have correctly spotted that the $$\frac{\rho}{2}\vec{v}\cdot\vec{v}$$ term is not Galilean-invariant. The challenge is that, neither is $$C$$! You can't perform a change of reference frame and use the old value of $$C$$ and the transformed value of $$\vec{v}$$ to calculate the pressures in the new frame of your system; you need additional information about how your system is set up in order to solve for the pressures, keeping in mind that external work is also not Galilean-invariant.