If one moves a plate through the air at a uniform velocity $v$, the pressure exerted on the area $A$ of the plate is:

$$p=\frac{1}{2}\rho_{air}v^2 \space .$$

This pressure is measured by the observer located on the plate, who feels a wind $-$ moving at $v$ $-$ hits the plate. I think the pressure measured by the observer at rest with respect to the air should be:

$$p^\prime=\frac{1}{2}\rho^\prime_{air}v^2 \space ,$$

where $\rho_{air}=\gamma \rho^\prime_{air}$. Is the second formula correct or there should be some other relativistic corrections?

I think there must be a problem with my calculation because, as far as I know, pressure is a Lorentz-invariant $(p=p^\prime)$ in relativity, whereas the dynamic pressure measured by two inertial observers are seemingly not the same $(p=\gamma p^\prime)$ using the above equations.

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    $\begingroup$ Check out the pressure stress energy tensor for a perfect fluid "en.wikipedia.org/wiki/Perfect_fluid" - which is a second order tensor, namely, $T=(\rho +p )v\otimes v+pg$ where $v$ is a $4$-velocity, $g$ is the metric tensor - and $\rho$ and $p$ are the density and pressure, respectively. $\endgroup$ – Cinaed Simson Nov 13 '19 at 8:51

There are two possible ways to define force in relativity, either as the four-vector $d\textbf{p}/d\tau$ or as the a three vector $d\textbf{p}_3/dt$, where $\textbf{p}_3$ is the momentum three-vector. The four-force has nicer transformation properties, but the three-force is what an observer actually measures. Depending on which of these you pick, you will get a different definition of pressure. Newton's second law doesn't have a simple form in terms of the three-force. I have a discussion of this sort of thing in section 4.5 of my SR book http://www.lightandmatter.com/sr/ .

If you go the four-force route, then the pressure of a perfect fluid is defined as the relevant space-space component of the stress-energy tensor, in the fluid's rest frame, which is the frame in which the stress-energy is diagonal. This definition is stated in terms of a particular frame of reference, so it's automatically frame-invariant. However, the components of the stress-energy are not frame-invariant. As an example, the stress-energy of dust, in its rest frame, has the form $\operatorname{diag}(\rho,0,0,0)$ ($c=1$), so the pressure is zero. However, if you do a boost along the $x$ axis, it becomes non-diagonal, and its xx component becomes $\gamma^2v^2\rho$.

If you take the three-force as your definition of force, then under a boost parallel to the force, the force doesn't change. Since the area also doesn't change under a boost perpendicular to the area, I think it is probably true that the pressure is invariant. However, I would be careful with this interpretation, because I think under a boost along some other axis, you will pick up shear forces, so it's no longer clear what is meant by a pressure.

I think your example of the plate moving through a fluid is not very well chosen. Your approximation for drag holds for a certain range of Reynolds numbers, with a drag coefficient that isn't constant as you claim but rather varies with Reynolds number. For a perfect fluid, the viscosity is zero, so the Reynolds number is infinite, violating your assumptions. So for your formula to be of any interest, you'll have to consider a fluid that is not a perfect fluid. Then you would have to figure out its equation of state and what its stress-energy tensor looks like. This seems like a mess. The transformation properties of force and pressure don't depend on how they occur, so if you want a physically motivated example to look at for concreteness, I would choose a perfect fluid instead.

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