# What is the relativistic form of the dynamic pressure of a fluid? Is it a Lorentz-invariant quantity?

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.

• 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. – 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 $$\gamma^2v^2\rho$$.