# Does pressure at a point in a fluid flow change with respect to different frames?

Let there be a cylindrical pipe through which a fluid is flowing.

Between 2 points at the same height: $$p_1+\frac12\rho(v_1)^2 = p_2+\frac12\rho(v_2)^2$$ Thus, $$p_1 - p_2 = \frac12\rho[(v_2)^2 - (v_1)^2] = \frac12\rho[(v_2+v_1)(v_2-v_1)]$$

Now, if I am using the same equation from the frame of an observer moving opposite to the fluid with speed $$v_3$$ m/s, then $$p_1 - p_2 = \frac12\rho[\{(v_2+v_3)+(v_1+v_3)\}\{(v_2+v_3)-(v_1+v_3)\}]$$ $$p_1 - p_2 = \frac12\rho[(v_2+v_1+2v_3)(v_2-v_1)]$$

This means that the pressures at those points are changing with respect to different observers, but fundamentally, they should remain same, as $$P=\frac{dF}{dA}$$ always, and neither force nor cross-sectional area changes.

• @ShubhamGoel suppose in your reference frame the velocity of the pipe is $u$, whereas the velocity of liquid is $w$. The relative velocity in the Bernoulli equation is $v = w-u$. So switching to another reference frame, $u'=u+U, w'=w+U$ does not change the relative velocity; $w'-u'=w-u=v$. Commented Feb 13 at 12:28