A very simple thought experiment has crossed my mind for which I cannot find a clear solution. Two U-tube manometers filled with oil ($\rho_0$) are located in a uniform gravitational field $g_0$. A solid sphere with mass $m_0$ is located on a mass-less piston with a base area of $A_0$ (See below figure-Left.), and another solid sphere with the same mass $m_0$ is located an arbitrary altitude away from the ground on a platform where is connected with a simple apparatus made of some fixed cogwheels and moving gearbars which are all mass-less. (See below figure-Right.) This apparatus changes the horizontal direction of the force – exerted by the column of the oil – into vertical without altering its magnitude. The vertical gearbar tends to push the mass $m_0$ upward. From the viewpoint of the observer at rest with respect to the manometers, the height of the oil for which the manometers are kept at equilibrium, is calculated to be:

$${\rho_0}{g_0}{\Delta h_0}=\frac{m_0 g_0}{A_0} \Rightarrow {\Delta h_0}=\frac{m_0}{\rho_0 A_0}$$

Manometers in SRT

Now consider the scenario from the viewpoint of an inertial observer who moves with a velocity $v$ in the $x$-direction towards the manometers. He believes that the lengths in the $y$-direction shall remain unchanged, and thus he anticipate that ${\Delta h}$, from his standpoint, must be equal to ${\Delta h_0}$. For the left manometer, he calculates:

$${\Delta h}=\frac{m}{\rho A}$$

However, the density of the oil and the area of the piston base – due to Lorentz contraction – are respectively changed into ${\rho=\gamma \rho_0}$ and ${A=\alpha A_0}$ where $\alpha=1/\gamma={\sqrt{1-v^2/c^2}}$. If mass is an invariant ($m=m_0$), for the left manometer, the moving observer finally calculates:

$${\Delta h}=\frac{m}{\rho A}=\frac{m_0}{\gamma \rho_0 \alpha A_0}={\Delta h_0}$$

Recall that the moving observer can also use the transverse mass $m=\gamma m_0$ to recalculate a correct result for the height of the oil in the left manometer. In this case, he should use ${\rho=\gamma^2 \rho_0}$ due to the oil transverse mass increase:

$${\Delta h}=\frac{m}{\rho A}=\frac{\gamma m_0}{\gamma^2 \rho_0 \alpha A_0}={\Delta h_0}$$

These are correct results. Nonetheless, how do the calculations go for the right manometer from the viewpoint of the moving observer with respect to the fact that $A$ is no longer Lorentz contracted ($A=A_0$)?

(Note that Einstein, Plank, Eddington, Balnusa, Ott and Landsberg were all in agreement with the fact that pressure is an invariant!)

  • $\begingroup$ Note that Einstein, Plank, Eddington, Balnusa, Ott and Landsberg were all in agreement with the fact that pressure is an invariant! I think you're confused here. Pressure is described by three of the diagonal components of the stress-energy tensor, so it isn't a Lorentz scalar. $\endgroup$ – Ben Crowell Mar 2 '18 at 17:00
  • $\begingroup$ @ Ben Crowell: Do you mean that, from the viewpoint of the moving observer, the pressure of the oil has different components inside the manometers? If so, I think that it is somewhat an odd claim/deduction! $\endgroup$ – Mohammad Javanshiry Mar 2 '18 at 17:37
  • $\begingroup$ ... That is, the observer who is stationary in the reference frame of the manometers, observes that the oil pressure is isotropic and scalar, whereas the moving observer claims vice versa! $\endgroup$ – Mohammad Javanshiry Mar 2 '18 at 17:44

You have to consider time dilation for the gears. Let's make this more clear. From the perspective of the observer at rest, there is a torque which makes the gears rotate. Although force makes things a bit complicated, this observer can find the rotation period of the gears. The period should be derived from the angular momentum and torque equation.

If you were to give boundary conditions like whether the period will be constant or not and other things, this question could be solved more easily. At any rate, the difference between these two frame is a $\gamma$ in the denominator. By considering the fact that the non-rest frame will see more torque (because the oil density is multiplied by $\gamma^2$, for example) and at the same time that, because of time dilation, the gear frequency is being decreased, these two effects will neutralize each other and as a result $\Delta h$ will be invariant for both observers.

  • $\begingroup$ The problem has nothing to do with time dilation because there is no motion for the toothed bars nor revolve the wheels around their axes. The system is at static equilibrium. Assume that only a tiny drop of oil in the right manometer can set the system in motion, but it lacks the drop indeed. $\endgroup$ – Mohammad Javanshiry Mar 6 '18 at 9:50
  • $\begingroup$ Of course it does. First before anything else there is a motion in x axis for wheels according to non-rest frame. Imagine gears as some rotating disk. You already know that $\delta h$ is derived by double integrating over F/m. That is, accelerating. by knowing this and considering this fact that time and force is relative, there won't be any problem. $\endgroup$ – Paradoxy Mar 6 '18 at 10:44

When the horizontal toothed bar is contracted to half of its rest length, it moves half the amount that the vertical toothed bar moves.

When the oil in the horizontal tube is contracted to half of its rest length, it moves half the amount that the oil in the vertical tube moves.

When the motion is halved, then the force is doubled. It's like a lever.

  • $\begingroup$ Sorry, but there is no motion in the apparatus; the system is at equilibrium, that is. Moreover, if there were any motion which is halved, the difference between the oil heights in the two manometers would be paradoxical per se. Indeed, both observers should agree with the equilibrium status of the system as well as the similar heights of the oils. $\endgroup$ – Mohammad Javanshiry Mar 4 '18 at 8:15
  • $\begingroup$ I tried to explain how relativistic plumbing and power transmission work. Somehow those two subjects seem important in this case. :) $\endgroup$ – stuffu Mar 5 '18 at 4:09

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