The solution to this paradox is highly informative as it gives a feel to tangible measurable effects of length contraction, time dilation, density increase and relativistic mass and how they are all inter related.
I will introduce the subject with 3 short thought experiments that will tell us how gravitational acceleration transforms with relative motion.
Thought Experiment 1) Consider a projectile that is fired horizontally from height h. Let's say the the object would normally take time (t) to fall to the ground from height (h), according to an observer standing on the surface the Earth. The equivalence principle tells us the horizontal velocity does not affect the rate it falls at.
We can conclude that the acceleration of gravity on an object with non zero velocity relative to the observer and the Earth is:
$$g_{EO} = g \tag{Eq1}$$
Thought Experiment 2) Now consider the same situation but the observer is comoving horizontally with the object. In this situation, the object only has motion relative to the Earth, so there is no gamma factor due to motion relative to the observer. Due to time dilation this observer will measure the time to fall as shorter than that measured by the ground observer by a factor of gamma and the acceleration to be greater by a factor of gamma$^2$.
$$g_{E} = g \ \gamma_E^2 = \frac{g}{(1-v_E^2/c^2)} \tag{Eq2}$$
Since there is no motion relative to the observer in this case, we have to conclude that the transformation of the acceleration is entirely due to the motion of the object relative to the gravitational body $(v_E)$.
Thought Experiment 3) Now consider the situation where the object has no horizontal motion relative to the Earth surface so it is falling straight down, but in this case, the Earth and the object are moving together relative to an observer in space. Now the object has motion relative to the observer but not to the Earth. This observer measures the time to fall as longer by a gamma factor proportional to the velocity of the object relative to himself and therefore sees the gravitational acceleration as slowed down by a factor of gamma$^2$:
$$g_{O} = g \ \gamma_O^{2} = g \ (1-v_O^2/c^2) \tag{Eq3}$$
When $|V_E| = |V_O| \ne 0$, as in the first scenario, multiplying the gamma factors in (2) and (3) gives us:
$$\gamma_E \gamma_O = \frac{\sqrt{1-v_O^2/c^2}}{\sqrt{1-v_E^2/c^2}} = 1 , $$
which is in complete agreement with conclusion (Eq1).
The above equations tell us that the gravitational acceleration experienced by an object moving horizontally in a weak gravitational field, is not only affected by the velocity relative to the observer, but also by the velocity relative to the gravitational body. When the object has motion relative to a gravitational body the acceleration it is subjected to, increases by a factor of gamma squared using the velocity relative to the gravitational body. See (Eq2) below. It is important to note that the gamma factors $\gamma_O$ and $\gamma_E$ derived below only apply to vertical acceleration and use the velocity of the object relative the observer ($v_O$) and relative to the Earth ($v_E$) respectively. The object can be any object with mass, e.g. a submarine or a block of water.
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Now we have a good idea how vertical acceleration due to gravity transforms when there is horizontal motion relative to a gravitational body, we are ready to tackle the submarine paradox itself.
Solution to the paradox:
By Archimedes' principle, the upward buoyant force on an object is equal to the weight of water that is displaced. We can write this as
$$F_{up} = m_{water}\ g$$
$$ \implies \rho_{water} \ V_{sub} \ g_{water},$$
where $(\rho_{water} )$ is the density of displaced water and $V_{sub}$ is the volume of the submarine (and therefore also the volume of the displace water). Density transforms as $(\rho')= \rho \gamma^2$ when the object is moving relative to the observer, due to relativistic mass increase and length contraction combined.
In the following equations ($\gamma$) without a subscript, is just the normal gamma factor of special relativity that is a function of the velocity of the object relative to the observer.
Initial Reference parameters: The submarine is neutrally buoyant when fully submerged and at rest with the water and the Earth, such that the downward force $F_{down} $ and the upward buoyancy force $F_{up}$ are equal in magnitude and both equal to the weight of the submarine so that $F_{down} = F_{up} = mg$, where m is the mass of the submarine and g is the regular acceleration of gravity.
Frame 1) When the sub is moving horizontally relative to a observer that is at rest with the Earth and the water, the upward buoyant force is:
$$F_{up1} = (\rho_{water} ) (V_{sub} \ \gamma^{-1}) \ g_{water} $$
$$\implies F_{up1} = \gamma^{-1} (\rho_{water} \ V_{sub} \ g_{water}) = \gamma^{-1} (m g) \tag{Eq4} $$
Note that in this frame the water is not moving relative to the observer so there is no change in the water density.
The downward force on the submarine in this scenario is:
$$F_{down1} = (m_{sub} \gamma)(g_{sub} \gamma_E^2 \gamma_O^2) = (m_{sub} \gamma)g_{sub}$$
$$\implies F_{down1} = \gamma (mg) \tag{Eq5}$$
The net force on the sub in this reference frame is:
$$F_1 = F_{up1} - F_{down1} = mg(\gamma^{-1} - \gamma) \tag{Eq6},$$
which is negative, so the sub sinks in this reference frame.
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Frame 2) When the observer is comoving horizontally with the sub, the sub has motion relative to the water and the Earth but not to the observer and the water has motion only relative to the observer, the upward force is:
$$F_{up2} = (\rho_{water} \gamma^2 ) (V_{sub}) (g_{water} \gamma_O^{2} ) = (\rho_{water} \gamma^2 ) (V_{sub}) (g_{water} \gamma^{-2} )$$
$$\implies F_{up2} = (\rho_{water} \ V_{sub} \ g_{water}) \tag{Eq7}$$
The downward force on the submarine according to the observer comoving horizontally with the sub in this scenario is:
$$F_{down2} = m_{sub} (g_{sub} \gamma_E^2) = m_{sub} (g_{sub}\gamma^2) $$
$$\implies F_{down2}= \gamma^2 (mg) \tag{Eq8}$$
The net force (Eq7) - (Eq8) in this reference frame is:
$$F_2 = F_{up2} - F_{down2} = mg(1 - \gamma^2) \tag{Eq9},$$
which is negative, so the sub also sinks in this reference frame and the paradox is resolved.
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Note that the ratio between the net force in scenario (S1) where the sub is moving relative to the observer and the net force in scenario (S2) where the sub is at rest with respect to the observer is:
$$\frac{(Eq6)}{(Eq9)} = \frac{F(\gamma^{-1} - \gamma)}{F(1 - \gamma^2)} = \frac{1}{\gamma},$$
which is in complete agreement with the Lorentz transformation of transverse force with relative motion.
Equations (6) and (9) are also in agreement with the equations derived by Supplee, Matsas and Vieira, but here we only used the equations of special relativity (and the equivalence principle) and arrived at the same result.
In conclusion, changes in density do have the effects assumed in the naïve interpretation of the submarine paradox, but the paradox is resolved by observing that the acceleration of gravity on an object is affected by the horizontal motion of the object relative to the gravitational body.
