# The Relativistic Boat

1. Dec 14, 2009

### genesic

This is a thought experiment that my friends and I (physics graduate students all) were pondering while we should have been doing Jackson homework. Consider a boat traveling at relativistic speeds. From the boat's frame, the water is contracted, and therefore is more dense. This would increase the buoyancy of the water and keep the boat afloat. However, from the water's frame, the boat is contracted, making it smaller and denser. Since a smaller amount of water is displaced, there is a smaller buoyant force, and so the boat sinks. So what actually happens?

2. Dec 15, 2009

### Jorrie

The standard solution is that when gravity is also considered, the 'relativistic boat' tends to sink, e.g.: http://arxiv.org/abs/gr-qc/0305106" [Broken].

IMO, another (easy) way to look at it is that the radial acceleration caused by Earth is larger than 1g when the boat is moving relativistically in a transverse direction, i.e., when it has an angular velocity relative to Earth's center of mass.

$$\frac{d^2 r}{d t^2} = \frac {3 m{{\it v_r}}^{2}}{ \left( r-2\,m \right) r} + \left( r-2\,m \right) \left( {{\it v_\phi}}^{2}-{\frac {m}{{r}^{3}}} \right)$$

as https://www.physicsforums.com/showpost.php?p=1046874&postcount=17", where $v_{\phi} =d\phi/dt$ (geometric units). It is clear that any non-zero $v_{\phi}$ causes an increase in ${d^2 r}/{d t^2}$.

[Edit: I erred when working out the limit when speed approaches c, so I took it out for now...]

[Edit2: I also realized that this can only work for a 'flat planet' scenario. For a normal, almost spherical planet, any relativistic speed will let the boat fly out of the water and possibly even reach escape velocity...]

Last edited by a moderator: May 4, 2017
3. Dec 15, 2009

### Jorrie

I would like to know if the following is a valid approach to the problem. In order to convert pervect's polar coordinate analysis:

$$\frac{d^2 r}{d t^2} = \frac {3 m{{\it v_r}}^{2}}{ \left( r-2\,m \right) r} + \left( r-2\,m \right) \left( {{\it v_\phi}}^{2}-{\frac {m}{{r}^{3}}} \right)$$

to a pseudo-Cartesian system for a 'flat planet' analysis, we can subtract the centrifugal acceleration $r v^2_{\phi}$ and also get rid of the angular velocity by replacing it with a horizontal (x) velocity: $v_x = r v_{\phi}$.

If we take the initial radial velocity $v_r = v_y = 0$, we get the initial vertical (Cartesian) acceleration of a free-falling submarine, moving at $v_x$ (with c=G=1):

$$\frac{d^2 y}{d t^2} = \left( r-2\,m \right) \left( \frac{{ v_x}^{2}}{r^2}-{\frac {m}{{r}^{3}}} \right) -\frac{{ v_x}^{2}}{r} = -\frac{m}{r^2}\left(1-\frac{2m}{r} + 2v_x^2\right)$$

In a weak field (1g), but high speed earth surface scenario, the vertical gravitational acceleration simply becomes: $a \approx (1 + 2v_x^2) [/tex] g, with g ~ -9.8 m/s[itex]^2$.

Last edited: Dec 15, 2009
4. Dec 15, 2009

### yuiop

This is basically the same as the "submarine paradox" that was investigated by George Matsas. See http://arxiv.org/abs/gr-qc/0305106

<edit> Oops, I just noticed that Jorrie has already posted that link.. sorry

P.S.

In an old thread https://www.physicsforums.com/showthread.php?t=225573&highlight=submarine&page=5 I worked out this aproximate weak field equation for the force acting on a particle moving horizontally in a gravitational field :

$$F' = \frac{GMm}{R^2} \frac{(1-Vv/c^2)^2}{(1-V^2/c^2)\sqrt{1-v^2/c^2}}$$

where:
V = horizontal velocity of the massive body (M) wrt the observer,
v = horizontal velocity of the test particle (m) wrt the observer,
M >> m.

This equation can be used to work out the buoyancy forces of the water or the force acting on the submarine, from the point of view of an observer at rest with the water or co-moving with the submarine.

Last edited: Dec 15, 2009
5. Dec 15, 2009

### Jorrie

I agree - both approximations 'resolve' the submarine paradox by increasing the gravitational force.