1. Oct 5, 2005

### skywolf

help!! supplies paradox was supposed to solve the problem of lorentz contraction when an object is flying through a fluid, but instead, it just replaces it with another. From the objects point of view it should be going up, but from the outside it should be going down. this happens wether you have supplies paradox or lorentz contraction paradox. Please help

thank you

2. Oct 5, 2005

### pervect

Staff Emeritus
What is "supplies" paradox? Do you have a link? I'm not familiar with it offhand, and Google gives too many false hits.

Offhand I'd expect that the properties of a fluid will change with your reference frame - a fluid that is nice and isotropic in its own reference frame won't maintain that isotropy if viewed from a relativistically moving frame.

3. Oct 5, 2005

4. Oct 5, 2005

### Jimmy Snyder

Last edited by a moderator: May 2, 2017
5. Oct 5, 2005

### skywolf

but it didnt explain why the bottom was not flat. what i got out of it was that the bullet was time contracted and therefore it did not accelerate as fast as the liquid. but if that was the case, then from the bullets frame, the liquid was time contracted and it did not accelerate as fast as the bullet, it seemed to trade one paradox for another. please clarify where i was mistaken.

thank you

6. Oct 6, 2005

### Jimmy Snyder

Actually, it does, but it is rather terse, I'll give you that. The key is here:

In other words: The bouyancy force causes an acceleration. It is transverse because it is perpendicular to the initial velocity, but that is not at issue here. By the equivalence principle this acceleration is equivalent to gravity. Gravity curves space and so curves the lake bottom.

7. Oct 6, 2005

### pervect

Staff Emeritus
I think this is a very interesting paradox, one that I haven't seen discussed before, and I thank you for posting it.

One thing I noticed right away that's false about the initial argument is that while the density of the fluid is proportional to gamma^2 as stated, the vertical component of pressure at any point (call it the z-component), which is more relevant, is unchanged by a boost. The buoyant force will in general be determined by the pressure, not the density of the fluid.

The indpendence of the z-component of pressure with respect to whether or not the fluid is flowing (in a direction perpendicualr to z, such as the x or y directions) is most easily seen using the stress-energy tensor, which is the way relativity deals with pressure directly. It can also be inferred by using a model in which the fluid consists of a "swarm of particles" with different velocities.

Basically, if one imagines a "particle clock", consisting of a particle moving up and down in the z direction, time dilation arguments indicate that the particle has a lower velocity in the z direction when it is moving in either the x or y direction. This causes the pressure to decrease. Other factors, such as the increase in particle density of the fluid, and the momentum/velocity relationship (relativistic mass, if you prefer) work to increase the pressure. The net result is again that the z component of the pressure stays constant.

Buoyant force should be $$\frac{\partial}{\partial z} T_{zz} V$$, where T_zz is the pressure component in the Z direction, and V is the volume of the "bullet" (which won't be the same in both frames).

However, I haven't untangled the entire paradox yet to my satisfaction.

8. Oct 27, 2005

### Jimmy Snyder

Subsequent to my last post in this thread, I have come to realize that my explanation is flawed. Interestingly enough, I also find that the wiki page has been substantially edited and the text to which I provided an incorrect explanation is no longer there. As Rosanna Rosannadanna would say, "never mind".

9. Oct 30, 2005

### pervect

Staff Emeritus
In a previous post, I talked about how the Z component of the pressure tranforms uncer a Lorentz boost (it remains constant, a very simple transform),.

Now I'll tackle what happens to the Christoffel symbols and equations of motion

The Rindler metric, which represents a uniform gravitational field oriented in the 'z' direction, is given by

$$ds1^2 = dt1^2 - (1+gz1)dx1^2+dy1^2+dz1^2$$

(Note: you may see slightly different forms of this metric from different authors - they are the same physical situation with a slightly different choice of coordinates).

The Christoffel symbols for this metric are (dropping the suffix 1 for ease of typing)

$$\Gamma^z{}_{tt} = g(1+gz) \hspace{.25 in} \Gamma^t{}_{tz} = \Gamma^t{}_{zt} = \frac{g}{1+gz}$$

The important Christoffel symbol here is $\Gamma^z{}_{tt}$. This is the sole symbol in the equation for the z component of motion, i.e.

$$m \left( \frac{d^2 z}{d\tau^2} + \Gamma^z{}_{tt} \left( \frac{dt}{d\tau} \right)^2 \right) = F_{ext} \frac{dt}{d\tau} = \gamma F_{ext}$$

Here F_ext is the z component of the external 3-force on the bullet, due to buoyant forces. The factor of dt/dtau multiplying F_ext converts the force component from a 3-force component dP/dt, to a 4-force, dP/$d\tau$.

If F_ext were zero, this would be the geodesic equation. Note m here is the invariant mass, not the relativistic mass.

We can re-write this as

$$\frac{d^2 z}{d t^2} = -\Gamma^z{}_{tt} + F_{ext} / (m \gamma)$$

We've previously calculated F_ext to be the constant pressure P_z of the fluid multiplied by the volume of the bullet in the moving frame, which is $P_z V_0 / \gamma$

If the bullet was neutrally buoyant in it's rest frame, $P_z V_0=m g$

Putting it all together, with the bullet at z=0, and the bullet being neutrally buoyant in its own rest frame, we get

$$\frac{d^2 z}{dt^2} = -g + \frac{g}{\gamma^2}$$

We thus see that the bullet sinks.

Now, let's redo this calculation in a boosted frame, the frame of the bullet. We make the variable substitutions

$$x = \frac{x'-\beta t'}{\sqrt{1-\beta^2}} \hspace{.5 in}t = \frac{t' - \beta x'}{\sqrt{1-\beta^2}}$$

with y = y' and z = z', we get the new metric

$$ds^2 = \frac{(1+gz+\beta)(1+gz-\beta)}{1-\beta^2}dt^2 + \frac{2\beta g z (2+gz)}{1-\beta^2}dx dt + \frac{(1 + \beta(gz+1))(1 - \beta(gz+1))}{1-\beta^2}dx^2 -dy^2 -dz^2$$

This generates numerous christoffel symbols, the ones we are interested in are
$$\Gamma^z{}_{tt} = \frac{g(gz+1)}{1-\beta^2} \hspace{.5in} \Gamma^z{}_{tx} = \Gamma^z{}_{xt} = \frac{\beta g (gz+1)}{1-\beta^2}$$

This gives us the equation of motion
$$m \left( \frac{d^2 z}{d \tau^2} + \Gamma^z{}_{tt} \left( \frac{dt}{d\tau} \right)^2 + 2 \Gamma^z{}_{tx} \left( \frac{dt}{d\tau} \right) \left( \frac{dx}{d\tau} \right) \right) = F_{ext}$$

We do not need to convert the 3-force into a 4-force in this case, because the bullet is stationary in its own frame. For the same reason, the velocity of the bullet dx/$d\tau$ = 0, and $dt/d\tau$=1.

Thus we can write

$$m (\frac{d^2 z}{d t^2} + \frac{g}{1-\beta^2} ) = F_{ext}$$

$$\frac{d^2 z}{d t^2} = -\frac{g}{1-\beta^2} + g$$

The buoyant 3-force in this frame is just m g, but the bullet still sinks, because of the way that $\Gamma^z{}_{tt}$ transforms. Effectively, the "felt" gravity in the frame of the bullet is increased by a factor of $\gamma^2$. There is also a velocity dependence of the "felt" gravity, but this factor is zero in the frame of the bullet (that's the christoffel symbol $\Gamma^z{}_{tx}$)

Last edited: Oct 30, 2005