Register to reply 
Supplee's submarine paradox 
Share this thread: 
#19
Oct2712, 08:21 PM

Emeritus
Sci Advisor
P: 7,620

Hmmm  I snarfed it from somewhere, it's labelled as the x axis. Oh well. That's a standard plot for accelerated motion. The 3d diagram adds an x axis, which you can't see in the 2 d diagram. The x axis for an observer at rest is the red line. The xaxis for a moving observer is the green line The time axis on the 3d diagram is the blue line  it's the time axis for an observer "at rest". There isn't any explicit line for the z axis on the 3d diagram, but it's orthogonal to t and x. 


#20
Oct2712, 10:40 PM

P: 1,162

In the accelerating system an inertial particle moving along the z axis at some initial positive x would be charted as a curve with motion also in the x direction. So from an inertial frame the accelerating car (?) across the floor would be charted as a curve with a component of motion in the +x direction. I was wondering if this was what you were referring to as curvature of the floor. Within a Rindler system the metric is static and such motion would be equivalent to driving a car with constant velocity some distance on a flat road on earth. Effectively inertial, disregarding the constant g . In that context would you say the road was curved? Since a line of simultaneity is limited to an instant of equal proper time by definition I am not sure how to interpret the green line that extends through a range of coordinate time per your blue line??? Thanks for your response ,,, 


#21
Oct2812, 12:02 PM

Emeritus
Sci Advisor
PF Gold
P: 5,597




#22
Oct2812, 12:40 PM

Emeritus
Sci Advisor
PF Gold
P: 5,597

Cool stuff, pervect!
I asked my students last week to email me examples of relativity paradoxes that we could discuss in class. I'd expected them to google and come up with straightorward, wellknown examples that I could try to lead them through with a little Socratic dialog. One of my students sent me this one, which was pretty challenging. In the time I had available before the next class, I tried to come up with some kind of explanation that would be understandable to these students, who know very little relativity at this point. The handout below is what I was able to work out. At step D, I'm unsure about the validity of the last step, since it's not clear to me how the force fourvector relates to the force 3vector and actual measurements with the bathroom scales. Step E shows the situation as a spacetime diagram (with t horizontal), in the idiom used by Takeuchi and Mermin, where you represent a frame of reference as a coordinate grid, and a Lorentz transformation as a distortion of the grid. I warned my students that this analysis was probably not 100% correct, but I thought it probably showed the most important factors in resolving the paradox. My own attempt at a simplified mathematical analysis of D (the two cars on the bathroom scales) was as follows. Let 1's frame be (t',x',y') and the elevator's frame (t,x,y). In the elevator's frame, 2's motion looks like x=vt, y=(1/2)at^{2}. Transforming to 1's frame, I get this: [tex]y'=\frac{1}{2}a\left(\frac{1}{1v^2}\right)\left(1v\cdot\frac{2v}{1+v^2}\right)t'^2[/tex] The first factor in parens equals γ^{2}, while the second factor can be interpreted as the nonsimultaneity effect described in E. The fact that the first factor isn't described in the handout implies that the handout is oversimplifying a little. The whole expression can be simplified to (1/2)a(1/(1+v^{2})t'^{2}. The distortion of the elevator seems to match up with pervect's conclusion, although it hadn't occurred to me that it could be interpreted in terms of boosts not commuting with rotations. It also seems to match up with the statement about the shape of the "container" in the WP article. 


#23
Oct3112, 04:18 AM

P: 1,162

Am i following you ?? Wouldn't this also mean the weight on the scales in the elevator would be increased by the gamma factor of their velocity relative to the elevator? I am still unclear how this leads to a perception in car 1 of curvature of the floor??? 


#24
Nov212, 02:43 AM

Emeritus
Sci Advisor
P: 7,620

I finally got the fermi transport approach to work  as a series in time, so that it's assumed gT << 1.
Start out by finding the normalized 4 velocity for one of the moving cars in Minkowskii coordinates: [itex]u^a = \left[ \gamma \cosh gT, \gamma \beta, 0, \gamma \sinh gT \right][/itex] Note the magnitude of the 4acceleration is [itex]\gamma g[/itex] Proceed to write and solve the Fermiwalker transport equations. The four velocity is of course transported to itself, the other three basis vectors are: [itex]\hat{x} = \left[(\gamma\,\beta1/2\,{\gamma}^{3}{g}^{2}\beta\,{T}^{2}+1/24\,{\gamma}^ {3}{g}^{4}\beta\, \left( 4+{\gamma}^{2} \right) {T}^{4}+ ... ),(\gamma1/2\,\gamma\,{g}^{2} \left( \gamma^2 \beta^2 \right) {T}^{2}+1/24\,{g}^{4} \left( \beta^2 \gamma^2 \right) ^{2} \gamma\,{T}^{4}+... ),0,(1/3\,{\gamma}^{3}{g}^{3 }\beta\,{T}^{3}+1/30\,{g}^{5}{\gamma}^{3} \left( {\gamma}^{2}2 \right) \beta\,{T}^{5}+...)\right] [/itex] [itex]\hat{y} = \left[0,0,1,0\right] [/itex] [itex] \hat{z} =\left[ [({\gamma}^{2}gT1/6\,{\gamma}^{2}{g}^{3} \left( {\gamma}^{2}2 \right) {T}^{3}+{\frac {1}{120}}\,{g}^{5}{\gamma}^{2} \left( {\gamma} ^{4}8\,{\gamma}^{2}+8 \right) {T}^{5}+...,({ \gamma}^{2}g\beta\,T1/6\,{g}^{3}{\gamma}^{4}{\beta}^{3}{T}^{3}+{ \frac {1}{120}}\,{g}^{5}{\gamma}^{6}{\beta}^{5}{T}^{5}+...),0,(1+1/2\,{\gamma}^{2}{g}^{2}{T}^{2}1/24\,{g}^{4}{\gamma} ^{2} \left( 3\,{\gamma}^{2}4 \right) {T}^{4}+...)\right] [/itex] Integrate the 4velocity to give the position P(T) (I also expanded it in a series). [itex]P(T) = \left[\gamma \sinh gT / g, \gamma \, \beta T, 0, \gamma \cosh gT / g \right] [/itex] Write the transform equations to Minkowski coordinates [itex] (t,x,y,z) = P(T)+X \hat{x} + Y \hat{y} + Z \hat{z} [/itex] Carried out to second order, this gives a coordinate transform from Fermi normal coordinates to Minkowskii coordinates (as a time series to second order in T). I have the results to higher order, they're just unwieldy, so I posted the secondorder results. [itex] t = X\beta\,\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) T\frac{1}{2} \, X\,\beta\,{g}^{2}\,\gamma^3\,{T}^{2}[/itex] [itex]x = X\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) \beta\,T\frac{1}{2} \,X\,{\beta}^{2}\,{g}^{2}\,\gamma^3\,{T}^{2}[/itex] [itex]y=Y[/itex] [itex]z=Z+ \left( \frac{1}{2}\,g\gamma+{\frac{1}{2} {Z{g}^{2}}\gamma^2} \right) {T }^{2}[/itex] (Note that z=0 when t=0, this was dropped when P(T) was expanded in a series in T). Transform the metric (using automated software) to confirm that it's Lorentzian to order T^2 Look at the interesting part  g_00, and note that [itex]\partial g_{00} / \partial X[/itex] is zero at T=0, but becomes nonzero as T advances. So we see something similar to Thomas precession here, though I don't think the formulae are quite the same. [itex]g_{00} =\left(1+\gamma\,g\,Z\right)^2+2\,X{g}^{2}\beta\, \left( 1+\gamma\,Zg \right) {\gamma}^{2}T+{g}^{2} \left( {\gamma}^{2} {Z}^{2}{g}^{2}\beta^2\,g^2\,X^2+1+2\,\gamma Z\,g \right) {\gamma}^{2}{T}^{2} [/itex] So  the less mathematical summary. In fermi normal coordinates of the cars, you have a nice diagonal metric, with an acceleration due entirely to g_00 that changes components (in the fermi normal basis) with time. This can probably be traced to the changing shape of the floor of the spaceship. The ferminormal description of events doesn't have any velocity dependent forces, just rotations of the basis vectors caused by combining Lorentz boosts in two differnt spatial directions. 


#25
Nov712, 02:00 AM

Emeritus
Sci Advisor
P: 7,620

Using the relations I derived earlier
[itex] t = X\beta\,\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) T\frac{1}{2} \, X\,\beta\,{g}^{2}\,\gamma^3\,{T}^{2}[/itex] [itex]x = X\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) \beta\,T\frac{1}{2} \,X\,{\beta}^{2}\,{g}^{2}\,\gamma^3\,{T}^{2}[/itex] [itex]y=Y[/itex] [itex]z=Z+ \left( \frac{1}{2}\,g\gamma+{\frac{1}{2} {Z{g}^{2}}\gamma^2} \right) {T }^{2}[/itex] and knowing that for car2, to second order, the minkowski coordinates are [itex]t = \gamma \lambda[/itex] [itex]x = \beta \gamma \lambda [/itex] [itex]z = (g/2) \gamma \lambda^2 [/itex] I get the following for the ferminormal coordinates (T,X,Z) of car2 (by assuming there's a second order series expansion in lambda for (T,X,Z) and using the method of undetermined coefficients) [itex] T = \gamma^2\left(1+\beta^2\right) \lambda [/itex] [itex] X = 2 \beta \gamma^2 \lambda [/itex] [itex] Z = 2g\left(\gamma^21\right)\gamma^3 \lambda^2 = 2g\beta^2\gamma^5 \lambda^2 [/itex] [itex]\lambda[/itex] above, should be interpreted as some affine parameter, not equal to proper time. Normalizing it to proper time will be tricky unless we approximate g_00 as being unity. The secondorder expression for the floor of the rocket, z = (1/2) g t^2 in minkowskii coordinates, is a mess in fermi coordinates. If we set T=0, though, it becomes managable: [itex] Z = (1/2) g \gamma^2 X^2 \beta^2 [/itex] I attribute the difference from Ben's diagrams as Ben's being drawn at some time T > 0. The product of the fermwalker transported basis vector [itex]\hat{x}[/itex] and the [0,1,0,0] minkowskii basis vector is to second order [itex]\gamma \left(1  (1/2)g^2\gamma^2 \beta^2 T^2 \right) [/itex]. The derivative of this with respect to T is zero, so the initial relative rotation rate between the two is zero, but the rotation become nonzero as T increases. This is becaue the dot product gives [itex]\sin \theta \approx \theta[/itex]. 


#26
Nov2612, 07:37 AM

Emeritus
Sci Advisor
P: 7,620

One last set of coordinate transforms, with slightly different notation
let (t,x,z) be the inertial coordinates. Then the hyperbolic coordinates [itex](T_h, X_h, Z_h)[/itex] of the accelerating spaceship (where the floor is flat and the middle car is stationary) are defined implicitly by the transformation to the inertial coordinates: [tex]t = \left(\frac{1}{g}+Z_h\right) \sinh gT_h \approx T_h + g\,Z_h\,T_h[/tex] [tex]x = X_h[/tex] [tex]z = \left(\frac{1}{g}+Z_h\right) \cosh gT_h\frac{1}{g} \approx Z_h + \frac{g\,T_h^2}{2}[/tex] The approximation here is to assume that [itex]T_h, X_h, Z_h[/itex] are all "small" and of the same order, and to include all terms of total degree less than three (i.e include terms of orders 0,1, and 2). We will use the same secondorder approximation for [T,X,Z], our fermi coordinates. Within this approximation, we can approximately write the fermi coordinates [T,X,Z] which we computed previously. These fermi coordinates are for an observer moving on the spaceship floor in the +X direction with velocity [itex]\beta[/itex] and [itex]\gamma = 1/\sqrt(1\beta^2)[/itex]. The process of doing this was not pretty, even with the approximations used. The results are: [tex]T = \gamma\,{\it T_h}\gamma\,\beta\,{\it X_h}+g\gamma\, \left( 1\gamma \right) {\it Z_h}\,{\it T_h}+g\beta\,{\gamma}^{2}{\it Z_h}\,{\it X_h} [/tex] [tex]X = \gamma\, \left( {\it X_h}\beta\,{\it T_h}\beta\,g\,{\it Z_h}{\it T_h} \right) [/tex] [tex]Z = {\it Z_h}+\frac{1}{2}\, \left( 1{\gamma}^{3} \right) g{{\it T_h}}^{2}+g{\gamma} ^{3}\beta\,{\it X_h}\,{\it T_h}\frac{1}{2}\,g{\gamma}^{3}{\beta}^{2}{{\it X_h}}^ {2} [/tex] The results vaguely resemble a Lorentz transform, with the presence of other terms. One of these other terms is rather interesting, a quadratic in [itex]T_h[/itex] which will give a nonzero coordinate acceleration for Z even when [itex]Z_h[/itex] is constant. 


Register to reply 
Related Discussions  
Total pressure of water against submarine  Introductory Physics Homework  7  
Submarine Fires Torpedo, Submarine's Recoil?  Introductory Physics Homework  16  
Submarine down  General Discussion  20  
What kinds of engineers usually build submarine?  General Engineering  8  
Make a miniature submarine  Introductory Physics Homework  1 