Supplee's submarine paradox


by bcrowell
Tags: paradox, submarine, supplee
pervect
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#19
Oct27-12, 08:21 PM
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Quote Quote by Austin0 View Post
Very interesting but hard to visualize.
Let me see if I am getting it at all.

If there is a vertical line for the x axis with a horizontal line crossing it being the z axis and the floor. With the y axis projecting from the intersection towards us.
no explicit time axis. Time being the floor moving up the page.
Then a particle moving horizontally from some z point across the floor with constant motion must describe a non-linear curve over any time interval due to acceleration. SO in this sense the floor is curved as described by this motion. Is this at all what you are talking about???
This 2d diagram has an explicit time axis (time for an observer at rest) going up the page, and an explicit z axis, going to the right.

Hmmm - I snarfed it from somewhere, it's labelled as the x axis. Oh well. That's a standard plot for accelerated motion.

The 3-d 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 x-axis for a moving observer is the green line

The time axis on the 3-d 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 3-d diagram, but it's orthogonal to t and x.
Austin0
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Oct27-12, 10:40 PM
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Quote Quote by pervect View Post
This 2d diagram has an explicit time axis (time for an observer at rest) going up the page, and an explicit z axis, going to the right.

Hmmm - I snarfed it from somewhere, it's labelled as the x axis. Oh well. That's a standard plot for accelerated motion.

The 3-d 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 x-axis for a moving observer is the green line

The time axis on the 3-d 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 3-d diagram, but it's orthogonal to t and x.
Sorry if I was unclear. I was not describing your diagram. I was describing a simpler orthographic view from an inertial frame where the idea was a series of superposed time snapshots of the floor moving up the page with the object moving across the floor.
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 ,,,
bcrowell
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Oct28-12, 12:02 PM
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Quote Quote by greswd View Post
i think the modern consensus is that this paradox is outdated.
This would seem to be contradicted by the fact that the Matsas paper was published by Phys.Rev. D in 2003.

Quote Quote by greswd View Post
Another way to look at it is that gravitational mass is unaffected by inertial mass.
Well, not really.
bcrowell
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Oct28-12, 12:40 PM
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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, well-known 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 four-vector relates to the force 3-vector 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)at2. Transforming to 1's frame, I get this:

[tex]y'=\frac{1}{2}a\left(\frac{1}{1-v^2}\right)\left(1-v\cdot\frac{2v}{1+v^2}\right)t'^2[/tex]

The first factor in parens equals γ2, 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+v2)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.
Austin0
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Oct31-12, 04:18 AM
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Quote Quote by bcrowell View Post
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, well-known 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 four-vector relates to the force 3-vector 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)at2. Transforming to 1's frame, I get this:

[tex]y'=\frac{1}{2}a\left(\frac{1}{1-v^2}\right)\left(1-v\cdot\frac{2v}{1+v^2}\right)t'^2[/tex]

The first factor in parens equals γ2, 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+v2)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.
As I understand your analysis at any given moment from car 1 , the position of car 2 is closer to the fulcrum than car 1 ,correct? In fact the ratio of the two distances from the center is gamma. SO if they balance this would imply that the inertial mass/energy increase by gamma of car 2 was directly equivalent to gravitational mass increase.
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???
pervect
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Nov2-12, 02:43 AM
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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 4-acceleration is [itex]\gamma g[/itex]

Proceed to write and solve the Fermi-walker transport equations. The four velocity is of course transported to itself, the other three basis vectors are:

[itex]\hat{x} = \left[(\gamma\,\beta-1/2\,{\gamma}^{3}{g}^{2}\beta\,{T}^{2}+1/24\,{\gamma}^
{3}{g}^{4}\beta\, \left( -4+{\gamma}^{2} \right) {T}^{4}+ ... ),(\gamma-1/2\,\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}gT-1/6\,{\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\,T-1/6\,{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 4-velocity 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 second-order 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 fermi-normal 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.
pervect
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#25
Nov7-12, 02:00 AM
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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 fermi-normal 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^2-1\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 second-order 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 ferm-walker 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 non-zero
as T increases. This is becaue the dot product gives [itex]\sin \theta \approx \theta[/itex].
pervect
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#26
Nov26-12, 07:37 AM
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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 second-order 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 non-zero coordinate acceleration for Z even when [itex]Z_h[/itex] is constant.


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