1. Oct 24, 2012

### bcrowell

Staff Emeritus
Supplee's submarine paradox arises when you consider a submarine moving horizontally at relativistic speeds. (Never mind that this would obviously destroy the sub.) When at rest, the sub was neutrally buoyant. In the moving sub's frame, the water is more dense, so buoyancy is increased, and the sub feels a net upward force. In the water's frame, the moving sub's volume is decreased, so the sub sinks.

References:
J. M. Supplee, Am. J. Phys. 57, 75 (1989).
http://physics.aps.org/story/v12/st4
Matsas, 2008, http://arxiv.org/abs/gr-qc/0305106
http://en.wikipedia.org/w/index.php?title=Supplee's_paradox&oldid=55814645 -- a version blessed by Chris Hillman

Matsas does a complicated GR calculation. In the sub's frame, he gets

$$F_{tot}=-mg\gamma(\gamma-1/\gamma)$$

and interprets the first term as a gravitational force, the second as a buoyant force. Transforming the force into the water's frame, he gets

$$F_{tot}=-mg(\gamma-1/\gamma)$$.

Both observers agree that the sub sinks.

The one in the water's frame makes sense to me. The sub's mass-energy is higher than its rest mass, so it gets heavier. The sub's volume decreases, so the buoyant force is decreased.

In the sub's frame, I can easily believe that the gravitational force is increased. But why is the buoyant force *decreased* in the sub's frame? I'd think it would be increased, both because of the increase in the water's weight in proportion to its rest mass, and because of the increase in the water's density. Of course we can just say that it's because Matsas did a big, complicated GR calculation, and that's the result. But I'd like to get some understanding of why the calculation comes out as it does.

CH, in the WP article, summarizes Matsas by saying that this all has to do with the shape of the sub and the shape of the container. I don't see this anywhere in Matsas. Matsas does discuss the shape of the sub, but not the shape of the container. Matsas goes through a long discussion of the process by which the sub accelerates from rest, but it's not clear to me why it's necessary to model this process.

Last edited: Oct 24, 2012
2. Oct 24, 2012

### atyy

The link you give says Supplee's version does have the buoyant force increasing in the sub's frame.

Matsas seems to say the same too on his p4: "The apparently contradictory conclusion reached in the submarine rest frame by the mariners, who would witness a density increase of the liquid volume elements is resolved by recalling that the gravitational ﬁeld is not going to “appear” the same to them as to the observers at rest with the ﬂuid."

3. Oct 25, 2012

### pervect

Staff Emeritus
The relevant section seems to be:

The point is apparently that you'd expect the volume of the sub to decrease by gamma due to Lorentz contraction if you had a Minkowskii metric. But the metric is the Rindler metric, so the volume change is more complex than a simple Lorentz contraction for "thick" subs. I haven't figured out any more details than that it's "different", I'm not even quite sure whether it changes more or less.

4. Oct 25, 2012

### Fredrik

Staff Emeritus
Is there a SR version of this problem? First of all, there's no such thing as gravity in SR, and therefore no such thing as buoyancy. This completely eliminates the problem, unless we specify forces acting on the sub in some other way. If we do that, don't we have to deliberately contradict ourselves (e.g. by specifying the same forces in two different coordinate systems instead of specifying them in one and transforming to the other) to get any kind of paradox?

5. Oct 25, 2012

### bcrowell

Staff Emeritus
Re the shape of the container, the aps.org article does seem to say something about this, although Matsas doesn't. The apr.org article is describing the Am J Phys paper by Supplee, which I don't have access to. But this interpretation doesn't seem to appear anywhere in the Matsas paper.

The aps.org article describes the Supplee paper as doing it using SR in a frame where there's no gravitational field, and the ocean is accelerating upward.

The problem is that Matsas gives $$F_{tot}=-mg\gamma(\gamma-1/\gamma)$$ and interprets the first term as a gravitational force, the second as a buoyant force. This doesn't seem consistent with the statement that the buoyant force increases. Distributing the factor of gamma, the term interpreted as the buoyant force just comes out to be mg, the same as when the sub is at rest. (My statement in #1 that the equation showed buoyancy as being smaller was based on just the 1/γ part, which probably didn't make sense.) I don't understand what physical mechanism would result in the buoyancy staying the same. I would expect it to go up by a factor of γ due to the water's higher density of mass-energy, and up by another factor of γ because the gravitational field is perceived as being stronger.

If you look at fig. 1, the volume clearly does decrease as the sub accelerates, although I don't know if it's by a factor of exactly γ. I think you're right, and the point of describing the sub's acceleration in such excruciating detail is that he wants to make sure and correctly describe the sense in which the accelerated sub is "the same" as the sub at rest, because otherwise there'd be no way to determine how much water it displaced. It's pretty complex, because he's using something like Born rigidity, but he claims that the condition for Born rigidity is not just that the expansion scalar vanishes (which is what I'd seen previously for SR) but that in addition, some other complicated differential equation is satisfied.

It seems to me that it might be better to approach the problem from a fresh perspective. What's essentially happening is that we have object A, the sub, and object B, the water displaced by the sub. When they're at rest relative to one another, they have the same mass-energy, so they balance in a gravitational field. When they're in motion, we get contradictory results in the two frames about which way the balance is.

So how about this simpler version of the paradox. We have an elevator, and inside the elevator we have a balance with object A on the left and B on the right. When the elevator's motion is inertial, there are no fictitious forces on A or B, so they balance. The balance's arms experience zero force from either object, and everything just floats freely in neutral equilibrium.

Now we place the elevator in a state of constant proper acceleration along the line perpendicular to the arms of the balance. By symmetry, the balance still balances. If it's constructed like a real balance, with its center of mass below its point of support, then we have a stable equilibrium.

Now we let A and B be little cars that drive symmetrically along the beams of the balance. Each car moves outward, away from the fulcrum. They have velocities -v and v. By symmetry, there must still be an equilibrium. How does the driver of car A explain this equilibrium? In his frame, car B has a greater mass-energy than car A. Therefore a greater force is required from the balance's arm in order to accelerate B "upward." A would expect the balance to tip. But in B's frame, we expect the opposite.

The situation also seems strange because in a frame instantaneously comoving with the elevator, we have a four-force FA acting on A, and an equal four-force FB on B. Since these four-forces are equal, transforming into a different frame such as A's leaves them equal. But how does A explain their equality? Suppose both cars were simply driving on the floor of the elevator. Shouldn't A expect that more force is required from the floor to accelerate B, since B's mass-energy is higher?

Last edited: Oct 25, 2012
6. Oct 25, 2012

Staff Emeritus
I'm not even sure I agree with Archimedes' principle here.

One way to think about buoyancy is that I am replacing a volume with potential energy -m(sub) g h with one of -m(water) g h. The force is the position derivative of this change in energy, which works out to Δmg directed vertically, just like we know and love.

In relativity, in the sub's frame, I am replacing a volume with potential energy -m(sub) g h with one of -m(water) γ3 g h. Two factors of gamma come about because of the increased density and the third factor is because it's moving. Now, to get the force, you need to take the gradient of this.

Without even doing the math, you can see that you'll have forces in both the y and z directions. So the "ignore drag" instruction is going to cause problems.

Put in simpler, but more mathematical terms, this problem asks you to calculate a four-force in two different frames, compare their y-components, and then marvel that they are not equal.

7. Oct 25, 2012

### bcrowell

Staff Emeritus
I agree with V50's point in #6 that Archimedes' principle is suspect, although I'm not sure I agree with the discussion in terms of a gradient.

It's clearly reminiscent of the garage paradox (does the moving bus fit in the small garage due to length contraction?), which is resolved by taking into account the frame-dependence of simultaneity. The mariners might expect that when they started accelerating, water would have to be inside the sub because the cavity created by the sub would be length-contracted. You get similar issues in Purcell's classic exposition of the relativistic link between electricity and magnetism, in which you consider an infinite current-carrying wire, and length contraction affects the positive and negative charges unequally.

But I don't think drag necessarily has to be an issue. You can just let the sub move through a pre-excavated cavity a<z<b in the water, where a is the height of the bottom of the sub, and b the height of the top. (This would also help to keep the sub from being instantly destroyed, which would be nice from the sailors' point of view.)

8. Oct 25, 2012

### greswd

i think the modern consensus is that this paradox is outdated.

Because applying time dilation, relativistic mass has been "repackaged" into proper velocity. The formula for relativistic momentum is still the same, but the gamma coefficient is "attached" to velocity instead of mass.

i think the reason for this repackaging was to distinguish between the often confusingly intertwined concepts of "kinetic energy converted to relativistic mass" and rest-mass energy.

Another way to look at it is that gravitational mass is unaffected by inertial mass.

9. Oct 25, 2012

### pervect

Staff Emeritus
I'm deleting this while I think it over,,,

Last edited: Oct 26, 2012
10. Oct 25, 2012

### Austin0

Is it possible that in this context that fluid dynamics would override factors of gravity and buoyancy.

There must be a density gradient in the liquid. At normal velocities this is insignificant.
But even assuming perfect axial symmetry this gradient ,infinitesimally higher density below than the density over the surface above resisting motion. At relativistic velocities this could result in a form of subsurface hydroplaning off a differential.
This would pertain from either frame and would not be affected by any scalar value derived from gamma.
just a thought.

11. Oct 25, 2012

### bahamagreen

Does relativity assume that the sub's dynamics would be subject only to its local situation, or that the sub is existentially subject to the dynamics implied by those distant observations?

In other words, is the paradox based on the idea that remote observations have dynamic influence, if only local dynamics effect the sub?

12. Oct 26, 2012

Staff Emeritus
I thought about this a little more. Since the sub is neutrally buoyant, it has the density of water. We could replace it with an equivalent amount of water, perhaps dyed so we could somehow track it. So this question is equivalent to asking "where are the streamlines of this water volume"?

Except that the water won't have streamlines. The flow will be turbulent.

So I don't think there even is such a thing as a "relativistic Archimedes' principle".

13. Oct 26, 2012

### pervect

Staff Emeritus
It's pretty easy to determine that the proper accleration of car A and car B is increased by gamma^2.

What appears to be happening, though, is that if you put a little gyroscope on car A, and point it initally so it points parallel to the floor of the rocketship, it doesn't keep pointing that way. "gravitomagnetic" forces twist it around.

Trying to actually do the fermi-walker transport is giving me an absolutely incredible mess.

If we call the direction of acceleration the 'z' direction, and the direction of the floor of the rocketship the 'x' direction, it's easy to see that in the frame of the rocketship, A and B are at simultaneously at the same z-value. But if you go to a local fermi frame for A, the relativity of simultaneity means that A and B are no longer at the same z-value.

Tyring to get an analytical expression means fermi-walker transporting the x' basis vector, and so far this has been intracdtable. It might be possible and enlightening to solve it in the low-velocity limit, though.

14. Oct 26, 2012

### Mentz114

I've done a kinematic analysis of the x-boosted static frame field for this metric
$$ds^2=-{dt}^{2}\,{e}^{2\,a\,z}+{dz}^{2}+{dy}^{2}+{dx}^{2}$$
with the following results for the expansion coefficient, the acceleration vector, the vorticity vector and the non-zero components of the shear tensor S.
\begin{align} \Theta&=0\\ \dot{U}&=\frac{a}{1-{v}^{2}}\ \partial_z\\ \vec{\omega}&=-\frac{a\,v}{1-{v}^{2}}\ \partial_y\\ S_{xz}&=S_{zx}=\frac{a\,v}{2-2\,{v}^{2}} \end{align}\\
The shear and vorticity were unexpected. Rotation around the y-axis rather complicates the picture.

15. Oct 26, 2012

### pervect

Staff Emeritus
I had sort of an inisght. We can do the boosts cleanly in the Minkowskii frame (where boosts make sense). But -I don't think the resulting motion is going to represent the same motion as constant X velocity in the accelerating frame, because the coordinate clocks tick at a different rate than the accelerating clocks....

16. Oct 27, 2012

### pervect

Staff Emeritus
Some space-time diagrams to help visualzie what's going on since the math is so messy. The first is a standard 2d rindler diagram, with time (going up the page) and the z axis shown, which represents a single accelerating observer.

But we need t,z, and also x for the issue under discussion, so we need a 3-d diagram. We need to stack a bunch of these diagrams on top of each other. Hence, the next diagram, the 3d one.

The blue line is time. The black lines are plots of z vs time for an ensemble of rockets. The red line shows the "x" axis, a line of simultaneity for someone with dx/dt = 0. The lines all have the same z coordinate, the rocket floor is level for someone with no x motion.

[url]https://www.physicsforums.com/attachment.php?attachmentid=52344&d=1351336667[/url]
The green line shows the line of simultaneity along the x axis for someone who is not stationary. You can see from the diagram, hopefully, why the rocket floor isn't level anymore when we start to move across it.

#### Attached Files:

File size:
23.4 KB
Views:
710
• ###### rindler2.png
File size:
52.6 KB
Views:
650
Last edited: Oct 27, 2012
17. Oct 27, 2012

### pervect

Staff Emeritus
I'm puzzled why the second attachment isn't showing in-line in my original post - I'm going to try to get it to display here, and talk some more about its significance.

If I haven't put everyone to sleep - perhaps I've gone off on a tangent, but I think it's an interesting one.

So lets try again to get the snapshot of the secondlife model (the second attachment) to display.

The "top view" , shown in the previous post, is the hyperbola of motion of the accelerating rocket in the t-z plane, t being the Minkowskii time of an unaccelerated observer.

If you extend it in the x direction this becomes the "world sheet" of the accelerating rocket depicted by the snapshot,.

The snapshot allows us "to see the t-x plane, with t being the blue line , and x the red line. You can and should imagine these lines extending in the 'z' plane, surfaces of constant t are planes now, the same with surfaces of constant x.

A simple Lorentz transform defines how T and x mix together You can draw this on the tx plane, extend the resulting line into a plane in the z direction, and see where it intersects the worldsheet of the rocket. The result is the set of points that are simultaneous to the observer "moving" in the accelerating rocket in the Einsteinian sense.

Having the clocks be Einstei-synchornized is necessary if we are to try to understand the physics in a quasi-Newtonian way.

As can be seen, to this moving observer, the rocket floor doesn't appear to be flat anymore!

The way I'd describe why the rocket floor, supposedly "rigid", became distorted is that Born rigid objects can't rotate, and we "snuck in" a rotation by doing a sequence of boosts that was equivalent to a rotation. It's semi-well known that when you combine two Lorentz boosts in a different direction, you get a component that isn't a boost, but a rotation.

Last edited: Oct 27, 2012
18. Oct 27, 2012

### Austin0

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???

Last edited: Oct 27, 2012
19. Oct 27, 2012

### pervect

Staff Emeritus
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.

20. Oct 27, 2012

### Austin0

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???

21. Oct 28, 2012

### bcrowell

Staff Emeritus
This would seem to be contradicted by the fact that the Matsas paper was published by Phys.Rev. D in 2003.

Well, not really.

22. Oct 28, 2012

### bcrowell

Staff Emeritus
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.

http://lightandmatter.com/delete_later/sub2.jpg [Broken]

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:

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

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.

Last edited by a moderator: May 6, 2017
23. Oct 31, 2012

### Austin0

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???

Last edited by a moderator: May 6, 2017
24. Nov 2, 2012

### pervect

Staff Emeritus
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:

$u^a = \left[ \gamma \cosh gT, \gamma \beta, 0, \gamma \sinh gT \right]$

Note the magnitude of the 4-acceleration is $\gamma g$

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:

$\hat{x} = \left[(\gamma\,\beta-1/2\,{\gamma}^{3}{g}^{2}\beta\,{T}^{2}+1/24\,{\gamma}^ {3}{g}^{4}\beta\, \left( -4+{\gamma}^{2} \right) {T}^{4}+ ... ),(\gamma-1/2\,\gamma\,{g}^{2} \left( \gamma^2 \beta^2 \right) {T}^{2}+1/24\,{g}^{4} \left( \beta^2 \gamma^2 \right) ^{2} \gamma\,{T}^{4}+... ),0,(-1/3\,{\gamma}^{3}{g}^{3 }\beta\,{T}^{3}+1/30\,{g}^{5}{\gamma}^{3} \left( {\gamma}^{2}-2 \right) \beta\,{T}^{5}+...)\right]$

$\hat{y} = \left[0,0,1,0\right]$

$\hat{z} =\left[ [({\gamma}^{2}gT-1/6\,{\gamma}^{2}{g}^{3} \left( {\gamma}^{2}-2 \right) {T}^{3}+{\frac {1}{120}}\,{g}^{5}{\gamma}^{2} \left( {\gamma} ^{4}-8\,{\gamma}^{2}+8 \right) {T}^{5}+...,({ \gamma}^{2}g\beta\,T-1/6\,{g}^{3}{\gamma}^{4}{\beta}^{3}{T}^{3}+{ \frac {1}{120}}\,{g}^{5}{\gamma}^{6}{\beta}^{5}{T}^{5}+...),0,(1+1/2\,{\gamma}^{2}{g}^{2}{T}^{2}-1/24\,{g}^{4}{\gamma} ^{2} \left( 3\,{\gamma}^{2}-4 \right) {T}^{4}+...)\right]$

Integrate the 4-velocity to give the position P(T) (I also expanded it in a series).
$P(T) = \left[\gamma \sinh gT / g, \gamma \, \beta T, 0, \gamma \cosh gT / g \right]$

Write the transform equations to Minkowski coordinates

$(t,x,y,z) = P(T)+X \hat{x} + Y \hat{y} + Z \hat{z}$

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.

$t = X\beta\,\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) T-\frac{1}{2} \, X\,\beta\,{g}^{2}\,\gamma^3\,{T}^{2}$
$x = X\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) \beta\,T-\frac{1}{2} \,X\,{\beta}^{2}\,{g}^{2}\,\gamma^3\,{T}^{2}$
$y=Y$
$z=Z+ \left( \frac{1}{2}\,g\gamma+{\frac{1}{2} {Z{g}^{2}}\gamma^2} \right) {T }^{2}$

(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 $\partial g_{00} / \partial X$ 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.

$g_{00} =-\left(1+\gamma\,g\,Z\right)^2+2\,X{g}^{2}\beta\, \left( 1+\gamma\,Zg \right) {\gamma}^{2}T+{g}^{2} \left( {\gamma}^{2} {Z}^{2}{g}^{2}-\beta^2\,g^2\,X^2+1+2\,\gamma Z\,g \right) {\gamma}^{2}{T}^{2}$

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.

Last edited: Nov 2, 2012
25. Nov 7, 2012

### pervect

Staff Emeritus
Using the relations I derived earlier

$t = X\beta\,\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) T-\frac{1}{2} \, X\,\beta\,{g}^{2}\,\gamma^3\,{T}^{2}$
$x = X\gamma+ \left( \gamma+Z\,g\,\gamma^2 \right) \beta\,T-\frac{1}{2} \,X\,{\beta}^{2}\,{g}^{2}\,\gamma^3\,{T}^{2}$
$y=Y$
$z=Z+ \left( \frac{1}{2}\,g\gamma+{\frac{1}{2} {Z{g}^{2}}\gamma^2} \right) {T }^{2}$

and knowing that for car2, to second order, the minkowski coordinates are

$t = \gamma \lambda$
$x = -\beta \gamma \lambda$
$z = (g/2) \gamma \lambda^2$

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)

$T = \gamma^2\left(1+\beta^2\right) \lambda$
$X = -2 \beta \gamma^2 \lambda$
$Z = -2g\left(\gamma^2-1\right)\gamma^3 \lambda^2 = -2g\beta^2\gamma^5 \lambda^2$

$\lambda$ 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:

$Z = (1/2) g \gamma^2 X^2 \beta^2$

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 $\hat{x}$ and the [0,1,0,0] minkowskii basis vector is to second order

$\gamma \left(1 - (1/2)g^2\gamma^2 \beta^2 T^2 \right)$.

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 $\sin \theta \approx \theta$.

Last edited: Nov 7, 2012