SR "Paradox" Thread Continuation: Tilt of Plates & Simultaneity

[PAllen]

TL;DR Summary

A recent thread posed an interesting SR false paradox, that got closed. Here I continue discussion.

The thread:
https://www.physicsforums.com/threa...tivity-paradox-than-the-twin-paradox.1009978/
proposed an interesting SR scenario. The thread could have led to instructive discussion of exactly how relativity of simultaneity resolved a "false paradox" that considered only time dilation. Instead, the OP of that thread rejected the obvious effects of simultaneity (the tilt of plates varies by reference frame) despite clear logical argument by @PeterDonis and others, and video provided by @A.T. . This required the thread to be closed. Since it is interesting to see how the change in tilt of plates required by relativity of simultaneity resolves the false paradox, both quantitatively and with detailed description, and because the techniques of dealing with extended bodies are skipped in elementary SR introductions, I will present the material in this thread. The next post will be the substance after this introduction.

 

[PAllen]

The original thread proposed a somewhat over complicated scenario, which only got worse later in the thread. Looking at the diagram in the OP of that thread, it is much simpler to just consider 3 plates: A, B1, and B2 and not have bifurcation of B into B1 and B2. Relating to the initial picture, we will have the +x direction be into the paper, -x be out of the paper, +y be towards the top of the paper, and +z be towards the right of the paper. This happens to ensure we have right handed spatial coordinates.

In reference frame C, plates B1 and B2 are moving in the -x direction at speed $\beta$, and also in the +y direction at speed v (of course I prefer to use symbols throughout rather than specific numbers). Plate A is moving in the +x direction at speed $\beta$, and in the +y direction at speed v.

To represent the plates, we describe a world tube (more accurately a congruence) for each. Each plate will be characterized by two parameters (u,s) representing 'elements' of the plate, as well as by t of frame C, representing time parameterization along each world line specified by a given (u,s) value. Then we have, for plate A:

$z=u$, $u ~\epsilon ~(-L,D)$
$x=s+\beta t$
$y=vt$

and for plate B2:

$z=u$, $u ~\epsilon ~(0,L)$
$x=s-\beta t$
$y=vt-h$

and for plate B1:

$z=u$, $u ~\epsilon ~(0,L)$
$x=s-\beta t$
$y=vt+h$

A collision is obviously some value of (x,y,z,t) in both A and either B1 or B2. This is obviously impossible because the y coordinates can never coincide for a given t. The sets of events described above for each plate are disjoint sets. Describing things in another frame, or even wild general coordinates, cannot change this. A change to arbitrary coordinates is a continuous bijection, which inherently maps disjoint sets to disjoint sets.

Be that as it may, it is often interesting to see how the description, or 'story' of what happens changes in different frames. In this case, the OP claims about time dilation are generally true, and length contraction is irrelevant if the plates are considered to extend infinitely in the +x,-x direction. What is interesting is to see exactly how the tilting of the plates (forced by relativity of simultaneity) avoids the collision implied by considering time dilation alone (per the OP argument).

I will consider only the change to a frame in which plate A has no motion in the x direction. Symmetry guarantees that considering the frame in which B1 and B2 have no x motion is equivalent, with various quantities reversed. It is unnecessarily complicated to consider a frame in which plate A has no motion at all (and the OP of the other thread did not do that).

To Lorentz transform the plate representation, one uses the reverse transform to substitute expressions of the primed coordinates into the unprimed coordinate references above, i.e. replace x with $\gamma(x'+\beta t')$. Then, you rearrange and solve for primed coordinates. The result is:

For plate A:

$z'=u$, $u ~\epsilon ~(-L,D)$
$x'=\gamma s$
$y'=\gamma vt'+v\gamma^2\beta s$

Let's notice a few things about this. First, the x' formula shows the effect of length contraction - two element separated by $\Delta s$ in frame C (for which the plate is contracted) are separated by $\gamma \Delta s$ in frame A. We also see the impact of time dilation. Since frame C measures y speed as distance v per second, and the distances in y are not affected by the Lorentz transform, and clocks on A run slow per C, then per A, the distance v is covered in less measured time. Thus y' speed is $\gamma v$. Finally, relativity of simultaneity means the plate slopes upward in y' over x', for a given t' by $\gamma \beta v$.

For plate B1:

$$z'=u,~u ~\epsilon ~(0,L)$$
$$x'=\frac s {\gamma(1+\beta^2)}-\frac {2\beta} {1+\beta^2} t'$$
$$y'=\frac {vt'} {\gamma(1+\beta^2)}+\frac {\beta v s} {1+\beta^2} +h$$

and for plate B2, all is the same except for -h in the last formula.

Notice that the second term in the x' formula just represents velocity addition for $\beta \oplus \beta$. Note that $\gamma$ for this composite velocity is (in terms of $\gamma$ for $\beta$) given by $\gamma^2(1+\beta^2)$. Thus noting that 'length' in the B frame is given by $\gamma s$, the first term in the x' formula above is just length contraction for the composite velocity. Similarly, you have time dilation by the composite velocity for the first term in the y' formula above. Finally, and crucially, note that the slope of y' over x' for given t' is $\gamma \beta v$. Thus the plates A, B1, and B2 are all parallel (but not horizontal), and displaced from each other.

The motion of the plates is then that all three plates are tilted in the positive y direction relative to the positive x direction. As the B plates move in the -x direction, any element of B1 or B2 would have a negative y component velocity. However, the positive y motion of all 3 plates together leads to a B plate element having a small positive y velocity (as shown in the formulas above), while the A plate, which has no motion in the x direction, is moving faster in the y direction. You can see from this analysis, that the slower motion of a B1 plate element compared to an A plate element, in no way implies collision because of the downward angled sliding of the B1 plate compared to the A plate.

A final exercise is to note that if one sets up the equation that for some s1 for the A plate, and some s2 value for the B plate, and any common value of t', you get both x' values being the same and both y' values being the same (thus producing a collision in the z overlap region of the plates), the resulting equations yield the result that this is possible if, and only if, h=0 (Duh!). Thus it can be shown by brute force that collision cannot occur in the problem as described in the A frame, any more than it can in the C frame.

 

[PAllen]

Actually, by reparameterizing y’ in terms of x’ and t’ for each plate, with a little messy algebra, you find that the difference in y’ values is always simply h, for all x’ and t’, consistent with the prior demonstration that they are parallel (using that their slopes are the same).

 

[Ibix]

This may also be helpful - it's a (2+1)d Minkowski diagram showing the worldtube of one cart and its rising plate in C's frame.

The cart is green and the rising plate is red. Some planes of simultaneity are shown in translucent white, and you can see the red plate is further and further above (in the +y direction) the green plate as time increases. Note that the lines where the red and green worldsheets intersect the simultaneity planes are parallel, so the rising platform is horizontal in this frame.

Now here's a diagram, still drawn in C's frame, but showing the simultaneity planes of A.

Here we see the exact same thing, but of course the relativity of simultaneity means that these planes of simultaneity are slanted with respect to the original ones. This time, we can see that the lines of intersection between the worldsheets and the simultaneity planes are not parallel. That might be more obvious if I add some lines parallel to the intersection of the green line and the simultaneity planes:

So the front of the rising platform is higher (larger y coordinate) than the rear of it using this definition of simultaneity. That change of definition of simultaneity is the origin of the slope of the platform - the cart length contracts but doesn't slope because its worldsheet lies in the $t-x$ and $t'-x'$ plane, while the platform with its worldsheet with extent in the $y$ direction does.

 

[PAllen]

I ignored the whole cart complication and simplified the original scenario a bit - that is all that is needed to address the issues.