Special relativity and the rod/slot paradox – III – Lorentz transformation causes inclination angles

In the 2nd post of this series about a variant of the rod/slot paradox in Einstein’s Special Theory of Relativity [STR] we defined a collision scenario between a vertically moving rod and a horizontally moving slot in a special reference frame A. In this 3rd post we apply the Lorentz transformation to some key events. Our objective is to get respective space and time coordinates in a frame B, in which the slot is at rest. We will find that a proper mapping of event coordinates leads to an inclination of the rod with respect to the x-axis and the slot in frame B.

For the initial conditions of the collision scenario see the previous 2nd post and a summary below.

Summary of initial conditions

The initial conditions of our collision scenario can be taken from the following drawing (and illustration 2 further below).

Initial conditions of a collision scenario for the rod/slot paradox

Illustration 1: Initial conditions of the collision scenario

I use sub-scripts A and B below to indicate in which of the frames certain coordinates and quantities are measured.

Summary of the initial conditions of the collision scenario
Seen from the slot, the rod approaches diagonally with constant components of its velocity.

Both the rod and the slot have a proper length L.
In frame A the rod rests with respect to its horizontal position. But the rod falls vertically there with a constant velocity v_rod,_A = -v_y.
In frame B the plate and thus also the slot are at rest (at a vertical position of y_B = y_A = 0). Frame B is co-moving with the slot.
Relative to frame A, frame B moves to the left with a horizontal velocity –ν_x.
At time t_B = t_A = 0 the origins of the coordinate systems of frames A and B coincide.
At time t_start,A = – v_y / h in frame A, the rod is located at height y_A = h. The rod will reach the vertical position of the slot (y_A = 0) at time t_A = 0.
The slot is placed in frame B with its left end at x_B = 0. The horizontal distance to the origin of frame A is chosen such that the slot’s left end reaches the rod’s left end at t_B = t_A = 0.
Due to the length contraction effect an observer measures the length of the slot in frame A to be L_slot,A = 1/γ_x • L. (For a definition of the usual γ-factor see below.)
In frame B the rod moves diagonally, but with a constant horizontal velocity ν_x and a constant vertical velocity v_y,B towards the slot.

These conditions lead to the following encounter of rod and slot at t_B = t_A = 0 as observed in frame A:

Illustration 2: Conditions at collision as seen in frame A

Due to the relativistic length contraction an observer in frame A predicts a collision of the rods right part with the plate. A potential paradox would exist if an observer in frame B would come to a different conclusion because he/she expects the length contraction to affect the moving rod.

Basic key events in our scenario are defined by their coordinates in frame A. For a further analysis we need to determine respective coordinates in frame B by applying the Lorentz transformation.

Lorentz transformation and the transformation of velocities

For our scenario we describe events in a reference frame A by respective x_A-, y_A-, t_A-coordinates (measured in A):

\[ \mbox{Event}\, \pmb{\operatorname{E}} \,\, \mbox{in}\, \pmb{\operatorname{A}} \ : \left[ \, x_A, \, y_A, \, t_A \, \right]_{E,A} \,. \]

Besides these coordinates events must be described by other characteristic information. We are in particular interested in events marking the (moving) left and right ends of both the slot and the rod at their encounter in frames A and B.

But let us first write down the required formulas of the Lorentz transformation [LT] of general event coordinates between our two reference frames A and B (and their coordinate systems) :

\[ \begin{align} x_B \,&=\, \gamma_x \, (\,x_A + \nu_x * t_A\, ), &\quad x_A \,&=\, \gamma_x \, (\,x_B – \nu_x * t_B\, ) \\[8pt] y_B \,&=\, y_A &\quad y_A \,&=\, y_B \\[8 pt] t_B \,&=\, \gamma_x \, (\, t_A + \nu_x / c^2 * x_A \,) &\quad t_A \,&=\, \gamma_x \, (\, t_B – \nu_x / c^2 * x_B \,) \end{align} \]

\[ \mbox{with : } \, \beta_x = {\nu_x \over c} \lt 1 \,, \quad \gamma_x = {1 \over \sqrt{ 1 – { \nu_x² / c²}\, }} \gt 1 \]

We have implicitly made the standard assumption that the origins of the coordinate systems coincide at t_A = t_B = 0. c is the the vacuum velocity of light. As we later will also need factors for relative velocities between frames in y-direction, I introduced the sub-scripts “x” and “y” for β and γ. Without sub-script we refer to factors related to the horizontal relative motions in x-direction.

The “+” and “-” signs result from the defined movement of frame B against frame A in negative x-direction.

The above formulas refer to space and time coordinates. An important point in our forthcoming analysis will be based on the Lorentz transformation of velocity components of moving objects. Let us assume that we measure the velocity components (v_x,A, v_y,A) [in x– and y-direction] of such a moving object in A, then the LT tells us that the respective velocity components (v_x,B, v_y,B) in B are given by:

\[ \begin{align} v_{x,B} \,&=\, { \,v_{x,A} + \nu_x \, \over 1 + v_{x,A} \, \nu_x / c^2} \, &\quad v_{x,A} \,&=\, { \,v_{x,B} – \nu_x \, \over 1 – v_{x,B} \, \nu_x / c^2} \\[10pt] v_{y,B} \,&=\, { \,v_{y,A} * \sqrt{\,1 – \beta_x^2 \, } \over 1 + v_{x,A} \, \nu_x / c^2} \, &\quad v_{y,A} \,&=\, { \,v_{y,B} * \sqrt{\,1 – \beta_x^2 \, } \over 1 – v_{x,B} \, \nu_x / c^2} \end{align} \]

Note the difference between the velocity components v_x,A and v_x,B measured for the moving object in frames A and B, respectively, on the one side and the value of the relative horizontal velocity ν_x between the frames A and B on the other side.

You may wonder why the velocity component in y-direction experiences a modification, whereas the spatial coordinates in y-direction remain unchanged by the LT. The reason is the time dilation effect experienced between observers in B and observers in A. We shall look at this effect in more detail in a minute.

Note two important consequences of the LT formulas:

An unchanged y-coordinate does NOT mean that two events measured in A to occur at the same time t_A with the same y-coordinate y_A, but different x_A-coordinates, will be transformed to get the same y-coordinate in B at a common identical time t_B in B. Simultaneous events in A may not occur simultaneously in B. This may have an impact on the y-coordinates!
A pure vertical movement of an object in frame A will lead to a diagonal one in frame B – with a horizontal velocity component v_x,A = ν_x, but a different vertical velocity v_y,B ≠ v_y,A .

Side remark: An “observer” is in a certain sense represented by a multitude of synchronized clocks placed at all relevant positions in the spatial coordinate system of a reference frame. The “observer” has furthermore means to register measured space and time coordinates of all interesting key events. For practical and experimental purposes you may assume that measured data at distant points are saved in some local protocols which can be evaluated later on.

The rod falls in frame B with a lower vertical velocity than in frame A

Let us first see what the vertical velocity component of the rod object looks like in frame B. To relate a velocity component to the rod, we use a leading sub-script “r“. From the LT formulas above we get (with v_r,x,A = 0) :

\[ v_{r,y,B} = v_{r,y,A} * \sqrt{ \, 1 – \beta_x^2 \,} = {1 \over \gamma_x} * v_{r,y,A} = – {1 \over \gamma_x}\, v_y = const. \]

This is consistent with the fact that due to time dilation any time difference measured in a frame A, which moves with velocity ν_x relative to frame B, appears stretched in frame B. So, when e.g. the rod falls from y_A = h down to y_A = 0 in a time interval Δt_A = h / v_y, the corresponding time interval in frame B will be measured to be longer, Δt_B = γ • Δt_A :

\[ \pmb{\operatorname{ A : }}\,\, \Delta t_A = h / v_y \quad \Rightarrow \quad \pmb{\operatorname{ B : }} \,\,\Delta t_B = \gamma_x * \Delta t_A \]

So, any of the rod’s end points will cover the vertical distance y_B = h in frame B with a properly reduced vertical velocity exactly in the prolonged time interval there:

\[ \begin{align} \pmb{\operatorname{ A : }}\,\, &{\Delta y}_A = h = v_{y,A} * h / v_y \quad \Rightarrow \\[10pt] \pmb{\operatorname{ B : }}\,\, & {\Delta y}_B = v_{y.B} * \gamma_x \, {\Delta y}_A \,= \, { 1 \over \gamma_x} \, v_{y} \,*\, \gamma_x \, {h \over v_y} = h \, . \end{align}\]

This ensures that our conditions at t_B = t_A = 0 get fulfilled in B, too: The particular event when the left end of the slot meets the left end of the rod in frame A at t_A = 0 occurs in frame B also at t_B = 0.

Key events occurring in frame A at t_A = 0

Let us now look at some events in frame A characterizing the collision event at t_A = 0 = t_B. We pick the following events Event 1 (E1) and Event 2 (E2), which occur when the left and the right end of the rod pass the line y_A = 0 in frame A.

We relate a coordinate like x, y or t

to the rod’s left end by a leading sub-script “r” and by a second sub-script “l “,
to the rod’s right end by a second sub-script “r”
and to the frame by a third sub-script A or B.

We indicate to which event “En” the coordinates belong by “|_En“.

The coordinates of the particular events E1 and E2 are given by:

\[ \begin{align} &\mbox{rod’s left end at} \, t_A=0 \quad \rightarrow\, \pmb{ \operatorname{E1}} \quad [0, 0, 0]_{E1,A} : \\[8pt] &\quad x_{r,l,A}|_{E1} = 0\,, \quad y_{r,l,A}|_{E1} = 0 \,, \quad t_{r,l,A}|_{E1} = 0 \\[10pt] &\mbox{rod’s right end at}\, \, t_A=0 \,\,\rightarrow\,\pmb{ \operatorname{E2}}\,\, \,\, [L, 0, 0]_{E2,A} :\\[8pt] &\quad x_{r,r,A}|_{E2} =L , \quad y_{r,r,A}|_{E2} = 0 , \quad t_{r,r,A}|_{E2} = 0 \end{align} \]

We have distinguished event E1 for the left end of the rod from event E2 for the rod’s right end because these events may transform differently, in particular with respect to the time coordinate. Due to our scenario’s specific setup (horizontal orientation of the rod in A), these two events do happen at the same time t_A = 0 in frame A.

Let us use the Lorentz transformation to find the coordinates of these events in frame B:

\[ \begin{eqnarray*} \pmb{ \operatorname{E1}}: \quad &x_{r,l,A}|_{E1} &= 0 &\,\,\Rightarrow \,\, x_{r,l,B}|_{E1} &= \gamma_x \,(0 +\nu_x * 0) & &= 0, \\ \quad &y_{r,l,A}|_{E1} &= 0 &\,\,\Rightarrow \,\, \,\,y_{r,l,B}|_{E1} &= 0 \, & &\, \\ \quad &t_{r,l,A}|_{E1} &= 0 &\,\, \Rightarrow \,\, \,\, t_{r,l,B}|_{E1} &= \gamma_x \,( 0 + \nu_x/c^2 * 0) & &= 0 \\[10pt] \pmb{ \operatorname{E2}}: \quad &x_{r,r,A}|_{E2} &= L &\,\,\Rightarrow \,\, x_{r,r,B}|_{E2} &= \gamma_x \,(L +\nu_x * 0) & &= \gamma * L, \\ \quad &y_{r,r,A}|_{E2} &= 0 &\,\,\Rightarrow \,\, \,\,y_{r,r,B}|_{E2} &= 0 \, & &\, \\ \quad &t_{r,r,A}|_{E2} &= 0 &\,\, \Rightarrow \,\, \,\, t_{r,r,B}|_{E2} &= \gamma_x \,( 0 + \nu_x/c^2 * L) & &= \gamma_x *\nu_x/c^2 * L \end{eqnarray*}\]

This means:

\[ \begin{eqnarray*} \pmb{ \operatorname{E1}}:\,\, &[\,0,\,0,\,0\,]_{E1,A} &\,\, \Rightarrow \,\, &[\,0,\,&0,\,&0 &\,]_{E1,B}, \\ \pmb{ \operatorname{E2}}:\,\, &[\,L,\,0,\,0\,]_{E2,A} &\,\, \Rightarrow \,\, &[\,\gamma_x \,L,\, &0,\, &\gamma_x \, \nu_x/c^2 \, L &\, ]_{E2,B} \end{eqnarray*}\]

We get something similar for events E3 (left end of rod passing y_A = h) and E4 (right end of rod passing y_A = h).

\[ \begin{eqnarray*} \pmb{ \operatorname{E3}}:\,\, &[\,0,\,h,\,0\,]_{E3,A} &\,\, \Rightarrow \,\, &[\,0,\,&h,\,&0 &\,]_{E3,B}, \\ \pmb{ \operatorname{E4}}:\,\, &[\,L,\,h,\,0\,]_{E4,A} &\,\, \Rightarrow \,\, &[\,\gamma_x \,L,\, &h,\, &\gamma_x \, t_{start,A} + \gamma_x \, \nu_x/c^2 \, L &\, ]_{E4,B} \end{eqnarray*}\]

That was easy! But let us try to understand these results a bit better …

The rod’s ends do not pass the line y_A = 0 at one and the same point in time in frame B

We assume that the rod has been falling with v_r,y = – v_y already for a while before it passes the lines y_A = h at time t_start,A and line y_A = 0 at t_A = 0 in frame A. The first strange consequence of the above results is:

\[ \begin{align} \Delta x_{rod, B} &\, = \, x_{r,r,B}|_{E2} – x_{r,l,B}|_{E1} \, =\, x_{r,r,B}|_{E4} – x_{r,l,B}|_{E3} \\[8pt] &\, = \, \gamma_x * L – 0 \,=\, \gamma_x * L \,\gt\, L \,\, . \end{align} \]

But, as the rod is moving with velocity ν_x relative to frame A in our scenario: Should we not expect a length contraction effect for the rod-length in B, i.e. a value Δx_rod,B < L?

This is an important point, which deserves some discussion. Any length measurement process is defined such that we should perform coordinate measurements at different spatial positions, but at the very same point in time in a chosen frame of reference (here at some t_B in B). The time value must be taken, of course, from synchronized clocks at the positions in question. So, length measurements in x-direction must be based on coordinates of two events with different x_B-coordinates, but with one and the same time coordinate t_B.

However, our events E1 and E2, which occur at the same time t_A = 0 in frame A, transform into two non-simultaneous events occurring at different times in frame B! Something analogous holds for E3 and E4.

Remember a basic consequence of the STR: What happens at the same time in one reference frame, does not necessarily happen at the same time in another frame. This is exactly what we find here:

While the left and the right sides of the rod pass y_A= 0 at the same time t_A = 0 in A, an observer in frame B finds

B: The rod’s left edge passes the geometrical point (x_B, y_B) = (0, 0)_B at t_B = 0 .
B: The rod’s right end passes (γ_x • L, 0)_B at a later time t_B = γ_x • ν_x /c² • L > 0 .

Again, something analogous is true for E3, E4. The difference in time is in both cases:

\[ \pmb{\operatorname{B:}} \quad \Delta t_{E2,E1}|_B \,=\, \Delta t_{E4,E3}|_B \,=\, \gamma_x \, {\nu_x \over c^2 } \, L \,. \]

So, determining the length difference between the spatial coordinates of these events is not a proper measurement of the length of the rod in frame B.

Consistency Check: So, what would happen if we measured the length of the rod at t_B = 0, instead? We then would have to look at a somewhat different event (instead of E2) in frame A. We can use the Lorentz transformation again to get the required corrected point in time for the rod’s right end in frame A:

\[ t_B = 0 \,\,\Rightarrow t_B = \gamma_x \,( \, t_A \,+\, {\nu_x \over c^2} * L\,) = 0 \,\,\Rightarrow \,\, t_A = – {\nu_x \over c^2} * L \lt 0 \]

This tells us that instead of picking E2 ∼ [L, 0, 0]_A, we would have to look at an earlier event (for the rod’s right end) in A:

\[ \pmb{\operatorname{Frame\, A :} \quad \operatorname{E2_R}} \,:\, [\,L,\, \, y_{r,r,A}|_{E2R},\,\, – \nu_x/c^2 \, L\,]_A\,, \quad {\small \mbox{with }} \,\, y_{r,r,A}|_{E2R} = v_y * {\nu_x \over c^2} \, L \,. \]

E2_R ∼ [ L, ν_x*v_y / c² • L, -ν_x/c² • L ]_A in A. The y-coordinate of this event has to be adapted because the rod’s right end has not yet reached y_A = 0 at this earlier time

\[ \begin{align} t_A|_{E2R} \, &= \, t_{r,r,A}|_{E2R} \,=\, – {\nu_x \over c^2 } * L \, ,\\[8pt] y_A|_{E2R} \, &= \, y_{r,r,A}|_{E2R} \,=\, {\nu_x * v_y \over c^2 } * L \,. \end{align} \]

Our 2-dimensional motion enforces more than just a correction of the time coordinate! We need to adapt the y-coordinate, too.

We now transform our particular event E2R from frame A to B to get the following coordinates in B:

\[ \begin{align} \pmb{\operatorname{ B :}} \quad \pmb{\operatorname{E2R}} \,:\, x_{r,r,B}|_{E2R} \, &= \, x_{r,r,B\,}(t_B=0) \\ &= \gamma_x \,L \,(1 – {\nu_x^2 \over c^2} ) = {1 \over \gamma_x} \, L \,,\\[10pt] y_{r,r,B}|_{E2R} \, &= \, y_{r,r,B\,}(t_B=0) \,, \\ &=\, y_{r,r,A\,}(t_A=0) \, = \, {\nu_x * v_y \over c^2 } * L \\[10pt] t_{r,r,B}|_{E2R} \, &= 0 \end{align} \]

We now can directly conclude from the resulting x_B-coordinate: This procedure indeed gives us the expected length contraction in x-direction!

Shifts in space and in time – and “perceptions”

The Lorentz transformation tells us that observers in frame B watch events E1 and E2 at different time coordinates. We should, however, also look at events which happen at the same time in B. This is a very natural approach: An observer analyzing the registered events is typically interested in what happened at different places at the very same point in time.

This is what we refer to as an observer’s “perception” of a situation in space: We cut a slice through space-time at a specific time-coordinate.

We have derived that the events E1 and E2_R in frame A do support a perception of simultaneous events in frame B. Respective distances in the spatial coordinates of the rod’s end points are:

\[ \begin{align} \pmb{\operatorname{Frame\, B :}} \quad \Delta_{x,R,B} \, &= \, x_{r,r,B}|_{E2R} – x_{r,l,B}|_{E1} = {1 \over \gamma_x} \, L \,, \\[8pt] \Delta_{y,R,B} \, &= \,y_{r,r,B}|_{E2R} – y_{r,l,B}|_{E1} = {\nu_x * v_y \over c^2 } * L \,. \end{align}\]

Consistency check: Note that between t_B|_E2R = 0 and t_B|_E2 = γ • ν_x/c² • L the right end of the rod consistently moves in B to the following x_B position

\[ \begin{align} &x_{r,r,B}|_{E2R} + \nu_x * \left(\,t_B|_{E2} -t_B|_{E2R} \,\right)\, =\, {1\over \gamma_x} \,L + \nu_x * \gamma_x \, {\nu_x \over c^2} \, L \\[8pt] &\quad=\, L \, (\,1/\gamma -\gamma*\beta^2\,) \,=\, \gamma \, L \,=\, x_{r,r,B}|_{E2}\,. \end{align} \]

This is indeed the x_B-coordinate of event E2.

Respective considerations for respective E3 and E4R lead to completely analogous results. What do our derived shifts in space and time coordinates for the right end point of the rod mean geometrically?

In frame B the rod moves with a constant inclination angle Ψ relative to the x-axis and thus also relative to the slot

In frame B any point on the rod to the right of its left end x_r,l,B crosses the horizontal lines given by y_B = h and y_B = 0 later then the rod’s left edge. This means that observers in B, who measure positions on the rod at a given equal point in time, register a rod which not only is moving diagonally towards the slot, but which is also rotated against the x_B-axis:

Frame B: The rod shows a constant inclination angle Ψ with a horizontal line through its left end point.

I have schematically depicted the situation as seen in frame B at t_B = t_start,B

\[ t_{start,B} = \gamma_x * t_{start,A} = – \gamma_x * {h \over v_y} \]

in the next illustration.
Note: Scaling, velocity component relations and angles are only schematic and may not reflect realistic conditions.

collision scenario: rod seen by slot at initial time

Illustration 3: The rod according to the initial conditions as seen in frame B at t_B = t_start,B

At collision time t_B = t_A = 0, we get the following resulting situation (again depicted schematically):

collision scenario: rod seen by slot at encounter

Illustration 4: Rod and slot seen at collision time t_B = 0 in frame B. An observer in B registers an inclination angle Ψ between the rod and the slot

From the perspective of B we deal with an object, the rod, which is moving diagonally. The rod’s horizontal velocity in B is given by the relative velocity ν_x of frame A versus frame B. In frame A the rod is oriented parallel to the x-axis. In frame B, however, the rod’s orientation exhibits a positive and constant inclination angle Ψ with the x-axis and parallel horizontal lines.

The inclination angle Ψ of the rod with the x-axis is given by

\[ \begin{align} \pmb{\operatorname{A, \, }} {\normalsize \text{inclination of rod }}: \quad \tan\,(\,\Psi\,) \,&= \, {\Delta_{y,R,B} \over \Delta_{x,R,B} } \,=\, { v_{y} * \nu_x / c^2 * L \over L / \gamma } \,, \\[10pt] \tan\,(\,\Psi\,) \, &=\, \gamma \, { v_{y} * \nu_x \over c^2 } \, . \end{align} \]

The angle’s sign obviously depends on the signs of the components of the object’s velocity components in frame B:

A falling object moving in positive x-direction in B would show a positive inclination angle. A falling object moving in negative x-direction in A would show a negative inclination angle. But an object moving in positive y-direction and in negative x-direction would again show a positive inclination angle.

Side remark: Our result is, by the way, directly related to the so called aberration effect in the STR. You may replace v_y with c (for a vertically propagating light wave front in A coming from a star above) and interpret –v_x as the observer’s movement (i.e. of B) relative to the star. Any telescope would have to be oriented vertically towards the incoming wave front in B, which shows the inclination Ψ (for v_y = c, in this case). One can show with a bit of geometry that the angle for the telescope’s inclination against the y-axis is given by |Ψ|, with tan(|Ψ|) ≈ β (neglecting the deviation of γ from 1.0). This is indeed the correct classical result for v_x << c.

Due to the squared velocity of light in the denominator, Ψ is a relative small angle. The angle Φ of the rod’s velocity vector v_rod in B versus the x-axis (!) is negative in our case and given by :

\[ \pmb{\operatorname{B, \,\,}} {\normalsize \text{incl. of rod velocity}}: \quad \tan\,(\,\Phi\,) \,= \, { – v_{y} / \gamma \over \nu_x } \,=\, – {1 \over \gamma} \, { v_{y} \over \nu_x} \]

The absolute value of this angle Φ can become significantly bigger than Ψ.

Side remark on time phase

The attentive reader has of course noticed that the so called general time shift or time phase seen in a moving frame B for events which happen at the very same time t_A in A was of major importance for our derivation:

\[ |\Delta t_{shift}| = \gamma \, {\nu_x \over c^2 }\, x_A \]

This time shift or time phase is obviously proportional to the distance of events in A from the origin of A. This corresponds to the statement that synchronized clocks along the x-direction in A appear out of synchronization in frame B. I admit, that this distance dependency is a bit hard to swallow. But it is a basic consequence of the Lorentz Transformation. In addition, depending on the sign of the relative velocity between the frames in question, the shift effect can go in both directions. I will use this a forthcoming post to build a scenario in which the rod actually moves through the slot without touching the plate.

The events marking the end-points of the rod define its movement and orientation completely

A reader has asked me why it is sufficient to regard events related to the end points of the rod. The reason is that we deal with linear transformations. Also the correction for time shift is a linear procedure. They transform straight lines or line segments into straight lines or line segments – if all points are measured simultaneously in both frames.

You can test this by transforming a point located somewhere on the rod from frame A to B – and correcting for simultaneity in the target frame. You will find that the point in B will reside on the line connecting the transformed end points of the rod – and the the ratio of the distances to the end points of the rod are the same in B as in A.

Conclusion

In this post we have looked at two reference frames

a frame A with a thin and lengthy rod (with proper length L) oriented in parallel to the x-axis and falling vertically downwards,
a frame B moving horizontally towards frame A in negative direction with a constant relative velocity –ν_x and containing a slot at a y-level of y_B = 0 = y_A. The slot also has proper length L and its left end is fixed at x_B = 0.

We have studied some first consequences of the Lorentz transformation for the description of the rod’s movement by an observer with synchronized clocks in frame B. A major result was that an observer in B registers the movement of the rod not only as a diagonal movement, but measures a constant small inclination angle Ψ between the rod and the slot.

We have also seen that events happening at the same point in time in A exhibit a time difference in frame B.

In addition, the transformation results indicated that the right end of the rod may touch the plate later and at a larger distance to the right of the slot’s right end in B than expected. In the next post

Special relativity and the rod/slot paradox – IV – consequences of an inclination angle in the slot’s frame

I will look closer at this distance and discuss a consequence – namely that in our “collision scenario” the rod will not move through the slot unharmed – neither in frame A, nor in frame B.

Stay tuned …