In 2nd post of this series about a variant of the rod/slot paradox in Einstein’s Special Theory of Relativity [STR] we have in detail defined a scenario leading to a collision 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 occurring in our reference frame 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 the slot** from the perspective of observers in **B**.

For the initial conditions see the previous 2nd post for details and a summary below.

Previous posts:

**Post I:**Special relativity and the rod/slot paradox – I – seeming contradictions between reference frames**Post II:**Special relativity and the rod/slot paradox – II – setup of a collision scenario

## Summary of initial conditions

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

**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* orthogonal components of its velocity.

- Both the rod and the slot have a
**proper****length***L*. - In frame
**A**the**rod***rests*with respect to itsposition. But the rod falls vertically there with a**horizontal***constant*velocity*v*,_{rod}_{A}=*-v*._{y} - In frame
**B**the plate (and thus also the**slot**) are at rest (at vertical position*y*=_{B}*y*= 0)._{A} - 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*in frame_{y}/ h**A**, the rod is located at height*y*=_{A}*h*there**.**The rod will reach the vertical position of the slot (*y*= 0) at time_{A}*t*=_{A}*0*. - The slot is placed in frame
**B**with its left end at*x*= 0. The horizontal distance to the origin of frame_{B}**A**are 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*= 1/_{slot,A}*γ**L*. (For a definition of the usual*γ*-factor see below.) - In frame
**B**the rod moves diagonally, but with constant horizontal velocity*ν*and a constant vertical velocity_{x}*v*towards the slot._{y,B}

These conditions lead to the following situation at *t*_{B} = *t*_{A} = 0:

**Illustration 2:** Conditions at collision as seen in frame **A**

Basic key events in our scenario are defined with 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**):

For the given setup let us 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) :

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-script “*x*” for *β* and *γ*.

The “+” and “-” signs result from the defined movement of frame **B** against **A** in negative x-direction. An important point in our forthcoming analysis will be based on the Lorentz transformation of velocity components (*v _{x,A}*,

*v*) of an object in A to their counterparts (

_{y,A}*v*,

_{x,B}*v*) in

_{y,B}**B**:

Note the difference between a velocity component *v _{x,A}* measured for an

*object*in frame

**A**and the value of the relative horizontal velocity

*ν*between the frames

_{x}**A**and

**B**.

You may wonder why the velocity component in *y*-direction experiences a modification, whereas the spatial coordinates in *y*-direction remain unchanged. The reason is the * time dilation* effect experienced between observers in

**B**versus observers in

**A.**We shall look at this in detail in a minute.

Note two points here:

- The denominator in the velocity formulas is important in our scenario as a pure vertical movement of an object in a frame
**A**will lead to a diagonal one in frame**B**– with a horizontal velocity component*v*=_{x,A}*ν*, but a different vertical velocity_{x}*v*≠_{y,B}*v*._{y,A} - An unchanged
*y*-coordinate does**NOT**mean that*two*events in**A**measured to occur at the same time*t*with the same_{A}*y*-coordinate*y*, but_{A}*different**x*-coordinates, will be transformed to get the same_{A}*y*-coordinate in**B**at apoint in time*common identical**t*in_{B}**B**. Simultaneous events in**A**may**not**occur simultaneously in**B**.

**Side remark:** An “observer” is represented by a gigantic multitude of synchronized clocks placed at all relevant positions in a frame to register (= measure) space and time coordinates of all interesting key events.

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

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

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*down to

_{A}= h*y*0 in a time interval Δ

_{A}=*t*=

_{A}*h / v*, the corresponding time interval in frame

_{y}**B**will be measured to be longer, Δ

*t*=

_{B}*γ*• Δ

*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:

This ensures that our conditions at *t*_{B} = *t*_{A} = 0 get fulfilled in **B**, too: The particular **event** that the left end of the slot meets the left end of the rod in frame **A** at *t*_{A} = 0 is reproduced in frame **B** 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*. We pick the following events

_{B}**Event 1**(

**E1**) and

**Event 2**(

**E2**), which occur when the left and the right end of the rod pass the line

*y*= 0 in frame

_{A}**A**.

We refer a coordinate

- 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 “**E**n” the coordinates belong by “|_{En}“.

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

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. Due to our scenario’s specific setup (horizontal orientation of the rod in **A**), these two events for the rod’s two ends 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**:

This means:

We get something similar for events **E3** (left end of rod passing *y _{A} = h*) and

**E4**(right end of rod passing

*y*).

_{A}= hThat was easy! 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

_{A}

We assume that the rod has been falling with *v _{r,y}* = –

*v*already for a while before it passes the lines

_{y}*y*at time

_{A}= h*t*and line

_{start,A}*y*= 0 at

_{A}*t*= 0 in frame

_{A}**A**. The first strange consequence of the above results is:

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 Δ

*x*<

_{rod,B}*L*?

This is an important point, which deserves some discussion. Any length measurement process is defined such that we should perform 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*-coordinates, but with one and the

_{B}**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 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*= 0 in

_{A}**A**, we find in frame

**B**

**B:**The rod’s*left*edge passes the geometrical point (*x*,_{B}*y*) = (_{B}*0, 0*)_{B }at_{ }*t*= 0 ._{B}**B:**The rod’s*right*end passes (*γ*•_{x}*L, 0*)_{B}_{ }at a later point in time*t*=_{B}*γ*_{x}*• ν*> 0 ._{x}/c^{2}• L

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

So, determining the length difference between these events is **not** a measurement of the length of the rod in **B**.

**Consistency Check:** So, what would happen if we measured the length of the rod at *t _{B}* = 0, instead? We can use the Lorentz transformation again to get the corresponding point in time for the rod’s

*right*end in frame

**A**:

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**:

**E2 _{R}** ∼ [

*L, y*]

_{R,A}, -ν_{x }/_{ }c^{2}• L*in*

_{A}**A**. The y-coordinate of this event has to be adapted because the rod’s right end has

**not**yet reached

*y*at this earlier time

_{A}= 0We now transform this particular event **E2R** from frame **A** to **B** to get the following coordinates in **B**:

We now can directly conclude from the resulting *x _{B}*-coordinate: This would indeed have given us the expected length contraction in x-direction!

**Shifts in space and in time**

The Lorentz transformation tells us that observers in frame **B** watch event **E2** at coordinates with different values in comparison to what observers in **A** measure. In particular the time coordinate changed.

However, we should also look at events which happen at the *same* time in **B**. We have derived that such events are **E1** and **E2 _{R}**. Respective distances in the spatial coordinates of the rod’s end points are:

**Consistency check:** Note that between *t _{B}|_{E2R}* = 0 and

*t*=

_{B}|_{E2}*γ • ν*the right end of the rod consistently moves in

_{x }/_{ }c^{2}• L**B**to the following

*x*position

_{B}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*and

_{B}= h*y*= 0

_{B}**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 falling diagonally, but whose lengthy symmetry axis is also

*rotated*against the

*x*-axis:

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

**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*= 0, we get the following resulting situation (again depicted

_{A}*schematically*):

**Illustration 4:** Rod and slot seen at collision time *t _{B}* = 0 in frame

**B**. An observer in

**B**register 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 in parallel to the

*x*-axis. In

**B**, however, the rod’s orientation exhibits by 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

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*with

_{y}*c*(for a vertically propagating light wave front in

**A**coming from a star above) and interpret –

*v*as the observer’s movement (i.e. of

_{x}**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*<< c.

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

The absolute value of this angle Φ is typically much bigger than Ψ.

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:

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. 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.

But, enough for today …

## 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*and containing a slot at a y-level of_{x}*y*= 0 =_{B}*y*. The slot also has_{A}*proper*length L and its left end is fixed at*x*= 0._{B}

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**. The transformation results indicated in addition 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

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 …