Problem analyzing rod-slot 2-D Lorentz contraction paradox

1. Jul 8, 2014

nixed

I have encountered a difficulty which arises from my attempt to combine non collinear Lorentz transformations in analyzing the following problem:

A thin rod is cut from a metal plate leaving a slot of exactly the same size. The two are separated and set in motion thus: the rod lies along x and moves in the x direction with velocity Vx . The plate is oriented along x but at position -y, is set in motion along y direction with velocity Vy. The center of the slot and the center of the rod will coincide at time tc. Will the rod pass through the slot? Consider the point of view of an observer on the ground G, and one on the rod R, and one on the plate P.

The observers P and G both agree the rod should pass as it is Lorentz contracted so is smaller than the slot but the observer R will say the slot is contracted so how can the rod pass?

To resolve the apparent paradox I attempted to relate the space time coordinates of events in each frame as follows:

Coords of an event in G = (t, X, Y) = { Xi } .
Coords of same event in R = {Xir} = L(Vx). {Xi} ie a Lorentz boost in the direction of x by the rod’s speed. In this frame the rod is stationary and the plate moves with velocity components –Vx and Vy’= Vy . √( 1 – (Vx/c)2 ), so the correct relative velocity of the plate and rod are

Vr = Vx2 + Vy2 – (Vx2 )( Vy2)/c2 and the angle this makes to x axis is tan Θr = Vy’/Vx

Now comes the problem! To get the coordinates of the same event in frame P , so all the motion is carried by the rod, I could either perform the transformation from the rod’s frame by the relative velocity vector Vr relating the rod and the plate so

Coords of event in P = L (Vr) . {Xir} = L(Vr). L(Vx). {Xi}

Alternativley I could go from frame G by transforming along y by L(Vy) to give

Coords of event in P = L(Vy) . {Xi}

However these two methods give different results! Despite both methods leaving the plate stationary and the rod moving- so apparently the same physical situation.

i.e. L(Vr). L(Vx) is not equal to L(Vy). The second method gives the same magnitude to the relative velocity between rod and plate but a different angle of this velocity to the x axis (call it Θp) such that

tan Θp = tan Θr /√( 1- (Vr/c)2 )
So which of these two methods gives the correct relation of the coordinates of an event in frame P to that in Frame R and to G?

2. Jul 8, 2014

WannabeNewton

The second half of your post is almost incomprehensible to me because of your choice of notation and lack of LaTeX.

Let $L$ be the rest length of the rod, $(-x_0, 0)$ the initial position of its center, and $(0,-y_0)$ the initial position of the center of the slot in the plate with the length of the plate aligned along the $x$ axis. Let the plate move up along $y$ with velocity $v_y$ at time $-t_0$ and let the rod move right along $x$ with velocity $v_x$ also at time $-t_0$. Then $\frac{x_0}{v_x}- t_0$ is the time at which the center of the rod reaches the origin and $\frac{y_0}{v_y} - t_0$ is the time at which the center of the plate reaches the origin. In order for these to coincide, we need $\frac{x_0}{v_x}=\frac{y_0}{v_y}$; furthermore choose $t_0 = \frac{x_0}{v_x}$ so that the centers of the plate and rod coincide at $t = 0$; with these initial conditions, the front end of the plate reaches the $x$ axis at the event $(0,L/2,0)$ and the back end of the plate reaches the $x$ axis at the event $(0,-L/2,0)$. At this instant, the ends of the rod will be located at $(\frac{L}{2\gamma}, 0)$ and $(-\frac{L}{2\gamma},0)$. This is all in the ground frame.

Now boost to the frame $(t',x',y')$ of the rod when its center passes the origin. The centers still coincide at $(0,0,0)$ but the front end of the plate reaches the $x'$ axis at $(-\gamma v_x L/2, \gamma L/2, 0)$ and the back end of the plate reaches the $x'$ axis at $(\gamma v_x L/2, -\gamma L/2,0)$; the ends of the rod are always located at $(L/2,0)$ and $(-L/2,0)$. In particular, this means the front end of the plate passes through the rod before the centers meet and the centers meet before the back end of the plate passes through the rod; this is of course because the plate is rotated in the rod frame by an angle relative to the $x'$ axis given by $\tan\theta = \gamma v_x v_y$.

Therefore, due to the rotation of the plate in the rod frame, the plate still completely passes through the rod even though it is length contracted along the $x'$ direction. I'll leave it to you to work out any further details.

3. Jul 8, 2014

4. Jul 9, 2014

nixed

Im sorry it wasn't clear: This is what I meant: The same physical event , call it i, has coordinates t, X, Y which are different in the three reference frames G (ground), P (plate), R (rod). The {Xi} is a 3x1 column matrix of the coordinates of the event in the G frame. {Xir} is the 3x1 column matrix of the same event in the R frame , {Xip} in the P frame. These coordinates are related to each other by the appropriate Lorentz transformations denoted L(V) etc. These L(V) are 3x3 matrices eg

L(Vx)=

$$\begin{pmatrix} γx & -γxVx & 0\\ xVx & γx & 0\\ 0 & 0 & 1 \end{pmatrix}$$

for the Lorentz boost by velocity Vx along the positive x axis.

γx= 1/√(1-(Vx/c)2)

i hope that clarifies the situation.

The problem then is that the apparent equivalence of the two possible routes from frame G to frame P Ie 1) a boost by Vx then by Vr or 2) a boost by Vy do not lead the initial coordinates {xi} in frame G being transformed into the same point in frame P because L(Vr).L(Vx)≠ L(Vy) So which of the two routes gives the correct coordinates of the event in frame P? Why do the two routes differ?

(im not sure why the latex did not formulate the matix -ill try and sort this out)

Last edited: Jul 9, 2014
5. Jul 9, 2014

DrGreg

To do subscripts within LaTeX you use V_x not [noparse]Vx[/noparse].

Also, within LaTeX, use \gamma, not γ.

6. Jul 9, 2014

nixed

Ill try the matrices again

{Xi}=$$\begin{pmatrix} t_i\\X_i\\Y_i\end{pmatrix}$$

L(Vx)=
$$\begin{pmatrix} \gamma_x & -\gamma_x V_x & 0\\ -\gamma_x V_x & \gamma_x & 0\\ 0 & 0 & 1 \end{pmatrix}$$

L(Vy)=

$$\begin{pmatrix} \gamma_y & 0 & -\gamma_y V_y\\ 0 & 1 & 0\\ -\gamma_y V_y & 0 & \gamma_y \end{pmatrix}$$

L(Vr)= R-1(ø). Lx(Vr) R(ø) =

$$\gamma\begin{pmatrix} \ 1 & - CV & -SV\\ -CV & C^2+S^2/\gamma & SC(\gamma-1)/\gamma\\ -SV & SC(\gamma-1)/\gamma & S^2 + C^2/\gamma \end{pmatrix}$$

where C=Cos (ø) S= Sin(ø ) where ø is the angle between the +ive Vr direction and the x direction.

γ = 1/√(1-(V/c)2) and V= Vr and γ = γx γy and γx=1/√(1-(Vx/c)2) γy=1/√(1-(Vy/c)2)

Last edited: Jul 9, 2014
7. Jul 9, 2014

pervect

Staff Emeritus
I don't see what additional benefits making the problem 2D has. if you consider the 1d problem, in the same detail, and resolve it first you might not even see the need to do the 2d problem. If you decide to go through with the 2d analysis, it would probably help you spot your mistake. This will be helped by the fact that you should be able to find an analysis in a textbook.

The 1d problem would omit the plate being set in motion in the y direction, instead have the rod directly over the plate and focussing on two events, the back of the rod (you can consider this to be the end of the rod with the most negative x) being just above the back of the slot, let's call this the "Back" event, and the front of the rod being just above the front of the slot, lets call this the "Front" event.

The resolution for the 1d case will be that in the plate frame, the back event will occur before the front event, this will be interpreted as "the rod being shorter than the plate", while in the rod frame, the front event will come before the back event, this will be interpreted as "the hole being shorter than the rod".

This isn't a problem because the front and back events are spacelike separated.

8. Jul 10, 2014

ghwellsjr

If we consider the 2D case but only look at the 1D part along the x axis, it makes it much simpler to analyze and we can draw a couple of conventional spacetime diagrams to illustrate what happens first in the frame where the rod is traveling along the x axis and then in the frame where the rod is stationary.

I show the two ends of the rod, which are separated by 8 feet, in red and black, and the panel in blue with its 8-foot long gap. I will have the rod traveling at 0.6c in the first frame which means that its length will be contracted to 6.4 feet. This can be calculated by taking the Proper Length of the rod, 8 feet, and dividing by gamma which in this case is 1.25. Keep in mind that the red and black lines are worldlines but the blue line segments are not. They are merely showing the instant in time when the y location of the panel with its gap equals zero:

Now if we transform to the rest frame of the rod, we get this spacetime diagram:

Note that the Proper Length of the rod is 8 feet.

But also note that the blue line segments show the instants in time when the y locations of the panel equal zero. As you can see, because of the motion of the panel (in both the y and -x directions) it appears at different times in this diagram and the gap passes across the rod with no problem.

I hope this helps you find the mistake in your calculations.

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9. Jul 10, 2014

WannabeNewton

I still have no idea what you're trying to do. Could you take a look at post #2 and use the notation, initial conditions, and overarching framework therein to perform your calculations?

10. Jul 11, 2014

nixed

Thanks for the help and the comments on the 1-d problem (the pole in a barn paradox). I appreciate how to analyse that now and how to use space time diagrams.

It was my attempt to create a similar diagram for a 2d situation comparing the three frames (ground, rod, plate) space time coordinates of events that led me into this problem of relating the coordinates in each of the three frames (G,R,P).

I assumed that if I had the space time coordinates with respect to the ground (frame G), I could find them with respect to the plate (by a Lorentz transformation L(Vx) from G) or with respect to the rod (by a Lorentz transformation L(Vy) from G). That is the case, -no problem.

If I had the coordinates with respect to the rod I could get the coordinates with respect to the plate by a Lorentz transformation along the direction of the relative velocity of plate to rod L(Vr). - again no problem (just a bit of a nasty matrix for L(Vr) if the direction is not collinear with x or y).

I thought the coordinates of events with respect to the plate, compared to the coordinates with respect to the ground say, would be the same, independent of how I generated them ( by boost along Vy, or alternatively by boost along Vx followed by boost along Vr ; since both lead to apparently the same physical situation of the plate being stationary and the rod moving with velocity Vr). After all if A is equal to B and B is equal to C then A is equal to C surely? - Well NO! not in relativity it seems!

My puzzlement arose when I discovered that the result of the x boost followed by the Vr boost ( represented by matrix multiplication
L(Vr).L(Vx)) is not the same as the boost along y (represented by matrix L(Vy)). After a lot of algebraic slog I discover that generally

L(Vr).L(Vx) = R(θpr). L(Vy)

Where R is a matrix which describes a rotation of the frame by the angle

Ω= θpr

where θp is the angle between the plate-rod relative velocity vector and the x axis in the plate reference frame, θr is the angle between Vr and x in the rod reference frame. (my mistake was to assume both these angles must be the same as they are at low speeds, but at speeds close to light they are not the same!)

It turns out that in general the combination of two non co-linear Lorentz boosts is never equal to a single Lorentz boost, but is always a combination of a boost and a rotation.

What I've stumbled into here is well known I discover, and is called Thomas rotation and the angle Ω is known as the Wigner angle.

11. Jul 11, 2014

WannabeNewton

It's awesome that you came upon the effect of Thomas rotation by yourself but you didn't need to go through all that pain! You could have recognized the non-equivalence of the two boost sequences right from the start. Indeed if you boost from the ground frame to the plate frame then you just boost straight up and that's the end of that. If you boost from the ground frame to the rod frame and then boost to the plate frame then you would know that in the rod frame, the plate would be rotated relative to the $x'$ axis by the angle $\tan\theta = \gamma v_x v_y$ due to relativity of simultaneity so a simply boost in the downwards direction in the rod frame will certainly not get you back to the plate frame; you have to rotate the coordinate system in the rod frame by the appropriate angle so as to align the $y'$ axis with the length of the plate, so that the $x'$ axis is aligned with the direction of the plate, and then boost in the $x'$ direction by the appropriate velocity.

12. Jul 11, 2014

nixed

Yes, I missed that. I think I started out thinking that boosts could commute (which they cannot if they are not co linear) and assuming that a boost along say Vr = Vy-Vx would be equal to a boost L(Vy)L(-Vx) which also is not true!

I have attached a file which summarises the coordinate transformations in a diagram, but im not sure how to get a picture to display in this message window.

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13. Jul 12, 2014

nixed

2I thought I would give my analysis of the rod slot paradox using the event coordinates I have been banging on about... (it is of course in agreement with earlier comments but I think the view point is helpful and it is much simpler than the previous discussion of combination of Lorentz boosts!...)

Imagine the plate is ridged and infinitely extended but has a slot length L0=2δ in its rest frame P. The plate moves along y axis at velocity Vy. Imagine the rod as a dust of particles of zero rigidity of length 2Δ in its rest frame R (which for generality can be any length) and moving along x with velocity Vx. From the perspective of the ground frame G the center point of the slot and of the rod collide at ti=o and at Xi=0 Yi=0. This point can set the origin of all the coordinate frames G,P R. In the ground frame the collision of plate and rod occurs simultaneously along the whole extent of the rod. The condition that at collision particles of the rod will pass through the slot is that the particles have coordinates in frame G :

$\begin{pmatrix} 0\\-δ\\0\end{pmatrix}$≤$\begin{pmatrix} t_i\\X_i\\Y_i\end{pmatrix}$≤$\begin{pmatrix} 0\\δ\\0\end{pmatrix}$

This set of points have the coordinates {Xir} in the rod frame of reference {Xir} =L(Vx) {Xi} and all these (and only these) points will pass the slot. So in the rod frame the points that pass are (in units where c=1):

$\begin{pmatrix} \gamma_xV_x\delta\\-\gamma_x\delta\\0\end{pmatrix}$≤$\begin{pmatrix} t_ir\\X_ir\\Y_ir\end{pmatrix}$≤$\begin{pmatrix} -\gamma_xV_x\delta\\\gamma_x\delta\\0\end{pmatrix}$

Notice the points do not all have the same time coordinate for the moment at which they pass through the slot. The negative sign for tir indicates that point passes before the center of slot and rod meet. We see that a rod of length 2 γxδ measured in the rod's frame can pass through the smaller slot (length 2δ in the plate frame) and it does so in a time Δt= 2γx Vx δ rather than in one instant as it did in the ground frame (relativity of simultaneity). This implies that the rod and plate are no longer parallel but that the slot is rotated through an angle Θ compared to the rod. (this is distinct from the Thomas rotation and is due to the relativity of simultaneity). To find the angle Θ is a simple matter of trigonometry: the center of the slot approaches the center of the rod at speed Vr and along a line at an angle θr to the rod. In time Δt= 2γx Vx δ the slot moves a distance VrΔt and the rod of length 2γxδ sweeps through the slot so we have a triangle in which we know two sides and the included angel and we want to find the opposite angle. We discover tanΘ = γxVxVy in units of c=1. (or in SI unit tanΘ = γxVxVy/c2)

Last edited: Jul 12, 2014