A moving rod; two Lorentz boosts compared with one

[Hiero]

Consider a rod oriented along and moving along the x direction at a speed v in frame A. In it's turn, frame A is moving at speed u in the y direction relative to frame B.

By my understanding of special relativity, the x component of the rod's velocity in frame B will be v/γ_u, and the y component is just u.
Also, the rod remains oriented along x (& same length) in frame B, because the ends are not displaced along the boost direction from A.
Also, in the rod's proper frame it is still oriented along x.As far as I can tell all of that is fine, but another perspective confuses me. I was thinking we should be able to look at frame B as being a single boost away from the rod's proper frame. This would be a boost with <x , y> velocity components being <v/γ_u , u>.
The point is, from the proper frame, we are boosting off at some angle with the rod's length, (not zero and not perpendicular) and as far as I can tell this necessarily causes the rod to be rotated when changing frames (whereas the analysis with an intermediate frame A told me that the rod remains along x).

Not sure where the problem lies. Thanks.

 

[Nugatory]

Not sure where the problem lies. Thanks.

Consider the two worldlines of the two ends of the rod, and consider two events on those worldlines that are simultaneous in frame A; both events lie on the x-axis so we conclude that the rod is oriented along the x-axis. In frame B both events lie on frame B's x-axis - but do they happen at the same time?

 

[Hiero]

Consider the two worldlines of the two ends of the rod, and consider two events on those worldlines that are simultaneous in frame A; both events lie on the x-axis so we conclude that the rod is oriented along the x-axis. In frame B both events lie on frame B's x-axis - but do they happen at the same time?

Yes that is part of the reasoning why it's along x in B, because simultaneous events on the endpoints in A are still simultaneous in B, because there is no displacement along the boost. This comes from the Lorentz transform Δt_B = γ_u(Δt_A+uΔy_A/c²) since Δt_A and Δy_A are both zero.

Am I misunderstanding?

 

[Dale]

I was thinking we should be able to look at frame B as being a single boost away from the rod's proper frame. This would be a boost with <x , y> velocity components being <v/γu , u>.

It turns out that this is not correct. The composition of two non colinear boosts cannot be represented as a single boost. It is a boost and a rotation:

https://en.m.wikipedia.org/wiki/Wigner_rotation

as far as I can tell this necessarily causes the rod to be rotated when changing frames

Yes, when boosting twice, but not colinear boosts.

 

[Hiero]

It turns out that this is not correct. The composition of two non colinear boosts cannot be represented as a single boost. It is a boost and a rotation:

https://en.m.wikipedia.org/wiki/Wigner_rotation

Yes, when boosting twice, but not colinear boosts.

Thanks for the link I will try to study up on this effect but I have to say immediately I am confused. So the two boost reasoning is correct and the rod maintains its orientation?
The reason it feels so bizarre is that I see no necessity to decompose the single boost into two; if I just see a rod flying off at some angle with its length, shouldn't I be able to make a single boost into its proper frame without worrying about "Wigner rotations"?

Edit:
Or maybe the point is that both ways are correct but that the proper frame's x-axis and frame B's x-axis are no longer parallel if we make two boosts instead of one? I'm not understanding the idea

 

[Dale]

if I just see a rod flying off at some angle with its length, shouldn't I be able to make a single boost into its proper frame without worrying about "Wigner rotations"?

Nope. It turns out that if you just do the single boost it will wind up rotated.

This is also why the restricted Lorentz group necessarily includes rotations. Without rotations (ie just boosts) it doesn’t form a group because the composition of two boosts is not a boost.

https://en.m.wikipedia.org/wiki/Lorentz_group

 

[Hiero]

Nope. It turns out that if you just do the single boost it will wind up rotated.

I understand that the rod will end up rotated by a single boost, (because one component is contracted and the other isn't,) but surely that's not what's meant by "Wigner rotation"?

 

[Dale]

surely that's not what's meant by "Wigner rotation"?

Umm, why not? This is an odd objection. Why wouldn’t it be called that?

 

[Hiero]

Umm, why not? This is an odd objection. Why wouldn’t it be called that?

Well the rod rotating is a consequence of the boost. I thought you were saying we have to boost and do some additional rotation? Sorry I've never heard of Wigner rotations before and am still confused as to if the rod is along the x-axis or not in frame B. One boost says it will rotate, two boosts says it will not rotate; which is correct?

 

[PeterDonis]

One boost says it will rotate, two boosts says it will not rotate

No, two boosts says it will rotate.

More precisely, the transformation you are calling "one boost" is not a pure boost; it's a single transformation that combines properties of both a boost and a rotation. Try constructing the transformation matrix for this transformation explicitly; you will see that it contains both "time-space" and "space-space" off-diagonal components (the former is what indicates "boost" when it appears by itself, and the latter is what indicates "rotation" when it appears by itself). Further, this transformation matrix is what you get when you compose (i.e., multiply) the matrices for each of the two boosts--which is why two boosts also says it will rotate.

 

[Hiero]

No, two boosts says it will rotate.

I'm still not seeing this even when I explicitly compose them.

Let (t,x,y) be the rods proper coordinates, (t',x',y') will be for frame A, and (t'',x'',y'') will be for frame B.

To go from the proper frame to frame A:
$$\begin{bmatrix} {Δt'} \\{Δx'} \\{Δy'}\end{bmatrix} =\begin{bmatrix} {γ_v} & {vγ_v/c^2} & {0} \\{vγ_v} & {γ_v} & {0}\\{0} & {0} & {1}\end{bmatrix}\begin{bmatrix} {Δt} \\{Δx} \\{Δy}\end{bmatrix}$$

To go from frame A to frame B:
$$\begin{bmatrix} {Δt''} \\{Δx''} \\{Δy''}\end{bmatrix} =\begin{bmatrix} {γ_u} & {0} & {uγ_u/c^2} \\{0} & {1} & {0}\\{uγ_u} & {0} & {γ_u}\end{bmatrix}\begin{bmatrix} {Δt'} \\{Δx'} \\{Δy'}\end{bmatrix}$$

So if I've composed these correctly it gives:
$$\begin{bmatrix} {Δt''} \\{Δx''} \\{Δy''}\end{bmatrix} =\begin{bmatrix} {γ_uγ_v} & {vγ_vγ_u/c^2} & {uγ_u/c^2} \\{vγ_v} & {γ_v} & {0}\\{uγ_uγ_v} & {uvγ_uγ_v} & {γ_u}\end{bmatrix}\begin{bmatrix} {Δt} \\{Δx} \\{Δy}\end{bmatrix}$$

(I apologize that u and v look very similar in subscript, bad choice.)

If we take two events located on the endpoints simultaneous in B, then in the proper frame there will be Δt, but that won't affect that Δx is still the proper length, say L, and Δy is still zero. So putting Δt''= 0 in the top line of the matrix gives Δt = -Lv/c². If we now use that value to find Δy'' from the bottom line, then we get Δy'' = 0.

That seems to say the rod will not have rotated?

 

[PeterDonis]

if I've composed these correctly

The matrix you computed looks transposed to me (i.e, I think the rows and columns should be exchanged). But I'm notoriously unreliable at properly ordering matrix computations.

 

[Hiero]

deleted

We choose two events with Δt''=0 located at the endpoints. These events will be separated in time in the proper frame, but again that is no problem. We can measure the endpoints at different times in the proper frame and will still get Δx = L and Δy = 0 because the rod is not moving in the proper frame.

The top line of the composite matrix (which says Δt'' = γ_uγ_vΔt + vγ_uγ_v/c^2Δx + (...)Δy) will give us the time displacement Δt which we will need. Putting in Δt'' = 0, Δx = L, Δy = 0, and rearranging we get Δt = -Lv/c^2 (the gammas drop off).

Now that we know Δt we can find Δy'' from the bottom line of the matrix by using Δx = L, Δy = 0, Δt = -Lv/c^2

Doing so yields Δy''=0. And so we see that Δt'' = 0 implies Δy'' = 0.

I'm not trying to wave hands I just don't which parts of the algebra to show or leave out (I have a bad habit of not writing algebra let alone typing it).

The matrix you computed looks transposed to me (i.e, I think the rows and columns should be exchanged). But I'm notoriously unreliable at properly ordering matrix computations.

It has indeed been a while since I've multiplied two matrices, but it looks ok to me. The top 3x3 matrix will multiply on the right side of the middle 3x3 matrix.

 

[PeterDonis]

I disagree

Note that I deleted this comment (and the next one you quoted).

Try comparing the matrix you obtained with the matrix you would write down for a single boost (just a pure boost in the direction of the rod's motion in frame B). Note that you will have to write down the single boost matrix in coordinates that are not aligned with its boost direction (either that or apply a rotation to the composed matrix you obtained).

 

[Hiero]

Note that I deleted this comment (and the next one you quoted).

Sorry about that, took me quite a while to type my response I guess.

Try comparing the matrix you obtained with the matrix you would write down for a single boost. Note that you will have to write down the single boost matrix in coordinates that are not aligned with its boost direction (either that or apply a rotation to the composed matrix you obtained).

I have been dreading doing this just because it will involve the gamma factor associated with sqrt(u²+(v/γ_u)² ) so I know things are going to get messy!

I'll have to take a stab at it tomorrow; it's getting late. I am curious though what are you saying I'll find, a different matrix? Do you think my composite matrix is right or wrong (ignoring the possible transpose issue)?

 

[Hiero]

The thing I can't get out of my head is that whatever that matrix comes out to be, it will cause the rod to rotate because any rod should rotate when boosted unless the boost is 0 or 90 degrees with the length. (It's just a consequence of shortening one nonzero component without changing the other nonzero component.) I guess I'll try to explicitly write that matrix tomorrow if you think it will be fruitful but I'm confused because you said the two boosts should rotate the rod so I thought the composition would be the path to understanding.

 

[jartsa]

Umm, why not? This is an odd objection. Why wouldn’t it be called that?

Well there seems to be two different phenomena that can cause a rod to rotate. To see that we just need to build a little bit more complex object out of rods:

So let's weld two rods together like this: V

Now if we boost vertically, the two rods rotate to opposite directions.

But if we boosts two times, the whole V rotates, causing the two rods to rotate to the same direction.

 

[Dale]

The thing I can't get out of my head is that whatever that matrix comes out to be, it will cause the rod to rotate because any rod should rotate when boosted unless the boost is 0 or 90 degrees with the length.

It is not actually about the rod. It is about the transformations. A pair of non-colinear boosts cannot be represented as a single boost. A pair of non-colinear boosts can be represented as a single boost and a rotation. So the boosts do not form a group, the boosts and the rotations together do form a group.

If you boost a rod in a direction off axis that is still just a boost.

 

[Dale]

Now if we boost vertically, the two rods rotate to opposite directions.

Which is a clear indication that the transformation is not a rotation.

 

[Ibix]

I think the relevant visualisation is three flat squares in a stack. The middle one is hinged to the top one along its north edge, and to the bottom one along its east edge. The bottom square represents the x-y plane. Open the hinge a bit and the middle square is tipped over, representing the x'-y' plane. Open the other hinge a bit and the top square represents the x''-y'' plane. I tried to sketch this freehand on my phone, so apologies for the dreadful draughtsmanship (edit: see #30 for a diagram that doesn't look like it was drawn by a six year old. Edit 2: also, apparently I don't know east from west):

The blue square is the x-y plane, the pink one is hinged to it at the left hand side and represents the x'-y' plane, and the green one is hinged to the pink one along the rear edge and represents the x''-y'' plane. You can certainly have just two squares joined by one hinge and have one square coplanar with the blue x-y and the other coplanar with the green x''-y''. But you cannot make the edges of your square line up with the edges of the green square - which is Wigner rotation. Or whatever its Euclidean equivalent is.

Interesting fact: it is possible to think about this by holding your hands together and hinging one hand about the blade of the palm and then the fingertips. If you do this on a crowded commuter train while frowning intently at your own hands then when you stop focussing on coordinate transforms you will realize that people are giving you very odd looks. Philistines.

 

[Dale]

We need something like a “post of the day” award here @Ibix

 

[Hiero]

So if I'm understanding correctly... when taking two boosts the rod does indeed stay along the x'' axis but not because it doesn't rotate, but instead because the x'' axis rotates with the rod? Whereas with a single boost the coordinates don't rotate and so the rod is rotated relative to x''?

So there isn't any contradiction or mistake it's just that two boosts leaves me in rotated coordinates compared to one boost?

The thing which bothers me is how should two vectors which are parallel to an intermediate vector not be parallel to each other? I suppose that's what Ibix's hinge picture is supposed to elucidate but I still find it strange.

 

[PeroK]

So if I'm understanding correctly... when taking two boosts the rod does indeed stay along the x'' axis but not because it doesn't rotate, but instead because the x'' axis rotates with the rod? Whereas with a single boost the coordinates don't rotate and so the rod is rotated relative to x''?

So there isn't any contradiction or mistake it's just that two boosts leaves me in rotated coordinates compared to one boost?

The thing which bothers me is how should two vectors which are parallel to an intermediate vector not be parallel to each other? I suppose that's what Ibix's hinge picture is supposed to elucidate but I still find it strange.

There is a similar paradox with time dilation. Let frame B be the reference frame. Frame A and C move in opposite directions at the same speed.

At time $t$ in frame B clocks At the origin of frames A and C both read the same $t'$.

But, at time $t'$ in either frame, the clock in the other frame reads something else.

In this Wigner case, that frames A and C have something in common in frame B does not mean they are symmetric with each other.

 

[Hiero]

that frames A and C have something in common in frame B does not mean they are symmetric with each other.

That is a good point. If it were three vectors measured in a single frame then parallel-ness must be transitive. This is different though, as we are comparing things in different frames.

I suppose the next step is take that matrix composed of two boosts and decompose it into a spatial rotation by some boost matrix. We could then find that boost matrix in terms of a rotation angle, (probably messy) but I'm not sure how the rotation angle would be found. Maybe we compare to a general boost matrix and find only one angle will give that form?

Wherever I learned SR from it skipped this effect now I feel I don't really know SR

 

[PeterDonis]

Wherever I learned SR from it skipped this effect

Don't feel bad, I had the same experience. In fact I have not seen a really in depth treatment of this in any textbook; they all seem to mention it quickly but not go into it.

 

[PeterDonis]

I'm not sure how the rotation angle would be found.

The Wikipedia article that @Dale linked to earlier discusses this. Note that you actually need to find two things: the rotation angle, and the axis about which the rotation occurs.

 

[Hiero]

Note that you actually need to find two things: the rotation angle, and the axis about which the rotation occurs.

Oh, for whatever reason I was assuming it was a z-axis rotation in this example (or in general, an axis perpendicular to both boosts). I can't justify that though, I guess it's a wrong assumption?

 

[PeterDonis]

for whatever reason I was assuming it was a z-axis rotation in this example (or in general, an axis perpendicular to both boosts)

You're right, it is, but that still means you had to figure that out.

 

[Hiero]

You're right, it is, but that still means you had to figure that out.

Yes true, but honestly I'm drawing a blank on how to justify that. All I have is the 'hand wavey' reasoning that the perpendicular space should not get mixed in with the plane of the boost. That is of course an unsatisfactory explanation though.

 

[Ibix]

My better version of the picture from post #20. This is actually a 2+1d Minkowski diagram; each plane is the t=0/t'=0/t''=0 plane

As a bonus, I've also generated (assuming I did the maths right!) a picture of what the green plane would look like if I did a single boost. The blue planes are identical; the green planes are co-planar but (Wigner) rotated in that plane. So the green plane below can be mapped down on to the blue plane by a single inverse transform; if I apply that same transform to the green plane above it'll lie in the blue plane but rotated with respect to it.

 

[Hiero]

@Ibix Wow! I thought you were giving an analogy in the first post, I didn't realize you were describing minkowski diagrams! That is quite a powerful visualization, very well made and captures the heart of the issue. I'm really impressed by this perspective, thank you.

 

[PeterDonis]

All I have is the 'hand wavey' reasoning that the perpendicular space should not get mixed in with the plane of the boost.

Yes, the rotation must be in the plane of the two boost directions, i.e., the x-y plane. (Why? Think about the composition of the two boosts; what transformation matrix components can be affected?) That means it must be about the z axis. The article @Dale linked to basically says the same thing, just with a lot more math.

 

[sweet springs]

The point is, from the proper frame, we are boosting off at some angle with the rod's length, (not zero and not perpendicular) and as far as I can tell this necessarily causes the rod to be rotated when changing frames (whereas the analysis with an intermediate frame A told me that the rod remains along x).

By boost of arbitrary direction, Lorentz contract works on boost direction of rod, so it has to rotate. Similarly if we put three rods assembled as XYZ axis or a cube box such an arbitrary direction boost, the cube becomes a moving parallelepiped.

Say rod proper frame is R, Lorentz transformations A→R and A→B keep rods vertical, but R→(A)→B does not.

 

[Dale]

The thing which bothers me is how should two vectors which are parallel to an intermediate vector not be parallel to each other?

The rod is not a vector. It is a worldsheet. Different frames select completely different lines on that worldsheet, and as a result the direction of the rod does not transform as a vector.

For flat spacetime and for actual vectors, two vectors which are parallel to an intermediate vector are indeed parallel to each other in all frames.

 

[Hiero]

The rod is not a vector.

I didn't mean that. The three vectors I was speaking of were the unit vectors in the x, x', and x'' directions; x vector is parallel to x' vector and x'' vector is parallel to x' (all this in the x' frame though).

I've come to grips with it though because all this parallel-ness is measured in the x' frame in which x'' and x are actually parallel. They only become non-parallel in the other two frames which doesn't violate that transitive rule of logic because they're no longer parallel to an intermediate vector in those frames.