Length contraction in diagonal motion and something strange?

[Ookke]

Object A goes horizontal line and object B diagonal line in 45 degrees angle (Fig 1). A and B have the same velocity $v$ in x-direction. In y-direction, B has velocity $v$, A has none. The magnitude of $v$ is not very important, but the total speed of B must be below $c$.

In their own rest frames, A and B are square-shaped and of same size. In A's rest frame (Fig 2), horizontal line is moving through A from right to left. In B's rest frame (Fig 3), diagonal line is moving through B from top-right corner to bottom-left corner.

The problem (Fig 4). We are in A's rest frame. I think the following three conditions must be met:
- B is length contracted in y-direction, but has its proper length in x-direction, because B is moving up with some velocity and there is no velocity in x-direction
- Diagonal line's angle is more than 45 degrees, due to length contraction, because the line itself is moving to left.
- Diagonal line must go through B so that it goes from top-right corner to bottom-left corner, as it does in B's rest frame, because it's absolute which parts of B meet with diagonal line and which not.

Putting these three together, I don't see how they all could be true. B is too shallow to give the diagonal line the room it needs.

There is no math yet, but before going into that, I'd like some comments if anyone finds this interesting and worth further effort. Of course if you see this is based on a simple mistake, please point that out, so I don't have to struggle with my misunderstandings anymore. Thanks.

 

[Nugatory]

You are going to have to be much more careful about how you define that diagonal line.

 

[DaveC426913]

How would you measure the parts of a moving line that is far enough away to have a relativistic delay in observing it (i.e the part of the line where B is)?

You draw the line as extending to infinity - and assume you can measure all parts of it at a single absolute instant in time. Which is forbidden by SR.

 

[Janus]

Remember, in the frame of B, the diagonal and horizontal lines are moving relative to it down and to the left (along the diagonal). This means B is measuring the length contracted values along this line of motion. You have to remove that contraction factor to get the proper lengths of the lines. When you do this, the angle between the diagonal and horizontal line become much less than 45 degrees in the rest frame of the lines themselves In fact it becomes less than the angle from corner to corner for B in figure 4. These are the lines that A applies length contraction to, which causes the diagonal line to pass from corner to corner of B in the rest frame of A.

What you did was to apply length contraction to the lines as they were measured in the rest frame of B, instead of as they were measured in the rest frame of the lines themselves.

 

[A.T.]

How would you measure the parts of a moving line that is far enough away to have a relativistic delay in observing it (i.e the part of the line where B is)?

What is a "relativistic delay"? That light speed is finite was known long before relativity.

You draw the line as extending to infinity - and assume you can measure all parts of it at a single absolute instant in time. Which is forbidden by SR.

After receiving the delayed signals one can certainly work out where the source was at a certain instant of time in your frame. That is "measuring" in SR, in contrast to "seeing". And it's the basis for inertial frames in SR, which do extend to infinity.

 

[A.T.]

When you do this, the angle between the diagonal and horizontal line become much less than 45 degrees in the rest frame of the lines themselves

The way I understand it is: The angle between the diagonal and horizontal rails is defined to be 45° in the rest frame of the rails (Fig 1)

 

[Dale]

Hi Ookke, why don't you try some of the techniques that we discussed previously. This will be easier than before because there is no acceleration involved.

 

[Janus]

The way I understand it is: The angle between the diagonal and horizontal rails is defined to be 45° in the rest frame of the rails (Fig 1)

But in figure # 3, he shows the the diagonal line to be 45 degrees in the rest frame of B. The point is that it can't be so in both rest frames and the angle will be less in the rest frame of the rails than in the rest frame of B.

 

[A.T.]

But in figure # 3, he shows the the diagonal line to be 45 degrees in the rest frame of B. The point is that it can't be so in both rest frames...

Why not? In the rest frame of B its rail is length contracted along the rail direction, but is still at 45° to the coordinate system axes.

 

[Ookke]

The way I understand it is: The angle between the diagonal and horizontal rails is defined to be 45° in the rest frame of the rails (Fig 1)

Yes, this is what I meant. Figure 1 is the rest frame of the rails, sorry I didn't state it explicitly.

 

[Ookke]

Hi Ookke, why don't you try some of the techniques that we discussed previously. This will be easier than before because there is no acceleration involved.

That crossed my mind too and I might well try it, just wanted to sketch this first and see what happens. I tend to dislike mathematical approach a little bit though, but nothing too serious.

The key issue seems to be the shape of B in the rest frame of A. I studied this a little and it could be that B is not necessarily rectangle-shaped, as it's drawn in Fig 4, but it could be something more twisted, parallelogram or worse. I'm not sure about this however. Anyway, I think that B must have some kind of shallow shape in that frame, due to length contraction in y-direction. This could be geometrically challenging to put together with the steep angle of diagonal line, but maybe if B is twisted enough, it can be done.

 

[Ookke]

But in figure # 3, he shows the the diagonal line to be 45 degrees in the rest frame of B. The point is that it can't be so in both rest frames and the angle will be less in the rest frame of the rails than in the rest frame of B.

In its own rest frame, B sees itself square-shaped and of proper size. It's just the diagonal rail that contracts in the direction of motion, as A.T. said, and I also think the angle must be 45 degrees. I checked also your advice in #4 and see what you meant. The angle would indeed be less than 45 degrees in that case.

 

[Dale]

The key issue seems to be the shape of B in the rest frame of A.

I agree. Guess how you determine the shape.

 

[georgir]

This is such an awesome question. Posting here just to subscribe to the following answers and discussion.
My guess is that the object "B" will seem skewed, not just contracted. Relativistic aberration, and all that. But no math from me, so take it with a grain of salt.

(EDIT: removed a claim about which way it should seems skewed, because every way i look at it, it still does not seem to make sense and my head hurts. i'll just wait for someone to explain it properly)

 

[Janus]

Why not? In the rest frame of B its rail is length contracted along the rail direction, but is still at 45° to the coordinate system axes.

Consider the following image. The top figure represents the rest frame of B, and the bottom figure the rest frame of the rails (I've added a third rail so the rails together form a isosceles triangle.)

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/zipline.gif
In the top figure, B is a square and the diagonal rail passes from corner to corner. B is at rest, so the three rails making up the triangle are moving in the direction of the arrow. Because they are moving relative to B, B measures these lines as length contracted along the line of the arrow. In other words, in the the top figure, the triangle is the length contracted version of the triangle. It is this length contracted triangle the B sees as a right isosceles triangle.

The bottom figure shows the rest frame of the rails. Now it is B that has a relative motion in the direction of the arrow. In this frame, it is not a square, but has length contracted along the diagonal. The triangle formed by the rails is now its proper, non-length contracted shape,and is not a right triangle. The angle between horizontal and diagonal rails has become less than 45 degrees.

It is this lower triangle that A would apply length contraction along the horizontal direction to in order to get the triangle shape in its rest frame.

 

[A.T.]

Consider the following image. The top figure represents the rest frame of B, and the bottom figure the rest frame of the rails (I've added a third rail so the rails together form a isosceles triangle.)

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/zipline.gif

1) Why is A's rail shown as horizontal in B's frame? This is not what OP's Figure shows. The OP's Figure shows A's rail as horizontal in the rail frame (Fig 1), and doesn't show it at all in B's frame (Fig 3).

2) Even if we take your B's frame picture as given, your transform to the rail frame still seems wrong: Why is only the vertical rail changed, but not A's rail? The boost is along the 45° line (B's rail). So only B's rail should preserve its orientation, while the other two rails should both change orientation. This would also show the inconsistency of your picture with the OP's scenario.

 

[Ookke]

I agree. Guess how you determine the shape.

I guess it already: world lines. :)

About the discussion in #15 and #16, here is my understanding of the scenario. This quick picture is not accurate in all things, but it's indended to show the following:
- In rails frame (left) A and B are length contracted in direction of motion. A is contracted in x-direction, B is somewhat "diamond shaped" due to contraction in direction of diagonal rail (as it was already drawn in previous posts). As a curiosity, because B conracts diagonally, it retains (at least in some sense) its original dimensions in x- and y-direction. A doesn't do that, but it contracts in x-direction and retains its height in y-direction.
- In B's frame (right) the diagonal rail is at 45 degrees angle relative to B, but the angle between the rails is larger due to length contraction in diagonal direction. The middle grey line represents what the direction of horizontal line would be, if the rails weren't moving, but they are. I'm not sure of A's shape in this frame (it could be skewed) but it seems to me that the horizontal rail must go through A as described in Fig 2 of #1. It's absolute which parts of A and B meet the rails and which do not. A surely has downward velocity in this frame, but it may or may not have velocity in x-direction.

 

[Dale]

. I'm not sure of A's shape in this frame (it could be skewed)

I agree. Qualitatively your sketch seems reasonable with that caveat.

 

[Ookke]

Here is some effort on world lines. In the rails rest frame, let's set $x=0$ and $y=0$ where rails cross and $t=0$ such moment when the center points of A and B both are on the crossing. The world lines for center points are

For A: $(t,x,y,z)=(t,vt,0,0)$
For B: $(t,x,y,z)=(t,vt,vt,0)$

The positions of corners of A and B can be calculated relative to their center points (see the picture). Let $\alpha=\sqrt{1-v^2}$, then A's length contracted width is just $\alpha$ itself and height is not contracted. Top-left corner coordinates, relative to center, are $(-\alpha/2,1/2)$, top-right $(\alpha/2,1/2)$, bottom-left $(-\alpha/2,-1/2)$ and bottom-right $(\alpha/2,-1/2)$.

Let $\beta=\sqrt{1-w^2}$, where $w$ is B's speed along the diagonal (not the same as $v$, but larger combined speed). The diagonal's proper length is $\sqrt{2}$ so its contracted length is $\beta\sqrt{2}$, so the length of $d$ that is in picture is half of that, $d=\beta/\sqrt{2}$. Solving $s$ that is in picture gives $s=\beta/2$. The coordinates: top-left $(-1/2,1/2)$, top-right $(\beta/2,\beta/2)$, bottom-left $(-\beta/2,-\beta/2)$, bottom-right $(1/2,-1/2)$.

If we want to look how things appear in A's rest frame, which moves at speed $v$ to left, we need to use formula
$(t',x',y',z') = (\gamma(t-vx),\gamma(x-vt),y,0)$ where $\gamma=1/\sqrt{1-v^2}$.

Substituting the world lines into this and simplifying:
A: $(\gamma t (1-v^2), 0, 0, 0)$
B: $(\gamma t (1-v^2),0, vt, 0)$

Then we need to put together the world lines and the corner coordinates, to find B's shape in A's frame... but this has been well enough for one post.

 

[Janus]

I decided to work this out for a particular example, and here are the results. Staring from the top figure working down, we have the rest frame of the rails, then the Rest frame of A and finally the rest frame of B. Each frame is shown as the A rail being horizontal, as if our observer moved from frame to frame while keeping his orientation the same relative to it.

http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/rails.gif
In the top figure we have A moving to the right at ~0.686c along its rail, while B moves at 0.8c upwards and to to right at an angle of ~30.96 degrees. A and B are squares in their own frames, and are length contracted along their respective lines of motion. The velocities of A and B are such that their centers remain vertical to each other at all times.

Middle figure is the Rest frame of A, which shows its proper shape of a square. The rails move to the left at ~0.686c, the angle between the A and B rail has increased to ~39.5 degree due to the length contraction of the rails in the horizontal direction. B moves upward at ~0.5657c and is length contracted in the vertical direction. The B rail passes from corner to corner as it did in the top diagram.

The ~0.5658c value can be worked out in couple of ways: you can apply length contraction to the rail's proper lengths from the top fig. to find the angle between the rails in the rest frame of A and then use trig and the relative velocity between A and the rails to work it out, or you can apply the rules for the relativistic velocity addition.

Finally, in the bottom figure, we have the rest frame of B. B is its proper shape of a square, the rails move down and to the left at a 45 degree angle at 0.8 c and A moves down at ~0.5657c The speeds follow from the previous fig. If B is moving at 0.8c relative to the rails in the rest frame of the rails, then the rails are moving at 0.8c relative to B in B's frame.The same argument holds for A and B.

The 45 degree angle between the rails comes from the fact that the rails are length contracted along their line of motion, which increases the ~30.9 degree angle measured in the rail rest frame to 45 degrees in the B frame, which causes the B rail to cross from corner to corner again.

 

[A.T.]

I decided to work this out for a particular example

Note that this is still not the OPs scenario. You have the rails at 45° in B's frame, while the OP has them at 45° in the rails' frame,

Each frame is shown as the A rail being horizontal, as if our observer moved from frame to frame while keeping his orientation the same relative to it.

How is that possible without rotating the coordinate axes? If the coordinate axes are rotated, then both shapes will also have a relative rotation between the frames, which is missing in your pictures. This makes things just more complicated, and has again nothing to do with the OP's scenario, which doesn't imply that A's rail is horizontal in B's frame.

 

[Janus]

Note that this is still not the OPs scenario. You have the rails at 45° in B's frame, while the OP has them at 45° in the rails' frame,

He then goes on to claim that in B's rest frame, B is a square and the line passes from corner to corner. Both conditions cannot be true in this scenario, unless B is rotated with respect to A However, he also shows B as not being rotated with respect to A in A's rest frame. All these things cannot be true. At least one of these assertions is contradictory to the others. I just decided to use the premises that B is a square in its own frame, the B rail passes from corner to corner of B and B is not rotated with respect to A to work out everything else.

How is that possible without rotating the coordinate axes? If the coordinate axes are rotated, then both shapes will also have a relative rotation between the frames, which is missing in your pictures. This makes things just more complicated, and has again nothing to do with the OP's scenario, which doesn't imply that A's rail is horizontal in B's frame.

Rotating the view in each frame has no effect on the shapes of the objects with respect to their lines of motion. If I am watching a train travel from left to right, It is contracted from engine to caboose. If I tilt my head so that the train now travels at a 45 degree angle in my field of view, the train does not change shape, it is still contracted from engine to caboose. The figures I showed are the same idea. The relative length contractions are applied in each frame and then each image is rotated so that the A frame appears horizontally. I haven't altered the transformations, I have just rotated the resultant figures so that the A rail is horizontal in the view shown.

As far as the A rail's orientation in the B frame goes, this is a consequence of the condition of the scenario that B remains directly above A in the rail frame( the horizontal component of each is equal) and B is not rotated with respect to A in the A frame.

In the rail frame, the A rail passes through the middle of A horizontally ( it passes the midpoint of one side of A, through the center and out the midpoint of the other side). If we imagine a third rail, passing vertically through the middle of A and at rest with respect to A, it also passes vertically through B (same rules apply. passing through the midpoint of the bottom, center and midpoint of the top). The vertical nature of this line is not changed by A and B's horizontal motion. In A's frame,the A rail moves horizontally While B moves upward on the third rail. If B is a square in its own frame, it is a rectangle in A's frame, contracted in the vertical direction. In B's frame, A and the third rail are moving downward, B is a square, and A is flattened in the vertical direction. The A rail must still pass from the midpoint of one side of A to the midpoint of the other. The top and bottom of A are still parallel to the top and bottom of B, This means that in the B frame, the A rail must be parallel to B's top and bottom. Ergo, if the B rail passes from corner to corner of B at a 45 degree angle to the bottom side of B, then the A rail is at a 45 degree angle to the B rail in the B frame.

The whole point of this exercise is to determine whether a contradiction arises in the given scenario. My conclusion is that the scenario itself contained contradictions (that the B rail crosses from corner to corner of a square B in B's frame, that the angle between rails is 45 degrees in the rail frame and that B is not rotated with respect to A in the A frame.) You can build a consistent scenario from any two but not all three. I chose a particular two.

Now you could choose another two, Say, 45 degree angle in the rail frame and B rail passing through corners of B. In which case, the angle between the rails in the B frame would be greater than 45 degrees (59 degrees with a 0.8c relative speed between rails and B) and B would be rotated with respect to A in A's frame. I guess it depends on which two the OP really meant to preserve.

 

[A.T.]

He then goes on to claim that in B's rest frame, B is a square and the line passes from corner to corner. Both conditions cannot be true in this scenario, unless B is rotated with respect to A .

So where exactly would such a rotation between A and B come in, if we realize the scenario step-by-step?
1) A & B are both at rest in the rail-frame, and perfectly aligned at the rail-crossing
2) A & B are both linearly accelerated to (v,0) & (v,v) respectively, during the same time period.
What happens in A's frame during that acceleration, so that B ends up rotated there? Is B merely accelerating upwards from (0,0) to (0,v) and contracting vertically, or is something else happening?

 

[Ookke]

Now you could choose another two, Say, 45 degree angle in the rail frame and B rail passing through corners of B. In which case, the angle between the rails in the B frame would be greater than 45 degrees (59 degrees with a 0.8c relative speed between rails and B) and B would be rotated with respect to A in A's frame. I guess it depends on which two the OP really meant to preserve.

I would like to keep these: angle is fixed to 45 degrees in rails frame, diagonal rail goes from corner to corner in B, and B is square-shaped in its own frame. However, I don't think that B is necessarily rectangle-shaped in A's frame for example, but could be quite skewed.

If 45 degrees angle is fixed in B's frame, the angle is smaller in A's frame and even smaller in rails frame. If we fix 45 degrees angle in rails frame, the angle is larger in A's frame and even larger in B's frame. We certainly can draw the horizontal rail always as horizontal, then the diagonal rail is the one that changes its direction depending on the angle. B would be square-shaped, but possibly rotated in its own frame, if the direction of horizontal rail is drawn as horizontal.

The confusion is pretty much about how B looks in A's frame. I'm quite confused, because if your technique works and everything is fine when 45 degrees is fixed in B's frame, why wouldn't it work if we fix 45 degrees in rails frame instead?

 

[Ookke]

So where exactly would such a rotation between A and B come in, if we realize the scenario step-by-step?
1) A & B are both at rest in the rail-frame, and perfectly aligned at the rail-crossing
2) A & B are both linearly accelerated to (v,0) & (v,v) respectively, during the same time period.
What happens in A's frame during that acceleration, so that B ends up rotated there? Is B merely accelerating upwards from (0,0) to (0,v) and contracting vertically, or is something else happening?

The rotation issue is very interesting. A and B could start from the same rest frame, with same orientation, and accelerate in their desired direction of motion (horizontal and 45 degrees) so that they retain their original shape in their own rest frames. To achieve that, both must accelerate so that the rear (looked relative to direction of motion) accelerates first, middle parts then and nose last. But the acceleration vector most definitely looks the same for each part of the object, so where could the rotation come from? Is this rotation something we could measure by gyroscope? If not, what is happening?

 

[Ookke]

I'm starting to get an idea of what B's shape in A's frame might be, but only quantitatively. I'd like to do this with numbers, but that must wait, it's harder than it sounds... Anyway, here is a picture where we imagine a measuring rod (blue in the picture) moving left along the horizontal rail (i.e. in A's frame), with x-velocity, position and contracted length that match B.

The rod's clocks are synced in its own frame, so in the rails frame, rear is ahead compared to nose. T, T/2 and 0 are times in rod's clocks. When we jump into rod's frame, when the rear of the rod is at the crossing of rails, we see rod longer than it was and all clocks show T.

Now, omitting quite a few details I would say that due to relativity of simultaneity, the parts of B (when looked in rod's frame) have moved upwards: the more right they are in x-position, the more they have moved up in y-position compared to the situation in rails frame. As a result, B is still diamond-shaped and somewhat skewed, but more horizontal position than it was in the rails frame. This is not to say that B retains its original shape and just rotates - most likely it will also change its shape. But the diagonal rail just might be able to go from corner to corner as it's supposed to. This is just an initial study, but I'm posting this anyway in case anyone finds these ideas useful.

 

[Ookke]

It seems clear enough that #26 is roughly correct and I don't have any particular need for exact solution (and there are even easier ways to practice world lines etc), so if there is no objection, I suppose the original problem is solved by declaring that rotation (among length contraction and the usual relativity stuff) resolves the problem so that there is no contradiction.

But this rotation itself contains problems that are yet not resolved.

So where exactly would such a rotation between A and B come in, if we realize the scenario step-by-step?
1) A & B are both at rest in the rail-frame, and perfectly aligned at the rail-crossing
2) A & B are both linearly accelerated to (v,0) & (v,v) respectively, during the same time period.

Let's modify the experiment the way A.T. suggested so that A and B are initially at rest where the rails cross, perfectly aligned and having the same orientation. Then let A and B accelerate briefly along the rails so that they get the same x-velocity, like the original setup was in #1. A and B accelerate so that they have their original size and shape in their own final rest frames.

Let's have an additional observer X that accelerates with A, rides some time with it and then jumps into B's frame by accelerating directly upwards starting from A's frame. When X becomes at rest with B, it can expect to find (as #26 suggests) that B is more or less standing on its corner instead of its side. This is somewhat surprising, because X and B started from the same point, at rest with each other and with the same orientation. Neither X's or B's accelerometer has recorded any rotation, only direct acceleration boosts to single direction. Or have they?

So which it would be:
- Either X or B (or both) have experienced rotation that shows in accelerometer, although they think they have accelerated only direct boosts that should not cause rotation.
- Neither X or B has experienced rotation that shows in accelerometer, but instead, the space itself has mysteriously rotated around them so that they do not have same orientation anymore in their final rest frame.
- Or then, this setup is somehow flawed. How?

This is my fourth post in a row in this thread, so if there is no discussion, that's OK and I will let this be. But if you find this interesting (I find) any insight, educated guesses or wild guesses from forum members are appreciated. Thanks.