# Length contraction in diagonal motion and something strange?

1. Oct 20, 2014

### 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.

2. Oct 20, 2014

### Staff: Mentor

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

3. Oct 20, 2014

### 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.

4. Oct 22, 2014

### Janus

Staff Emeritus
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.

5. Oct 22, 2014

### A.T.

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

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.

6. Oct 22, 2014

### A.T.

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)

7. Oct 22, 2014

### Staff: Mentor

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.

8. Oct 22, 2014

### Janus

Staff Emeritus
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.

9. Oct 22, 2014

### A.T.

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.

10. Oct 22, 2014

### Ookke

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

11. Oct 22, 2014

### Ookke

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.

12. Oct 22, 2014

### Ookke

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.

13. Oct 22, 2014

### Staff: Mentor

I agree. Guess how you determine the shape.

14. Oct 23, 2014

### 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)

Last edited: Oct 23, 2014
15. Oct 23, 2014

### Janus

Staff Emeritus
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.)

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.

Last edited by a moderator: May 7, 2017
16. Oct 23, 2014

### A.T.

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.

Last edited by a moderator: May 7, 2017
17. Oct 23, 2014

### Ookke

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.

18. Oct 23, 2014

### Staff: Mentor

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

19. Oct 23, 2014

### 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.

20. Oct 24, 2014

### Janus

Staff Emeritus
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.

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.

Last edited by a moderator: May 7, 2017