# Direction of length contraction in combined motion

1. Mar 12, 2015

### Ookke

Left picture: There are two moving rails that cross at center (red dot) which itself doesn't move. The rails speeds are close to c, so that they are length contracted by factor 10 (roughly 0.995c). There are markings in rails at interval 10 length units in rails rest frame, so the interval is only 1 length unit in this frame. Markings from both rails meet the center simultaneously.

Right picture: The same from horizontal rail's rest frame. The whole vertical rail and center are moving right at speed approx. 0.995c. The vertical rail has also some velocity in y-direction downwards, but considerably less than c, about 0.1c by my approximations. In right picture, the markings of horizontal rail are at interval 10 length units (not in scale, but should illustrate the idea) because this is horizontal rails own rest frame. But the markings of vertical rail are at interval 1 (or at least close) also in this frame, which can be seen e.g. by that y-velocity is only 0.1c and markings from both rails must meet center simultaneously also in this frame.

Maybe there is no problem, but I'm curious that the vertical rail seems to be strongly length contracted also in the frame of right picture, although the vertical rail's velocity is mostly horizontal and only slightly vertical in this frame. I would expect that length contraction happens only in the direction of the motion, which would mean that vertical rail shouldn't be much length contracted in y-direction.

2. Mar 12, 2015

### wabbit

Maybe I'm seeing this wrong, but are you sure your right hand diagram is equivalent to the left one? It seems to me the arrows representing the direction of the moving rail in the right hand picture should be pointing South-East, not East. As you draw it, the crossing point is stationary on the vertical rail frame, but it was moving in that frame in the lefthand picture.

(Hope this is clear...if not I'll try a better formulation)

3. Mar 12, 2015

### Staff: Mentor

Please show your work. This doesn't look right to me. There will be some effect on the vertical rail's y-velocity, but I think it's much less than you are estimating; the vertical rail will still be moving at a highly relativistic speed in the horizontal rail's rest frame. [Edit: no, actually the y-velocity is unchanged by a boost in the x-direction.] [Edit again: see follow-up posts.]

Last edited: Mar 13, 2015
4. Mar 12, 2015

### Ookke

Yes, that is the idea, but I "separated" the velocity components. So the movement is south-east, but mostly east.

I think the diagonal velocity of markings in vertical rail would be too much, if the vertical rail's y-velocity was also close to c.

The way I got 0.1c was that horizontally close to c (approx 0.995c, Lorenz factor 10) moving observer sends vertical light pulse, that reflects back at distance 1 units. The light pulse is vertical in this observer's own frame and takes 2 time units to two-way travel. But in outside frame, where this observer moves, the light pulse path is close to horizontal and takes 20 time units to get back to the sender. So the light pulse y-speed relative to the sender, which is moving horizontally, is quite much less than c looked from outside frame. I don't have anything more formal, but this was the idea.

5. Mar 12, 2015

### Ookke

Here is a picture about the value 0.1c. I don't draw all the arrows, because it would be messy, but the motion is the same as in previous picture.

Upper picture is in crossing's (red point) rest frame. Blue point is a clock that is stationary and synchronized with the crossing, the distance between crossing and clock is 1 length unit. Vertical rail markings are moving down and $\Delta t$ is the time one marking takes moving between crossing and clock. It's about 1 time unit, slightly more, because the speed is slightly below c.

Lower picture is in horizontal rail's rest frame, where crossing and clock are moving horizontally with approx. 0.995c. Red and blue clocks are still synchronized and their distance is not changed by motion, because horizontal motion does not affect either. Vertical rail marking takes time approx. 1 unit (slightly more) moving between red and blue clocks, according to these clocks, but as in the outside frame the clocks are slow by factor 10, one marking takes approx. 10 time units (slightly more) to travel from red clock to blue clock. That gives the speed quite close to 0.1c. I think this accuracy is enough, if this is otherwise correct.

6. Mar 12, 2015

### Staff: Mentor

This argument assumes that whatever is reflecting the light pulse is not moving vertically. More precisely, it only deals with one reflection, so the vertical motion of the reflector is irrelevant. Try looking at multiple reflections where the mirror is moving vertically, and seeing how the mirror's vertical velocity transforms. Or you could just look at the correct formula for relativistic velocity addition, in the case where the boost velocity (from one frame to the other) and the velocity being transformed are orthogonal:

Plug $U = 0.995c$ and $V = 0.995c$ into the formula and see what you get.

7. Mar 13, 2015

### jartsa

I think it's like this:

Boost in the x-direction decreases the y-velocity. (So that x-velocity plus y-velocity always stays smaller than c) (Pythagoras' formula is the correct formula to add the velocities)

Boost in the x-direction does not change lengths in y-direction. Because why would boost in the x-direction change lengths in y-direction?

Last edited: Mar 13, 2015
8. Mar 13, 2015

### pervect

Staff Emeritus
Are you sure? If the y velocity is .99c, and remains unchanged after a boost in the x-direction that makes the x-velocity equal to .99c, then both the x and y velocity will be .99c if the y-velocity is unchanged and the total velocity will be sqrt(.99^2 + .99^2), which is greater than c :(

I would think that the proper velocity would transform as a 4-vector, meaning the y-component of the proper velocity would remain unchanged (because the y-component of a 4-vector isn't changed by a boost in the x direction). I believe this implies that the y-component of the ordinary velocity would change - letting the 4 velocity be (dt/dtau, dx/dau, dy/dtau) we boost in the x direction, leaving dy/dtau unchanged but increasing dt/dtau. Then the ordinary velocity is dy/dt = (dy/dtau) / (dt/dtau) which should decrease if (dy/dtau) is unchanged and (dt/dtau) increases.

9. Mar 13, 2015

### Staff: Mentor

Hm, yes, you're right.

Yes, this looks right. Sorry all for the mixup on my part.

10. Mar 13, 2015

### pervect

Staff Emeritus
From the picture, the vertical rail is moving down, the horizontal rail is moving to the left.
That's about what I get, the drawing is confusing because you don't show the downward component, the arrows pointing to the right on the vertical rail should point to the right (mostly) but also down slightly.

I believe that the vertical rail remains vertical, and the vertical spacing remains the same, but it was a rather hasty calculation that I don't want to go into the details. Basically, though, you'd want to write the equations of motion of at least 4 dots, then lorentz-transform them appropriately.

yes
That's what I'm currently getting.

11. Mar 13, 2015

### Ookke

It can be a bit misleading, but the arrows in the first picture are intended to show that the vertical rail, as a whole, is moving to right, and the vertical rail also has some velocity in y-direction.

To be more accurate, the red arrow in the first picture would be strictly horizontal, as the center has only horizontal x-velocity in that frame, but the horizontal black arrows would be mostly right, but also slightly down, as you suggested.
I also believe that, because the motion of #1 right picture is horizontal. Why should it affect verticality? If the direction of motion was something else, e.g. down-right, I'm not sure about the vericality any more, but that would be a different case.

Any 4-vectors or advanced math stuff are out of my reach currently, sorry. I hope those are not really necessary to analyze the point in this exercise (as I see it) that is the direction of length contraction, which seems a bit strange.

12. Mar 13, 2015

### jartsa

If you calculate the velocity of one point with x-speed 0.995 and y-speed 0.0995 it will be something like 0.99996 c. If you calculate what Lorentz factor corresponds to that velocity, it's about 100.

So the vertical rail will have relativistic mass 100 * rest mass, when the rail has speed 0.99996 c. Horizontal motion and vertical motion contribute equally to the relativistic mass.

13. Mar 13, 2015

### Ookke

Both x- and y-velocity cannot be close to c, as you already suggested in #7, because the diagonal speed would exceed c. I don't know how to solve this thought experiment, but it's interesting.

14. Mar 13, 2015

### jartsa

If you look carefully, y-speed has a zero after the decimal point, it's about 0.1 c.

By the way I tried it myself. I got Lorentz-factor 111. For results closer to 100, change 0.995 c to a value that more accurately corresponds to Lorentz factor 10.

Y-velocity is of course 1/10 of original y-velocity, when Lorentz-factor is 10. Rail moves ten times slower, when time dilation makes things ten times slower.

Last edited: Mar 13, 2015
15. Mar 14, 2015

### Ookke

I agree. Now the funny part is that the vertical rail must be length contracted in y-direction by factor 10, although the y-velocity is only 0.1c. If it had its proper length, even approximately, this combined with well below c speed would result that the markings in horizontal and vertical rail would be out of sync. They would not hit the center simultaneously. That would be a contradiction, because if they hit the center simultaneously in one frame (like that of #1 left picture) they must do that in every frame. Different frames can see lots of things differently, but they must agree about events that happen at specific place and time i.e. at specific point of spacetime. Analogy: if two cars collide on a highway, they do that in every frame. But the events that led to the collision can be seen differently in different frames, or in court.

16. Mar 14, 2015

### jartsa

Yes. Thank you. I have learned something new. So let's see now ...

If a runner is running at speed 0.99 c, what happens when there's a slight uphill?

If the runner keeps his horizontal speed at 0.99 c, when going up the hill, he will length contract a lot. Also he will have to accelerate a lot. I mean he has to work hard to increase his kinetic energy a lot.

If a marble is rolling at speed 0.99 c, what happens when there's a slight uphill? Marble gains some vertical speed, loses some horizontal speed, no great length contraction.

17. Mar 15, 2015

### pervect

Staff Emeritus
Now I'm not sure if I analyzed the same problem you did :(. Though I think it's similar.

In some circumstances, doing a pair of Lorentz boosts causes a rotation. So I wouldn't rule it out a-priori, though it doesn't appear to happen in this case from my analysis.

You need high school algebra to fully analyze the problem - length contraction is not the only thing going on. Specifically, there are often effects due to the relativity of simultaneity. Ignoring the effect because you are not familiar with it will possibly generate false answers. I am not aware of any good way of explaining the effect without algebra, though you can try for instance http://www.bartleby.com/173/9.html ("Einstein's train").

The approach I would suggest would be to write the equations of motion of each point on the vertical rail. There are N points, we will just number them N, and write the equations of motion generically. Sorry if this is "too advanced", or "too hard", I don't think there is a relialble way of working relativity problems without at least this much math.

What I get (see my comments about not being sure I understand the way you set up the problem)

$x(t) = 0$
$y(t) = -\beta t + L N / \gamma$

where v is the velocity of the rail, L is the proper length between points (10, in your example), $\beta = v/c$ and $\gamma = 1 / \sqrt{1 - \beta^2}$

There are actually "N" equations, one for each of the N points on the vertical rail. Letting N=0 gives you the center point, N=1 is the point above the center, N=-1 is the point below the center, etc.

Then you use the well-known "Lorentz transform" to convert the equations of motion from (t,x,y) rest frame into the moving (t', x' y') frame. You'll initially get x' and y' as a function of t. You need to use the equation for t' as a function of t to get x' and y' as functions of t', rather than t, because t' is not the same as t, due to the aforementioned relativity of simultaneity.

You can see the basics of the Lorentz transform in the wiki at http://en.wikipedia.org/wiki/Lorentz_transformation, there are a lot of other online references as well.

I'd suggest doing a pair of such transforms, one to confirm that the spacing between points is indeed L in the rest frame, the other to determine what happens when you look at it from the frame moving to the right.

It is also helpful to set c=1 by the proper choice of units, it makes the algebra simpler.

If this approach is "too mathematical" for you, I apologize, but it is probably the lest sophisticated approach that captures all of the subtle nuances of relativity.

18. Mar 15, 2015

### Ookke

"I understood it until it was explained to me" Maybe I didn't explain it very well, but the key ideas I had in mind in #1 right picture
- vertical rail, as a whole, is moving to right, in the rest frame of horizontal rail
- vertical rail is strictly vertical i.e. perpendicular to horizontal rail, not rotated
- the crossing (red dot) has velocity only to right, it has no y-velocity. We can think the crossing as physical, e.g. that there is an observer at crossing, if we wish.
- each individual point of vertical rail (take e.g. any black dot there) has velocity to right, due to the whole vertical rail is moving, and velocity also to down, as it was defined in #1 left picture that the vertical rail has down-velocity relative to the crossing.

I'm somewhat aware of this rotation, it's surprising to me and can be used to construct thought experiments I find interesting. Maybe it's not an issue here, though, but we can keep this possibility in mind.

It seems that I have reached a kind of dead end in my relativity hobby, as the thought experiments and things to analyze are getting so complicated that mathematics cannot be avoided anymore. That's too bad, because math really isn't my cup of tea, although physics is interesting. But let's see how this goes. Thanks for the pointers, I need to dig into these at least a little bit. I suppose taking two successive dots on vertical rail, and equations to them, would be enough, because the main issue is length contraction and the distance between points.

19. Mar 15, 2015

### Ookke

Here is some effort on the mathematical approach. Luckily I got some practice in this forum last autumn, here is the link if someone wants to take a look:

Looking at #1 left picture, let's fix the origin at crossing (red dot). Fix $t=0$ so that some vertical rail marking (black dot) is at the crossing, name that marking A. Name the nearest vertical rail marking above crossing B. The distance to it is 1.

World line for A would be $(t,x,y)=(t,0,-vt)$ and for B $(t,x,y)=(t,0,1-vt)$. $v$ is in the units where $c=1$ and $v$ is defined to correspond Lorenz-factor $\gamma=10$, which is roughly $v=0.995c$.

If we want to look things from the horizontal rail's rest frame (#1 right picture) we need to boost left. Equation for Lorenz-boost to left (negative x-direction) is $(t',x',y')=(\gamma(t+vx),\gamma(x+vt),y)$. Substituting the world lines into boost equation, using $\gamma=10$, gives

$(t',x',y')=(10t,10vt,-vt)$ for A
$(t',x',y')=(10t,10vt,1-vt)$ for B
As we see, $t'=10t$, so using that
$(t',x',y')=(t',vt',-v(t'/10))$ for A
$(t',x',y')=(t',vt',1-v(t'/10))$ for B

So at $t'=0$ the y-coordinate of A is 0 and y-coordinate of B is 1. The y-coordinate of B would be 0 (i.e. B would be at the center) at approximately $t'=10$, slightly more. This is no new information, but at least matches what I already thought. I have a feeling of circular reasoning, but at least this is something to begin with.

20. Mar 18, 2015

### Ookke

Here is a simpler version demonstrating the same problem, or maybe there is no problem, just that the direction of length contraction is not as straightforward as someone (at least I) might expect.

The upper picture is in black rail's rest frame. Black rail's proper length is 1. The red rail is moving down relative to the black rail at v that corresponds Lorenz-factor 10, approx. 0.995c. Red rail's proper length is 10, so it's contracted length in this frame is 1. The black and red rail actually have the same x-position, but they are just drawn side by side to get both visible. Let's fix the origin at the center of black rail and t=0 so that the ends of the rails meet. There is also an observer (black dot) that moves at the same speed v to left. The world lines for the rails in this frame are:

$(t,x,y) = (t,0,0.5)$ black rail's upper end
$(t,x,y) = (t,0,-0.5)$ lower end
$(t,x,y) = (t,0,0.5-vt)$ red rail's upper end
$(t,x,y) = (t,0,-0.5-vt)$ lower end

The lower picture is in the black dot's (i.e. observer's) rest frame. The thick grey arrow indicates that the system of rails is moving to right at v relative to the black dot, and there is also red arrow saying that red rail has some down-velocity relative to the black rail. We use equation for Lorenz-boost to left, which is $(t',x',y')=(\gamma(t+vx),\gamma(x+vt),y)$. We need to substitute the world lines into the boost equation and use $\gamma=10$, as it was defined that v corresponds to this value, and get the world lines in this frame:

$(t',x',y') = (10t,10vt,0.5)$ (black upper)
$(t',x',y') = (10t,10vt,-0.5)$ (black lower)
$(t',x',y') = (10t,10vt,0.5-vt)$ (red upper)
$(t',x',y') = (10t,10vt,-0.5-vt)$ (red lower)
And by using $t'=10t$ get
$(t',x',y') = (t',vt',0.5)$ (black upper)
$(t',x',y') = (t',vt',-0.5)$ (black lower)
$(t',x',y') = (t',vt',0.5-0.1vt')$ (red upper)
$(t',x',y') = (t',vt',-0.5-0.1vt')$ (red lower)

By using t'=0 we see that that the black and red rails are equally long in this frame, which means that the red rail is length contracted by factor 10 also in this frame. And from y-coordinates of the red rail we see that its y-velocity is only approx. 0.1c. So the question is why the red rail is length contracted by factor 10 in y-direction, although its y-velocity is only 0.1c.

Last edited: Mar 18, 2015
21. Mar 18, 2015

### Ookke

I suppose post #20 gives some idea of solving this question too. If runners have the same proper length, slightly uphill going runner is length contracted and much shorter than the runner that is moving horizontally.

22. Mar 18, 2015

### Staff: Mentor

Careful! That equation is not a valid Lorentz transformation; it only holds for an event at the spatial origin, i.e., if $x = y = 0$. That is certainly not true for the events you are transforming.

23. Mar 19, 2015

### Ookke

If that is not ok, I don't know what to do. Actually the two rails system of #20 (forgetting the observer, the black dot) doesn't have any length in $x$-direction, so that dimension shouldn't have effect on relativity of simultaneity, but maybe the equations are still wrong.

I suppose that world lines, Lorentz boost and using $t'$ instead of $t$ are quite standard procedure, but I didn't find any universal formula that would give relation between $t'$ and $t$. And working on the boost formulas like $t=\gamma(t'-vx')$ and $x'=\gamma(x-vt)$ didn't seem to lead anywhere. If anyone can e.g. link some study material, it's appreciated.

24. Mar 19, 2015

### Staff: Mentor

Sure there is; it's the Lorentz transformation formula, just as you wrote it. But that formula transforms events, not worldlines; to look at worldlines, you have to find two events on a given worldline in a given frame, and then find the equation of the unique straight line, in that frame, that passes through those two events. Any pair of events on a given worldline will do.

Alternately, if you know the coordinates of a single event on a worldline in a given frame, and you know the worldline's velocity vector in that frame, you can write the equation for the entire worldline, since the velocity vector tells you how the spatial coordinates change with time and you have a set of spatial coordinates at one particular time (the coordinates of the event you already know).

25. Mar 23, 2015

### Ookke

Ok, let's try. Hopefully I understand your advice correctly. So the worldlines in #20 are the following, and we need to use t' instead of t. (It's the setup of #20 I'm considering from now on, and we can forget the original #1 setup.)
(t',x',y') = (10t,10vt,0.5)
(t',x',y') = (10t,10vt,-0.5)
(t',x',y') = (10t,10vt,0.5-vt)
(t',x',y') = (10t,10vt,-0.5-vt)

Let t'=1 => t=0.1 and we get the events
(1,v,0.5)
(1,v,-0.5)
(1,v,0.5-0.1v)
(1,v,-0.5-0.1v)

For t'=0 => t=0 we get
(0,0,0.5)
(0,0,-0.5)
(0,0,0.5)
(0,0,-0.5)

Then, get "event at t'=1" minus "event at t'=0", respectively for each of these four, and we get velocity vectors (I guess that's the term)
(1,v,0)
(1,v,0)
(1,v,-0.1v)
(1,v,-0.1v)

Then construct the lines. Use the information that line goes through the event at t'=0, and use the velocity vector multiplied by arbitrary $a \in \mathbb{R}$. The lines are
(0,0,0.5) + a(1,v,0) = (a, va, 0.5)
(0,0,-0.5) + a(1,v,0) = (a, va, -0.5)
(0,0,0.5) + a(1,v,-0.1v) = (a, va, 0.5 - 0.1va)
(0,0,-0.5) + a(1,v,-0.1v) = (a, va, -0.5 - 0.1va)
and using symbol t' instead of "a" we see that these are in fact the same as we already had in #20. So there is not really difference in this case, but I tried the same setting x=1 instead of x=0 in the frame of rails (frame F) and checking how it looks in the moving frame F', and the time component was indeed more complicated than t'=10t.