# Length contraction and Minkowski diagram

1. Oct 31, 2012

### 71GA

Hello,

in my other topic members on the forum helped me to understand time dilation by using Minkowski spacetime diagram. Now i would also need some assistance with explaining length contraction in Minkowski spacetime diagram.

I have observer Žiga in his frame $x,ct$ and observer Ranja, which is moving relative to Žiga with her frame $x',ct'$. In my Minkowski diagram i have drawn both frames as Žiga sees them. I also added a meter sticks $\Delta x$ and $\Delta x'$ for each frame.

Q1: What do i have to add in the picture so i can show that to Žiga it seems Ranja's meter stick is shorter?

Q2: What do i have to add in the picture so i can show that to Ranja it seems Žiga's meter stick is shorter?

http://shrani.si/f/1M/12l/5VLNB8Y/minkowski.png [Broken]

Last edited by a moderator: May 6, 2017
2. Oct 31, 2012

### Muphrid

Draw a line from point 1 parallel to the $ct$ axis (call this line $x = d$). Then, the $ct$ axis and this new line represent worldlines of a literal "ruler" that Ziga has.

All that remains is to show that Ranja will measure the distance between the worldlines according to $x'$ axis, and this distance will be shorter than the distance $Ziga$ measure. It will look longer because we're in Euclidean space, but when you compute the invariant interval between the two points of intersection of Ranja's $x'$ axis with the $x=d$ line, it is shorter.

3. Oct 31, 2012

### Staff: Mentor

Q1: Draw the world lines of both ends of Ranja's meter stick. Where do they intersect one of Ziga's lines of simultaneity (that is, a line parallel to to the x' axis, on which all points will have the same t' coordinate)? That's where the ends of Ranja's meter stick will be at the same time in Ziga's frame... And of course the length in Ziga's frame is just the distance between the where the two ends are at the same time.

Q2: Same answer, just the other way around. Draw the world lines of both ends of Ziga's meter stick, where the intersect one of Ranga's lines of simultaneity.

4. Nov 1, 2012

### 71GA

I did draw 2 pictures accordingly. Please confirm if this is correct.

Picture a:
Žiga thinks that Ranja's meter stick is shorter, because to him $\Delta x$ (his meter stick) seems longer than distance from origin to point $\textrm{G}$ (which is Ranja's meter stick how he sees it).

Picture b:
Ranja thinks that Žiga's meter stick is shorter, because to her $\Delta x'$ (her meter stick) seems longer than distance from origin to point $\textrm{H}'$ (which is Žiga's meter stick how she sees it).

http://shrani.si/f/J/bz/4kG04Wv/capture.png [Broken]

Last edited by a moderator: May 6, 2017
5. Nov 1, 2012

### bobc2

Here is a summary of the picture for time dilation, including a derivation of the Lorentz transformation. I'm using the symmetric Loedel diagram so that you don't have to use the hyperbolic curves to calibrate distances for the two different coordinate systems (blue and red).

You can directly visualize the length contractions. We show 3-D cross-section views (X2 and X3 supressed) across the 4-dimensional universe for a blue guy and a red guy, each traveling in opposite directions along the rest black X1 axis at relativistic speeds. Each is carrying along a steel beam, which, before they started their trips, had identical measurements in the black rest system.

Looking along the blue guys instantaneous 3-D space, along the blue line labeled "Simultaneous Blue", you can tell that blue "sees" red's beam length to be shorter than his own. Likewise, looking along the "Simultaneous Red" line you can see that the red guy "sees" the blue's beam length to be shorter than his.

There is something deeply mysterious about the way nature has contrived this scheme of having observers moving at different speeds experience different continuous sequences of 3-D cross-section "NOW" views of the 4-dimensional universe, as they ('...their consciousnesses', as Hermann Weyl put it) move along their various X4 axes at the speed of light). It is perhaps because it was the only way nature could work it out so that the laws of physics would be the same for all observers.

Last edited: Nov 1, 2012
6. Nov 1, 2012

### Staff: Mentor

Looks good to me.

7. Nov 1, 2012

### bobc2

Yes, what you have done is certainly correct and tells the story quite well. Also, here is an additional consideration with a slightly different emphasis. We will just consider the meter stick carried by Ranja. The front and rear boundaries of the stick are shown in heavy blue slanted lines. We show the stick as a 4-dimensional object. So, Ziga and Ranja view the same object through two different 3-D cross-sections. Ziga (Red circle) and Ranja (blue circle) are shown at two positions along their respective X4 axes, first as they pass each other at the origin, then later when Ziga is at position 1 and Ranja is at position 1'.

Last edited: Nov 1, 2012
8. Nov 1, 2012

### Jeronimus

What do you use to draw those diagrams?

While you did it the right way, just from the looks of it, the drawings don't seem to be accurate.
To both common is vrel. Noone's frame is special. This is why if a ruler of 1 meter at rest relative to Ziga would be measured as shorter by Ranja by a given factor, the same would be the case the other way around.
A ruler at rest relative to Ranja, Ranja measures to have a length of 1 meter, would be measured to be shorter by the same factor by Ziga.

In the drawings however, this does not seem to be the case. To me it looks like the ruler in Ranja's frame is shortened by a higher factor when measured by Ziga than Ziga's ruler when measured by Ranja.

How did you decide at which distance to place 1' for x'/t' in Ranja's diagram?

Last edited by a moderator: May 6, 2017
9. Nov 1, 2012

### bobc2

You can't compare line lengths directly between the two coordinate systems. The numerical distance values on the axes for the primed Ranja inertial frame are obtained using the hyperbolic calibration curves. That's why I did the sketch in post #5. In that sketch you can compare line lengths directly between red and blue since their coordinate axis slopes are the same (except red X4 is negative of blue X4, etc.). Notice the calibration curve that 71GA used in post #1.

10. Nov 1, 2012

### Jeronimus

I am not comparing the lengths but the factor by which they seem to have been contracted.

I am comparing the lines between 0/1 and G/1. G/1 seems to fit about 5 times into 0/1.
While H'/1' seems to fit about 10 times into 0'/1'.

If i think correctly, the factor should be the same, in both cases.

Also, i would have labeled the lengths differently if you want to apply the standard formulas for length contraction. Something like that (lines are probably not parallel as i only use paint to draw it)

edit: Δx*1/γ = Δx' and Δx2 * 1/γ = Δx2'

since Δx2 measured by Ranja is of the same length as Δx measured by Ziga, same goes for Δx' measured by Ranja and Δx2' measured by Ziga.

So Δx2/Δx' should be equal to Δx/Δx2' which is clearly not the case in the drawing (by OP). A bit surprising because it is no hand drawing but computer generated. The fault could only be the coordinates 1', 2' not being at the correct distance.

Last edited: Nov 1, 2012
11. Nov 2, 2012

### 71GA

I use Inkscape.

You observe very good. Distance really cant be measured, but procentual it should be the same and IT IS NOT. This is because my picture isnt fully geometricaly correct. Maybee i will use gnuplot to draw real hyperbola and fix this.

It is simple. If i take point $1' (0'1')$ and use Lorentz transformation i get point $1(\gamma, \beta\gamma)$, so i can draw $x'$ axis. It goes same for axis $ct'$ but you have to use Lorentz transformation on point $1'(1'0')$ and you will get point $1(\beta\gamma,\gamma)$.

12. Nov 2, 2012

### bobc2

I haven't yet computed the graph for the space-like view, but here is a graph using actual Lorentz transformation computations for plotting the hyperbolic (proper time) calibration curves. You can see that the line distances on screen are not the same for the corresponding times shown in the rest frame as compared to the moving frame. Notice that when the blue observer is located in time (along his X4 axis) at his position labeled 30 that his view along his instantaneous 3-D world (represented by his blue X1 axis) intersects the rest frame black time axis (black X4) at a point much less than a value of 30. In other words, the blue guy has, say 30 hours showing on his own clock while he "sees" something like 27 minutes displayed on black's rest frame clock (neglecting time delay for light travel, etc.).

We have exactly the same situation with the space-like plots.

Last edited: Nov 2, 2012
13. Nov 4, 2012

### Jeronimus

Also, it is noteworthy maybe, that you do not measure space-interval(length) contraction in an equivalent way as you measure time-interval expansion(dilation).

In fact, if you did the equivalent steps, you would get space-interval expasion(dilation).

Ziga measures following coordinates for two events E1(x1/t1) E2(x2/t2).
Ranja measures the two events to happen at different coordinates at E1'(x1'/t1') E2'(x2'/t2').

In the case of time dilation, we measure t2-t1 = Δt measured by Ziga, and then compare it to t2'-t1' = Δt'

In the case of length contraction however, we measure the distance between any two events which are at the front and end of an object and happen simultaneously within the frame we are at rest in.

If we did the equivalent steps. Therefore calculate the distance between x2-x1 = Δx and then compared that to x2'-x1' = Δx', we would get Δx' = Δx/γ length dilation.

here is a drawing showing this

Δx2 = 1m (measured by Ranja as x2-x1) expands to the length of the light-blue line which is >1m (measured by Ziga as x2'-x1'), while
Δx = 1m expands to the purple line > 1m

Last edited: Nov 4, 2012