# Teaching SR without simultaneity

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• Orodruin
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But seriously, can it ever fit with both doors closed simultaneously?
Yes and no.
• In the rest frame of the garage, yes.
• In the rest frame of the ladder, no.

vanhees71
Yes and no.
• In the rest frame of the garage, yes.
• In the rest frame of the ladder, no.
Wait, what? I was being facetious. But now I'm confused. The ladder can't be contained by the garage, at least that's what I thought.

Wait, what? I was being facetious. But now I'm confused. The ladder can't be contained by the garage, at least that's what I thought.
If it's moving fast enough and is length contracted enough then it will fit as viewed in the garage frame. Briefly. Then it'll slam into the end of the garage at a large fraction of lightspeed and leave a sizeable crater.

vanhees71
If it's moving fast enough and is length contracted enough then it will fit as viewed in the garage frame. Briefly. Then it'll slam into the end of the garage at a large fraction of lightspeed and leave a sizeable crater.
Ugh... It's late here. I'm still conceptually struggling with the idea that length contraction is "real" TBH. The non-simultaneity of the garage doors feels more natural to me. But I guess the fact that in one frame the ladder is entirely in the garage and observes both doors to be closed... It hurts my brain. I guess that's why I brought it up as an example.

Wait, what? I was being facetious. But now I'm confused. The ladder can't be contained by the garage, at least that's what I thought.

Maybe it helps, if you look at the animated Minkowski diagram in above posting #67.

Imagine, the garage is at rest in the unprimed frame (with black coordinate-axes) and has the rest-length equal to ##\overline{OC}## in this frame. In this frame, the events ##O## and ##C## happen simultaneously.

vanhees71
Wait, what? I was being facetious. But now I'm confused. The ladder can't be contained by the garage, at least that's what I thought.
Suppose that your front door is three feet wide and four feet high. How can you fit a 4.5 foot wide sheet of plywood into the house?

Maybe it helps, if you look at the animated Minkowski diagram in above posting #67.

Imagine, the garage has the rest-length ##\overline{OC}## and is at rest in the unprimed frame (with black coordinate-axes).
Its just that in my mind (however misguided), the garage and the ladder have the same scale of measure in either frame (so the garage is always smaller than the ladder). I'm just saying, that in my mind, it is more a matter of an observers perception of time. But I could bet well be wrong!?

Its just that in my mind (however misguided), the garage and the ladder have the same scale of measure in either frame (so the garage is always smaller than the ladder).

No.
• Invariant is the spacetime interval between two events.
• The spatial distance and the temporal distance of events are not invariant.

vanhees71
No.
• Invariant is the spacetime interval between two events.
• The spatial distance and the temporal distance are not invariant.
Yeah... I know. Its challenging to think about and doesn't follow daily common sense

No.
• Invariant is the spacetime interval between two events.
• The spatial distance and the temporal distance of events are not invariant.
And apologies, I didn't mean to mistake the situation. I do understand how it works, and the animation is a great illustration. I was just commenting on how my intuition conflicts with the "paradox".

And apologies, I didn't mean to mistake the situation. I do understand how it works, and the animation is a great illustration. I was just commenting on how my intuition conflicts with the "paradox".

Imagine an intuitive situation: You are in a lab and measure the length of a moving rod with a stationary ruler. You get the correct result only, if you capture the numbers on the scale of the ruler at both ends of the (moving) rod "simultaneously".

vanhees71
Imagine an intuitive situation: You are in a lab and measure the length of a moving rod with a stationary ruler. You get the correct result only, if you capture the numbers on the scale of the ruler at both ends of the (moving) rod "simultaneously".
Hey now! If the rod isn't moving relativistic, my lab partner and I should be able to share information fast enough to make an accurate observation.

But I get your point

Hey now! If the rod isn't moving relativistic, my lab partner and I should be able to share information fast enough to make an accurate observation.

But I get your point

That's not the point. You could use two cameras which are triggered by two local, synchronized clocks. It is not about observation. It's about the relativity of "simultaneously" between different frames, as you can see in the Minkowski diagram.

vanhees71
That's not the point. You could use two cameras which are triggered by two local, synchronized clocks. It is not about observation. It's about the relativity of "simultaneously" between different frames, as you can see in the Minkowski diagram.
I really do understand, I just don't find it intuitive. That's why I brought it up. The example challenges everyday experience. I don't want to derail the topic any further.

Sagittarius A-Star
Ugh... It's late here. I'm still conceptually struggling with the idea that length contraction is "real" TBH. The non-simultaneity of the garage doors feels more natural to me. But I guess the fact that in one frame the ladder is entirely in the garage and observes both doors to be closed... It hurts my brain. I guess that's why I brought it up as an example.
The easiest way to think of it is to note that the ladder has extent in the time direction. So if you simplify the ladder to a 1d line in space, it's a 2d sheet in spacetime - one dimension is length, the other is duration. But different frames differ in which direction they call length and which duration, so different frames take different lines across the sheet and call that the length, and the different lines have different lengths.

That's what's being illustrated in the Minkowski diagram in #67. The shaded area is the 2d rod. According to the unprimed frame (black axes) the x-direction is a horizontal line and the t-direction is a vertical line, so OC is the length of the rod. But according to the primed frame (red axes) the x'-direction is sloped upwards and the t-direction is sloped to the right, so OB is the length of the rod. They're different lengths.

Whether or not length contraction is "real" is up to you. I would say that it's similar to me slicing a sausage perpendicular to its length while you slice it on the diagonal. I get a circular cross-section, you get an elliptical one. That's a Euclidean analogy to length contraction - but is it "real" that the sausage is elliptical according to you and circular according to me? Or are we just measuring different things?

vanhees71
Summary:: De-emphasizing simultaneity in SR curriculum. Thoughts? Experiences?

focus on spacelike separation between events, differential ageing and clocks measuring worldlines, and the geometry of spacetime.

Such an approach is described in the following not published paper:
Special and General Relativity based on the Physical Meaning of the Spacetime Interval

Alan Macdonald
...
In Sec. 2 we discuss the interval of special relativity. The interval has a simple physical meaning as something measured by light, a clock, or a rod. It is thus a fundamental physical quantity. Its meaning does not depend on the notion of inertial frames. In the approach to special relativity described in Sec. 2, we define ##\Delta s## physically, without reference to inertial frames. We then prove that if inertial frames are introduced, then ##\Delta s^2 = \Delta t^2 - \Delta x^2##. But the physical meaning of ##\Delta s## as a directly measureable quantity is more important than the mathematical formula for it in terms of coordinate differences (as important as the formula is). The Lorentz transformation is even less important. Note however, that since ##\Delta s## has a physical meaning independent of coordinates, it is automatically invariant under a Lorentz transformation.
Source:
http://www.faculty.luther.edu/~macdonal/Interval.pdf

via:
http://www.faculty.luther.edu/~macdonal/

Orodruin
As I teach the high-school-teacher students, it's easy to discuss such didactical issues with them, and after giving the intro to SR for the 3rd time now, the way to construct Minkowski diagrams seems indeed to help them a lot.
...

(Shameless plug)
You may be interested in a recent book that was published.

Teaching Einsteinian Physics in Schools
Kersting and Blair, Routledge 2021, https://doi.org/10.4324/9781003161721
(has supplementary material at the bottom: https://www.routledge.com/Teaching-...-Teachers/Kersting-Blair/p/book/9781003161721 )
Chapter 7 is my contribution: Introducing relativity on rotated graph paper

In https://physics.stackexchange.com/a/689291/148184 ,
I describe an argument I used in that chapter
(which was left out of my AJP article because the article was already too long).

When I have time, I'll try to write up aspects as an Insight.

The story line goes like this:
• The Speed of Light Principle and the Velocity give the shape of moving observer Bob's light-clock diamond (with edges parallel to the light cone, to the rotated graph paper).
• The Relativity Principle determines the size (the scaling) of Bob's light-clock diamond.
Two inertial observers meet at an event O.
2 seconds after they meet, they send a signal to the other.
We expect they have the same results, in accord with the relativity principle.
• Take v=(3/5)c for simplicity.
The attempts:
• Assuming absolute-time (same-height diamonds) fails the Relativity Principle.
Alice receives at 3.2, but Bob receives at 5.
Bob's clock-diamonds have to scale up (take longer to tick than Alice's clock)
so that Alice receives later (greater than 3.2) on her clock
and Bob receives earlier (less than 5) on his clock.
This implies time-dilation... but by how much?

• Assuming absolute-space (same cross-section length) fails the Relativity Principle.
Alice receives at 5, but Bob receives at 3.2.
Bob's clock-diamonds have to scale down (so the cross-section length is shorter than Alice's)
so that Alice receives earlier (less than 5) on her clock
and Bob receives later (greater than 3.2) on his clock.
This implies length-contraction... but by how much?

• By trial and error (or some analysis), we are successful.
Alice and Bob receive at 4.
We see the Doppler Effect since k=(4 ticks)/(2 ticks)=2 (as expected for v=(3/5)c )
(and equality of light-clock areas [invariance of square-interval]), in accord with Special Relativity.
Time-Dilation
and Length-Contraction and Relativity of Simultaneity are consequences,
with the correct factors (by counting diamonds along the triangle legs).
Try this for v=(4/5)c.

Play with these using
https://www.geogebra.org/m/HYD7hB9v#material/UBXdQaz4 (make sure BOB's diamonds are shown)
https://www.geogebra.org/m/kvfsq664 (updated)... (make sure BOB's diamonds are shown)
[You can manually adjust Bob's velocity and "lengths [in the lab frame]" of the light-clocks.]

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vanhees71
The easiest way to think of it is to note that the ladder has extent in the time direction. So if you simplify the ladder to a 1d line in space, it's a 2d sheet in spacetime - one dimension is length, the other is duration. But different frames differ in which direction they call length and which duration, so different frames take different lines across the sheet and call that the length, and the different lines have different lengths.

That's what's being illustrated in the Minkowski diagram in #67. The shaded area is the 2d rod. According to the unprimed frame (black axes) the x-direction is a horizontal line and the t-direction is a vertical line, so OC is the length of the rod. But according to the primed frame (red axes) the x'-direction is sloped upwards and the t-direction is sloped to the right, so OB is the length of the rod. They're different lengths.

Whether or not length contraction is "real" is up to you. I would say that it's similar to me slicing a sausage perpendicular to its length while you slice it on the diagonal. I get a circular cross-section, you get an elliptical one. That's a Euclidean analogy to length contraction - but is it "real" that the sausage is elliptical according to you and circular according to me? Or are we just measuring different things?
You hit the nail on the head. I am getting more philosophical than scientific on the issue.

Im still trying to fit my 12 foot ladder in my 10 foot garage
Put it at an angle and rely on Pythagoras Theorem (assuming your garage is 7 foot wide).

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vanhees71
Put it at an angle and rely on Pytharoras Theorem (assuming your garage is 7 foot wide).
Not as much fun as accelerating the ladder to 55% of lightspeed. I get the impact on the back wall releasing about 260 megatons.

vanhees71