- #71
Sagittarius A-Star
Science Advisor
- 1,081
- 810
Yes and no.valenumr said:But seriously, can it ever fit with both doors closed simultaneously?
- In the rest frame of the garage, yes.
- In the rest frame of the ladder, no.
Yes and no.valenumr said:But seriously, can it ever fit with both doors closed simultaneously?
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.Sagittarius A-Star said:Yes and no.
- In the rest frame of the garage, yes.
- In the rest frame of the ladder, no.
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.valenumr said: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.
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.Ibix said: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.
valenumr said: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?valenumr said: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.
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!?Sagittarius A-Star said: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).
valenumr said: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).
Yeah... I know. Its challenging to think about and doesn't follow daily common senseSagittarius A-Star said:No.
- Invariant is the spacetime interval between two events.
- The spatial distance and the temporal distance 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".Sagittarius A-Star said:No.
- Invariant is the spacetime interval between two events.
- The spatial distance and the temporal distance of events are not invariant.
valenumr said: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".
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.Sagittarius A-Star said: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".
valenumr said: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
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 said: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.
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.valenumr said: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.
Orodruin said: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.
Source: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.
vanhees71 said: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.
...
You hit the nail on the head. I am getting more philosophical than scientific on the issue.Ibix said: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?
Put it at an angle and rely on Pythagoras Theorem (assuming your garage is 7 foot wide).valenumr said:Im still trying to fit my 12 foot ladder in my 10 foot garage
Not as much fun as accelerating the ladder to 55% of lightspeed. I get the impact on the back wall releasing about 260 megatons.PeroK said:Put it at an angle and rely on Pytharoras Theorem (assuming your garage is 7 foot wide).