Need help understanding the Ladder Paradox and space-time diagrams?
With the given details, An observer O at rest relative to the garage with measured length Lg= 6.10 m. The garage is equipped with doors that can open or close instantaneously at a chosen time. Observer O notices that a rocket car is traveling toward the garage at a speed of v=0.6c. Observe O also measures the length of the car while it is moving to be Lc= 4.88 m. Observer O′ is the driver of the car.
Testing whether the car will fit, Observer O rigs the doors of the garage so that the front garage door closes as soon as the back car enters the garage. The closing of the front door sends a light signal to the back door and causes it to open, allowing the car to pass through. What happens to the car when it passes throught the garage.
I calculated that L′g= 4.88 m and L′c= 6.10 m.
I am having trouble understanding space-time diagrams. The question is whether the car will car will succesfully enter and leave the garage. To test whether the car will fit, Observer O rigs the doors of the garage so that the front garage door closes as soon as the back end of the car enters the garage. The closing of the door sends a light signal to the back door and causes it to open, allowing the car to pass through without breaking the doors. According to Observer O, what would happen to the car? How can you tell from the space-time graph that the light will approach the backdoor, thus opening it?
- 1 month agoFavorite Answer
If Observer O says the garage is 6.10 m, that is the proper length because they are at rest relative to each other. Observer O' will see the garage as smaller than it actually is.
If Observer O sees the car coming towards it at a speed of 0.6c, Lc is gonna be smaller than the actual length. The formula for this is
L = L0 / y
where L0 is the proper length, L is the observed length and y is gamma, which is equal to 1 / sqrt( 1 - v^2 / c^2). v is the speed of the object, which here is the car
4.88 = L0 / 1 / sqrt( 1 - (0.6c)^2 / c^2) = 6.1 m
You don't need the length of the garage from O' point of view but you got both of those right
For the garage opener, I would say the car would crash into the front door because it fits exactly in the garage so if it moves at all it'll hit the other door. The light signal is not instantaneous so the car will move before the light gets to the other door
I'd have to see the graph to tell if it has anything useful but in my experience I try to avoid the graphs whenever possible they make me more confused, I try to use just the formulas
- nebLv 71 month ago
I don’t normally supply a link in my answers, but the following link has all of the spacetime diagrams you need to understand this version of the ladder paradox. It gives a far better description than we can supply with this limited forum.
- nyphdinmdLv 71 month ago
In the fraem of reference of the car, the garage has a length Lg' = Lg*sqrt((1 - (0.60^2)) = Lg*sqrt(0.36) = 0.6 Lg = 3.66m
In the same frame of reference, the car has a length Lc' = Lc/sqrt(1 - (0.6)^2) = Lc/0.6 = 8.133m
so the car will not fit in the garage. As the end of the car enters, the front of the car is crashing through the rear of the garage.