I Length contraction vs expansion

[freshman2013]

I'm coming across a contradiction that I can't resolve. Two points on the ground are separated by 30km. If you're in a moving frame at velocity +v relative to the ground and you view those two points on the ground, would those two points be greater or less than 30km apart? At first, I thought it would be less, since the proper length is 30km and the ground is moving at a velocity -v relative to you. So the separation would be Lo/lorentzfactor and thus be contracted. But when I saw the solutions and they used the lorentz transformations, the solutions yielded a result greater than 30km. The solutions make actually sense, but I couldn't resolve my original method and the method used in the solutions. Can anyone tell me what is wrong with my original approach? It's problem 2a on the link below. Thanks!

https://tbp.berkeley.edu/exams/459/download/ Problem 2a
https://tbp.berkeley.edu/exams/536/download/ the solutions

 

[Vitro]

I'm coming across a contradiction that I can't resolve. Two points on the ground are separated by 30km. If you're in a moving frame at velocity +v relative to the ground and you view those two points on the ground, would those two points be greater or less than 30km apart? At first, I thought it would be less, since the proper length is 30km and the ground is moving at a velocity -v relative to you. So the separation would be Lo/lorentzfactor and thus be contracted. But when I saw the solutions and they used the lorentz transformations, the solutions yielded a result greater than 30km. The solutions make actually sense, but I couldn't resolve my original method and the method used in the solutions. Can anyone tell me what is wrong with my original approach? It's problem 2a on the link below. Thanks!

https://tbp.berkeley.edu/exams/459/download/ Problem 2a
https://tbp.berkeley.edu/exams/536/download/ the solutions

Judging from the answer question 2a isn't asking for the separation between 2 points on the ground but the separation between the two flashing events. That's just a X coordinate difference after Lorentz transforming the coordinates of the two flashes to the spaceship frame.

If the question was asking about the distance between two marks on the ground then your initial intuition would have been correct.

 

[freshman2013]

Judging from the answer question 2a isn't asking for the separation between 2 points on the ground but the separation between the two flashing events. That's just a X coordinate difference after Lorentz transforming the coordinates of the two flashes to the spaceship frame.

If the question was asking about the distance between two marks on the ground then your initial intuition would have been correct.

But the flashes were simultaneous in the rest frame, so shouldn't that be equivalent to calculating the separation between 2 events? If you instead replaced the two lights with a ruler that is 30km long, then you would see the two ends simultaneously in the rest frame, but they won't be simulataneous in the moving frame.

 

[Vitro]

shouldn't that be equivalent to calculating the separation between 2 events?

It is the separation between 2 events, that's what I just said. The separation between 2 events is the difference of their X coordinates, regardless of simultaneity.

 

[PeterDonis]

At first, I thought it would be less, since the proper length is 30km and the ground is moving at a velocity -v relative to you. So the separation would be Lo/lorentzfactor and thus be contracted.

The separation between two events on the worldlines of the two points on the ground that are simultaneous in the spaceship frame will be less than 30 km. But the two events the question is asking about are simultaneous in the ground frame, not the spaceship frame.

 

[Nugatory]

But the flashes were simultaneous in the rest frame, so shouldn't that be equivalent to calculating the separation between 2 events?

The "separation" between two events (that is, the difference between their $x$ coordinates) doesn't tell you much of anything unless the events are simultaneous. One way of seeing this is to imagine that you are driving a car down the road at 100 km/hr. I am on the side of the road two kilometers behind you, and I clap my hands at noon (using your clock). One hour later my brother one kilometer further behind you claps his hands. The two hand-clap events have (x,t) coordinates (-2, 0) and (-103,1). The "separation" between these non-simultaneous events is 101 kilometers, which has nothing to do with anything.

Conversely, if the hand-clap events were simultaneous (using your clock) we could interpret the "separation" as the distance (in your frame) between me and my brother.

If you instead replaced the two lights with a ruler that is 30km long, then you would see the two ends simultaneously in the rest frame, but they won't be simulataneous in the moving frame.

The two ends of a ruler don't wink in and out of existence - they're around for as long as the ruler itself exists. In a space-time diagram you'd represent an end of the ruler with the line $x=x_0+vt$ ($x_0$ being the initial position), while you'd represent the emission of a flash of light as a single point.

Thus, with the ruler all observers can find simultaneous "near/far end of the ruler is here" events. They'll be different events for different observers moving at different speeds relative to one another; but each observer will interpret the separation between his unique pair of simultaneous events as the length of the ruler in his frame.

 

[Chestermiller]

Imagine 4 guys on the train, A, B, C, and D. A is further forward than B, B is further forward than C, and C is further forward than D. A and B are close together, and C and D are close together. All four guys are carrying clocks, and all four of these clocks are synchronized, so that they all read exactly the same time in the train frame of reference.

Guy A is directly opposite the location on the ground where and when the forward bolt of lightning strikes, and, at the same time (according to the clocks on the train), guy C is directly opposite the spot on the ground where the second bolt of lightning will strike. But, according to the guys on the train, the second bolt of lightning has not struck yet. The distance between guys A and C is less than 30 km.

Guy D is directly opposite the location on the ground where and when the rearward bolt of lightning strikes. According to the guys on the train, this second bolt of lightning strikes after the forward bolt. So the distance between guys D and A is greater than 30 km. Guy B is directly opposite the location on the ground where the forward bolt of lightning struck, but he is there at the time that the rearward bolt of lightning struck (according to the clocks on the train). The distance between guys D and B is less than 30 km (and equal to the distance between guys A and C).

 

[MikeLizzi]

If you just transform the 2 events from a reference frame in which they are at rest, to one in which they are moving, you will get 2 events whose spatial separation is larger. But, as has been explained by others, the events in the second reference frame will be for different times. Others have suggested how to proceed to get the spatial separation for the same time. Here is my method.

http://www.relativitysimulation.com/Documents/DeterminingTheLengthOfMovingObject.htm

 

[Herman Trivilino]

But the flashes were simultaneous in the rest frame, so shouldn't that be equivalent to calculating the separation between 2 events?

Yes, but as others have pointed out, that's not equivalent to calculating the distance between the events. The reason is because the frame in which the mesaurements are made is in motion.

Keep in mind, as you have already pointed out, that the $\Delta x$ and $\Delta x'$ that appear in the Lorentz transformation equations are not lengths, they are the differences between two positions. Even in the nonrelativistic case (that is, the Galilean transformation equations) the difference in positions is not necessarily a length, but it's trivial to see why because you don't have the conflating issue of relative simultaneity.

 

[Andrew Mason]

I'm coming across a contradiction that I can't resolve. Two points on the ground are separated by 30km. If you're in a moving frame at velocity +v relative to the ground and you view those two points on the ground, would those two points be greater or less than 30km apart? At first, I thought it would be less, since the proper length is 30km and the ground is moving at a velocity -v relative to you. So the separation would be Lo/lorentzfactor and thus be contracted. But when I saw the solutions and they used the lorentz transformations, the solutions yielded a result greater than 30km. The solutions make actually sense, but I couldn't resolve my original method and the method used in the solutions. Can anyone tell me what is wrong with my original approach? It's problem 2a on the link below. Thanks!

https://tbp.berkeley.edu/exams/459/download/ Problem 2a
https://tbp.berkeley.edu/exams/536/download/ the solutions

Just to add to what has been said, there is a subtle but important distinction between a "length" of an object at rest in a particular frame of reference (proper length) and the "distance" component of a spacetime interval (the difference in the spatial components between two points in spacetime).

With respect to the Lorentz transformations, it is sometimes said that moving clocks run slow and moving rulers contract. What this means with respect to rulers is that the number of local rulers (ie. rulers measuring a unit of distance and which are at rest in a particular frame) that fit between two spacetime events that are simultaneous in the "rest" frame (ie. the proper length) is less than the number of local rulers that fit between those same two spacetime events in the "moving" frame.

AM