# A Special Relativity Question

1. Jan 22, 2005

### grounded

Imagine a long straight train that is not moving. (Relative to the ground)

Each individual boxcar on the train is exactly the same length, which is 100 km.

All the boxcars are numerically numbered starting at the front. (Example 1,2,3,4,5…)

There are two observers, Chuck and Sam.

Chuck is standing at the front of the train (in front of boxcar #1) and is at rest compared to the train.

Sam is next to Chuck, but Sam is in a craft that can instantly accelerate to 150,000 km/sec and can instantly stop.

If Sam instantly accelerates to 150,000 km/sec and travels for exactly one second (according to Sam) in the direction of the train (Example 1,2,3,4,5…), what number boxcar will Sam stop at?

I am pretty sure (tell me if I’m wrong) that Special Relativity says that passing a 100 km long boxcar at the speed of 150,000 km/sec would cause Sam to measure the boxcar to only be 86 km long due to length contraction. This would mean that Sam should pass 1744 - 86km long boxcars in one second, which means Sam will stop on boxcar # 1744.

I am not a believer of SR, but I would like the opinions of people who have a deeper understanding of SR than myself. I believe Sam will stop at #1500 and any change in the length of the boxcars will be due to error, but that is for another discussion.

I would like to know which boxcar you think Sam will stop at and the perspective of both Sam and Chuck.

2. Jan 22, 2005

### Janus

Staff Emeritus
Sam stops at boxcar 1733 according to both Sam and Chuck(the cars are closer to 86.6 meters long according to Sam. If you are going to round off, you need to make sure that your answer comes within 1 boxcar of the correct one).

Since Sam's clock runs slow according to Chuck, and it Sam's clock that determines how long he travels, Sam will travel for longer than 1 sec by Chuck's clock (about 1/.866 or 1.15 sec). Sam therefore travels 173,210 km before stopping according to Chuck and stops at boxcar 1733.

3. Jan 22, 2005

### pervect

Staff Emeritus
I get 1732 and a half. The formujla is 1500 / sqrt(1 - (v/c)^2). Approximating c as 3*10^8 m/s gives 1732, using the exact value of c in google calculator gives 1732.45

As you've noted, Sam thinks the boxcars are shorter. Chuck thinks that Sam's clocks run slow.

Something very similar to this experiment has already been done, and the results are consistent with SR. Muons have a very short lifetime, which can be measured in the laboratory, so short that when they are generated in the Earth's upper atmosphere by cosmic rays, they could not survive long enough to reach the Earth's surface without relativity. However, they do survive. On Earth, we explain this by saying that the muon's lifetime is extended because of time dilation. The muon would explain this by saying that the distance from the upper Earth's atmosphere to the surface of the Earth is shorter.

4. Jan 22, 2005

### grounded

If Sam stops at boxcar #1733 then he will be 173,300 km away from Chuck.

Relative to Sam, if his ship is only capable of traveling 150,000 km/sec how does he travel for one second and end up 173,300 km from Chuck? Wouldn't Sam calculate his own speed to be 173,300 km/sec? After all, Sam knows where he started, knows where he finished, and knows how long it took him to get there.

I understand how SR says Chuck will see Sams clock running slower which explains the extra distance covered beyond 150,000 km, but how does Sam explain himself being further from Chuck than his ship is capable of bringing him in one second?

5. Jan 22, 2005

### Hurkyl

Staff Emeritus
No, he'd calculate his own speed to be zero. He observes the train cars (which are length contracted) passing by him at 150,000 km/sec, until he hits the brakes after one second (by his own clock).

6. Jan 22, 2005

### grounded

Come on now… Sam is the one who instantly accelerated from a declared rest position, besides it doesn’t matter which one you want to view as being in motion, so why do you have to view the train as moving?

Anyway, if you say Sam is traveling at 150,000 km/sec, then how does he end up being 173,300 km away from where he started?

7. Jan 22, 2005

### Janus

Staff Emeritus
Until Sam decelerates to a stop relative to train 1733, he measures the distance from Chuck as 173,000 km. Once he has decelerated, the train has uncontracted. The fact that he is next to car 1733 does not change, but the length of the train from Chuck to car 1733 has increased. This does not mean that Sam figures that he traveled at 173,000 km/sec.

8. Jan 22, 2005

### pervect

Staff Emeritus
Let's say that Sam is in a rocket ship that can accelerate at a trillion gravities, 1 gravity = 9.8 m/s^2

You can use the formulas at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

to determine that it takes Sam 16.8 microseconds, by his watch, to accelerate up to 150,000 km/sec, and that an observer on the train sees that Sam accelerates for 17.6 microseconds.

The same observer on the train sees Sam cover a distance of 1.4 km during his acceleration phase.

Sam, who is in a rocket ship that's a foot long (rather big for a rocket ship that can accelerate at a trillion gravities, but still well inside the ultimate physical limit of c^2/a), can then determine his speed relative to the train, by seeing how long it takes a specific point on the train to traverse the length of his spaceship. He picks a point at the beginning of the second boxcar, to give his ship time to settle down - it's a sturdy ship, but accelerating at a trillion gravities still causes it to deform. [edit -add. It's a very sturdy ship. Think of a structure a trillion feet high in a 1g field. A trillion feet is slightly over twice Earth's orbital radius around the sun.]

Observing that his foot long ship takes approximately  two nanoseconds [end edit] to traverse a fixed point on the leading edge of the boxcar, he computes that he is travelling at about half the speed of light. Doing a more detailed calculation with a more accurate clock readings over multiple points, he determines that he is indeed going exactly 150,000 km/sec as planned.

After the end of his accleration phase, Sam waits for one second by his watch, during which time he does not accelerate. At the end of a second, he deaccelerates for another 16.8 microseconds (his time) at a trillion gravities, coming to a stop somewhere on boxcar 1733. He overshoots very slightly due to the fact that he did not stop and start instantaneously, which is unpysical, but could "only" accelerate at a trillion gravities, during which time he covered 1.4 km starting up and 1.4 km stopping, plus the fact that the exact distance he travels in 1 second would put him in the middle of the boxcar anyway.

Last edited: Jan 23, 2005
9. Jan 22, 2005

### Hurkyl

Staff Emeritus
No it doesn't matter: that's why we get the same answer if we analyze the problem from Sam's viewpoint or from Chuck's viewpoint.

However, when you said:

You specifically asked for the analysis where Sam is stationary, because you're asking for measurements relative to Sam.

Just before hitting the brakes, Sam observes chuck exactly 150,000 km behind him. It's not until after the deceleration (which changes Sam's frame of reference) that the distance is observed to be more.

10. Jan 23, 2005

### grounded

Are you guys sure that is right? While traveling, Sam will measure his speed to be 150,000 km/sec. After traveling this speed for 1 second, and just before he stops, Sam will measure himself to be 150,000 km away from Chuck. But the instant he stops, Sam will measure himself to be 173,300 km away from Chuck?

So Sam will not travel 173,300 km per second, but he will travel 173,300 km in 1 second?

What number boxcar will Sam be at the instant before he stops?

Are you saying that if I get in a ship and travel away from Earth at a speed of 150,000 km/sec for exactly one second, when I stop I will be 173,300 km from Earth?

I’m sorry to keep repeating myself but I still believe this is wrong, and this is why:

You are saying that a ship traveling from point A to point B will reach point B faster if he continuously and instantly starts and stops multiple times during the trip rather than driving straight through at the same speed. The first time Sam stopped he was 173,300 km from Earth, if he didn’t stop he would only be 150,000 km from Earth. Do you see my confusion?

11. Jan 23, 2005

### Janus

Staff Emeritus
Yes. Because while 'traveling' he is in a different frame of reference than when he is 'stopped', and the distance between Chuck and car 1733 is different for these two frames.
No. From Chuck's frame he will have traveled 173,000 km in 1.15 sec. From Sams frame he will have 'traveled' 150,000 km in one sec. It is the "uncontracting" of Chuck's frame when Sam 'stops' that causes Sam to measure his distance to Chuck as 173,000 km once he is at rest with respect to the Train again.
1733
If that one sec is as measured by you, yes.
How in the world did you ever come to that conclusion?
He would would be 150,000 km from Earth as measured by himself, he would be 173,300 km from Earth as measured from the Earth. When he stops, that 150,000 km becomes 173,300 km because he is no longer measuring it from a frame that has a relative motion with respect to the Earth.

12. Jan 23, 2005

### Hurkyl

Staff Emeritus
No, Sam will measure his speed to be 0, but will measure the train cars passing by at 150,000 km/sec.

Yes.

Yes.

No.

Same one as after he stops.

If that second is mesaured by your clock, then yes.

Not true. If you travel for 2 seconds, then stop, you will measure your distance exactly twice as far as if you went 1 second then stopped.

Yes, and let me explain:

Those two numbers are not the same measurement. The 173,300 km is the "distance to Earth, according to the reference frame where Earth is stationary", and the 150,000 km is the "distance to Earth, according to this reference frame where the Earth is not stationary".

You seem to have gotten the idea that there's a difference between time as measured by Chuck's clock, and by Sam's clock... but you still haven't internalized the difference that there's a difference between distance as measured by Chuck's ruler, and by Sam's ruler, (and their respective perception of simultaneity).

In other words, instead of looking at Sam's reference frame, you've been looking at some strange hybrid that uses Sam's clock, but Chuck's rulers.

Also, there seems to be a common misconception that "instant acceleration" avoids all of the side effects of acceleration, but that's incorrect. The phase where Sam decelerates is the phase where his notion of distance, time, simultaneity, et cetera smoothly change from how they were when he was at full speed to match those of Chuck. Making the acceleration instant just makes this change instant as well.