1. Jun 26, 2013

### medwatt

Hello,
I am a bit confused about the length contraction formula. The formula is :
l=lo*sqrt(1-u^2/c^2), where lo is the distance measured in a frame at rest with the moving object.
Now I was looking at a problem where a spacecraft just as it touches the finish line (event 1) a message from the back (which is measured by the astronaut to be 300m) is simultaneously sent (event 2).
Now, using Lorentz trans. an observer standing just at the finish line will say that event 2 occurred at x=375m.

My confusion.
I would like to consider the spacecraft as a whole, as a single object with the front just at the finish line and the back where event 2 occurred. In that case, the astronaut will say the length of the space craft is 300m. By definition of proper length that should be the proper length because it is being measured by the astronaut who is at rest with the craft.
So if I plug that into the formula for length contraction I should expect the result to be shorter as measured by the observer standing at the finish line. That is x should be less than 300m.

I know I am wrong somewhere in my thinking and I hope someone will point that out.

Thanks.

2. Jun 26, 2013

### ghwellsjr

In order for you to see the correct Length Contraction using the Lorentz Transformation, you have to find two events at the end points of the object that are at the same time. The two events you picked were:

t1 = 0; x1 = 0
and
t2 = 0; x2 = 300

Then you transformed these to a speed of 0.6c and got:

t'1 = 0; x'1 = 0
and
t'2 = -225; x'2 = 375

You probably didn't bother to transform the time components and so maybe you didn't notice that t'1 did not equal t'2 or maybe you didn't know that it mattered. Well it does. So you can keep trying different values for t2 until t'2 equals 0. That's what I did and I found that 180 works. If you use this event:

t2 = 180; x2 = 300

you will find that it transforms to:

t'2 = 0; x'2 = 240

and now you get the correct Length Contraction.

The rationale for doing it this way is that it doesn't matter in the unprimed frame which two time components for the events at the endpoints of the object you use since the object isn't moving but it does matter in the primed frame where the object is moving.

A better way to understand this is to draw a spacetime diagram of the stationary object in its rest frame and then transform the entire spacetime diagram into a frame moving at 0.6c with respect to the first one. Then you can see that the length along any horizontal axis follows the formula. Maybe I'll do that in a subsequent post.

3. Jun 26, 2013

### medwatt

I dont understand why you said t2=180. t2 as you have defined it is the time of the second event as measured in the craft and that is already said to coincide with the craft touching the finish line meaning t2=0. So what do you mean to say t2=180 ?

4. Jun 26, 2013

### ghwellsjr

Well, like I said, a spacetime diagram is a better way to understand the situation. Here's one for the rest frame of the spacecraft. The finish line is shown in black approaching at 0.6c. The front of the spaceship is depicted by the blue line and rear by the red line:

If you transform the event at the rear of the spaceship at the Coordinate Time of 0, it ends up at a Coordinate Time of -225 in the rest frame of the finish line:

But if you transform the event at the rear of the spaceship at the Coordinate time of 180, it ends up at a Coordinate Time of 0 in the rest frame of the finish line and the length of the spaceship is 240 meters.

Actually, you can measure its length at any time. I show another example at the bottom.

Does this make it clear?

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5. Jun 29, 2013

### Staff: Mentor

What George is saying is that, according to stationary observers, the two events did not occur simultaneously. As reckoned by them, the message was sent first, and then the front of the spaceship reached the finish line. So, according to these stationary observers, the spaceship moved forward during the intervening time between the two events, and the distance between the two events was not only greater than the contracted length of the spaceship, but also greater than its rest length. If they had measured the length of the spaceship from the location of its front end and its tail end at the same time (as measured by the synchronized clocks in their own frame of reference), they would have obtained the contracted length.

Chet