How to retrieve the lenght-contraction from the LT?

[quasar987]

I am trying to retrieve the lenght-contration formula from the Lorentz transformations (and am failing miserably). I'd appreciate if someonew could tell me where I'm mistaken. Thanks.

Actually I just don't know how to represent a proper length in terms of events, but here's my latest "attempt"...

The Lorentz transformation tell us how space-time points transform from an inertial coordinate to another. So suppose for exemple, that S is the coordinate system at rest with respect to Earth and some space station, and suppose S' is the system moving with speed +v (purely in the x direction) relative to S (a spaceship for instance). Suppose also that at t = 0, t' = 0 also and that at that time the origins O and O' coincide. Next, we need two events. Event 1, (say, a flash of light), will happen at t = t' = 0 and will happen at x = x' = 0. Event 2 happens simultaneously but at the space station, located, say, a distance x from the origin O. The LT tell us what is the space coordinate of that flash in the primes coordinate:

x' = \gamma(x -v0) = \gamma x

So it's basically telling me that the distance Earth-Space Station is greater as seen from an observer in the spaceship. Evidently, this is not what I wanted to show. I would have liked to get the opposite, namely

x = \gamma x'

How do we do this ?!?

 

[jtbell]

Event 1, (say, a flash of light), will happen at t = t' = 0 and will happen at x = x' = 0. Event 2 happens simultaneously but at the space station, located, say, a distance x from the origin O. The LT tell us what is the space coordinate of that flash in the primes coordinate:

x' = \gamma(x -v0) = \gamma x

So it's basically telling me that the distance Earth-Space Station is greater as seen from an observer in the spaceship.

Hint: When you measure the distance between two moving objects, you need to locate their positions simultaneously. What are the times of events 1 and 2 in the spaceship frame?

 

[quasar987]

Event 1 we have defined to happen at t'_1 = 0 and event 2 will happen (before), at time

t'_2 = -\gamma \left(\frac{vx_2}{c^2}\right)

But what does it matter? The transformation of position from S to S' only makes use of x and t:

x'_2 = \gamma (x_2 - vt_2)

And t_2 is 0..


So as I understand your message, you're saying we cannot measure the distance between the two events in the moving coordinate system because we cannot measure their distance simultaneously. Then how do we get the lenght-contraction formula in terms of events and LT?

 

[JesseM]

You have to find two events in the object's own frame, one at the position of the back end and one at the position of the front end, which are not simultaneous in the object's frame, but which map to simultaneous events in the observer's frame.

 

[quasar987]

Oh, THAT's the trick.. thanks JesseM.

 

[pervect]

I am trying to retrieve the lenght-contration formula from the Lorentz transformations (and am failing miserably). I'd appreciate if someonew could tell me where I'm mistaken. Thanks.

Actually I just don't know how to represent a proper length in terms of events, but here's my latest "attempt"...

The length of an object in any frame is given by computing the distance between the ends of the object at the same time.

The proper length of an object is just the length of the object in its own reference frame.

One way of solving this problem is as follows:

Write down the parametric equations of motion of one end of the object Probably the easiest way to do this is to write

x1(lambda), y1lambda), z1(lambda), t1(lambda). Lambda can be any variable you like, the important thing is that as lambda progresses, it "traces out" the path that the object travels in space and time. If you don't know what variable to chose, make lambda the proper time of the point whose path through space-time you are tracing.

Repeat the procedure to write down the parametric equations of motion of the other end of the object.

x2(lambda), y2(lambda), z2(lambda), t2(lambda)

Now, all you need to do is to solve for a pair of lambda' so that t1(lambda1) = t2(lambda2), then you can compute the distance via the pythagorean theorem.

distance^2 = (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2 where you compute
x1,y1,z1 at lambda1 and x2,y2,z2 at lambda2.

lambda here is called an "affine parameter". You can use a seemingly more direct appraoch if you like, you can write the equations of motion as

x1(t), y1(t), z1(t) and x2(t), y2(t), z2(t)

if you like - but you'll have to be careful to use the same notion of time. In this case, when you go to a new frame, you have to transform two variables at once, and it's easy to get confused. By using an affine parameter lambda to describe the objects trajectory, it's easier to avoid being confused about which notion of time you are using at the moment.

Example:

In its own frame, the path throgh space-time of an object is very simple

x1=0, y1=0, z1=0, t1=lambda
x2=L, y1=0, z1=0, t2=lambda

In a moving frame, this becomes

x1 = gamma*(x1-v*t1) = -gamma*v*lambda1
y1=z1=0
t1 = gamma*(t1-v*x1/c^2) = gamma*lambda1

Repeat the procedure for x2,y2,z2,t2, solve for t1=t2, and compute the square root of the sum of the squares to get the distance. In this case, since y1=z1=0, the distance is just x2-x1.

 

[jtbell]

Or, go back to quasar987's original situation, let the Earth be at the origin of both reference frames. In the spaceship frame, S', measure the Earth's position at time 0 (call this event #1):

x_1' = 0 and t_1' = 0

Also in S', measure the location L of the space station at time 0 and call this event #2:

x_2' = L and t_2' = 0

These two events are not going to be at the same time in S, but that's OK because the Earth and the space station are both stationary in S, so they're always at the same locations in S.

x_1 = 0 and t_1 = 0

x_2 = L_0 (the proper distance between Earth and space station) and t_2 = unknown

Substitute the x's and t's into the Lorentz transformation equations, and you have two equations for the two unknowns L and t_2. Solve them together and you get

L = \frac {L_0} {\gamma}

as expected.