Simple special relativity mistake - find i can't

[n3xtras]

First: thanks for reading
Second: I'm not interested in other "experiments" that prove contraction of lengths (i know them and i know it works)
Third: Please find where is the problem in my simple thought

Let be a "rest frame" S with an object to measure its length and do it with a ray of light/body that move from A to the and B and come back to A
It needs a time T so that its length is d=c*T/2 (d=V*T/2 in case i use other body different to light )

Observe the same experiment in a frame S' that has a v' relative velocity in the same direction of the length of the object AB
Because of that the event doesn't happen in my frame i argue i can observe it in time T'=γT so turns to me that the object appears long d'=c*T'/2

The length is greater than the one in the rest frame. There must be an obvious mistake...
Where my thought fails and why?

 

[Nugatory]

The length is greater than the one in the rest frame. There must be an obvious mistake...
Where my thought fails and why?

Your second experiment is not measuring the length of the object in S' because the ends of the object are moving while the light is in flight. On the way to the mirror, the light travels a distance equal to the length of the object in S' plus the distance the mirror moved after the light was emitted but before it reached the mirror. On the way back, there's a smaller error.

To get the length in S' to come out right, you have to find the locations of both ends of the object at the same time in S' ("at the same time" - don't forget relativity of simultaneity!) then measure the distance between those two locations. A space-time diagram may help make this more clear.

 

[ghwellsjr]

First: thanks for reading
Second: I'm not interested in other "experiments" that prove contraction of lengths (i know them and i know it works)
Third: Please find where is the problem in my simple thought

Let be a "rest frame" S with an object to measure its length and do it with a ray of light/body that move from A to the and B and come back to A
It needs a time T so that its length is d=c*T/2 (d=V*T/2 in case i use other body different to light )

You are describing the process by which a Laser Rangefinder measures length and nowadays, that is more in line with the definition of the unit of length. Here is a spacetime diagram depicting the measurement of an object whose length measures to be 6 feet. The blue line represents the laser rangefinder with dots marking off 1-nanosecond increments of Proper Time and the green line represents the target at the far end of the object. The thin black line shows the path of the laser pulse:

Note that I am assuming the speed of light to be 1 foot per nanosecond so the time it takes for the light to progress from one end of the object to the other and back is 12 nanoseconds. That number is the only thing that matters for a laser rangefinder to determine that the length is 6 feet by applying your formula.

Note also that your term "T" refers to the Proper Time of the clock in the laser rangefinder. We can apply your formula T'=γT to mean that T' is the Coordinate Time and we see that since v=0, γ=1 and so T'=T=12 nsecs in this example. The light propagates according to the Coordinate Time, not according to the Proper Time.

Observe the same experiment in a frame S' that has a v' relative velocity in the same direction of the length of the object AB
Because of that the event doesn't happen in my frame i argue i can observe it in time T'=γT so turns to me that the object appears long d'=c*T'/2

The length is greater than the one in the rest frame. There must be an obvious mistake...
Where my thought fails and why?

The time that you are describing, T', is the Coordinate Time in this new frame. The Proper Time of the laser rangefinder is still T. This is describing the Time Dilation of the clock in the laser rangefinder as shown in this diagram transformed from the first one to a speed of 0.6c:

As you can see, the dots marking off the Proper Time of the laser rangefinder are stretched out (dilated) so that it takes longer in this frame for the clock to reach 12 nanoseconds. Your formula tells us by how much: T'=γT. Since v is 0.6c, γ=1.25 so T'=1.25*12=15 nanoseconds. And as you can see the total path length for the light is 15 feet which takes 15 nanoseconds. Note also that the length of the object is contracted at this speed so that the light takes the right amount of time to make its trip but the observer operating the laser rangefinder has no awareness of either the Contracted Length of the object or of the time it takes for the light to hit the target or of the round-trip time it takes for the light to propagate. As I stated before, the laser rangefinder only knows of the Proper Time of its own clock and so establishes the length of the object to be 6 feet, just as in its rest frame.

Does this all make sense to you? Any questions?

 

[n3xtras]

First of all thanks for both answers... even if i did not read well ghwellsjr for now (it needs more time)

Your second experiment is not measuring the length of the object in S' because the ends of the object are moving while the light is in flight. On the way to the mirror, the light travels a distance equal to the length of the object in S' plus the distance the mirror moved after the light was emitted but before it reached the mirror. On the way back, there's a smaller error.

Let's see using lorentz equations (c=1):
T1' = γ ( T1 + v d) in the first flight
T2' = γ ( T2 - v d) in the back flight

d=cT/2 i.e. T/2 and T1=T2=T/2

2d' = cT' = T1' + T2' = γ ( T/2 + v T/2) + γ ( T/2 - v T/2) = γT
d'=γd

no seems to help me

To get the length in S' to come out right, you have to find the locations of both ends of the object at the same time in S' ("at the same time" - don't forget relativity of simultaneity!) then measure the distance between those two locations. A space-time diagram may help make this more clear.

Of course you are right ! and is so simple in a generic space-time diagram to see, but doing math...

P.S. Normal "experiments" that show length contraction use a car or an object moving and measure length when it passes at a fixed point (not focusing on same time) then argue distance and they works... even with me. Maybe here is something more complicated by the fact that there is somehow three frames: the one in rest, the one moving, the one of the ray of light or some other worse misunderstanding

Don't want to get stuck - Thanks anyway

For anyone who needs a refresh (me sure I'm newbie) i found http://www.youtube.com/user/PhysicistMichael?feature=watch

 

[Dale]

There must be an obvious mistake...

Yes. The described experiment does not correctly measure the length of a moving object.

Given an object with length L in a frame and moving with velocity v, a pulse of light takes L/(c-v) to go from A to B and L/(c+v) to go from B to A. So the total time is 2γ²L/c. Multiplying this quantity by c/2 only gives L for γ=1.

So clearly, the proposed experiment only measures length for objects at rest.

 

[n3xtras]

I got. Thanks very much. Especially ghwellsjr , finally i could read.