Length contraction, train along a line, and a detector

In summary: Yes, the timer will record some fraction of a second of shadow. You can get this from the LT. The train is stationary in the primed frame, the front at x'=0 and the back at x'=-1 (measuring in light seconds). The front of the train is at the origin of the non-primed frame at t=0. The back of the train is at the origin when \begin{eqnarray}x'&=&\gamma(x-vt)\\-1&=&-\gamma vt\end{eqnarray}This is rather different from what you would see if you looked at the whole
  • #1
Nick666
168
7
I was reading a closed thread here about the length contraction being real or not. The best I could understand my self is that space is the one that contracts, the final post by an experienced forum member saying something like its the space-time that changes, it rotates.

But I still have to ask this. So we have a train that is 299.792.458 meters long, traveling along a line. We setup a timer on that line such that it begins to record when the train touches it, and stops recording when the train stops touching it. The faster the train moves the shorter the time recorded. But at the train speed of 299.792.000 m/s, will the timer record more or less than a second ?
 
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  • #2
Nick666 said:
So we have a train that is 299.792.458 meters long
In which frame?
 
  • #3
Its a train on a railroad line. Just that its a long train.
 
  • #4
Nick666 said:
I was reading a closed thread here about the length contraction being real or not. The best I could understand my self is that space is the one that contracts, the final post by an experienced forum member saying something like its the space-time that changes, it rotates.

But I still have to ask this. So we have a train that is 299.792.458 meters long, traveling along a line. We setup a timer on that line such that it begins to record when the train touches it, and stops recording when the train stops touching it. The faster the train moves the shorter the time recorded. But at the train speed of 299.792.000 m/s, will the timer record more or less than a second ?

I guess you meant that the this was the rest length of the train.

It's easier to see what's happening in the train's frame. The faster the timer moves towards to train, the less time it will take to pass. If the timer gets close to the speed of light it will be just over 1 second in the train's frame.

But the timer will be running very slow in the trains frame. So, the timer will record only a fraction of a second.

All observers will agree on what the timer records.
 
  • #5
Nick666 said:
So we have a train that is 299.792.458 meters long
A.T. said:
In which frame?
Nick666 said:
Its a train on a railroad line. Just that its a long train.
A.T. is asking you who made the measurement of 299792458 metres: someone on the train or someone on the line?
 
  • #6
I think the measurement is what perok said...at rest.
PeroK said:
So, the timer will record only a fraction of a second.

Ok, so from the stationary railroad line's point of view, the train's length contraction is real.
 
  • #7
Nick666 said:
I think the measurement is what perok said...at rest.
Ok, so from the stationary railroad line's point of view, the train's length contraction is real.
In the stationary railroad coordinates the train is still the same length. Everyone agrees on the reading on the timer clock for its passage of the train.

One might say that for the timer the train appeared much shorter than it does for the train observer.
 
  • #8
Mentz114 said:
One might say that for the timer the train appeared much shorter

Ok. So how about this ? if we have a timer, on a railroad line, that records for how long it doesn't detect sunlight photons, and a 299792458 meters long train moving at almost c, the timer will record for a fraction of a second ?
 
  • #9
Yes, the timer will record some fraction of a second of shadow. You can get this from the LT. The train is stationary in the primed frame, the front at x'=0 and the back at x'=-1 (measuring in light seconds). The front of the train is at the origin of the non-primed frame at t=0. The back of the train is at the origin when
[tex]\begin{eqnarray}
x'&=&\gamma(x-vt)\\
-1&=&-\gamma vt
\end{eqnarray}[/tex]You can see that t is a very small number.

This is rather different from what you would see if you looked at the whole train, which is what the Terrell rotation thread you have also posted on, is talking about. Here you are focussing on one point at a time, and arranging not to have to worry about lightspeed delay. If you factor both of those things in, life gets more complex. It's like the difference between trying to work out what a map would look like if you moved the origin (easy, just move the gridlines) and what a photo would like if you moved the camera (a very different problem).
 
  • #10
Nick666 said:
Ok. So how about this ? if we have a timer, on a railroad line, that records for how long it doesn't detect sunlight photons, and a 299792458 meters long train moving at almost c, the timer will record for a fraction of a second ?
Is that not the same as the original setup ? I do not wish to change my statement. No one disputes the timer will register much less than a second. How does this impact on the length of the train ?
 
  • #11
Length contraction is a coordinate effect. That's because length is a coordinate effect. It's not enough to say that from the stationary observer's point of view the length of the train is contracted just because his clock reads a fraction of a second. That observer must also measure the speed of the train and then calculate the length of the train based on the time it took to pass.

And in order for the stationary observer to measure the speed of the train, he must make two measurements of, say, the leading edge of the train at two different times, at least one of which must be a remote measurement. That remote measurement requires him to make an arbitrary assumption about the one-way speed of light (something that he cannot independently measure). Depending on that arbitrary assumption, he can get a wide range of speeds for the train and therefore a wide range of lengths for the train. None of these lengths can be considered any more real that any of the others, not even the one that he gets when he assumes that light takes the same amount of time to go from him to the leading edge of the train as it does for the reflection to get back to him.
 

FAQ: Length contraction, train along a line, and a detector

What is length contraction?

Length contraction is a phenomenon in physics where the length of an object appears shorter when it is moving at high speeds relative to an observer. This is due to the effects of special relativity, where space and time are perceived differently depending on one's frame of reference.

How does length contraction relate to a train moving along a line?

The concept of length contraction is often demonstrated with the example of a train moving along a line. As the train moves at high speeds, observers on the train will perceive it as having its normal length, but observers outside of the train will see the train as being shorter in length due to its movement.

What is the formula for calculating length contraction?

The formula for calculating length contraction is L' = L * √(1 - v^2/c^2), where L' is the contracted length, L is the original length, v is the velocity of the object, and c is the speed of light. This formula is derived from the principles of special relativity.

Can length contraction be observed in everyday life?

No, length contraction is only noticeable at extremely high speeds, close to the speed of light. In everyday life, objects are not typically moving at these speeds, so the effects of length contraction are negligible and cannot be observed.

How is length contraction measured in experiments?

Length contraction can be measured in experiments using particle accelerators, where particles are accelerated to very high speeds and their length is measured before and after acceleration. The difference in length is confirmed to be in accordance with the formula for length contraction.

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