Length contraction clarification

[Mentz114]

The common usage of the word 'interval' implies a separation in time. Such as the 'interval between cooking and eating'. So 'space-interval' can be confusing.

But if 'space-interval' has been defined, eg as simultaneous measurement

$x'_2-x'_1 = (\gamma t_1 + \gamma\beta x_2) - (\gamma t_1 + \gamma\beta x_1) = \gamma\beta(x_2-x_1)$

no confusion can arise whatever word is used.

It is not clear what the motives are for the OP's questions. It seems learning is not the priority.

 

[Dale]

seriously? you were confused by my first three posts, of four days ago (quoted just above)?

I thought these posts were good:
https://www.physicsforums.com/showpost.php?p=4587238&postcount=2
https://www.physicsforums.com/showpost.php?p=4587243&postcount=5

I was confused by this one:
https://www.physicsforums.com/showpost.php?p=4587276&postcount=7

And this one was simply wrong:
https://www.physicsforums.com/showpost.php?p=4591084&postcount=16

 

[Trojan666ru]

I think the other answerers were following the answer posted by tiny tim

What tiny tim said is wrong
The distance between the rods won't contact, because if you consider this thought experiment, you'll understand it

"Imagine you are the observer waiting to see a space racing. The track is exactly 1 Light second long. I am sitting in a rod A & There is another rod B infront of me which is kept 100km away from me.
YOUR STOPWATCH IS AIMING AT MY SPEED.

This setup is moving at 99.999999999% of c with respect to YOU. According to your prediction If the distance between my rod and the rod ahead of me contract, then I'll reach the end line before it ticks 1 second in YOUR STOPWATCH!

You can think in another way too
What if there is NO rod in front of the First Rod?

 

[GrayGhost]

well, that refers to taylor and wheeler's "spacetime physics",

which calls s² the interval squared, not the interval

in other words, taylor and wheeler agree with me, that that wikipedia page is wrong, and "interval" is one-dimensional

I must agree. WIKI's page is misleading, if not mistaken IMO. A spacetime interval has a length s, and its square is not the spacetime interval itself. The "spacetime interval" is that which possesses the length of s, and "that" is the line element (or path) between the 2 reference events in 4d spacetime. s² is just the distance squared, ie a squared spacetime interval. The metric being the function describing the spacetime interval's length s. So when WIKI says ...

The reason s² is called the interval and not s is that the sign of s² is indefinite.

While I understand the point they are making regarding the possible polarities of s², I do not see how that justifies labeling a "squared distance" the spacetime-interval itself.

Thank You,
GrayGhost

 

[Nugatory]

This setup is moving at 99.999999999% of c with respect to YOU. According to your prediction If the distance between my rod and the rod ahead of me contract, then I'll reach the end line before it ticks 1 second in YOUR STOPWATCH!

You can think in another way too
What if there is NO rod in front of the First Rod?

Try this:

1) Write down the coordinates of the two events "I am at the start line" and "I am at the finish line". Use the Lorentz transforms to find the coordinates of these two events for the observer at rest relative to the track. That covers the case in which there is no second rod in front of the first rod.

2) Now suppose that there is a second rod in front of the first. Do the same calculation for that one. Use this and the result from the previous step to calculate the distance between the rods according to the observer at rest relative to the track. You will see that distance is contracted.

 

[Trojan666ru]

this is the wrong interpretation of LT. It is measuring one
end of the rod at one place while the other end of rod at
another place.
This is wrong.
Because when you measure the end point, the front point
moves forward to different point.

 

[Trojan666ru]

The formula for spacetime interval is given by,
ds2 =- (cdt)2 + dl2
After a little bit of algebra we get
dt =|ds2|^1/2/c = dt[1
- (v/c)^2]^1/2.
The following
dt = |ds2|^1/2/c
is the expression for the proper time difference not
only for observers at rest in the coordinate system but
for all observers, however they move,

 

[Dale]

What tiny tim said is wrong
The distance between the rods won't contact

No, that part of what tiny tim said is correct (most of what he said is correct, only that one very small post wasn't).

You can write the worldline of the endpoint of any rod as $(t,x)=(t,p_i)$ where p_i is the position of the i'th endpoint in the rest frame. If you do a Lorentz transform on that then you can get the worldline in the primed frame, set t'=0 and see the distances. You will see that they all contract.

 

[Nugatory]

this is the wrong interpretation of LT. It is measuring one
end of the rod at one place while the other end of rod at
another place.
This is wrong.
Because when you measure the end point, the front point
moves forward to different point.

The at-rest observer can calculate the length of a rod by taking the time at which the front of the rod passed over the starting line of the track and subtracting that from the time at which the rear of the rod passed over the starting line. That gives us the time it took for the rod to pass that point, and we multiply that by the speed to get the length of the rod.

The same technique works to get the distance between the rods; we just use the times when the back of the leading rod and the front of the trailing rod pass the at-rest observer (or anything else that is at rest relative to him).

I really really strongly suggest that you try these calculations for yourself.

 

[Dale]

After a little bit of algebra we get
dt =|ds2|^1/2/c

You made a mistake with your algebra. This is not true in general, although it is true for dl=0.