Length contraction clarification

[Trojan666ru]

Consider two iron rods of about 10 metres separated by 10 metre. They are in series position and not connected in anyway. If the rods are passing nearby an observe with a velocity of 90% speed of light, then what would he observe?
1) does the separation between the rods (ie 10m) undergo length contraction?
2) What must the velocity of the rod if it appears half of its original length?
3) at that speed do every object appears half of its original length?
4) what will happen if these rods are connected with a tiny there's? Do the whole system appear contracted?

 

[tiny-tim]

Hi Trojan666ru!

1) 3) 4): the same contraction formula applies whether there is something there or not …

it is the space that appears to contract (in a different frame) ​

[2): you can work that out from the formula]

 

[Trojan666ru]

What i need to know is. Do the separation between the rods undergo length contraction?

 

[Doc Al]

What i need to know is. Do the separation between the rods undergo length contraction?

Yes. (That was already answered.)

 

[tiny-tim]

What i need to know is. Do the separation between the rods undergo length contraction?

yes

as i said, the space (the separation) contracts the same

 

[Trojan666ru]

But here the observer is at rest and the rods are moving. I thought space contracts only when the observer is in motion
From the below picture can you show me the answer?

 

[tiny-tim]

But here the observer is at rest and the rods are moving. I thought space contracts only when the observer is in motion

no

a space interval contracts whenever it and the observer are in relative motion

remember, the contraction is as measured by the observer …

if he measures an interval (or eg a rod in that interval) as being 1 metre when it's stationary, then he measures it as less than 1 metre when it's moving

 

[Naty1]

Hi Torjan...

a space interval contracts whenever it and the observer are in relative motion

yes. that's why it is called RELATIVITY.

An observer in your example sees the distant rods [and the distance between them] as contracted, but an observer riding on a rod will see your distant observer as contracted also...and will see the rods at their normally measured length...because they are stationary with respect to him...

With constant relative velocity observers, nobody knows who is moving and who is 'stationary'...such motion is RELATIVE not absolute.

Right now, from a galaxy far,far away, moving rapidly in a straight line, you will see their turkey narrower than they do and they will see your turkey narrower than you see it. Which turkey is moving and which is not is arbitrary.

Happy thanksgiving.

 

[Nugatory]

What i need to know is. Does the separation between the rods undergo length contraction?

You've already gotten several answers to this question ("yes"), but there are two other things you might want consider.

First, the length contraction formula is derived from the Lorentz transformations as a special case, so any time you find yourself even slightly unclear on whether the length contraction formula applies (you just asked "does X undergo length contraction?") you can go back to the Lorentz transforms, use them directly. In this situation, the question you're really asking is "does the difference between the x coordinates of the ends of the two rods change", and the Lorentz transforms will answer that question. So would a space-time diagram, which is basically a graphical representation of what the Lorentz transforms are telling you.

Second, the way you've asked the question suggests that you're falling into a trap - if you are, you're in good company, that trap has caught plenty of people over the years, but it's still a trap. Nothing ever "undergoes" length contraction, and phrasing it that way will tempt you into thinking of length contraction as something that happens to objects because they're moving. It's not - it's something that happens to observations of objects when the observer is moving relative to the object. Train yourself to frame relativity problems in these terms and you'll see that many of the most perplexing contradictions and paradoxes... just resolve themselves.

 

[Naty1]

Nothing ever "undergoes" length contraction... it's something that happens to observations of objects when the observer is moving relative to the object.

But it IS a physical effect with measureable consequences.

Maybe clarify it this way:

Nothing ever "undergoes" length contraction in it's own local frame of reference...
it's something that happens to observations of objects when the observer is moving relative to the object.

 

[ghwellsjr]

Consider two iron rods of about 10 metres separated by 10 metre. They are in series position and not connected in anyway. If the rods are passing nearby an observe with a velocity of 90% speed of light, then what would he observe?

Your question is ambiguous. When you specify the length of objects and the separation between objects, you have to state the frame in which the dimensions apply. There are three different ways your question could be interpreted:

A) You could mean that the two rods are 10 meters long and separated by 10 meters in the frame in which the observer is at rest and therefore they are already traveling at 90% speed of light.

B) You could mean that the two rods are 10 meters long and separated by 10 meters in the frame in which they are at rest and the observer is passing nearby with a velocity of 90% speed of light and then you transform to the rest frame of the observer. (This is the way all the previous responders assumed you meant.)

C) You could mean that the two rods start out at 10 meters long and separated by 10 meters in their own rest frame and then the two rods are independently and identically accelerated to a velocity of 90% speed of light and pass by an observer who remained at rest in the original frame.

So depending on which scenario you mean, the rest of your questions will have different answers.

1) does the separation between the rods (ie 10m) undergo length contraction?
2) What must the velocity of the rod if it appears half of its original length?
3) at that speed do every object appears half of its original length?
4) what will happen if these rods are connected with a tiny there's? Do the whole system appear contracted?

What is a "tiny there's"?

 

[nitsuj]

Consider two iron rods of about 10 metres separated by 10 metre. They are in series position and not connected in anyway. If the rods are passing nearby an observe with a velocity of 90% speed of light, then what would he observe?
1) does the separation between the rods (ie 10m) undergo length contraction?
2) What must the velocity of the rod if it appears half of its original length?
3) at that speed do every object appears half of its original length?
4) what will happen if these rods are connected with a tiny there's? Do the whole system appear contracted?

Another way to look at it is from the perspective of the constancy of c.

what is the difference in time it takes light to travel the space between the rods at rest and when in motion? What measurement must change for the constancy of c?

 

[yuiop]

What is a "tiny there's"?

I suspect he meant tiny threads connecting the rods.

 

[Nugatory]

I suspect he meant tiny threads connecting the rods.

My thought too, but guessing what question was intended has a fairly poor track record. Let's let trojan clarify his question for us - he's the only person who KNOWS what he meant.

 

[Trojan666ru]

no

a space interval contracts whenever it and the observer are in relative motion

remember, the contraction is as measured by the observer …

if he measures an interval (or eg a rod in that interval) as being 1 metre when it's stationary, then he measures it as less than 1 metre when it's moving

What is spacetime interval?

 

[tiny-tim]

What is spacetime interval?

no such thing

where did you get "spacetime interval" from?​

 

[Trojan666ru]

Go and read wikipedia
http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals
You made a mistake in the beginning

 

[tiny-tim]

http://en.wikipedia.org/wiki/Spacetime#Spacetime_intervals

hmm … I've never seen that use of "interval" anywhere else, and i don't like it

that wikipedia article says that a spacetime interval (between two events) is s²

everyone else calls s² the separation (between two events)

this is because "interval" normally means a one-dimensional measurement (metres or seconds), while s² (obviously) is two-dimensional (square metres or square seconds), and so a completely different word is necessary

calling it an "interval" is misleading, and likely to lead to misunderstanding

 

[Trojan666ru]

no

a space interval contracts whenever it and the observer are in relative motion

remember, the contraction is as measured by the observer …

if he measures an interval (or eg a rod in that interval) as being 1 metre when it's stationary, then he measures it as less than 1 metre when it's moving

But that's not true if you go through This link, read what it says

http://www.physics.fsu.edu/users/ProsperH/AST3033/relativity/Interval.htm

http://www.askamathematician.com/2013/07/q-how-do-you-prove-that-the-spacetime-interval-is-always-the-same/

 

[tiny-tim]

But that's not true if you go through This link, read what it says

http://www.physics.fsu.edu/users/ProsperH/AST3033/relativity/Interval.htm

http://www.askamathematician.com/20...at-the-spacetime-interval-is-always-the-same/

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

 

[Trojan666ru]

It's not only from wiki, you can google it. Can you give me any link supporting your "variable space interval"

 

[Dale]

tiny-tim, the spacetime interval is a very common and very important concept in SR. The spacetime interval is defined as $ds^2=-c^2dt^2+dx^2+dy^2+dz^2$. This quantity is frame invariant as you can see simply by doing a Lorentz transform on it and simplifying. In fact, you can define the Lorentz transform as the transform which leaves this invariant and make the invariance of the spacetime interval your single postulate. From the spacetime interval you can figure out if a reference frame is inertial or not, calculate out the fictitious forces in any frame, and even determine if your spacetime is curved or flat. It is hugely important.

What you are calling a "space interval" is always called a "distance" or "length", to avoid just this confusion with the spacetime interval.

However, your original point, that it is only relative motion which is important for determining length contraction, is correct. It was just problematically worded with the use of the word "interval" instead of "distance".

 

[nitsuj]

while s² (obviously) is two-dimensional (square metres or square seconds), and so a completely different word is necessary

Why do you say length OR time, the formula (obviously) includes length AND time, in turn is a spacetime interval, I love the term finally an interval we all can agree on lol

Spacetime interval
time-like interval
light-like (null) interval
space-like interval

looks pretty straight forward

wiki
"which takes into account not only the spatial separation between the events, but also their temporal separation."

Looks right to me and seems to agree with your terminology too, less the understanding of what the spacetime interval calculation includes...

 

[Trojan666ru]

Dalespam: I have to clear some doubts. How can i ask you doubts personally?

 

[tiny-tim]

tiny-tim, the spacetime interval is a very common and very important concept in SR. The spacetime interval is defined as $ds^2=-c^2dt^2+dx^2+dy^2+dz^2$. This quantity is frame invariant…

yes i know!

but

i] that's ds², not s² … it isn't what we were talking about

(ds² is the metric at a particular event, s² is a measurement between two events)

ii] I've never seen s² called "interval" before (and eg taylor and wheeler don't call it "interval")

 

[Dale]

I guess then that you are talking about the quantity $s=\int \sqrt{ds^2/d\lambda^2} d\lambda$ where $\lambda$ is an affine parameter along the path in question.

 

[tiny-tim]

I guess then that you are talking about …

me?

it's not my word

i'm trying not to talk about "interval"

i say "separation" for s²

i only used "interval" in its ordinary english sense, in the phrase "space interval" …

a space interval contracts whenever it and the observer are in relative motion

 

[dvf]

hmm … I've never seen that use of "interval" anywhere else...

It's pretty common jargon: https://www.google.com/search?q="spacetime+interval"&tbm=bks

 

[Dale]

i only used "interval" in its ordinary english sense, in the phrase "space interval" …

I wouldn't use it like that. It confused me, so it is likely to confuse others also.

 

[tiny-tim]

i only used it …

Hi Trojan666ru!

1) 3) 4): the same contraction formula applies whether there is something there or not …

it is the space that appears to contract (in a different frame) ​

a space interval contracts whenever it and the observer are in relative motion

… to describe a 1D region of space in which there might be a rod or string, or in which there might be nothing

i had already used the word "separation" …

… the space (the separation) contracts the same

… but apparently that was not clear enough, so i tried "interval" as an alternative

It confused me, so it is likely to confuse others also.

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

 

[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.