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Length contraction clarification

  1. Nov 29, 2013 #1
    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???
     
  2. jcsd
  3. Nov 29, 2013 #2

    tiny-tim

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    Hi Trojan666ru! :smile:

    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) :wink:

    [2): you can work that out from the formula]
     
  4. Nov 29, 2013 #3
    What i need to know is. Do the separation between the rods undergo length contraction?
     
  5. Nov 29, 2013 #4

    Doc Al

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    Yes. (That was already answered.)
     
  6. Nov 29, 2013 #5

    tiny-tim

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    yes

    as i said, the space (the separation) contracts the same
     
  7. Nov 29, 2013 #6
    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?
     

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  8. Nov 29, 2013 #7

    tiny-tim

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    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
     
  9. Nov 29, 2013 #8
    Hi Torjan.......

    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.
     
  10. Nov 29, 2013 #9

    Nugatory

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    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.
     
  11. Nov 29, 2013 #10
    But it IS a physical effect with measureable consequences.

    Maybe clarify it this way:

     
  12. Nov 29, 2013 #11

    ghwellsjr

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    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.

    What is a "tiny there's"?
     
  13. Nov 29, 2013 #12
    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?
     
  14. Nov 29, 2013 #13
    I suspect he meant tiny threads connecting the rods.
     
  15. Nov 29, 2013 #14

    Nugatory

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    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.
     
  16. Dec 3, 2013 #15
    What is spacetime interval?
     
  17. Dec 3, 2013 #16

    tiny-tim

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    no such thing :confused:

    where did you get "spacetime interval" from?​
     
  18. Dec 3, 2013 #17
  19. Dec 3, 2013 #18

    tiny-tim

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

    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
     
  20. Dec 3, 2013 #19
    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 [Broken]

    http://www.askamathematician.com/2013/07/q-how-do-you-prove-that-the-spacetime-interval-is-always-the-same/
     
    Last edited by a moderator: May 6, 2017
  21. Dec 3, 2013 #20

    tiny-tim

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    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
     
    Last edited by a moderator: May 6, 2017
  22. Dec 3, 2013 #21
    It's not only from wiki, you can google it. Can you give me any link supporting your "variable space interval"
     
  23. Dec 3, 2013 #22

    Dale

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    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".
     
  24. Dec 3, 2013 #23
    Why do you say length OR time, the formula (obviously) includes length AND time, in turn is a spacetime interval, I love the term :smile: 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...
     
  25. Dec 3, 2013 #24
    Dalespam: I have to clear some doubts. How can i ask you doubts personally?
     
  26. Dec 3, 2013 #25

    tiny-tim

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    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")
     
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