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I Consequences of length contraction

  1. May 20, 2016 #1
    If you are in a space ship, traveling near light speed, will length contraction enable you to se astronomical events lightyears away, before it is possible to see them from earth?

    This because the distance is shorter for you in the spaceship and light speed is constant.

    Example: A star 100 ly. away from earth will be 50 ly. away from a spaceship traveling 87% of speed of light towards or from it, even if the spaceship are traveling past the earth when this is observed. So the spaceship can se 50 years "in to the future" on this star compered to observed from the earth. (On earth we can see what happened 100 years ago, and from the spaceship we can see what happened 50 year ago on the same star.)

    Is this a valid conclusion and correct understanding of the theory of relativity?

    regards
    Vidar
     
  2. jcsd
  3. May 20, 2016 #2

    stevendaryl

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    Whether you are in the rocket or on the Earth, you're seeing the same events. It's just that for the rocket's frame, those events only took place 50 years ago, while for the Earth's frame, those events took place 100 years ago.
     
  4. May 20, 2016 #3
    Thanks a lot :smile:
    I suspected it must be something like this, because I won't dare to thing about the consequences if my conclusion where true :woot:

    I will try to find the math to understand why this is so.
     
  5. May 20, 2016 #4
    The math is the Lorentz transformation. An event that happened at (x = 100ly, t = -100y) in Earth's frame transforms to (x' = 373.2ly, t' = -373.2y) for a spaceship heading in the +x direction at v = 0.866c.

    If you say the event happened 100y ago on a star which is 100ly away from Earth (and stationary wrt Earth) then it happened 373.2 years ago on the star which is 50ly away in the spaceship frame. But don't forget that the star is not stationary wrt the spaceship, so when the event happened it was 373.2ly away (if I did the math right :-).
     
  6. May 20, 2016 #5

    pervect

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    OK, let's call the earth observer E, and the star P is 100 light years away from E. And we have a spaceship S, moving towards the star, flying right by earth, who sees the star as only 50 light years away.

    A light beam L emitted by the star P at some event of omission (I won't give that a letter) is seen by (i.e. arrives at the loaction of) both E and S at the same time, since both E and S are at the same location.

    The explanation for this involves a third result of relativity, one other than length contraction, or time dilation. This result is called "the relativity of simultaneity". "At the same time" doesn't mean the same thing to observers A and E. To really understand this, you probably need to draw a space-time diagram, but that's the basic explanation of how it's possible.
     
  7. May 21, 2016 #6
    Thanks a lot for your responses.

    I found a formula that gave the same result.

    Is this the right formula?:

    t = lambda (t’- (vx/c2 ))

    if we put in the numbers: Can we use ly and c=1 together in the same formula?
    I så the:

    lambda = 2

    c = 1

    v=0.87 (87% speed of light)

    x = 100 (100 ly away from earth)

    t’ = -100 (100 years ago from earth)

    then t= 2(-100-(0.87 * 100/12)) = -374 years


    Im not sure I understand this, because if this is true then the clock on the star will have gone bacwords in time.

    So lets take a journey in space to see what I mean:

    Lets say that our spaceship started on earth and gradually increased the speed with 1g ( 9.8 m/s2) then in about a year or so it will reach 0.87 c. Let’s sync the clock on the star, the earth and the spaceship when it starts.

    As the spaceship accelerate the star will come gradually closer to it is 50 ly away from the spaceship ( the spaceship will then be approx. 1-2 ly away from earth). How will the clock on the star behave to us on the spaceship during this time?

    It must move forward in time in my mind ( gradually slower to ½ speed when we are 50 ly (or 48ly away)), the clock can’t move backwards (373ly??). So the event on the star must have taken place later in time.

    Since the spaceship have been on the journey only for a year or so, the clock on the star have been slower by approx. ½ year.

    I mean the event must have happened later than 100 y ago to us on the spaceship.

    This must be closer to 48 years or so if light from distant objects reaches us at the same time if we are in different timeframes, but on the same place in space.

    Is my assumption correct here?

    What is the math to calculate this?
     
  8. May 23, 2016 #7

    Erland

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    Yes, this is the same calculation as Vitro did. The difference (373.2 years for Vitro) is due to roundoff error: you use sqrt(3)/2 = 0.87 while Vitro uses 0.866.

    The clock you see at the star at this instant, whether you are on the Earth or at the spaceship, shows -100 years.

    But if there is another spaceship moving with the same velocity w.r.t. the Earth (and the star) as the first spaceship, with a clock synchronized with the clock at the first spaceship, passing next to the star when the clock on the star shows -100, then the clock on this second spaceship would show -373.2 at this instant, so the observers at the Earth and the first spaceship would see the clock on the star showing -100 and the clock of the second spaceship next to the star showing -373.2.

    No clocks go backwards in time.

    I think you need to look up the relativity of simultaneity. There are plenty of posts here about that.
     
  9. May 23, 2016 #8
    Yes, I agree. Not to confuse other reders. The numbers in the formula should also be: t=2(-100-(0.87*100/1)) berceuse i set light speed to 1 and the formula is c2 so 12 shuld be 12 = 1.

    Yes i realise this now because light from distant objects reaches us at the same time if we are in different timeframes, but on the same place in space. And
    if we are in the same place ( but diferent time frames) we are agread of what is simulanesly localy where we are. Am I correct here?

    So both the spaceship and earth, se this as it happend 100 years ago even if the spaceship are only 50 ly away ( but in the same place as the earth). and that is because the clock on the star goes on half speed.
    And eaven if the spaceship has accelrated ( the star dosent know that :wink:) and the light travled from the star must then have been traveling for 100 years because as far as the spacesip knows (in that timeframe). the stars clock has always been running on half speed. And then the numbers add up.

    Is this one way to look at it and a valid conclution?

    And thanks again for you responses and help me figure this things out. I think I am cloaser to an understanding now.
     
  10. May 24, 2016 #9
    No, it's much more simple: Clock synchronization means that clocks strike 5 simultaneously, at the same time. A distance of 50 light years causes a 50 years delay of news transmission. Therefore time synchronized spaceships 50 light years apart, watching each others TV programs, will see the clocks on their TV screens to be 50 years behind the clocks hanging on their walls.

    When clocks are synchronized, then any apparent difference in clock readings is caused by the information propagation delay.

    EDIT: Or is there perhaps a 373.2 light years distance between the two spaceships, and a 373.2 years apparent time difference between the clocks of the two spaceships?
     
    Last edited: May 24, 2016
  11. May 24, 2016 #10
    Not quite. In the rocket's frame the events currently seen in the rocket took place much longer than 50 or even 100 years ago:

    The distance between the light and the star is now 50 ly. The distance increased at rate 1 c - 0.87 c, so it took this much time: 50 ly / (1 c - 0.87 c) = 384 years, for the light to get 50 ly away from the star.

    The light was emitted from 384 ly distance from the rocket according to rocket, not from 50 ly distance from the rocket according to rocket.
     
  12. May 24, 2016 #11

    Dale

    Staff: Mentor

    The spaceship's frame is non inertial. The idea of simply stitching together the momentarily comoving inertial frames results in exactly the problem you identified. The result is mathematically invalid.

    Here is a paper that addresses one simple solution.
    http://arxiv.org/abs/gr-qc/0104077
     
  13. May 25, 2016 #12
    If I understand this correctly:
    Is it safe to say that you bouth are right because you accually are talking abaut to differents eventes in time.

    1. When the spaceship sees the event. ( it happend 50 y ago to the spaceship)
    2. When the event happend on the star ( The spaceship was (or shoud have been in thath timeframe) 384 ly away)

    Is this a correct understanding?
     
  14. May 25, 2016 #13
    No, that's not correct, so I will rephrase post #10:

    A scientist on the rocket might be pondering like this:

    This light beam I am seeing now left that star 384 years ago, the star was 384 light years away from me at that time. Then during 384 years the light traveled 384 light years towards me, and it's now here. During the 384 years the star traveled 335 light years towards me, and it's now 50 light years away from me.


    velocity of the star towards the rocket = 0.87 c
    0.87 c * 384 years = 335 light years
    384 light years - 335 light years = 50 light years
     
    Last edited: May 25, 2016
  15. May 25, 2016 #14

    stevendaryl

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    No, I was wrong. In the frame of the rocket, the star is 50 light-years away, but the light from the star took more than 50 years to get to the rocket.
     
  16. May 26, 2016 #15

    Let's say the aforementioned light beam contains a message: "It's year 0 now here on the star".
    Seeing that message the aforementioned scientist thinks to herself: "That light has been traveling for 384 years, so it must be the year 384 on the star now ... oh but clocks run at half speed on the star, so it's actually year 192 on the star now".

    Now let's say the rocket stops in one minute, and by that I mean that the rocket's speed goes from 0.87 c to 0 c. Distance to star becomes 100 light years, and the aforementioned scientist is now saying: "That light beam containing the message that it's year 0 on the star has been traveling for 100 years, so it must be year 100 on the star now".

    During one minutes time the scientist's idea about the current calendar time on the star went from year 192 to year 100.


    (A year on the star is measured by a clock on the star, that clock is same kind of clock that is used on the rocket)
     
  17. May 26, 2016 #16
    Disregarding cosmological expansion, any photon that arrives at the rocket would have arrived at that discrete location in space at that discrete time, regardless of whether the rocket was actually there or not. i.e. all arriving photons travel at c from their source to an observer.

    In SR based relativistically rolling wheel solutions the length contraction of a particular spoke can be used to determine the location of a point on the circumference of the wheel at the time of an emission while the emitted photons themselves travel in a direct, and non length contracted, straight line at c towards the observer.

    If points A and B both transmitted an encoded "the time on the third stroke will be X" type (synchronized between A and B) series of pulses, with B transmitting a reversed 'count down' pulse, a rocket traveling relativistically between these 2 points would always be able to determine a consistent location from pulse B regardless of its velocity etc.
     
  18. May 26, 2016 #17

    Erland

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    For quite a long time, I thought there was an error in your reasoning, since your figure, 384 ly, differs significantly from mine and Vitro's: 373.2 ly, obtained by a different method.

    But in fact, there is nothing wrong with your reasoning (nor with ours), but instead another roundoff error, which is surprisingly large!

    We want the factor √(1- v2/c2) to be 1/2. Then we must have v/c = (√3)/2 = 0.8660254037844386 (using http://web2.0calc.com/). Using this, we obtain the answer 373.205087568875990858 ly, where last ditits of course are highly uncertain (they differ slightly if we change the form of the input), but we can be quite certain that the answer with six correct decimals is 373.205088 ly.
    If we use the approximation 0.87 for (√3)/2, we obtain the answer 384.615384153846154 ly, with a difference of about 11.4 ly from the correct answer. A surprisingly large roundoff error, indeed, but nonetheless a roundoff error.

    So, can't we agree to use as many decimals as possible in our calculations in this thread...?
     
    Last edited: May 26, 2016
  19. May 26, 2016 #18
    You have to be careful with the notion of "now" for events that occur at a distant location. The notion of "now" is in these cases relative.

    Because of the relative nature of "now".
     
  20. May 26, 2016 #19
    I doubt you'll get a consensus on that! It's better to follow the usual rules about the use of significant figures. When you express the value of ##\frac{\sqrt 3}{2}## rounded to two significant figures you get 0.87. You should therefore not be surprised that 384.6 and 373.2 do not agree, but are close to agreeing, in the 2nd sig fig.

    ##2+2=5## for sufficiently large values of ##2##.
     
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