Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Length Contraction question.

  1. May 24, 2005 #1
    In the SR section of my book there is a question like this.

    An astronout is traveling at .9C.
    The spacecraft has a proper legth of 100m.
    a)what is the length observed by observer? I get this as I devide 100 by gamma which is about 2.29
    b)Astronouts view is also effected by the time it takes for the light to reach him. What length spacecraft does he see when the spacecrafts tail is next to him?
    I got the answer to be about 39m but in the answers section they have 23m.

    Could someone help me with this please.

    This is what I have done, is it correct?

    43.6/C=1.433333E^-7 Sec. (Time it take the light in front of space craft to reach observer)

    1.43333E^-7*.9C=39.24m (The length of space craft while the tail is next to him.)
     
  2. jcsd
  3. May 24, 2005 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi.

    I've joined a picture, because with words, it would be too confusing.

    We can get two distinct relations for t*:

    [tex]t^*=\frac{x}{c}[/tex]

    and

    [tex]t^*=\frac{43.6-x}{0.9c}[/tex]

    Putting them together gives the answer: 22.95m.
     

    Attached Files:

  4. May 24, 2005 #3
    [tex]t^*=\frac{x}{c}[/tex]
    Is the x 43m?

    Can someone please explain it through so I can understand it better.

    So far what I understand is first find how long it take light to reach the tail from head.

    Then you find how long it has traveled by

    [tex]{t^*}{.9C}={43.6-x}[/tex] then

    [tex]{t^*}{.9C}-{43.6}={-x}[/tex]

    but I can't seem to get 23m.

    [tex]t^*=\frac{x}{c}[/tex]

    [tex]t^*=1.4533^-7[/tex]

    [tex]{t^*}{.9C}=39.24[/tex]

    [tex]{39.24}-{43.6}=-x[/tex]

    [tex]x=4.36[/tex]

    Can someone please help.
     
    Last edited: May 24, 2005
  5. May 25, 2005 #4

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    x is not 43m, x is the unknown, i.e. the lenght of the rocket as seen by the observer.


    By "putting them together", I meant

    [tex]\frac{x}{c} = \frac{43.6-x}{0.9c}[/tex]

    Then solve for x.


    Or is your problem that you don't know where the two equation for t* come from?
     
    Last edited: May 25, 2005
  6. May 25, 2005 #5
    I kind of get it now.

    Is it that [tex]t^*[/tex] is the time it takes for like to reach the observer and science [tex]t^*[/tex] is [tex]\frac {x}{C}[/tex] we can change it to [tex]{1.9x}={43.6}[/tex] thus [tex]x=22.95[/tex].

    I would really appriciate if you could run through it as well.

    thanx
     
  7. May 25, 2005 #6

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There exists a photon that parted from the nose some time ago, and arrived at the eye of the observer when the tail was right next to him. This is undeniable. This is the photon illustrated in the picture. Now, we defined the distance it traveled as 'x', since it is in fact, the lenght of the rocket that the observer will observe. At the time of emission of that photon, say t=0, the tail is a distance 43-x from the observer, and following the definition we gave of that photon, it reaches the observer at the same time as the tail. Suppose this time is t=t*. Then between t=0 and t=t*, light has traveled a distance x (at speed c) and the tail has traveled a distance 43-x (at speed 0.9c). This translate into the equations

    [tex]t^*=\frac{x}{c}[/tex]

    and

    [tex]t^*=\frac{43.6-x}{0.9c}[/tex]


    (I've included a newer version of the drawing, where I corrected a little mistake in the path of the photon)
     

    Attached Files:

  8. May 26, 2005 #7
    This is an interesting problem....Clearly the "austronaut" can not be on the ship since the length would be 100 meters. However, if he is on the platform then he must stand a slight distance at a right angle to the direction of flight and so there is, possibly, a hypotenuse matter here, which can theoretically, I guess, be ignored if he is close enough.

    What is interesting is that the observed contraction is taken as real, but the observation of the astronaut involves an illusion....right?
     
  9. May 27, 2005 #8

    Doc Al

    User Avatar

    Staff: Mentor

    I wouldn't phrase it quite like that. All measurements and observations made by the astronaut on his platform will corroborate the fact that (according to him) the ship's length has contracted. Even the light that reaches him from the front of the ship at the instant that the rear of the ship passes him is consistent with that, once the speed of light is taken into account.

    I agree that this is an interesting problem. To better appreciate the second part I would recommend to bayan that he imagine that the astronaut has built a huge ruler in space parallel to the direction of the spaceship. As the ship moves by, he (or another observer in his frame) can see exactly where the front of the ship is because the ruler is right next to it. (Of course, this is just a thought experiment--lots of luck "seeing" anything as the ship zooms by. Pretend you have a super-fast camera that can capture the light at an "instant".)

    So, what the astronaut "sees" at the instant that he passes the rear of the ship is an image of the front of the ship passing the "23 m" ruler mark. His is not misled into thinking that the ship is only 23 m long, of course, since he knows that the image does not reflect where the front of the ship is now (according to him) but where it was some time ago. He knows that now (according to him) the front of the ship is at ruler mark "44 m".
     
  10. May 27, 2005 #9
    thanx for all your efforts.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Length Contraction question.
Loading...