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Help me to Understand: I am traveling at 90% of the speed of light.

  1. Sep 25, 2009 #1
    I am traveling at 90% of the speed of light in a given direction. I generate a new beam in the direction that I am traveling. If I was to measure the speed of the new beam in reference with the speed that I am currently traveling would that new beam of light be moving forward at 10% of the speed of light?



    ------>------>------>----->(me*)_____>____________>_____________>

    ^ new beam of light ^
    measure ref. speed is the same speed I am moving at.
    (90% the speed of light)

    I ask this to clarify my understanding of the limit the speed of light imposes.
    Another question....
    If I was traveling at 50 percent of the speed of light, and I shine a light perpendicular to myself as I am moving forward what effect would this have on the new beam of light?
    How would this effect be observed by me and a observer?


    ---->-------->------->------->---->(me*)---------->----> 50% speed of light ​

    l
    l
    l
    l
    \/​
    New beam 100% speed of light. ​




    I promise this isn't a homework question. I am just trying to fully understand what effects are produced by the speed of light in relation to a individuals position.
     
  2. jcsd
  3. Sep 25, 2009 #2

    Nabeshin

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    Science Advisor

    One of the two fundamental postulates of special relativity is that the speed of light is constant in all inertial frames of reference. Therefore, if I am moving at .9c with respect to some observer and I shine a beam of light in the direction of my motion, I measure its speed as c. The observer I am moving .9c with respect to also measures c.

    Similarly for a beam propagating perpendicular, you will always measure c.

    Note: while velocity of light is constant, its frequency is not. There will be an observed doppler shift.
     
  4. Sep 25, 2009 #3
    No the light will be moving at the speed of light.

    No effects.

    [
     
  5. Sep 25, 2009 #4
    In relation to the second question. So to an outside observer on a parallel plane, but a sufficient distance away the light leaving me while I travel forward will curve. To me as I shine the light away from myself it will appear in a straight line?

    In relation to first question. I think this statement is the source of confusion. If you think of the speed of light as a "sound barrier*" this would imply that if you were moving at 90% the speed of light the new beam would surpass the speed of light. If you shined the light while moving at 90% of c would not the light move to [your position+10%] and remain?

    *I know that the sound barrier is totally different than the speed of light I only use it as an example of something that is incapable of traveling faster than a certain speed.
     
  6. Sep 25, 2009 #5
    i do not really understand what your question is here or if there is one. What I am saying is that if you measure the speed of light in any inertial frame of reference you would get the same value. if you are in a ship moving at 90%c with respect to a given point P and you have a beam of light inside the ship that is directed from the back to the front.If measure the speed of that beam with respect to the ship you will get 100%c not 10%c. You will also get 100%c if you measure the speed of that beam with respect to the point P.This is not very intuitive but it is the way things are. It took Einstein to figure out how this happens and is the basic idea behind special relativity.
     
  7. Sep 25, 2009 #6

    russ_watters

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    Staff: Mentor

    Light just plain doesn't work that way.

    Imagine you have a laser and three accurate clocks in the middle of your spaceship. You synchronize the clocks and move two to opposite ends of the ship. Then you fire a laser forward from the middle clock to the first clock, then turn around and fire one backwards to the second clock. By this method, you can calculate the speed of light, since you know the distance and have measured the time. As long as your spaceship is moving at constant speed (whether 0 or any other), you will always measure the speed to be C.

    Now, if you fire that laser out a window in any direction and someone somewhere breaks off a piece and measures its speed, they will also get C regardless of how you are moving relative to each other.

    This is the central concept of SR.
     
  8. Sep 25, 2009 #7

    Doc Al

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    The observer who sees you moving at 0.5c will see the light as being fired at some forward angle, not at 90 degrees to your direction of motion. (About 60 degrees from your direction of motion.) All observers will measure the light to be moving at speed c, of course.
     
  9. Sep 25, 2009 #8
    I guess the problem comes down to another question.
    Perhaps this will help me to understand.
    Is it a property of light that even if you are traveling at nearly the speed of light that if you then shoot another light ahead of you it will not increase in speed with the additional speed that your traveling at. At the same time measure the speed of this light as it approaches this barrier (c) from outside the craft will it always measure c?

    I guess in relation to your above example this would be like one of those clocks flying at the other clock at 90% the speed of light. If that beam of light was still sent from one clock to the other would the light still reach the clock at the sime time. As the above example. I understand that the distance is decreasing, but imagine that the far clock is in fact moving away from the first at speed of light.

    .9c Clock{1obs}-------->1.0c Clock{2obs}
    l
    l
    l
    \/
    {3obs}

    Perhaps telling me what is happening in this example for the observer at point 1,2 and 3 could help. I understand that this is difficult to explain, but I really want to understand the principle that yall are trying to get across. I refuse to think that there is some principle out there that is beyond my understanding. I'm just looking for that "Oh, now it makes since moment."


    I'm trying to figure out how light behaves once it is at c.
     
  10. Sep 25, 2009 #9
    I think I've figured it out. In trying to figure out how to lay out the last example to illustrate my last question I think I started to understand how it works.
     
  11. Sep 25, 2009 #10
    Holy profanity, I just realized that If you were to travel to a distant star instantaneously that the universe from your point of view would look totally different depending on how far away you went. I mean completely different. In fact, you couldn't even really predict where your target star might be unless you knew exactly how far away it was and in what direction it was moving. You would also have to know how old the star was if you were to have any hope of actually getting there before it was dead...or reborn....or a black hole.

    On a lighter note I guess you could be fairly certain to actually arrive at a black hole if you already knew its exact location.
     
  12. Sep 25, 2009 #11
    Thanks guys, I guess any further discussion would be reserved for a new topic in a new thread.
     
  13. Sep 25, 2009 #12

    russ_watters

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    Yes! Repeat it over and over until you accept it: the speed of light is always C to an inertial observer. No iffs, ands or buts.
    If the light beam leaves from the same distance, it gets to the other clock in the same time. But note that now that the clocks are moving wrt each other, they won't stay synchronized. So it requires significantly more calculation to figure out what is going on.

    As the above example. I understand that the distance is decreasing, but imagine that the far clock is in fact moving away from the first at speed of light.
    It is probably better not to bring acceleration into the mix until you first grasp the concept from an inertial (non accelerating) perspective. But anyway, an accelerating observer may measure the speed of light to be variable, depending on how he makes the measurement. Observer #3, however, is presumably not accelerating and thus will measure the speed of light to be C.
    Ok: Light travels at C to an inertial observer. Period. Nothing else to say. No "once it is at....": if it is light, it is traveling at C.
     
  14. Sep 25, 2009 #13

    russ_watters

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    If you know where a star is now and how fast it is moving with respect to us, there isn't any problem with jumping in a windowless spaceship and traveling there, nor is there any problem figuring out how much the star has aged since you saw it in your telescope to when you got there. The equations that govern motion and time are quite well understood.

    Just keep in mind that that star you want to travel to that is 4.5 light years away will take less than 4.5 years (by your clock) to get to at high speed.
     
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