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Travelling faster than the speed of light - SR

  1. Apr 23, 2012 #1
    Picture yourself accelerating to 3/4 the speed of light, taking your girlfriend with you, and leaving your best mate behind, back on earth (with no speed!). At this speed you stop accelerating and continue to travel through space at 0.75c.
    From this position, you see light travelling at the speed of light and would not know any better than to ASSUME you are stationary, with speed = 0 (even though you are moving away from your stationary friend back on earth).
    You then send your girlfriend on her way, accelerating her in the same direction you began accelerating from your mate. Relative to you, she accelerates to 0.75c and keeps travelling.
    But then your friend stops and thinks for a bit. He is not the brightest boy but does a quick calculation and finds that:
    0.75c + 0.75c = 1.25c
    "Wow" he says, that dudes girlfriend is travelling faster than the speed of light. "Is she?"
    But faster than light travel requires infinite energy! It Isn't possible!

    Really I can see nothing wrong with this, and if this is a problem, then it raises another question:

    Couldn't we calculate how fast WE, on earth, and the solar system are travelling through space? As time dilation is not linear (from lorentz factor), we could put a clock on a plane (going really, really fast), and one on earth and calculate the difference in time and from that work out how fast we are actually going! (obviously direction is going to be a big factor here, the earths spin on its axis, earths rotation around the sun, and suns rotation around milky way galaxy). But wouldn't this be possible???
     
  2. jcsd
  3. Apr 23, 2012 #2
    First of all, that's 1.5c :)

    Secondly, that's not how you move from one coordinate system to another. Space and time work differently in special theory of relativity, that's the whole point of it. In SR, you would get

    [tex] v = 1.5c/(1+0.75^2) = 0.96c [/tex]

    (see http://en.wikipedia.org/wiki/Velocity-addition_formula)
     
    Last edited: Apr 23, 2012
  4. Apr 23, 2012 #3
    thats embarassing :(.

    well thats very interesting. I'll have to look into it a bit more. Thankyou!
     
  5. Apr 23, 2012 #4

    phinds

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    When you do, spent some time learning about frames of reference. You seem to believe that there is some absolute velocity that we could measure. This is nonsense, as you will find. Velocity is ALWAYS relative to something; it has no other meaning. What would you use as the basis for the speed of the earth? You can choose ANY reference point you like (and get different speeds accordingly).
     
  6. Apr 23, 2012 #5
    Just a quick, un-related topic. (I've been having fun with SR lately).
    I have been trying to find the optimum speed to get from point a to point b in the fastest time possible, from an observers point of view.
    I used the common time delation equation:
    t- = t/√(1 - (v2/c2)
    and replaced t with the distance formula, distance/velocity.
    now, when I pick a distance (won't matter what distance, lets say 1000m) and do this sum multiple times with different velocities (a spreadsheet and graph in excel helped here), I get the optimum speed is around 0.707 times the speed of light.

    Obviously this is because any slower and you would just take longer to get there, simple! And any longer and time dilation would have a profound effect on the observers point of views time.

    Is this correct? Or have I made some errors?
     
  7. Apr 23, 2012 #6

    Ich

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    The observer's time is distance/velocity, obviously. Time dilation means that the measured rocket time is shorter by that factor. So there is no optimum velocity, you go as fast as you can and get as the lowest limit a time of distance/c (observer) or 0 s (rocket).
     
  8. Apr 23, 2012 #7
    no matter what you do, no matter how fast you accelerate, no matter how fast your velocity, light still leaves you at a velocity 'c'... sad, perhaps, but true.
     
  9. Apr 23, 2012 #8
    I don't think im making myself clear!!!

    I'm saying there is a dude standing at point B, with a timer and he can see point A. When the rocket leaves point A (dude starts the timer) it will need to travel at around 0.7c to optimize time FROM THE OBSERVERS REFERENCE FRAME (dude at point B). Any faster than that and the rocket would be travelling forward in time, and making the trip longer to complete from the observers reference frame but not from the reference frame of the rocket.
     
  10. Apr 23, 2012 #9

    DrGreg

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    Your time dilation formula is the wrong way round. It should be [tex]
    t_{rocket} = t_B \cdot \sqrt{1-\frac{v^2}{c^2}} = \frac{d_B}{v} \sqrt{1-\frac{v^2}{c^2}}
    [/tex]Both [itex]t_{rocket}[/itex] and [itex]t_B[/itex] get smaller as v approaches c. Ich's answer (#6) is correct.
     
  11. Apr 24, 2012 #10
    This is the bit that I don't understand:
    Shouldn't that be the other way around (rockets time is distance / velocity). I'm trying to find the optimum speed to get somewhere according to the observers reference frames time, not the rockets reference frame. (In the rockets reference frame, of course optimum time is going to occur at maximum velocity, but he will arrive at his destination at a later date, according to the observer).

    Just so I know I have the formula correct:

    tobservers reference frame = trocket/√(1-v2/c2)

    So the observers time will always be greater than the rockets time for speeds between 0-c.
     
  12. Apr 24, 2012 #11

    Ich

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    Right.
    And the observer's time is definitely distance/velocity. Think about it. It could not be different, as that is the definition of velocity.
    That means that the rocket time is smaller than distance/velocity.
     
  13. Apr 24, 2012 #12
    Exactly! The rocket time is smaller by a factor of "the lorentz factor?" (if thats how you say it :smile:). And that factor slowly increases as v increases.
    ie. the faster the rocket travels, the more time passes (relative to the rocket) on earth. By that logic you wouldnt want to travel from point A to point B at near the speed of light, because although from the rockets reference frame less time passes, on earth more time passes.
    That is why you need the balance, and I graphed it and found optimum to be approximately 0.707c.
     
  14. Apr 24, 2012 #13

    Doc Al

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    Optimum for what? What do you mean by 'balance'?
     
  15. Apr 24, 2012 #14
    Optimum for getting from point A to point B in the least time possible (from the rest reference frame)
    By balance I mean getting the least time with
    t = distance/velocity
    while keeping the effect of time dilation (as the faster the rocket travels, the more time is going to pass in the rest frame) to a minimum.
     
  16. Apr 24, 2012 #15

    Doc Al

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    Well, what criteria did you use? (Obviously not minimum time. For that you just go as fast as possible.)
     
  17. Apr 24, 2012 #16
    to get minimum time (according to the rest frame) I used this formula:
    where γ is the lorentz factor:

    t' = tγ
    t' = (distance/velocity)γ

    graphing that you get a minimum at v = 0.707c
     
  18. Apr 24, 2012 #17

    Ich

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    t' is smaller than t by a factor of γ.
    Try
    t' = (distance/velocity)/γ
    instead.
     
  19. Apr 24, 2012 #18

    Doc Al

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    I understand this one:
    Earth time = spaceship time * gamma
    But what's this mean? Whose distance/velocity did you use?
     
  20. Apr 24, 2012 #19
    I used the spaceship's distance/velocity
    Is that incorrect?
    Please explain why, I really dont understand.
     
  21. Apr 24, 2012 #20

    Doc Al

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    If you used the spaceship's distance/velocity that would be correct. But then you'd end up with the same equation as the first one, since that distance/velocity would equal t:
    t' = (distance/velocity)γ = tγ

    And you're right back where you started.
     
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