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Does time essentially stop if you could travel a C?

  1. Nov 7, 2004 #1
    I was explaining to one of my friends about how if you could travel the speed of light then you could get to any destination, no matter how far away, instantaneously. And if you were going a bit slower then the speed of light (lets say 98% C) then you could reach a destination, lets say 100 light years away, in just a few years (or you would just age a couple of years even though you traveled a distance of 100 light years)

    Would someone here mind to elaborate on this and give a more technical explanation? My friend is very skeptical on what I am saying.

  2. jcsd
  3. Nov 7, 2004 #2
    Okay, basically, there are three things that happens to objects that move at speeds near to c (relative to an inertial observer): masses increase, time slows down, and the length of the object contracts in the direction of motion. All of these quantities are equal to the rest values either multiplied or divided by the Lorentz factor, [itex]\sqrt{1-(\frac{v}{c})^2}[/itex]. For the object, all of those properties are normal for it. However, the inertial observer now moves relative to it, and so all of those things happen to the observer (according to the object, mind you).

    Now, imagine that you live in a city where speed of light is only 5 mph. You want to travel some place that is 1 mile down the street. So, you take off running at 4.99 mph (very near c). Now, how long does it take you to get there? You can think of it in two ways:

    1. From your point of view, everything else is moving at a speed of 4.99 mph, right? Therefore, everything else seems "squished" in your direction of motion. Though you still seem to be moving at 4.99 mph, the distance between you and your destination is multiplied by the Lorentz factor, .063 in this case. So, the distance you must travel becomes a paltry .063 miles, and you get there in about 45.6 seconds. However, this is according to you. According to everybody else, though, it still took you 12 minutes to reach your destination. If you were able to move even faster, say at 4.99999999 mph, though your actual speed is only .00999999 mph faster, the Lorentz factor would be about 6.3e-5, and the 1 mile becomes about 4 in. Once again, for everyone else, you took 12 min to travel, but for you, you only took .04 seconds. You can probably see where this is going.

    2. You could also think of it from the point of view of the rest of the world, by using time dilation. For everybody else, it takes 12 mins for you to get to your destination, no matter what. Therefore, they simply have to multiply that time by the Lorentz factor for your speed to figure out the amount of time it took for you. If you're moving at 4.99 mph, you get 12 min * .063 = 45.6 s. If you move at 4.99999999 mph, you get 12 min * 6.3e-5 = .04 s.

    Notice that the result is the same no matter what method we choose: the first one or thie second. They're both expressions of the same thing: special relativity.

    By the way, though the text is not word for word, I borrowed heavily from George Gamow's Mr. Tompkins in Wonderland . It's a great book, written in the forties, which explains relativity wonderfully. (The whole idea of a city with a "slow" c came from this book).
    Last edited: Nov 7, 2004
  4. Nov 8, 2004 #3


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    As you approach c relative to some inertial observer, the time it takes (in your frame) to cross a distance (in the observer's frame) approaches 0. Also, the time (in the observer's frame) to cross some distance (in your frame) approaches infinity. (duh, the inverse)

    You could technicly say it takes a photon 0 seconds in your frame to cross any distance in its frame AND that it takes an infinite amount of time in the photon's frame to cross any distance in your frame. Makes no sense. Don't make measurements at c.

    Oh, always remember to state what frame your measurements are in.
  5. Nov 8, 2004 #4


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    The only technical correction that comes to mind is to expand on your last remark, and to talk about the limit as one approaches 'c' rather than talking about travelling at c. One can definitely say that in the limit as v->c, the proper time elapsed for an observer moving a finite proper distance 'd' with a velocity v goes to zero.

    The proof is simple. For the observer at rest, we can assign coordinates to the events of the start and end of the trip as follows:

    start of trip: t=0, x=0
    end of trip: t = d/v, x=d

    The invariant lorentz interval is

    [tex]L = c^2 (\Delta time)^2 - (\Delta distance)^2 = [/tex]
    c^2*(d/v)^2 - d^2 =

    For the observer in motion, the coordinates will be:

    start of trip: t=0, x=0
    end of trip: t=tau, x=0

    The invariant lorentz inteval will be

    L = c^2*tau^2

    Because the Lorentz interval is invaraint for all inertial observers, we write:

    c^2*tau^2 = d^2*(c^2/v^2-1)

    tau = d *sqrt(1/v^2 - 1/c^2) =
    \frac{d}{v} \sqrt{1-(v/c)^2}

    This result can be interpreted from the moving observer's POV as the fact that the distance he travels is shorter than the proper distance 'd' by the factor of sqrt(1-(v/c)^2) due to Lorentz contraction.

    From the stationary obsever's POV, the moving observer's clocks run slow by a factor of 1/sqrt(1-(v/c)^2)

    In any event, in the limit as v->c, the value of tau approaches the limit zero, because sqrt(1-(v/c)^2) approaches zero while d/v approaches a finite value d/c.

    Your friend might also want to look at the sci.physics.faq on the relativistic rocket http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    This handles the case of an observer travelling at a constant acceleration, rather than a constant velocity, however.
    Last edited by a moderator: May 1, 2017
  6. Mar 29, 2009 #5
    Re: Does time essentially stop if you could travel at C?

    Time never stops. The mistaken impression is with the time keeping mechanism and with the observations. All lights travel at the speed of light. Even if their on board clocks appear to have stopped, it still takes them 1 year to travel 1 light-year distance.
  7. Mar 29, 2009 #6
    Your time yes. Time no.

    Basically, if you travelled somewhere at c (assuming for the sake of the discussion that you could) then, for a person at rest with both your point of departure and your destination, a time of x/c will have passed, where x is the separation between your point of departure and your destination according to someone at rest with respect to both.

    I'm assuming your point of departure and your destination are spaceports, stars, train stations, what have you which can be at rest relative to each other.

    So, you can get somewhere instantaneously for you but you will miss out on everything that happened while you were travelling. So, you can't forget about your beloved's 30th birthday on Alpha Centauri until the day before and travel to be with her instantaneously. You will arrive closer to her 34th, during which time she will have got tired of waiting, met someone else, got married and now be pregnant with her second child.

    Which goes to show that very long distance relationships just don't work.


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