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

Why does time slow down as you approach C?

  1. Sep 23, 2011 #1
    I was talking with my friend and the best answer i could come up with is "It does." But, i could not explain why. Could you guys help me out?


  2. jcsd
  3. Sep 23, 2011 #2


    User Avatar
    Science Advisor

  4. Sep 23, 2011 #3
    On a more simplistic level, you are experiencing time at your own rate. If you traveled above c ( currently thought to be impossible, sans new CERN reports in the news about neutrinos ) everything else in the universe would be standing still relative to you. Time is never truly at a fixed rate of passage for anything, it's completely dependant on your velocity relative to everything else.

    At least that's how I'll look at it until shown to be incorrect.:uhh:
  5. Sep 23, 2011 #4
    because the speed of light could not be a constant if time was the same for everyone in all frames of reference.

    but c is a constant therefore time must be different in inertial frames at different speeds.
  6. Sep 23, 2011 #5


    User Avatar
    Gold Member

    WHOA HERE ... unless I have this totally wrong, time does NOT slow down for you when you get near the speed of light. It APPEARS to other inertial frames of reference to have slowed down but to YOU it doesn't seem to slow down at all. To you it seems like things in the OTHER frames of reference have slowed down (and of course, they would disagree).
  7. Sep 23, 2011 #6


    User Avatar
    Gold Member


    Also, the explanation to 'why' is pretty complicated and can't be summed up in 1 line. Experimentally it has been shown that the speed of light for any observer moving at any velocity relative to anything else will be constant.

    The question relates to how one measures distance and velocities and times. For example, if an observer at rest sees a train moving 100mph to the right and another observer is driving by the train going 60mph, the person in the car will only measure the train going at 40mph. So typically, observers measure objects traveling at different speeds depending on their own velocities. Experimentally, however, it has been shown that the speed of light is measured as the same velocity for all observers, regardless of their velocity!

    Since any velocity measurement is simply a change in distance during some change in time, for two observers to measure something at the same velocity yet the observers themselves are moving at different velocities during those measurements, something has to be different with how they measure time and distance. The details of that, given correctly by the Lorentz transformations, tell us that people will measure time changing differently and specifically, the observer moving relative to the guy initially at rest will experience a slower time change.
  8. Sep 23, 2011 #7
    I was listening recently to an audio book by Brian Greene 'the fabric of the cosmos' and he used a very simple analogy concerning spacetime. If you travel north east, then it would take you longer to get to a line that was north of your starting point than if you simply travelled north, because some of the distance component has been transferred to the easterly direction.

    Similarly seeing as space and time are two aspects of the one spacetime, then if you are not moving time is going as fast as it can. But once you start moving in space then the time component slows down because some of the time component of spacetime is being used up by travelling through space.

    So if you travel at the speed of light then you are travelling through space as fast as you can so you are using all the time component of your spacetime motion on travelling through space. Therefore time stands still.

    This is how I understood what was being said, anyway.
  9. Sep 23, 2011 #8


    User Avatar
    Gold Member

    Hm ... I wonder if that's really the right way to say it. As I understand it, the guy moving will not EXPERIENCE a slower time change but he will have UNDERGONE a slower time change even though not experiencing it. I think this is more than semantic nitpicking.

    The guy who travels off at 99.99% of the speed of light (forget about getting squashed by initial G force in this thought experiment) and travels in a huge arc back to the starting point will feel that he has been traveling through time at a perfectly normal rate, but when he gets back he'll see that a clock on the ground that was synchronized with his to start with is now far into the future of his clock even though he EXPERIENCED time as normal.
  10. Sep 23, 2011 #9
    Time slows down for the observers only, like when entering the event horizon of a black hole to an observer outside the object entering will appear to take forever to fall in, however if you are in the object you would pass right through it in no time at all.
  11. Sep 23, 2011 #10


    User Avatar
    Gold Member

    Yes that was a poor way of stating it. He will experience time like usual, but upon returning to compare, his clock will have advanced much less than the original one.
  12. Sep 25, 2011 #11
    So suppose I start a stopwatch and I start travelling at the speed of light C for 100 seconds and then returned to my original stopwatch. how many seconds will have lapsed on my stopwatch?
  13. Sep 25, 2011 #12
    Depends on how close to the speed of light you go. You cant go c... But you can get closer to closer to c and as you get closer and closer you can make the stopwatch lapse as much as you want.
  14. Sep 25, 2011 #13
    well, I was wondering how you'd solve that numerically. Say I went 99% of C for 100 s, assuming C is 3x10^8 m/s. How much time would have elapsed on my stopwatch?
  15. Sep 25, 2011 #14
    If the stopwatch you are using to measure out that 100s is on your body when you travel that fast, 100s.

    If the stopwatch is not moving at all,

    [itex]\gamma = \frac{1}{\sqrt{1-\beta^2}} [/itex]
    [itex]\beta = \frac{v}{c} [/itex]
    [itex]\Delta t = \frac{\tau}{\gamma}[/itex]

    [itex]t = 14.1067 s [/itex]
    Now, if you went 0.9999c,
    [itex]t = 1.41418 s [/itex]
    And if you were only going 0.5c
    [itex]t = 96.8246 s [/itex]

    So, notice that you really have to be traveling VERY close to the speed of light to see an effect.


    This is the time you would experience out of the stationary frame's 100s. If you want to see how long the other would experience just multiply 100s by gamma.

    Secondly, I fixed my numbers, the were a bit off.
    Last edited: Sep 25, 2011
  16. Sep 25, 2011 #15
    Gravity has the same effect on time. I.e. Flying a clock in a plane can cause a difference.
    Would flying around the sun at 1/2 the speed of light in a gravitational field that was immensely stronger than that on earth cause the clock to differ in the same manner as they would if there was no gravity but you were going very close to the speed of light? I know the answer is yes, just curious if anyone knows how the gravitational effect changes the difference in the clocks in comparison to the speed concept? Then we will know which way might be achievable before the other. Or is the combo the best approach? Sending yourself into the future might be useful if you were gravely sick for instance, or just currious :)
  17. Sep 26, 2011 #16
    Then why does one person age more than the other?
  18. Sep 26, 2011 #17


    User Avatar
    Gold Member

    Have you read the thread? Read posts 8 & 10
  19. Sep 26, 2011 #18
    I really like the 'photon clock' explanation. Imagine you have a photon clock which measures one second as the time it takes a photon to go up a distance d, bounce off a mirror and go back down a distance d. Now imagine you have this photon clock on a train moving at speed v.

    Relative to you, the photon clock isn't moving, so the time it takes for the photon to go up and down is simply 2d/c, where c is the speed of light. However, I am outside of the train looking in. I see the photon moving to the side as well as up and down, because the train is moving at speed v. Since the speed of light is a constant, I measure the time for the photon to go up and down as a value greater than 2d/c because the distance travelled by the photon is further.

    Thus for me, a greater time has elapsed than for you, even though we have both seen the same event occurring. And so we conclude that the faster you move, the more time slows down.
  20. Sep 26, 2011 #19


    User Avatar
    Science Advisor
    Gold Member

    No matter how much you have accelerated in the past, you are still just as far from the speed of light as you were before you started. You will still measure the speed of light to be c just like before.

    Furthermore, even though an observer you left behind sees and measures time for you to be going slower, you see and measure his time to be going slower by the same amount, even though he never accelerated. You are both in the same boat as far as "time slowing down" or as far as "approaching c" goes. To you, he appears to be the one who is traveling at some high rate of speed and therefore "approaching c" and to him, you are the one who is moving at the same high rate of speed and therefore "approaching c".

    And to top it off, you don't even have to accelerate at all to have your "time slowing down" or "approaching c". All you have to do is transform a frame in which you are at rest to one in which you are "approaching c" as close as you want and in that frame your time will be slowed down as much as you want. Then, you can see what happens if you actually accelerate in the opposite direction approaching c and then, according to that frame, your time will be speeding up.
  21. Sep 26, 2011 #20
    I am out of touch in physics for more than a decade, so please excuse my obsolete and corrupt memory/understanding of the concepts...

    As per my understanding velocity is a vector quantity, and if the traveler indeed comes back to the starting point - while travelling in the arc - the velocity vector relative to the stationary clock will not always be 0.9999c, half of the journey time should dilate, and the other half, it would be other wise... so when he does comes back and stops, there should not be any time difference.

    The time dilation effect should come into picture, when the person travelling at 99.99%c, tries to read the stationary clock - since that information from stationary clock can at best be sent at c... so the 1 second ticks sent by stationary clock to the person travelling at 99.99% c, will not be 1 second ticks read as per the clock with the travelling man.... and vice versa ( if the stationary person tries to read the clock travelling at 99.99%c)

    So, if the traveller continue travelling at 99.99%c, in the direction away from the stationary clock - the time ticks sent by stationary clock when read on traveller's clock will appear dilated - relatively.

    However, I am not sure what happens, when the traveller takes a U-turn and approaches the stationary clock.... and if it must come to rest to read the clock, there would be a process of deceleration also - and it would no more be non-accelerating frames of observation....the traveller must read the stationary clock, while maintaining the constant speed.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook