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Relative aging equation

  1. Sep 26, 2008 #1

    I was searching online for an equation which would reveal the effects on aging as velocities approach light speed.

    I found this

    t = squareroot of (1-(v^2/c^2))

    Pardon my math ignorance (I am not a physics student) but when I entered in a velocity of 80% the speed of light the result I got was 0.6 and I didn't know how to interpret this. Would my answer mean that I have aged 60% less than a stationary observer if i'm moving at 80% the speed of light? That seems a bit high to me and I am puzzled as to why distance didn't play a roll in this equation. I also assumed that how long you traveled at this speed would additionally play a roll.

    If im trying to dumb down this math too much, please say so. I found this equation and it looked like something I could tackle.



  2. jcsd
  3. Sep 26, 2008 #2


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    First of all, you need to know that if it's possible to describe you as "moving at 0.8c" and the other guy as "stationary", it's equally correct to describe you as stationary and the other guy as moving at -0.8c.

    The formula says that the other guy's aging rate is 60% of your own, so if you have aged 10 years, he has aged 6 years (which by the way is 40% less than you). But when he has aged 6 years, he can use the formula and conclude that you have only aged 3.6 years, and he would be right! This is the "twin paradox". There are lots of threads about that, so you might want to check out one of those.

    The distance is only relevant when you try to consider the point of view of someone who changes their velocity. This is discussed in most of the threads about the twin paradox.

    How long you travel is obviously relevant, since e.g. 60% of 10 is 6 and 60% of 20 is 14. If you travel for 20 years, the difference is twice as large as if you travel for 10 years.
  4. Sep 26, 2008 #3

    Jonathan Scott

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    If you travel at 80% of the speed of light relative to some other observer's frame of reference, then according to that observer you appear to age at only 60% of the rate shown by his own clocks.

    That rate simply applies to the interval for which you travel at that speed, so if the observer sees 10 seconds elapsing, then the apparent elapsed time for you during the same interval as seen from the observer's frame of reference is only 6 seconds. For example, if the observer sets up two cameras along your path at two points which you pass 10 seconds apart, and uses those cameras to take a picture of your watch as you pass, from the same distance in each case, those two pictures will show a separation of only 6 seconds.

    This is a very confusing area, as can be seen by the number of threads on the "twin paradox" and similarly confused ideas.

    The maths of paths between events in space-time is surprisingly similar to the maths of bits of string in space, where the displacement between the ends of a piece of string corresponds to displacement in space and the length of the string corresponds to the time, both as measured in the observer's frame, and the slack in the string (measured for example as the maximum width that it can be waggled from side to side, given end points at fixed events) gives the amount by which something has aged along the path assuming it moved in a straight line (equivalent to constant velocity in space-time). This means that for a given amount of time, the greater the distance between the end points, the less ageing occurs along the path. For a string where both endpoints are at the same point in space (corresponding to being at rest), the slack at the midpoint allows it to be moved exactly as far from one side to the other as the original length of the string, so the ageing is equal to the time. At the other extreme, a string stretched out straight corresponds to travelling at the speed of light, with no slack at all, and no ageing.
  5. Sep 26, 2008 #4
    Ok, great!

    So to apply this to time travel, could one say that if I left earth in year 2000 and traveled at 80% the speed of light away from earth for 6 years (according to me) and then (ignoring decelerations and accelerations) traveling 6 years back (according to me), I would return to Earth at earthtime year 2020. (that's 12 years to me 20 for stationary earth)

    Did I do this right?

  6. Sep 26, 2008 #5

    Jonathan Scott

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    Yes, that calculation sounds right to me, if somewhat impractical at present!
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