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Time Dilation Problem

  1. Jul 16, 2010 #1
    I was reading The Elegant Universe by Brian Greene and I wanted to see how to do one of his examples with math. Here's an excerpt:
    I believe I've included all the relevant information. If not let me know what I'm missing. I would like to understand the math behind this.
     
  2. jcsd
  3. Jul 16, 2010 #2
    Yes, you've included everything we need to know! Here is a description of the scenario and two formulas you can use to make the calculation that Greene does. I haven't explained the reasoning behind the formulas, how they're derived or the more general formulas that these are special cases of, but maybe this will give you at least a taste for what's going on.

    I've drawn a Minkowski diagram (a kind of spacetime diagram) representing your example. The yellow axes are a coordinate system in which Gracie doesn't move. The horizontal axis represents space; we only need to consider one dimension of space for this problem. The vertical axis represents time. In the yellow coordinate system, Gracie's position in space doesn't change. She only moves in time, from event A (where/when George crosses her path at the beginning) to event E (where/when George meets her again at the end). A coordinate system in which an object doesn't move is called it's "rest frame" because the object is at rest in that coordinate system (reference frame).

    George travels from A to F. F represents the event of him turning round. Then he goes from from F to E.

    If you can draw a horizontal line between two events on this diagram, they happen at the same time as each other in Gracie's rest frame. For example, in Gracie's rest frame, George turns around (event F) simultaneously with whatever is happening in Gracie's life at event C. It's natural for Gracie to think of F as simultaneous with C, although, being wise in the ways of relativity, she knows that C and F won't be simultaneous in all coordinate systems.

    Those light blue lines at 45 degrees to the yellow axes are the path that a pulse of light would take through spacetime. I've drawn it to a scale (such as years of time and light years of space) where the speed of light is one, so the slope of the blue lines is 1. The closer George's path is to a blue line, the closer his speed to light speed. Imagine that the slope of his paths are 1/0.995. (I haven't quite drawn them accurately; that would be a bit hard to see.)

    I've drawn extra red lines (FB and FD) that make the same angle as George's path, but on the other side of the blue lines representing light speed. (The top set of three slanting lines should really be the same angles apart from each other as the bottom three, but my drawing is a bit rubbish...) These other red lines are lines of simultaneity in George's "rest frames" (those are coordinate systems that moves along with George; coordinate systems where George always has the same space coordinate). There are two of them here: one that moves along with George from A to F, and another that moves with George from F to E. There are two of them because the simplest kind of coordinate systems, called an "inertial reference frame", is a coordinate system that has a constant velocity. George has a constant velocity on his way from A to F, and from F to E, but when he turns round, his velocity changes direction, so there's no single inertial reference frame in which he's at rest for the whole journey.

    All events that lie on one of these red lines (FB and FD and any lines parallel to them) happen at the same time as each other in George's current rest frame. Before he turns around at F, the event he's experiencing is (from his perspective) simultaneous with event B that Gracie is experiencing. But after he turns around, his experience is simultaneous with Gracie's experience of event D, according to his new rest frame. Bizarre, but that's how it is.

    By symmetry, we conclude that Gracie lives three years between A and B, and three years between D and F. But the total amount of time she has to wait to see George again is much longer. The number of years Gracie waits from A to the midpoint C is...

    [EDIT: Oopsh, I made a mistake here, which DrGreg corrected: see #5 for the answer.]
     

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  4. Jul 16, 2010 #3

    DrGreg

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    Actually [tex]
    \frac{0.995}{\sqrt{1-(0.995)^2}} = \sinh(\tanh^{-1}(0.995))
    [/tex]
     
  5. Jul 16, 2010 #4
    @Rasalhague - which is the application you used to draw the Minkowski diagram?
     
  6. Jul 16, 2010 #5
    Thanks for the correction, DrGreg, and congratulations on your upcoming 1000th post! I should have written

    [tex]\frac{3}{\sqrt{1-0.995^2}} = 3\cosh(\tanh^{-1}(0.995)) = 30.0376...[/tex]

    about 30 years.
     
    Last edited: Jul 16, 2010
  7. Jul 16, 2010 #6
    Photoshop. I judged the angles by sight, which is why they're a bit wonky ;-)
     
  8. Jul 19, 2010 #7
    Can someone explain why it isn't so that Gracie experiences 6 years and George experiences 60 years? If you take George's frame of reference to be static, isn't Gracie then the one who should come back in George's future?
     
  9. Jul 20, 2010 #8
    George's rest frame is not an inertial reference frame, it's an accelerated reference frame. The simple time dilation equations are valid only in inertial reference frames.

    You can use George's rest frame as reference, but it's more complicated. There are many easy to find threads about this in this forum.
     
  10. Jul 20, 2010 #9
    Just to clarify: While the two are moving inertially they WILL both will see the other aging less when taking their own frame as being static. This is not an illusion or anything. Both are ageing quicker RELATIVE to the other when considered static.

    When one turns round though this symetry is broken which is why there should be no confusion when the traveller returns much younger.

    We can be certain about which person is to be considered the traveller because only one of them would have experienced accelerating forces.
     
  11. Jul 20, 2010 #10
    I understand this. Does this mean that if you take any two bodies travelling along distinct paths through the universe, at any point that they meet, you can tell which one has undergone the most acceleration by finding out which one has 'aged' the least?
     
  12. Jul 21, 2010 #11
    This acceleration thing is where I get confused. Sure, George knows that he is the one accelerating, unlike with constant velocity where neither travelers know if they are moving. But lets say that George passes Gracie going .995 the speed of light, but then he slows down to a stop. In this case, he still would have traveled into Gracie's future, correct? But if we consider the relative velocity between the two at each moment, wouldn't there still be a symmetry? Wouldn't we mathematically be able to show for each relative velocity that if Gracie experiences X years and George experience Y years, then when George experiences Y years Gracie will experience X years? It seems like even though one is decelerating, the equations of time dilation should hold. So as George decelerates, his time begins to match Gracie's until their relative velocities are equal and they do match. I understand this is not what happens but I don't see why.
     
  13. Jul 21, 2010 #12
    True - the relative velocities are the same. Even so, only one of them accelerates - i.e. change inertial frame. At least in SR, a change of inertial frame is an absolute, not a relative, matter. In this sense, there isn't in Special relativity, the complete relativisation of all motion that relationists had hoped for.

    It doesn't quite work like this in SR. Roughly, as one changes frame, the `lines of simultaneity' change too and this introduces an effect which more than compensates for the effects of time dilation.

    Putting it vividly: As George slows down he moves through frames which disagree about what is happening `now' with Gracie. In particular, as George (say) drinks a coke during the deceleration process, the Gracie who is simultaneous with George when he starts drinking the coke (in one frame) is a much younger Gracie who is simultaneous with George when he finishes drinking the coke (in a different frame).
     
  14. Jul 21, 2010 #13

    Look at "relativity of simultaneity" on wikipedia. Check out the explanation of the light hitting each end of the train simultaneously in one frame and not so in another. Then apply that phenomena to this situation.

    As one person slows (accelerates) the agreement of what is considered simultaneous changes as the relative velocities change. What happens is that this experience of simultaneity will change enough to compensate for the moving persons measuring of the stationary persons clock running slow. Basically as the moving person slows down he will, assuming its possible, witness the other persons clock start to run faster until they stop. At this point the clock will have run faster enough to be in the 'future' even though it was running slower all the time before it strated accelerating to a stop.
     
  15. Jul 22, 2010 #14
    I kind of get it.. Can you further explain how the simultaneity "compensates" for the other person's clock running slow?
     
  16. Jul 22, 2010 #15
    Look at the beautiful(!) Minkowski diagram attached (least I hope it's attached - never done this before...).

    That big black bar up the middle is the stationary twin's worldline. The up axis is time. The horizontal line at the bottom is a line of simultaneity in the stationary twin's frame. All horizontal lines, all lines parallel to it, are also lines of simultaneity for this twin. Any two points parallel to the x axis represents two events at the same time, according to stationary twin.

    That faint bluish line that goes to the right then goes to the left - that's the moving twin's world line. He changes frame at space-time point A.
    Those faint reddish angled lines - they are the lines of simultaneity for the moving twin. THe lower three are lines for the outward going frame, the upper three lines for the inward bound frame. You can add more - as long as they're parallel.

    Look at the red lines just above and just below point A. From moving twin's point of view, there's little time between them, they're just a click away. But from stationary twin's point of view, there's a long time between them. Sure, during outward journey, moving twin thinks stationary twin's clock ticks slower; suring, during inward journey, moving twin thinks stationary twin's clock ticks slower. But he mustn't calculate total ticks of stationary twins' clock by simply adding up the number of ticks during the two phases of the journey - that would miss out all the ticks of the clock of stationary twin's clock during t.

    All that extra time at t on stationary twins worldline more than makes up for the effects of time dilation. This qualitative explanation should be enough to see *how* it's possible - it's not enough to show that it more than compensates - for that, we'll need to start working through the actual formula
     

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