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Interaction in 4-D spacetime

  1. Jan 10, 2012 #1
    As time is considered the 4th dimension can we say the 3-d world to be floating(advancing forward in time) in the 4D. If so is every inertial frame has is own velocity rate(moving in the time dimension) in the 4-D. When you have 3-D world if you want to bring 2 objects in contact you can bring them to the same point. But when dealing with 4-d consider this let A and B be start at the same point at t=0 now let B go for a trip and come back. As you can see A and B have their own 3-d layers advancing in time. Because of time dilation after B's return its 3-d layer would be below that of A's isn't it?how can they ever interact again?as they will continue to maintain a distance in the 4-d spacetime(as time rate is the same)?

    i m just trying to imagine 4-d spacetime. apologize if the question is non-sensical.hopefully someone can clarify me?
  2. jcsd
  3. Jan 10, 2012 #2


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    The four dimensions of spacetime are three normal ordinary orthogonal coordinates of space usually designated by x, y and z and on normal ordinary coordinate of time designated by t. All four coordinates specify an event. If any one of the coordinates is different, it is a different event. In order for two objects to be in contact, they have to have all four coordinates the same. So you shouldn't say A and B start at the same point at t=0, you should say they start at the same event. But that event is gone in an instant and they can never return, but they can still continue to occupy the same events, just with t increasing. If one or both of them change locations and reunite, meaning they continue to share the same events again, then you have the Twin Paradox situation. Keeping track of their instantaneous speeds defines their instantaneous Proper Times which is how you determine their relative ages when they return.

    All Frames are identical and none is preferred. You can define any scenario in any single Frame and then transform all the coordinates for the relevant events into another Frame, if you want. Twins that start out at the same event and end up together at a different event will have the same accumulated age difference no matter what Frame you use.
  4. Jan 10, 2012 #3
    Herman Weyl (a colleague of Einstein's and one of the foremost mathematicians and physicists of the 20th century) put it this way:

    "The objective world merely exists, it does not happen; as a whole it has no history. Only before the eye of the consciousness climbing up in the world line of my body..."

    Weyl's description would be basically consistent with what I think you are saying, but Weyl's picture would have each observer moving along his own world line at the speed of light.

    I think I know what you are driving at (correct me if I'm wrong). You're saying that if each twin is moving along his own world line at the same speed (c), and the traveling twin is taking a different path through 4-D, then you can't really bring them back together if one is lagging behind the other (in proper time) after they get back on the same 4-D path (world line). This situation is depicted in the sketch below.


    Weyl's description implies that the whole 4-D world line is there ("...it merely exists, it does not happen"). That means that even though red is present after 10yrs at his 10Yr event, there also exists his 13yr event, available for the encounter with his 10yr-old blue twin.
  5. Jan 11, 2012 #4
    Snip3r, ghwellsjr is furnishing some good perspective here. I've tried to present his perspective with the use of a space-time diagram. I've added two new observers (dark brown and light brown observers) to the space-time diagram shown in the previous post. The simultaneous spaces for dark brown and light brown at the proper times along their respective world lines that include the red-blue event (event where red and blue reunite) are shown. It's just like ghwellsjr pointed out; here we see you can always find a transformation for which any given observer's world will include the red-blue reunion event. As ghwellsjr indicated they will all agree that sure enough there is an event at which red and blue unite, and further, the proper times shown on red's and blue's world line clocks at that event will be the same for any of the other frames, as is shown explicitly for dark brown and light brown in this example. That is to say, the dark brown guy "sees" red's clock reads 13 years while blue's clock reads 10 years. And the light brown guy sees exactly the same clock times on red's and blue's clocks.
    Last edited: Jan 11, 2012
  6. Jan 13, 2012 #5
    thats spooky!but on this i would say we can never ask the question "why" in this model because you will always answer "because thats the 4-d object in the path you just traversed"

    also does spacetime diagram really depict whats happening in the 4-d world. i mean if there is only one time axis for all the observers they would age the same after separating and re-uniting isn't it?you can easily prove this by calculating the resultant vector along the time axis and it will turn out to be the same for both twins however this is not the case in reality.

    so how will the picture be in the 4-d world??
    yes i agree with that. all observers agree on simultaneous events happening at the same place

    please point out any errors in what i said above and thanks for your replies :)
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