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Who does time really slow down for?

  1. Mar 29, 2005 #1
    Hi. I was thinking about SR and the theory that you can use its predictions to travel into the future (well...in less time than stationary observers). But I just though it this, I was hopeing someone can clear this up for me.

    I originally though...
    If I travel at .9c, time runs slow for me from a statoinary reference frame. And in my frame of reference, time is running fast around me. Hence, I can travel into the future.

    Now I think...
    I travel at .9c, time runs slow for me from a stationary reference frame. But from my frame of reference, I am stationary and everything around me is traveling at .9c. So when my body of motion comes to rest. more time would have passed up for me.

    The second one makes little sense at all, but can somebody please clear this up for me.

    Thanks in advance :smile:
  2. jcsd
  3. Mar 29, 2005 #2


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    Two observers with a relative velocity of 0.9c will both see the other's clocks as running slowly, while their own appears normal. The only way to compare the actual elapsed times on the clocks is to bring the clocks back together again, which necessarily involves some acceleration (at least one of the travellers has to turn around.)

    If you calculate the proper time along each of the two worldlines, you'll find that the observer who turns around and comes back has experienced a shorter proper time. This is the famous "twin paradox."

    - Warren
  4. Mar 29, 2005 #3


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    If you are travelling at 0.9c, you will indeed see all the clocks around you running slow, but if a series of clocks is lying on the ground you'll also see them out-of-sync, and this is what explains how your observations can be consistent with those of someone on the ground who sees your clock running slow. For example, suppose there is a long rod that's one light year long in its own rest frame, and there are clocks mounted on either end that are synchronized in this frame. If you travel at 0.9c from one end of the rod to the other, then if both your clock and the clock at the end you start from read 2005.00 when you depart, then when you arrive at the clock on the other end, the clock there will read 2006.11 and your clock will read 2005.48. From the point of view of an observer on the rod, this is because it takes you (1 light year)/(0.9 light years/year) = 1.11 years to go from one end of the rod to the other, but your clock is slowed down by a factor of 0.436, so on your clock only 0.436*1.11 years = 0.48 years will pass. From your point of view, the length of the rod is shrunk down to 0.436 light years, and the clocks on either end are out-of-sync--at the same moment that the clock on the end you leave from reads 2005.00, the clock on the other end already reads 2005.90. It will take you (0.436 light years)/(0.9 light years/year) = 0.48 years to get to the other end, but the clock on that end is ticking slower, so only 0.436*0.48 years = 0.21 years will pass on that clock. But since it started out reading 2005.90, by the time you reach it the clock will read 2005.90 + 0.21 = 2006.11 years. So both frames give the same prediction, for different reasons.
  5. Mar 30, 2005 #4
    Does this mean that SR does NOT imply that one can travel into the future (in less time than a stationary observer)?
  6. Mar 30, 2005 #5


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    If one observer accelerates as in the twin paradox, then if twin A departs from twin B and later returns to find twin B has aged more than him, that definitely seems like it would qualify as travelling into the future. If no acceleration is involved, as in my scenario where one observer is moving past a row of clocks at constant velocity, it's tougher to say what "travelling into the future" means. If you set out from earth in 2005 AD to a planet on the other end of the galaxy 100,000 light years away (in the galaxy's rest frame) at 99.99999% light speed, then in your frame at the moment you leave earth the planet on the other end is already at the date 102004.99 AD (according to a calendar designed for the galaxy's rest frame), and in your frame the journey take 44.7 years while on this distant planet only about 0.02 years pass, so when you arrive the date is a little after 102005.1 AD...have you "travelled into the future" here since the planet's date is so much later, or have you not travelled into the future since in your frame less time elapsed on the planet between the time you left earth and the time you reached it than the amount of time that passed on your ship? Note also that if aliens on this planet have been keeping track of radio transmissions from earth, the most recent one they'll have got when you arrive was sent in 2005.1 AD, very shortly after you left earth, so you won't be getting any news about earth's distant future (unless you turn around and return to earth).
  7. Mar 30, 2005 #6


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    You certainly can travel into the future. Spend enough time at a high enough speed and you can travel arbitrarily far into the future in a short time.

    In fact, get near enough to c (or get too close to a black hole) and you could watch the entire universe age and die from the comfort of your pilot's chair.

    The problem is, you can't return to the present again. You might as well have gone into suspended animation.

    fiction based on this principle
    'A World Out of Time' - Larry Niven
    A spaceship pilot loops around the centre of the galaxy at relativistic speeds and returns to Earth 3 million years in it future.

    'The Forever Wars' - Joe Haldeman
    An interstellar war is fought over centuries Earth-time but the combatants, at relativistic speeds, experience only decades subjective-time. This leaves two lovers out of sync by the end of the war. The one who comes home early waits on a "time shuttle" - which loops at relativistic speeds, slowing down her time - and waits for her lover to come home.
  8. Mar 30, 2005 #7
    This question also goes for DaveC426913. You are saying that if you travel near c, time will slow down (durrr), but you just stated earlier that from the travelers frame of reference frame, he is stationary and the universe i traveling near c past him, hence, time runs normal for the traveler and time runs slow for the universe around him. How does this allow time to run slow for you (or, time running fast around you)?

    I really want to get this cleared up, thanks a lot.
  9. Mar 30, 2005 #8


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    Well, in my example, at the moment the ship departs the earth, in the earth's frame the date on earth's clocks is 2005 AD and the date on the destination planet's clocks is also 2005 AD, but in the ship's frame simultaneity is defined differently, so at the moment the ship departs the date on earth's clocks is 2005 AD but the date on the destination planet's clocks is already 102004.99 AD. In the earth's frame, the journey takes 100,000.01 years, so at the moment the ship arrives at the destination planet, in the earth's frame the date on earth is 2005 + 100,000.01 = 102005.01 AD and the date on the destination planet is also 2005 + 100,000.01 = 102005.01 AD. But in the earth's frame, time is running slower on the ship, so only 44.7 years pass according to onboard clocks.

    In the ship's frame, clocks on both the earth and the distant planet are running slow, so although the journey takes 44.7 years according to the ship's clock, only 0.02 years pass according to clocks on earth and on the destination planet. So, at the moment the ship arrives at that planet, in the ship's frame the date according to clocks on earth is 2005 + 0.02 = 2005.02 AD, and the date according to clocks on the destination planet is 102004.99 + 0.02 = 102005.01 AD. So although clocks on both the earth and the destination planet ran slow throughout the journey in the ship's frame, because of the way the two frames define simultaneity differently, both predict that if the ship leaves earth in 2005 AD, the journey to the distant planet takes 44.7 years in the ship's frame, but the date on the destination planet when it arrives will be 102005.01 AD according to clocks on the planet. From the ship's point of view, this is just because clocks on the planet were out-of-sync with clocks on earth to begin with, since clocks on the planet already read 102004.99 AD "at the same moment" that clocks on earth read 2005 AD. Since different frames define simultaneity differently, they disagree about what clocks on the planet read "at the same moment" that clocks on earth read 2005 AD, that's the key here.
  10. Mar 30, 2005 #9
    What do you mean by "simultaneity differently"?
  11. Mar 30, 2005 #10


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    I said that different frames define simultaneity differently, which means they don't agree on whether two events happening at different locations happened at the same time or not (that's what 'simultaneity' means, the question of whether two events happened simultaneously or not). If one frame assigns the events the same time-coordinate, then another frame will assign them two different time-coordinates.
  12. Mar 30, 2005 #11


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    He had to accelerate. While accelerating, he's not in an inertial frame of reference and relativity does not apply.
  13. Mar 30, 2005 #12


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    But eNathan's question also applies to cases where two observers are moving inertially relative to each other--unless you understand about the two frames having different definitions of simultaneity, it might be hard to understand how their observations could be consistent if each one said the other one's clocks were running slow.
  14. Mar 30, 2005 #13
    Are you saying that if I accelerate to .9c and then back to 0c, my clock will still be in sync with another stationary observer (IE no relativistic effects have occurred). This makes no sense because it would be almost impossible to maintain a constant velocity all the time. If my speedometer in my car says I am going 55 MPH, that is probably still in a margin of error of .01 MPH, at least.
  15. Mar 30, 2005 #14


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    When he said "relativity does not apply", he didn't mean no relativistic effects would occur, just that you can't use the usual equations of relativity in a non-inertial frame. You can predict how much the accelerating observer's clock will fall behind by looking at the situation from the point of view of a separate inertial observer, though.
  16. Apr 3, 2005 #15
    Ok, so we have Earth which I'll define as E we have the ship which will be S and the distant planet will be X.

    S is traveling to X from E. Current year on E is 2005 C.E and current year on X relative to E is 2005 C.E.

    Now S takes 44.7 years to get to X because S is traveling at 99.99999c. The reason it takes only 44.7 years for S is because this speed allows him to travel this distance (100,000 light-years) in a short amount of time.

    Now, If this is correct so far I do not understand a few things. If S was able to reach X in 44.7 years, why did X only age .02? And for that matter, why did E age 100,000 years?

    I made a http://www.pixelpolice.com/upd1/XESgra.jpg [Broken] to explain myself better.

    Ok, so once you gather that information I have provided on the graphic (could be false, I do not understand this stuff, thats why I made this). So if it was correct, which I'll be amazed if it was, go on to section A. If the information was not accurate, please explain why and go to section B. And if you want to just check out both sections, feel free.

    Section A:
    S takes 44.7 years relative to S.
    If S is traveling away from E does E appear slower or faster? If it is slower (which I think it is) is it because S is traveling at 99.9999c so S is traveling with the "image" of E (in 2005 [at the exact moment he left] which is traveling the speed of light). So the reason S has a slow appearance of E is because time for E relative to S is only increasing .0001% a second relative to E's actual second?

    Ok, now X would be viewed as faster or slower to S? If it is faster (which for some reason I think it is but feel i am most likely wrong) is it because X's "images" are rushing towards S as it approaches X's year relative to X. Every second S travels closer into X's future relative to S because in fact it is 102004.99 C.E on X relative to X, therefore since S's view of X from E was seeing X as being in the year 2005 C.E, this would make S "catch up" to X's actual time?

    Now, if it only takes S 44.7 years to travel this distance of 100,000 light-years, why is it that X doesn't actually age 44.7 years? And why is it that E doesn't only age 44.7 years relative to E during the duration of 44.7 years that S took to travel. Why does S's speed affect the relative time of E and X for that matter? If you were to omit S and just turn the clock 44.7 years ahead, then X and S relative to themselves would have aged 44.7 years, correct? So I do not see why S can drastically change the time of either X or E.

    If S lands on X and looks back at E wouldn't he simply be seeing it 44.7 years into the future?

    Umm, I had questions for Section B, but I guess I'll leave this for now.
    Last edited by a moderator: May 2, 2017
  17. Apr 3, 2005 #16


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    X didn't age only .02 years. X and E aged the same amount, 100,000 years, since you said that "it was the same time" on X as it was on E when S left.

    Note that determining this is a matter of convention, but since X and E are moving at roughly the same velocity, they both agree on what "the same time" is. S, however, doesn't share the same definition of "at the same time" as E and X do. This is a very important point!

    I can't figure out what you mean by "year relative to" which occurs at a lot of places on the diagram.

    It's easiest to analyze the problem when every observer (X,S,E) has a coordinate system which defines exactly one time coordinate and up to three space coordiantes for each event. For this simple problem, we can get away with one space coordinate. So we might say:

    Event: S reaches X

    E's coordinates: t=102,005 CE, x=100,000 light years
    X's coordiantes: t=102,005 CE, x=0 light years
    S's coordinates t=2049 CE, x=0 light years
    Last edited by a moderator: May 2, 2017
  18. Apr 3, 2005 #17

    By year relative to I was referring to actual time. When I say relative to, it means its refering to the actual time of (X,E,S). Therefore, E's relative time to E = E's real time which is 2005 C.E. X's relative time to E refers to E's view of X from E, since X's relative time to X is 102005 C.E the view of X from E would be seeing X at the year 2005 C.E because X is 100,000 away from E.

    I actually messed up on the graphic, for earth it says "current year relative to X 102005 C.E." When in fact the view of E from X would be 2005 C.E minus 100,000 years.

    Perhaps this is what I am not understanding, can you explain this more?

    Also, since S has traveled 100,000 light years away (I hate the term light year becuase it is misleading-- well to me) So I'll use "the large numbers." So S traveling at 99.99999c travels 5.865698 x10^17 in 44.7 years. Now, lets drop these notions and I'll just simply get to the bare bone. We will assume that E cannot see X and X cannot see E so they dont even know what year it is. Therefore, the year on X and E when S departs from E is 0 for both. Now, S travels this distance in 44.7 years, why is it not year 44.7 on X and E?
  19. Apr 3, 2005 #18


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    When you say "current year on X relative to E", you mean the current year in E's reference frame, right?
    In E's frame, it didn't take 44.7 years -- it took (100,000 light years)/(0.9999999 light years/year) = 100,000.01 years. But in E's frame, S's clock is slowed down significantly, by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex], which for (v/c)=0.9999999 works out to be around 0.000447. So, in 100,000.01 years, S's clock will only have ticked (100,000.01 year)*(0.000447) = 44.7 years.

    In S's frame, the reason it takes 44.7 years is different. For S, his own clock ticks at the normal rate, but the distance between E and X is shrunk down to 0.000447 times the distance as seen in E's frame. So he only sees the distance as 44.7 light years, and X is moving towards him at 0.9999999c, so he gets there in 44.7 years.
    In S's frame, both E and X only aged 0.02 years. This is because S sees E and X moving at 0.9999999c, and he also sees clocks on those planets slowed down by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex], or 0.000447. So, in 44.7 years he sees their clocks tick by only (44.7 years)*(0.000447) = 0.02 years. But remember, in S's frame X's clock doesn't read 2005 AD at the moment S departs E, it reads 102004.99 AD, which means after 0.02 years it will read 102005.01 AD.

    But in E's frame, both E's and X's clocks tick by the same amount as S goes between the two planets, or 100,000.01 years. And in E's frame, X's clock did read 2005 AD at the moment S left E, which means that E should also predict that X's clock will read 102005.01 years at the moment S arrives.
    There are a few things wrong with that diagram. For one, it's not E's frame vs. X's frame you should be looking at, but E's frame vs. S's frame (E and X share the same frame, since they are at rest relative to one another, so they won't disagree about anything). Also, in both S's frame and E's frame, S leaves E in 2005 AD. But they disagree about what the date on planet X is at the "same moment" that S leaves E--in E's frame, X's clock also reads 2005 AD at that moment, but in S's frame, X's clock already reads 102004.99 AD at that moment.
    Last edited by a moderator: May 2, 2017
  20. Apr 3, 2005 #19
    Thank you, I finally understand it now!
  21. Apr 5, 2005 #20
    Two twins, A and B are born on the planet X, which has a constant inertial rate. At the time of B's departure from X both A and B had an age of 0. Twin B travels 3 light years at a speed of 99c. Once reaching the destination of 3 light years, twin B returns to X again at 99c. What age will A and B be and why are the ages different?

    Ok, so here goes.

    A experiences time [tex]t_0[/tex]
    B experiences time [tex] t_{dil}[/tex]
    B travels at .99c (184140 mps)
    A is stationary and has an age of t(0) = 0 (not sure what this means).

    To find how long B is gone we take the distance and divide it by velocity.

    [tex]t_0 = \frac{d}{v}[/tex]


    [tex]t_0 = \frac{1116000}{184140} = 6.060606061 [/tex]
    is this light years becuase it involves twin B who is traveling in light years? And if he only felt 6 light years, is that really 6 years for him even though they are light years?

    To find how long B was gone in reference to A we must find the time dilation for the equation

    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]


    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{(184140)^2}{(186000)^2}}} = .1410673598[/tex]


    [tex]t_{dil} = \frac{t_0}{0.1410673598} = \frac{6.060606061}{0.1410673598} = 42.96249728 [/tex]
    this is years correct? because this involves twin A who is stationary so its years?

    Thus, we conclude that if A and B were to meet again A = 42.96249728 years of age and B = 6.060606061 years of age.
    Last edited: Apr 5, 2005
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