A question about stationary reference frame

  1. Is this correct?

    There are two clocks on Earth that are synchronized. One clock goes out on a spaceship at .8c.

    Now according to the stationary or Earth FOR one year goes by for its clock but he sees that only .6 of a year has gone by on the clock of the spaceship.

    Now the spaceship returns home after another year has passed on my clock.

    So this means that 2 years has passed on my stationary clock and only 1.2 years has passed on the moving clock according to my FOR?

    Correct?
     
  2. jcsd
  3. This is assuming that at the 1 year mark on my clock I see the spaceship turn around and head back to Earth.
     
  4. anyone can get that value by plugging in the values in time dilation equation. What exactly you want to know by asking is this correct?
     
  5. He wants to know whether or not it is correct. And yes (assuming perfect conditions) it is correct.
     
  6. What I don't understand is where the paradox sets in.

    To me it does not seem like a paradox that one clock has passed two years and the moving clock which now returns home has only passed 1.2 years.

    The other thing I want to understand is what is happening in the FOR of the moving clock.

    How do I use the lorentz transformation to figure this out?
     
    Last edited: Dec 24, 2011
  7. I think I am confusing what I mean by FOR.

    In my reference frame one year has passed and in the moving observers reference frame .6 years has passed.

    There can only be one clock in each reference frame, correct? So that .6 years which has passed after the 1 year passed in my own reference frame is the lorentz transformation for time? Correct?
     
  8. how did you arrive at the number 0.6 for moving clock?that is lorentz transformation
    This is where the paradox steps in. Relativity states that any inertial frame is equally valid. So if i am in that ship i can very well say i am stationary and you are moving at the velocity of 0.8c away from me. So when i calculate 1 year from my FOR it will be just 0.6 in yours
    yes in your frame you have only 1 clock for you
     
  9. ghwellsjr

    ghwellsjr 5,058
    Science Advisor
    Gold Member

    This is very similar to the scenario you asked about in your thread called Twin paradox question except that there you had the traveling twin go out for one year according to his clock and then return the next year whereas now you have the twin go out for one year according to the Earth clock and return in another similar year. I worked out in detail on post #22 how you use the Lorentz Transform to transform events from one frame to another frame.

    You asked about how you use the Lorentz Transform to figure out what happens in the FoR of the moving clock. I pointed out there that the moving clock is not stationary in a single inertial FoR and that you need two FoRs, one for the outbound portion of the trip and one for the inbound portion of the trip and I showed a way to do this without incurring any jumps in time. But I also pointed out that you should transform the entire scenario into one of these two FoRs and then do it again for the other FoR.

    It is not correct that there can be only one clock in each reference frame--every clock is in every reference frame. A reference frame is a coordinate system--we use it to describe where everything is at different times. You described this very well in your first post where you talked about what happened to both clocks in the Earth FoR. Why did you then feel the need to question your good start?
     
  10. atyy

    atyy 9,774
    Science Advisor

    There is, as you say, no paradox.

    The time elapsed on each clock is indpendent of the frame of reference, ie. it is the same in every frame of reference.

    However, the time dilation formula only works in a Lorentz inertial frame (ie. the frame of a non-accelerating observer). An accelerating observer (such as the spaceship twin who undergoes two acelerations) has to use a different formula. A paradox only arises if one erroneously uses the time dilation formula for the accelerating observer.

    You have correctly used the time dilation formula for both clocks (non-accelerating and accelerating) described from a Lorentz inertial frame.
     
    Last edited: Dec 24, 2011
  11. ghwellsjr,

    I was under the impression that each reference frame has its own time. You are making it confusing for me.

    Let me give you an example.

    I am the stationary observer and I see the spaceship moving at .8c.

    Now one year passes for me so that would be the clock in my reference frame correct?

    Isn't it correct to say that .6 of a year elapses for the moving ship or in other words for that FOR relative to mine?

    How is it possible to say that a frame of reference can have multiple clocks? That does not make sense. I say in my FOR one year has passed and in the moving ships FOR relative to me .6 of a year has passed.

    It would be strange to say that the moving ship is part of my FOR and that his clock has moved .6 and mine has moved one year. I need to say in my stationary FOR one year has passed and the moving FOR has elapsed .6 of a year relative to me.
     
    Last edited: Dec 24, 2011
  12. Alright,

    So how do we flip this around?

    We have to say there are two synchronized clocks on earth. One leaves Earth on a space ship and sees 1 year on its clock and it sees .6 of a year on the Earth clock. Now he returns home and his clock says 2 years has passed but only 1.2 has passed on Earth.

    Now I'm getting confused.
     
  13. HallsofIvy

    HallsofIvy 40,223
    Staff Emeritus
    Science Advisor

    No- the situation is not symmertric. In this scenario, the clock was stationary on earth, then had to accelerate to .8c, then had to accelerate to turn around, then had to decellerate to rest on earth again. There was no acceleration or deceleration for the clock that stayed on earth.
     
  14. DaleSpam

    Staff: Mentor

    You cannot just flip it around. As has been pointed out multiple times the traveling twin is not at rest in any inertial frame during the whole journey. The traveling twin is a non-inerital observer. If you want to use the standard formulas you can only apply them in inertial frames. However, you can look at different inertial frames.

    For instance, consider the inertial frame where the traveller is at rest during the first half of the journey. In that frame, the earth travels at .8c for 5/3 2 years = 3.33 years until they are reunited. The travelling twin starts at rest for .6 years during which time he experiences no time dilation. Then, to catch up with the Earth twin he must move at .98c experiencing a time dilation factor of 41/9 for 2.73 years = .6 years. So, in that inertial frame you also wind up with 2 years on the earth clock and 1.2 years on the travelling clock at the reunion.
     
    Last edited: Dec 24, 2011
  15. Well, is there really such things as inertial frames in reality?

    Because even in both my examples one clock still needs to leave the Earth so it had to make an acceleration at some point.
     
  16. DaleSpam

    Staff: Mentor

    There is nothing wrong with acceleration. You can use all of the standard formulas in an inertial frame regardless of the non-inertial acceleration of the various objects wrt that frame. The important thing is that the reference frame itself be inertial. As long as your reference frame is inertial you can have any non-inertial objects that you like.

    Does that distinction between inertial frames and non-inertial objects make sense? Remember, a reference frame is basically just a coordinate system.
     
  17. There is no symmetry even if you dont know the word "acceleration". Or even if the ship blames the earth as accelerating or decelerating
     
  18. ghwellsjr

    ghwellsjr 5,058
    Science Advisor
    Gold Member

    Sorry to be confusing you, I will try to make it clear.

    First, I recommend you study How did Einstein Define Time, especially towards the end where I explain the difference between Proper Time and Coordinate Time.

    But I'll give you a synopsis here. A Frame of Reference in Special Relativity is defined as a coordinate system with the normal kind of x, y, and z components to specify a 3-dimensional distance for any point from a predefined origin and with synchronized clocks at every location which keep what is called Coordinate Time. These are normal clocks but they are not allowed to accelerate, that is, they must remain at the locations where they were synchronized to the clock at the origin. When they are synchronized, they all read zero. So we now add a fourth component to the coordinate system which is the time. The four components making up the location and time are called an "event", which doesn't have to imply that any thing actually happened there at that time, just like we can talk about the location of a point in a normal coordinate system, even if there isn't anything there. (Actually, we just imagine that these clocks exist at every location just as we imagine that there are grids of measuring rulers laid out in all directions from the origin.)

    Now if we want to describe a scenario in this Frame of Reference, we specify the events for each real clock that we want to consider. Since these clocks are allowed to change locations and therefore experience acceleration and therefore experience speed and therefore experience time dilation with respect to the Coordinate Times on the clocks that are fixed to the Coordinate Locations, we have a different term to refer to the times on these real clocks which is called Proper Time. We can determine how fast any clock is ticking by knowing its speed in the Frame of Reference and applying the reciprocal of the Lorentz Factor to it. Adding up all the ticks allows us to keep track of the time after any period of Coordinate Time we want to consider.

    So let's see how this works for the scenario you described in your first post. You started with two synchronized clocks on Earth. Let's assume that they are located at the origin so that the x, y, z and the time components are all zero and that the Proper Time on these two clocks is also zero. I like to use the nomenclature of [t,x,y,z] to specify events and since y and z are always zero in a simple scenario, I like to omit them and use [t,x] to specify each event. Remember that the "t" in this event is the Coordinate Time on the clock that is located at x=0, even though we have two more real clocks at this location at the start of the scenario, the two clocks you specified.

    So let's refer to the "stationary" clock as clock S and the "traveling" clock as clock T. At the start of your scenario, both clocks are specified by the event [0,0] and they both read 0. Then T accelerates to a speed of 0.8c (instantly, to keep things simple) for one year. Where will it be after one year? That's easy, it will be at location x=0.8 (we are using units of time in years and distance in light years.) The corresponding event is [1,0.8] for T and for S it is [1,0]. Now the first question is what time will be displayed on these two clocks? Well T has been at a speed of 0.8c so its time will be dilated by a factor of 0.6 (you already know how to do this calculation so I won't repeat it). So keep in mind that there are two times related to clock T at the point of turn-around, one is the Coordinate Time of the event of the turn-around which is 1 year and the other is the Proper Time displayed on the clock which is 0.6 years. Meanwhile, the Proper Time on clock S is 1 year and the Coordinate Time for the corresponding event is also 1 year.

    Now clock T returns and it takes another year of Coordinate time for it to get back. The event describing its reuniting with clock S is [2,0]. This event applies to both clocks but the Proper Time on clock T has advanced by another 0.6 years so it now reads 1.2 year while clock S has advanced by 1 year so that its Proper Time reads 2 years, again the same as the Coordinate Time, since it never moved.

    Now I know that you know all the details with regard to the Proper Times because you described them in your first post but you may not have been aware of the concept of Coordinate Time or how they applied to events.

    Then you asked about how to view things from the Frame of Reference of clock T. As I pointed out in post #8, clock T is not stationary in any Frame of Reference so you cannot use the Lorentz Transform to answer your question. It's actually stationary in three frames, the one it shares with clock S at the beginning and ending of the scenario, the one it is stationary in during the outbound portion of the trip and the one it is stationary in during the inbound portion of the trip.

    So let's start by transforming three of the four events we have already specified from the first FoR into the FoR in which clock T is stationary during the first half of its trip. I'm not going to bother to transform the event for clock S corresponding to the time of the turn around for clock T because it has no bearing on anything.

    Here at the three events in FoR 1:
    [0,0] start of scenario for both clocks
    [1,0.8] turn around event for clock T
    [2,0] end of scenario for both clocks

    The first event [0,0] will remain the same no matter what other FoR we transform to so I won't go through its calculation.

    For the other events, the first thing we need to do to use the Lorentz Transform is calculate gamma, γ, where the speed as a fraction of c, is β = 0.8 according to:
    γ = 1/√(1-β2)
    γ = 1/√(1-0.82)
    γ = 1/√(1-0.64)
    γ = 1/√(0.36)
    γ = 1/0.6
    γ = 1.667

    The formulas for the Lorentz Transform are:
    t' = γ(t-βx)
    x' = γ(x-βt)

    For the outbound portion of the trip, β=0.8, and the event of the traveling clock at the end is [1,0.8]:

    t = 1
    x = 0.8

    t' = γ(t-βx)
    t' = 1.667(1-(0.8*0.8))
    t' = 1.667(1-0.64)
    t' = 1.667(0.36)
    t' = 0.6

    x' = γ(x-βt)
    x' = 1.667(0.8-(0.8*1)
    x' = 1.667(0.8-0.8)
    x' = 1.667(0)
    x' = 0

    So the turn-around event for clock T is:
    [0.6,0]

    We can calculate its speed by taking the difference in its starting and ending locations divided by the difference in its starting and ending times. But the starting and ending locations are both 0 so it is stationary in this FoR. So all we have to do to see how much time has advanced on clock T during the first leg of the trip is take the difference in the Coordinate Times for these two events which is 0.6-0 or 0.6 years.

    Now let's jump to the last event [2,0].

    t = 2
    x = 0

    t' = γ(t-βx)
    t' = 1.667(2-(0.8*0))
    t' = 1.667(2-0)
    t' = 1.667(2)
    t' = 3.333

    x' = γ(x-βt)
    x' = 1.667(0-(0.8*2)
    x' = 1.667(0-1.6)
    x' = 1.667(-1.6)
    x' = -2.667

    So the ending event for both clocks is:
    [3.333,-2.667]

    Now in order to calculate the speed of clock T during the inbound portion of the trip we have to do the difference thing I mentioned earlier. The coordinate location difference for clock T between the start and end of his inbound trip is -2.667-0 = -2.667 light years. The corresponding coordinate time difference for clock T is 3.333-0.6 = 2.733 years. Dividing these we get a speed for clock T of 0.9756c. Now we apply the reciprocal of the Lorentz Factor on this speed to see what time dilation clock T experiences:

    1/γ = √(1-β2)
    1/γ = √(1-0.97562)
    1/γ = √(1-0.9518)
    1/γ = √(0.0482)
    1/γ = 0.2159

    Now we multiply this factor by the Coordinate Time difference of 2.733 years to get 0.6 years as the Proper Time difference for the clock T during the inbound portion of the trip (just like we got in the first FoR). Finally, we need to add the Proper Time at the end of the outbound portion of the trip to the additional time accumulated during the inbound portion of the trip to get the final Proper Time on clock T as 1.2 years.

    Now let's see how much clock S has advanced in this FoR. We need to do the difference thing again but for this clock and we will do it from the beginning of the scenario to the end. The location difference is -2.667-0 or -2.667 light years. The time difference is 3.333-0 or 3.333 years. The speed is 2.667/3.333 or 0.8c so the reciprocal of the Lorentz Factor is 0.6. So the Proper Time on clock S has advanced by 0.6 times 3.333 or 2 years.

    So as you can see, even in the FoR for the outbound portion of the traveling's clock trip, we still calculate the Proper Times for both clocks to be the same as they were in the original FoR. In fact, it won't matter what FoR we use, we will always get the same answers. If you want to do the transform for the inbound portion of the trip, you will see that it is similar to the outbound and ends up with the same Proper Times.
     
  19. Isn't coordinate time pretty much equal to what my clock says in my FOR?

    Let's say there is a train in my coordinate system that is going at .5c and I want to know how much time has passed for this train after 5 years in my own coordinate system. Am I correct in saying that 5 years is the coordinate time?

    Now is Proper time equal to the amount of the time that has passed for the train moving at .5c after 5 years of my coordinate time?

    So the Proper Time for the train moving at .5c relative to my FOR is 4.33 years. Correct?

    To figure this out I just took my Coordinate Time/The Lorentz Factor of .5c.

    I still have not figured out what the Lorentz Factor exactly is. According to the wikipedia it says it's the change in coordinate time/change in proper time.

    It's saying the change in proper time is equal to the square root of 1 - Bsquared. I don't really understand what that means. It is also saying that equals to the square root of c2-v2. Not really sure what that means. The speed of light - the velocity something is going = proper time?
     
    Last edited: Dec 25, 2011
  20. atyy

    atyy 9,774
    Science Advisor

    Coordinate time is not clock time.

    Clock time is "proper time", and is the same in every reference frame. It is a property of the spacetime trajectory of a clock.

    Coordinate time is the "time" coordinate assigned when "space" and "time" coordinates are assigned within one reference frame to all conceivable events in spacetime.

    Proper time coincides with the coordinate time of a Lorentz inertial frame for a clock that is stationary with respect to that Lorentz inertial frame.

    For clocks that are moving relative to a Lorentz inertial frame, the proper time can be calculated from the coordinate time of that Lorentz inertial frame and the "Lorentz factor".
     
  21. ghwellsjr

    ghwellsjr 5,058
    Science Advisor
    Gold Member

    We're talking about inertial Frames of Reference which means they don't accelerate which means they don't change their speed or direction. So if you remain stationary in what you call "my FOR", then yes, as long as your clock was synchronized to the previously synchronized coordinate clocks defining your FOR, and you also never accelerate, then the Proper Time on your clock will remain synchronized to the Coordinate Time in your inertial FOR. But the "clocks" that we are talking about that keep Coordinate Time are usually "imaginary" clocks that would behave exactly like real clocks at each location that we want to consider. This is why atyy said, "Coordinate time is not clock time." We use the term "Proper Time" to refer to the time on your real clock in your FOR even if it never accelerates and so it keeps the same time as the Coordinate Time in your FOR.
    Yes.
    Yes.
    Yes.
    Yes, the Lorentz Factor of .5c is 1.1547 so it looks like you did this correctly.
    That is correct and you did everything correctly both in this example and in your first post so I don't understand why you say you have not figured it out.
    You've left some things out here. The Lorentz Factor, gamma, is given by:

    γ = 1/√(1-β2)

    And beta, β, is the ratio of the speed that the clock is traveling at divided by the speed of light:

    β = v/c

    So the other way to express gamma is:

    γ = c/√(c2-v2)

    But I like to use the first formula for gamma since it uses speeds as a ratio of the speed of light which is what you used also.

    You said the Lorentz Factor is equal to the change in coordinate time/change in proper time. So that means the change in proper time is equal to the change in coordinate time divided by the Lorentz Factor, correct? So we could express this as:

    Δτ = Δt/γ
    Δτ = Δt * √(1-β2)

    where Δτ (delta tau) is the change in proper time and Δt is the change in coordinate time.

    So given a speed as a factor of the speed of light as in your example of β=.5, did you actually do this calculation for the Lorentz Factor?

    γ = 1/√(1-β2)
    γ = 1/√(1-0.52)
    γ = 1/√(1-0.25)
    γ = 1/√(0.75)
    γ = 1/0.866
    γ = 1.1547

    Now you said there was a 5-year change in the coordinate time so the change in the proper time for the train is equal to 5/1.1547 which equals 4.33 years.

    Now since you got the correct answer, how did you get it if you didn't understand how to do the calculations? Where are you still confused? It looks to me like you have perfect understanding (except for the incomplete quotes from wikipedia).

    Also, have you studied the calculations I gave you for doing the Lorentz Transform in my previous post? Does it all make sense? Did I answer your question about how to do the Lorentz Transform?
     
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