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SR Simultaneity Question

  1. May 29, 2006 #1
    OK, can someone answer this for me?

    Let's say we have two spaceships, Ship A and Ship B, traveling toward earth at v= 0.866c relative to earth. They are at rest relative to each other, not accelerating, and Ship B is at a distance of 5 light years behind Ship A (as observed by them). They have planned ahead of time to each accelerate to v = -0.866c relative to earth at the same time (as observed by them). They plan these events to occur 5.77 years after Ship A passes earth (according to them). So, according to the ships, ship A will pass earth at t=0, ship B reaches earth at t=5.77 years, and at t=5.77 years, both ships will accelerate to v = -0.866c. The ships turn around (relative to earth) simultaneously (as observed by the ships). Now for the earth observer. Earth will observe the distance between the ships as 2.5 light years (due to length contraction), earth will observe ship A pass by earth at t=0, ship B turn around at earth at t=2.89 years. But earth will not observe ship A turn around simultaneously with ship B. When and where will ship A turn around according to an observer on earth?

    I don't really need diagrams, but an explanation using Lorentz transformation equations would be great.

    Thanks,
    Alan
     
  2. jcsd
  3. May 29, 2006 #2

    JesseM

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    They turn around simultaneously in the frame where they were originally at rest, but not in their new rest frame after the turnaround.
    Yes, assuming you define the event of ship A passing the earth as happening at time-coordinate t'=0 in the earth's frame.
    OK, the Lorentz transformation equations are:

    [tex]x' = \gamma (x - vt)[/tex]
    [tex]t' = \gamma (t - vx/c^2)[/tex]

    So let's have the x,t frame be the rest frame of the ships before the turnaround, and the x',t', frame be the frame of the earth. Here v=-0.866c (the velocity of the earth in the x,t frame), and [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex] = 2. If you want to define things so that ship A passes earth at coordinates (0,0) in both frames (the equations I gave for the Lorentz transformation assume the origins of the two coordinate systems coincide, although there is a more general form where this doesn't have to be true), then in this frame ship A must be at rest at x=0 while ship B would be at rest 5 light years behind it at x=-5, so the event of ship B's turnaround has the coordinates x=-5, t=5.77, and the event of ship A's turnaround has coordinates x=0, t=5.77 (because it turns around at the same moment in this frame). So plugging the event of ship B's turnaround into the Lorentz transformation gives:

    x' = 2(-5 - (-0.866)*5.77) = 0 light years
    t' = 2(5.77 - (-0.866)*(-5)) = 2.89 years

    And plugging in the event of ship A's turnaround gives:

    x' = 2(0 - (-0.866)*5.77) = 10 light years
    t' = 2(5.77 - 0) = 11.55 years

    Assuming I haven't made any math errors, these should be the coordinates of both events in the earth's frame. Note that in the earth's frame ship A turned around 11.55-2.89=8.66 years after ship B, meaning that it had travelled 8.66*0.866 = 7.5 light years since the moment ship B turned around, so it makes sense that it'd be at position x'=10 light years at this moment, since the separation between them was 2.5 light years in this frame which means ship A must have been at position x'=2.5 light years at the moment ship B turned around at position x'=0 light years.
     
    Last edited: May 29, 2006
  4. May 29, 2006 #3

    pervect

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    This is unfortunately a potentially ambiguous question.

    Let us set up the problem with one possible interpretation of what this could mean, using the Lorentz transforms.

    First, some notation. We will introuduce two coordinate systems for the problem. The first will be the coordinate system of the Earth.

    Let Axe be defined as the x coordinate of ship A in the Earth frame of reference.

    Let Ate be defined as the t coordinate of ship A in the Earth frame of reference.

    Then we can write the trajectory of ship A as follows.

    [tex]
    Axe = \beta \, c \, \lambda_1 \hspace{.5 in} Ate = \lambda_1
    [/tex]

    In your example

    [tex]
    Ax_e = .866 \lambda_1 \hspace{.5 in} Ate = \lambda_1
    [/tex]

    Here [itex]\lambda_1[/itex] is an arbitrary parameter, representing some particular "point" that the ship is at along its journey. [itex]\beta = v/c[/itex] is the normalized velocity of the ship (.866 in your example).

    Similarly, for ship B, we can write:

    [tex]
    Bxe = \beta \, c \, \lambda_2 + K \hspace{.5 in} Bte = \lambda_2
    [/tex]

    We will write down the details of this in your example after some discussion.

    At this point, K is an undetermined constant, one that we will get from the conditions of the problem. So far, what we've said boils down to a statement that ship A is moving with a normalized velocity [itex]\beta[/itex], that ship A reaches x=0 at t=0, and that ship B reaches x=0 at some other time which is a function of our parameter K.

    Now let's introduce the Lorentz transforms.

    We will introduce a frame that is moving with a velocity [itex]+ \beta[/itex], the same velocity that the ships are intially moving at.

    We will call this frame s.

    Then we can write down the Lorentz transforms to convert from the Earth frame to the "S" frame. Note that the "S" frame does not work after the ships accelerate! We'd need another frame to deal with the results after acaceleration occurs. This is why your question was potentially ambiguous.

    Here are the Lorentz transforms.

    [tex]
    ts = \frac{(te - \beta \, xe/c)}{\sqrt{1-\beta^2}}
    [/tex]
    [tex]
    xs = \frac{(xe - \beta \, c \, te)}{\sqrt{1-\beta^2}}
    [/tex]

    Note that I am including the factor 'c' for completness, I would generally recommend that one set c=1 and work in geometric units for these sorts of problems. In your example, we can easily do this - distances are measured in light years, and times in years, therefore we can set c=1 and forget about it.

    Now, let us talk about simultaneity. Two points [itex]A:(Axe,ate), B:(Bxe,Bte)[/itex] are simultaneous in the Earth frame when the time coordiantes are the same, i.e. when [itex]Ate=Bte[/itex].

    Let us apply that to our example. Looking at our expressions for Ate and Bte, we see that simultaneous events occur when [itex]\lambda_1 = \lambda_2[/itex].

    We can then deduce that K is the length of the ship in the Earth frame.

    Thus, in your problem, to be specific
    K=2.5
    [itex]\beta=.866[/itex]
    c=1

    and we write

    [tex]
    Bxe = .866 \lambda_2 + 2.5 \hspace{.5 in} Bte = \lambda_2
    [/tex]

    Now, let us look at what hapens when we apply the Lorentz transforms to the above equations.

    We know from the Lorentz transform defintion that

    [tex]
    Axs := \frac{(Axe - \beta \, c \, te)}{\sqrt{1-\beta^2}}
    [/tex]


    Substituting and simplifying, we find
    [tex]
    Axs = 0
    [/tex]

    Similarly, we find that

    [tex]
    Ats = \lambda_1 \sqrt{1-\beta^2}
    [/tex]

    I.e the ship A is stationary at the origin in the "S" frame, and it's time coordinate. is lower in the S frame than it is in the E frame.

    Specifically, in your example, Ats = .5 [itex]\lambda_1[/itex].

    Similarly, we find that

    [tex]
    Bxs = \frac{K}{\sqrt{1-\beta^2}}
    [/tex]
    [tex]
    Bts = \sqrt{1-\beta^2} \lambda_2 + \frac{\beta}{\sqrt{1-beta^2}}\frac{K}{c}
    [/tex]

    in your example

    [tex]
    Bx_s = 5 \hspace{.5 in} Bt_s = .5 \lambda_2 + 5*.866
    [/tex]

    Now the length in the ship frame occurs between two points where At_s = B_ts. This is a rather messy calculation, we find that

    [tex]
    \lambda_2 = \lambda_1 + \frac{\beta}{\sqrt{1-\beta^2}}\frac{K}{c}
    [/tex]
    or that
    [tex]
    \lambda_1 = \lambda_2 - \frac{\beta}{\sqrt{1-\beta^2}}\frac{K}{c}
    [/tex]

    in your example
    [tex]
    \lambda_2 = \lambda_1 + 5*.866
    [/tex]
    [tex]
    \lambda_1 = \lambda_2 - 5*.866
    [/tex]


    and we get the length in the ship frame as [itex]K/\sqrt{1-\beta^2}[/itex] as one would expect (5 in your example).

    This is just background, now we come to your original question.

    If you can specify the parameter [itex]\lambda_2[/itex] where ship "B" turns around - (I'm afraid I couldn't figure that out from your verbal description of the problem, as your assumptions were not clear at all), we can solve for [itex] \lambda_1[/itex] = [itex]\lambda_2 - 4.33[/itex], and then easily find the coordinates for event A (in either or both of the coordinate systems) from the other equations.

    Similarly, you could specify [itex]\lambda_1[/itex], calculate [itex]\lambda_2[/itex], and find the coordinates of event B (in either or both of the coordinate systems)

    Note that after the turnaround occurs, a different defintion of simultaneity will apply. If the ships turn around "simultaneously" in one particular frame, in the frame after the turnaround occurs, they will no longer consider the turn-around "simultaneous".
     
    Last edited: May 29, 2006
  5. May 29, 2006 #4

    pervect

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    The following might be a simpler way to address the overall issues.

    Consider a coordinate system (x,t).

    Then events with the same value of t are simultaneous in the stationary frame.

    The lorentz transforms are just (x,t) -> (x',t') via the following eq:

    [tex]
    t' = \frac{(t - \beta \, x/c)}{\sqrt{1-\beta^2}} \hspace{.5 in} x' = \frac{(x - \beta \, c \, t)}{\sqrt{1-\beta^2}}
    [/tex]

    Events that are simultaneous in the transformed (primed) frame have identical values of t'.

    If we consider the set of events that are simultaneous with the origin in the moving frame (t'=0), we find that the equation of this in the non-moving frame is

    [tex]
    t = \frac{\beta x}{c}
    [/tex]

    This can be easily graphed on a space-time diagram, for instance.

    [add]
    Notice that for any arbitrary event (not just the origin), t'=constant implies that
    [tex]t = \frac{\beta x}{c} + K[/tex], i.e. a line through any point with a constant t', a "line of simultaneity in the primed frame", will have a slope dx/dt = [itex]c/\beta[/itex].
     
    Last edited: May 29, 2006
  6. May 30, 2006 #5
    Jesse,

    I assume then that ship A will pass back by earth at t' = 23.1 years according to earth. Now, after Ship A turns around, I assume ship A and ship B will again be at rest relative to each other, but will not agree that they turned around simultaneously in this new frame. So, how far apart will they be now? It looks like earth will observe them to be 10 light years apart. So will they observe themselves to be 20 light years apart in their own new frame, since earth will observe this distance as length contracted?
    And if the ships separated by 15 light years as observed by them, then in his new frame, Ship A will observe that Ship A turned around 17.32 years after ship B turned around. So according to ship A, he turned around at t = 23.1 years, and passes back by earth at t = 46.2 years.
    Is this right?

    Thanks,
    Alan
     
    Last edited: May 30, 2006
  7. May 30, 2006 #6

    robphy

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    convenient arithmetic with regard to the Lorentz Transformations...

    Just a thought concerning convenient arithmetic with regard to the Lorentz Transformations...

    Of course, if you are given specific values in your problem you've gotta work with them. However, if you are trying to get the conceptual idea behind a certain type of scenario, it might be wise to make your numerical calculations as simple as possible.

    A Twin Paradox type situation with the same departure and return speeds involves an obvious factor of 2. Choosing a speed so that gamma=2 (i.e. v=0.866=sqrt(3)/2) introduces another factor of 2. For some (like me), when doing arithmetic (rather than algebra) it's easy to lose track of how things are related with there are coincidental symmetries... not to mention my eyes glaze over looking at too many decimal values (which is one reason I hate grading).

    Instead, choosing v=4/5 yields other nice fractions...
    gamma=5/3, beta*gamma=4/3 and k=3... which is much nicer when doing calculations.

    My $0.02.... i mean 2/100 dollars.
     
  8. May 30, 2006 #7
    Good point, robphy. I could have made the arithmetic simpler. But, if I start changing things now, I would really get confused.

    Thanks,
    Alan
     
  9. May 30, 2006 #8

    JesseM

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    Yup.
    No, in the earth's frame they are 2.5 light years apart up until the moment B turned around, after that the distance between them increases by 2*0.866 = 1.73 light years per year until A also turns around. Ship A turns around 8.66 years after ship B, so the distance between them has increased by 8.66*1.73 = 15 light years in this time. Add that to the initial separation of 2.5 light years at the moment B turned around, and you see that in the earth's frame they are 17.5 light years apart at the moment A turns around. After this they are again at rest with respect to each other, so that distance stays constant.
    They will observe themselves to be 17.5 * 2 = 35 light years apart in their new frame after the turnaround.
    Where did the 15 come frome?
    Again, I don't know where you're getting these numbers.
    the 23.1 years was in the earth's frame, not the two ship's new rest frame.
    Well, let's check what the coordinates of the turnaround in their new rest frame x'',t'' are, by transforming from the coordinates in earth's frame x',t'. We assume that in this frame, x''=0, t''=0 is the event of A passing the earth, just as it is in earth's frame. In earth's frame B's turnaround had coordinates x'=0, t'=2.89, so:

    x'' = 2(0 - (-0.866)2.89) = 5
    t'' = 2(2.89 - (-0.866)0) = 5.77

    And in earth's frame A's turnaround had coordinates x'=10, t'=11.55, so:

    x'' = 2(10 - (-0.866)11.55) = 40
    t'' = 2(11.55 - (-0.866)10) = 40.41

    After each one turns around, they will remain at rest at these positions while earth moves at 0.866c in the direction of A. It's 35 light years from x''=5 to x''=40, so earth will take 35/0.866 = 40.41 years to travel this distance. Since it was at x''=5 at time t''=5.77 when B turned around next to it, it will reach A's position at x''=40 at time t'' = 5.77 + 40.41 = 46.19 years.

    An alternate way to get this conclusion would just be to do a transformation on the time that A returns to earth, which in earth's frame has coordinates x'=0, t''=23.09. So the coordinates in the two ship's new frame must be:

    x'' = 2(0 - (-0.866)23.09) = 40
    t'' = 2(23.09 - (-0.866)0) = 46.19

    So, you get the same answer whichever way you look at it.
     
    Last edited: May 30, 2006
  10. May 30, 2006 #9

    pervect

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    Re-reading your question, it appears that you say in my notation that you specify the location of ship B by the fact that

    Bts = 5.77

    I found some sign errors and a typo in my post, the corrected equations are:

    K = -2.5 (B is behind A)

    a sign error:
    [tex]
    Bts = \sqrt{1-\beta^2} \lambda_2 - \frac{\beta}{\sqrt{1-\beta^2}}\frac{K}{c}
    [/tex]


    there is no square root in the expression for lambda1

    [tex]
    \lambda_1 = \lambda_2 - \frac{\beta}{1-\beta^2}\frac{K}{c}
    [/tex]

    This gives the revised set of corrected equations in your example of
    [tex]Ate = \lambda_1 \hspace{.5 in} Axe = .866 \lambda_1[/tex]
    [tex]Ats = .5 \lambda_1 \hspace{.5 in} Axs = 0[/tex]
    [tex]Bte = \lambda_2 \hspace{.5 in} Bxe = .866 \lambda_2 - 2.5[/tex]
    [tex]Bts = .5 \lambda_2 + 4.33 \hspace{.5 in} Bxs = -5[/tex]

    The "s" frame equations are just the Lorentz transformed versions of the "t" frame equations.

    Using the corrected equations and Bts=5.77

    [tex]Bts = .5 \lambda_2 + 4.33[/tex]

    implies that [itex]\lambda_2[/itex] = 2*(5.77 - 4.33) = 2.88

    Thus we have

    Bt_s = 5.77, Bx_s = 5 (specified) -- coordinates of B in S frame
    Bt_e = 2.88, Bx_e = 0 -- coordinates of B in E frame.

    Now for the event at A to be simultaneous in the 's' frame, Ats = Bts. And we know Bts=5.77, thus Ats must also be 5.77

    We find [itex]\lambda1=11.54[/itex] (it's easiest to do that from Ats = .5 lambda1).

    Then we find

    At_e = 11.54, Ax_e = 9.99
    At_s = 5.77, Ax_s = 0

    Thus At_s = Bt_s, as desired - the events are simultaneous in the S frame, which is the frame before acceleration.

    After acceleration, these events will not be simultaneous, and the distance between the ships in the new frame will not be equal to 5.

    As Jesse noted, when A eventually turns around at Earth coordinates Axe=10, Ate=11.54, B has already been moving for some (11.54-2.88) seconds at a velocity of -.866 c, so it's X coordiante Bxe is -7.5 at Bte=11.54, placing the ships 17.5 light years apart.

    What is it that you have against diagrams, anyway? This would be a lot clearer on a diagram, I think.....
     
    Last edited: May 30, 2006
  11. May 30, 2006 #10
    Thanks, guys, I got it.

    Jesse, I got all those wrong numbers from my first math error. Then that led to getting all the subsequent numbers wrong.

    Pervect, I've got nothing against diagrams, I just didn't want everyone to go to that much trouble. At least it looks like a lot of trouble to post them.
    Actually, these latex graphics look like a lot of trouble, too, but maybe that's just because I have never tried it.

    Thanks,
    Alan
     
  12. May 30, 2006 #11

    pervect

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    Now let's look at the easy way to solve this problem.

    First, put B's turnaround at the origin of the coordinate system - it makes the problem simpler. Thus, at t=0, Bx_e = Bt_e = 0.

    Now we know that the equation for A's motion is

    Ax_e = [itex]\beta[/itex] At_e + 2.5, because at t=0, B is at 0, and A is 2.5 light years ahead of B in the Earth's coordinate system.

    We also know, from previous remarks, that the "equation of simultaneity" is just

    Ax_e = c At_e /[itex]\beta[/itex] = At_e /[itex]\beta[/itex]

    see my remarks about why the slope of a line of simutaneity is always c/beta.

    This gives a much simpler set of equations to solve, for the same results.
    To solve, just substitute the second in the first:

    Ate / [itex]\beta[/itex] = [itex]\beta[/itex] Ate + 2.5

    Ate = 2.5 / ([itex]1/\beta - \beta[/itex]) = 8.66
    Xte = 8.66 / .866 = 10

    If you compare them, you will see that they are identical. X is still at (very near?) 10 at turnaround, and 8.66 = 11.54 - 2.88

    Note that on a space-time diagram, the equation for the slope of a "line of simultaneity" has a simple geometric interpretation.

    [tex]
    \]
    \unitlength 1mm
    \begin{picture}(97.5,92.5)(0,0)
    \linethickness{0.3mm}
    \put(20,10){\line(0,1){80}}
    \linethickness{0.3mm}
    \put(20,10){\line(1,0){70}}
    \linethickness{0.3mm}
    \multiput(20,10)(0.12,0.36){167}{\line(0,1){0.36}}
    \linethickness{0.3mm}
    \multiput(20,10)(0.36,0.12){167}{\line(1,0){0.36}}
    \put(22.5,25){\makebox(0,0)[cc]{$\theta$}}

    \put(40,12.5){\makebox(0,0)[cc]{$\theta$}}

    \linethickness{0.3mm}
    \multiput(50,10)(0.12,0.36){167}{\line(0,1){0.36}}
    \put(20,92.5){\makebox(0,0)[cc]{Time}}

    \put(97.5,10){\makebox(0,0)[cc]{Space}}

    \end{picture}

    \[
    [/tex]

    the geometric interpreation is that the angle [itex]\theta[/itex] in a space-time diagram that an object makes with the 't' axis is the same angle that the "line of simultaneity" makes with the 'x' axis.

    If the t axis is vertical and the x axis is horizontas (as is standard), the two nearly vertical lines on the diagram above are worldinles of particles. The more nearly horizontal line is a "line of simultaneity" that consists of events regarded as "simultaneous" by either of the two previous worldlines.

    Note that the rightmost, almost vertical line would be observer "A" in the problem we just solved.

    Note that this solution process was much simpler and less error-prone than the previous method.
     
    Last edited: May 30, 2006
  13. May 30, 2006 #12

    pervect

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    Heh, I was just fixing up my post on the easy way to do it, so as not to make all the errors.

    As far as the diagrams go

    http://jpicedt.sourceforge.net/ makes it pretty easy to draw graphs. You need to know to include the following additional text.

    Code (Text):

    tex]
    \]
    ......paste code from jpicedt here
    \[
    /tex]
     
    Another fine point is how to draw greek letters (use $\theta$ to draw theta).
     
  14. May 31, 2006 #13
    Jesse,

    The first thing I notice here is that in earth's frame, ship A passes earth at t' = 0, turns around at x' = 10 light years and t' = 11.55 years. And ship A passes back by earth at t' = 23.1 years.

    And in ship A's frame(s), ship A passed earth at t = 0, and ship A passes back by earth at t" = 46.2 years.

    And if we assume that earth had a space station 10 light years from earth (observed in earth's frame), that an earth observer would observe that ship A turned around at the space station. But an observer on ship A would initially observe that ship A turned around 5 light years from earth (at the space station), but after ship A turned around, ship A's observer would observe that ship A travelled a total distance of 40 light years from the time he first passed earth until the time he passes back by earth.

    Is this correct?

    And if ship A had a clock, and this clock was started at t = 0, when ship A first passed earth, what would this clock read (according to ship A's observer) when ship A passes back by earth?

    Thanks,
    Alan
     
    Last edited: May 31, 2006
  15. May 31, 2006 #14

    JesseM

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    These are just coordinates from two different frames, they don't represent what ship A "observes"...better to think in terms of the proper time ship A observes between first passing earth and returning to it, which will only be 11.55 years.
    I don't think you can use the word "observes" for non-inertial observers, unless you specifically define a non-inertial coordinate system for them (but you couldn't construct this system by assuming that at any given moment they will use the same definition of simultaneity as their current instantaneous inertial rest frame, because when they accelerate that could lead to the same event being simultaneous with multiple different points on their worldline). Remember, observing something in an inertial frame is a somewhat abstract notion, you don't observe an event at the time the light from the event actually reaches you, it's the coordinates you retroactively assign events after you see them under the assumption that the light from those events was travelling at c relative to you. In the outbound-leg rest frame, the time on earth's clock that is simultaneous with the event of A's turnaround is 2.89 years, but in the inbound-leg rest frame, the time on earth's clock that is simultaneous with the event of A's turnaround is 20.21 years...would you say that the events of the earth's clock ticking all the dates in between those were never "observed" by A, even though he did see the light from them?
    This is the "proper time" I referred to earlier, it'd be 11.55 years. And there's no need to specify "according to ship A's observer", all frames must agree on what a clock carried on ship A will read at the moment it returns to earth, otherwise you'd have an inconsistency in real physical predictions.
     
    Last edited: May 31, 2006
  16. May 31, 2006 #15
    Jesse,

    So, if each ship (A and B) had observers, and clocks, ship A's clock would read t = t" = 5.77 years when ship A turned around (according to ship A), and ship B's clock would read t = t" = 5.77 years when ship B turned around (according to ship B). And afterward, these clocks would be synchronized again. And these ships would be 35 light years apart (as observed by them) after they turn around.

    Is this correct?

    And if a real person was on ship A, would this person say that ship A turned around 5.77 years after passing earth, ship A turned around 40.41 years after passing earth, or would this person say that neither statement is objectively true?

    Thanks,
    Alan
     
  17. May 31, 2006 #16

    JesseM

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    You seem to have changed the meaning of the t'' notation, which was used earlier to denote the time-coordinate of the inertial frame where A was at rest during the inbound leg after the turnaround. Aside from that, yes, if ship A's clock read 0 at the time it passed earth, then it will read 5.77 years at the time it turns around, and if ship B's clock was set to be synchronized with ship A's clock in the inertial frame where they were both at rest during the outbound leg before the turnaround, then B's clock will also read 5.77 years at the time it turns around.
    You mean each clock would be reset to the time-coordinate of its turnaround in the outbound rest frame (which I assumed would still have t''=0 be the moment A passed the earth)? If the clocks were not manually reset, they would no longer be synchronized in their mutual rest frame after the turnaround.
    In my last post I discussed why I don't agree with your use of the term "observed". They would be 35 light years apart in the inertial frame where they are both at rest after the turnaround.
    The person could calculate how much time elapsed between the event of ship A passing earth and the event of ship A turning around according to different reference frames, but the person would say that according to their own clock (which shows the 'proper time' along their worldline), 5.77 years had passed between these events.
     
    Last edited: May 31, 2006
  18. May 31, 2006 #17
    Jesse,

    You're right, I misused the notations in my last post.

    OK, let's say that T refers to time in ship A's non inertial frame, and T' refers to time in ship B's non inertial frame. Clocks on both ships measure T and T' respectively. Then the non inertial observers on both ships would agree that ship A passed earth at T=T'=0, ship A turned around at T=T'=5.77 years, and ship B turned around at T=T'=5.77 years, and ship A passed back by earth at T=T'=11.55 years.

    And everyone would agree that ship A turned around at the space station.

    And both ship observers would say that the ships separated by a distance of 30 light years during their acceleration.

    Of course they would both realize that the ships did not accererate simultaneously in the SR sense.

    And both ships' clocks will read the same time, even though they each lost synch along the way.

    Is this all correct?

    And I guess it is safe to assume that if we did not assume instantaneous acceleration, the ships could not actually accelerate at a rate that would cause the ships to separate from each other at v>c.

    Thanks,
    Alan
     
  19. May 31, 2006 #18

    JesseM

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    But unlike with inertial observers, there is no standard or conventional way to define the "frame" of a non-inertial observers, you could make up a variety of different coordinate systems in which they stayed at rest throughout the journey which would assign different coordinates to distant events. If you just want to talk about time as measured by clocks on each ship, that's just the proper time, but if you want to talk about the space and time coordinates each one assigns to distant events then you need to define exactly what coordinate system each one is using.
    OK, but that's just the proper time measured by each, it won't help each one to assign time-coordinates to distant events.
    Ship A would say that he passed the earth when his clock read 0. But Ship B cannot say what time that A passed the earth without a coordinate system. Are you just assuming that if a distant event is simultaneous with the event of B's clock reading 0 in B's instantaneous inertial rest frame at that moment, then that must mean those events would also be simultaneous in whatever non-inertial coordinate system you make up? Unfortunately that would lead to badly-behaved coordinate systems where the same event can happen at two different time-coordinates, and other events are not assigned a time-coordinate at all...that was what I was talking about when I said in my second-to-last post:
    So you need to explicitly define a coordinate system that doesn't have these problems (perhaps by defining the coordinate transformation between an inertial frame and your non-inertial coordinate system), only once you have it can we figure out what time on B's clock is simultaneous with the event of A passing earth in whatever coordinate system you have defined.
    Both ship's own time of turnaround was 5.77 years according to their clocks, but if you invent a non-inertial coordinate system for each one, they won't necessarily assign the event of the other ship turning around the same time-coordinate.
    Yes, this is a definite physical event and should be true regardless of what coordinate system you come up with.
    Again, not without defining a coordinate system for each. Where did you get the 30? In the inertial frame where they are both at rest after the turnaround, their separation was 35 light years.
    They accelerated simultaneously in one inertial frame, non-simultaneously in others.
    In what frame will they read the same time? Or are you just referring to the fact that both read 5.77 at the moment they turn around?
    Not in any inertial frame, but if you're talking about a non-inertial coordinate system, coordinate velocities can exceed c in such coordinate systems, depending on how you define them. If you stick to inertial frames, then even with instantaneous acceleration nothing ever moves faster than c (although the distance between them can increase faster than c in an inertial frame if they are moving in opposite directions).
     
    Last edited: May 31, 2006
  20. May 31, 2006 #19
    Jesse,

    I've got it for now. The reason I said the ships separated by 30 light years is because they were 5 ly apart before the turnaround, and 35 ly after.

    And I didn't mean that the clocks stayed synched (in the SR sense) the whole time. Just that they read the same time for the same events.

    I'm not a physicist, so I'm not used to being so formal with these words. But I'm realizing how important it is.

    Thanks,
    Alan
     
    Last edited: May 31, 2006
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