# Non Twin Spaceflight Time Dilation

1. Nov 29, 2008

### Austin0

Having read a great number of twins threads I "think" I may have a handle on the concept but still a certain ,"uncertainty " remains so I hope you will bear with ONE more space voyage. Hopefully it is a simple one.

There are two space stations at rest in freespace near Earth and Alpha C ,,,,
E (St) and A (St) respectively , that have synched their clocks/calenders.

Two Spaceships E shp and A shp that are traveling at the same relativistic velocity on a course for E (St) by way of A (St) and are spatially separated by the same distance as the stations. They also have synched the clocks in their frame and they are physically identical to the stations clocks.

As E shp passes A St they exchange bursts of info. The ships current time and the current A station time = Earth time in E-A frame. They also send the info to E St and A shp allowing them to calculate the start time of the journey when they receive the transmissions.

Upon reaching Est and Ast respectively they once again exchange info in passing with NO acceleration in the entire journey.

So my question is: Am I correct in assuming that E ship and Earth station would agree on the elapsed time as measured by clocks in their respective frames?
NO net time dilation effects???
Likewise for A shp and A sta.??

That their world lines as drawn by either frame would be equivilent??

2. Nov 29, 2008

### Staff: Mentor

In which reference frame (ships' frame or stations' frame) are the distances the same? They can be the same in only one frame.

Note also that in the ships' frame, the station clocks are not synchronized with each other, and in the stations' frame, the ship clocks are not synchronized with each other, because of relativity of simultaneity.

3. Nov 29, 2008

### RandallB

you need to be more careful making your description clear. Streets do not cross streets; “by way of” means you’re traveling On A street to get to E street! You mean crossing A street before crossing E street.

A reference frame ?

[/QUOTE] As E shp passes A St ……
The ships current time and the current A station time = E-A frame.
Upon reaching Est and Ast respectively they once again ….

4. Nov 29, 2008

### JesseM

Pretty sure "St" was supposed to stand for "station", not "street", since Austin0 said:

5. Nov 29, 2008

### JesseM

Let's put some numbers on this scenario:
OK, let's say the stations are 4 light-years apart in their own rest frame.
Let's say the ships are moving at 0.8c relative to the stations. I assume when you say "spatially separated by the same distance", you mean the distance between the ships in their own rest frame is the same as the distance between the stations in their rest frame? If so, that means they're 4 ly apart in their rest frame, which means in the station rest frame the distance is shrunk by a factor of $$\sqrt{1 - 0.8^2}$$ = 0.6, so in the station frame the ships are only 2.4 ly apart. Also, if the ships' clocks are synchronized and a distance d apart in their own rest frame, then in a frame where they're moving at speed v, they'll be out-of-sync by vd/c^2; so with the numbers I gave, in the station frame the ship clocks will be out-of-sync by (0.8)*(4) = 3.2 years, with the trailing clock (A) ahead of the leading clock (E) by this amount.
OK, let's say that at the moment the E ship reaches A station, both clocks read 0 years. And in the station frame E is synchronized with A, so in this frame E station reads 0 years at the same moment. Finally, in the station frame the A clock reads 3.2 years at this moment.
In the station frame, the A ship is 2.4 light years behind E, and it's moving at 0.8c, so it'll reach the A station in 2.4/0.8 = 3 years. But the A ship's clock is slowed down by a factor of 0.6, so it only elapses 3 * 0.6 = 1.8 years. And it started out reading 3.2 years, so by the time it reaches the A station it reads 3.2 + 1.8 = 5 years, while the A station reads 3 years at that moment.

Meanwhile, in the station frame the E ship has 4 light years to cross to reach the E station, so at 0.8c it takes a time of 4/0.8 = 5 years to reach the E station in the station frame. But the E ship's clock is slowed down by a factor of 0.6, so it only elapses 5 * 0.6 = 3 years in this time. So, the E ship reads 3 years and the E station reads 5 years when they meet, the reverse of the readings when the A ship meets the A station.
I'm not sure what you mean by "agree on the elapsed time as measured by clocks in their respective frames"--agree on the elapsed time between what pair of events? You can look at my numbers above and see if they agree with your assumption. In the station frame the time between the E ship passing the A station and the E ship passing the E station was 5 years, while in the ship frame the time between the E ship passing the A station and the E ship passing the E station was 3 years, so if you were saying you thought the times would be the same in both frames you were incorrect. But note that in the ship frame the time between the A station passing the E ship and the A station passing the A ship was 5 years, while in the station frame the time between the A station passing the E ship and the A station passing the A ship was 3 years, so there is an equivalence there.

Last edited: Nov 30, 2008
6. Nov 30, 2008

### Austin0

________________________________________________________________________
Hi I understand the distance between the ships can be equivilent to the station distance in only one frame. I was not specific because in this question it is not relevant , it could be either frame or even just approximate.
I also understand that asynchronicity would be equivalent. I made no reference to synchronizing clocks between the two frames.
I made no reference to specific velocity because this also is not relevant.

My question was one of basic principle.
Comparing the elapsed time of the clock in the earth ships frame with the elapsed time as recorded in the earth station -alpha station reference frame.

Comparing the elapsed time on alpha ship with the elapsed time as recorded on the alpha station.

As far as I could see the situations were completely symmetrical with no acceleration involved to indicate a preferred frame.

So inflight dilation should be equivilent between earth ship and earth station.
That if they are both valid inertial frames then it seemed that their respective world lines would be identical as considered from either frame.
That therefore there would be no elapsed time differential. They might not agree on what the current time was but would agree on the elapsed time of the flight.

I am sorry if I did not express my original question clearly. Thanks

7. Nov 30, 2008

### Austin0

__________________________________________________________________________

Hi I am sorry if I made things overly complicated and not completely clear.

I am sure that if you think about it you will see that the relationship between the earth ship and earth station and between the alpha ship and alpha station are completely identical. The distance between the ships is not relevant at all. This would effect the quantitative elapsed time that it took alpha ship to reach alpha station but would have no bearing on whether or not there was any accrued elapsed time difference between alpha ships clock and alpha stations clock.

So according to your calculations there would be a residual , net time discrepancy between the different inertial frames.
Specifically between earth ship and earth -alpha station.
And between alpha ship and earth alphas station.

In the twins question I understood it was acceleration that determined a preferred inertial frame.
So in this situation I cannot understand why the frames would not be totally equivalent.
It seems like time dilation and any question of asynchronicity would be reciprocal and would not have any residual or anisotropic effect.
Thanks,, I will try to be more clear and definitive in the future.

8. Nov 30, 2008

### Staff: Mentor

it's still not quite clear to me exactly what you're asking. So I've made a diagram and you can decide whether it answers your question. It shows the two ships and the two stations both in the reference frame in which the stations A and E are stationary (top), and in the reference frame in which the ships A and E are stationary (bottom). In each case, I show the situation at two different times: when ship A passes station A, and when ship A passes station E.

The rectangles are the stations. The things that look like rockets are the ships. The numbers are the readings on clocks carried on each station or ship, in years.

Distances are in light-years. The speed of light is c = 1 light-year / year. (This simplifies calculations a bit.)

Unlike JesseM, I assumed the distance between the ships equals the distance between the stations, in the stations' reference frame.

I've omitted the details of the calculations. If you have questions about the calculations, ask!

In both reference frames, ship A passes station A when the clock on ship A reads 0.00 yr and the clock on station A reads 0.00 yr; and ship A passes station E when the clock on ship A reads 3.00 yr and the clock on station E reads 5.00 yr.

In the stations' reference frame, as ship A travels from station A to station E, 5.00 yr elapse on both the station A and station E clocks; and 3.00 yr elapse on both the ship A and ship E clocks. That is, the ship clocks run slower than the station clocks, in the stations' rest frame.

In the ships' reference frame, as ship A travels from station A to station E (actually better described as station E traveling from its initial location to ship A), 3.00 yr elapse on both the ship A and ship E clocks; and 1.8 yr elapse on both the station A and station E clocls. That is, the station clocks run slower than the ship clocks, in the ships' rest frame.

#### Attached Files:

• ###### Austin0.gif
File size:
30.8 KB
Views:
141
Last edited: Nov 30, 2008
9. Nov 30, 2008

### JesseM

No it's the other way around--the earth ship is analogous to the alpha station, and the alpha ship is analogous to the earth station. After all, the earth ship is "in front" in the rest frame of the stations, so it's the first ship to reach either station (the alpha station); similarly, the alpha station is "in front" in the rest frame of the ships, so it's the first station to reach either ship (the earth ship). You can also see this analogy in the numbers I gave:

earth ship and alpha station clocks both read 0 when they pass each other; then, when earth ship later passes earth station, earth ship clock reads 3 years while earth station clock reads 5 years.

Now, replace "alpha station" with "earth ship" and vice versa, and replace "earth station" with "alpha ship", and you get:

alpha station and earth ship clocks both read 0 when they pass each other; then, when alpha station later passes alpha ship, alpha station clock reads 3 years while alpha ship clock reads 5 years.

...and these are exactly the numbers I got in my previous calculation. The only thing I didn't calculate was the time on the alpha ship when it passes the earth station (the final passing-event), if you do you find that at that moment both the earth station clock and the alpha ship clock read 8 years, so again the symmetry is preserved.

10. Nov 30, 2008

### RandallB

Like I said - things like this need to be very clear; dosn't matter if Austin0 might be more than sure, we need to more than 'petty sure'

11. Dec 1, 2008

### Austin0

[
___________________________________________________________________________
Hi I am not talking about the numbers you derived. I am unclear regarding the basis of your choice of inertial frame to apply them from.
(1) Are not earthship and earth station both equally valid inertial frames?
(2) From the perspective of earth ship ,,earth station is moving towards the ships at rest inertial frame at their relative v ?
And vice versa?
(3) From the perspective of either frame, a world line diagram would be a vertical vector for their own frame with a straight vector for the other frame at some angle to the left?
That these diagrams would be identical?

(4) That all the above would apply equally to alpha ship and alpha station regarding each other?
(5) That from either frame the expectation would be that the other frames clocks were dilated by Lorentz factor???

So my question is: What frame of reference are you using to determine that one frames clocks would be more dilated?? And therefore there would be a cumulative difference at the end of the elapsed time.
I am clearly missing something here. With the twins question it appeared that that preferential determination was made on the basis of acceleration and the fact that it implied an invalid inertial frame. But that is not a factor here so I am confused. Thanks

Last edited: Dec 1, 2008
12. Dec 1, 2008

### Austin0

On what basis did you determine [from what frame] that it would be the ships clock that was actually dilated and thus showed a shorter elapsed time???

If it is the stations frame then what determined that it was those clocks that determined the cumulative results of dilation accurately?

From what frame or rational basis are you determining that between two equivalent valid inertial frames that there would be any cumulative difference in elapsed time between the frames?

In this case, the stations view of reality was accurate and the ship when it arrived conformed with the expected elapsed time but from the perspective of the ship the expected elapsed time for the station (1.8 yrs.) was widely divorced from the reality of (5 yrs).
What is the explanation for this asymmetry of perception??
From what I have learned regarding ,for instance , length contraction ,is that it applies reciprocally.
That two rulers that are equivalent when at rest, would be viewed as contracted equally from either frame when in relative motion.
But at the same time ,by applying the transformation, they could both accurately determine their relationship . But this does not seem to be the case here.

So this is my question and the sourse of my confusion.

Thanks for your help . I really appreciated the effort you put in the drawings and they were very clear. I can see I must start doing that myself.

Last edited by a moderator: Dec 1, 2008
13. Dec 1, 2008

### Austin0

/QUOTE]

Point taken.
I am just getting the hang of communicating in this realm and will try to be more lucid in the future.

14. Dec 1, 2008

### Staff: Mentor

I suspect that you may be confused by two different meanings of the word "time" in common usage. The first is "what a clock reads at a particular moment." The second is "the elapsed time (or time interval) between two events," which can be calculated by subtracting one clock reading from another. Time dilation refers to the second kind of "time" but not to the first.

The ships' clocks are "actually dilated" only in the stations' frame.

In the stations' frame, during the trip, the time interval on the stations' clocks is 5.00 - 0.00 yr = 5.00 yr. Using ship A's clock, the time interval in the ships' frame is 3.00 - 0.00 = 3.00 yr, which is less than the time interval in the stations' frame. Note that we can also calculate this using ship E's clock: (-2.33 yr) - (-5.33 yr) = 3.00 yr.

In the ships' frame, during the trip, the time interval on the ships' clocks is 3.00 - 0.00 yr = 3.00 yr. Using station A's clock, the time interval in the stations' frame is 1.80 - 0.00 yr = 1.80 yr, which is less than the time interval in the ships' frame. Note that we can also calculate this using station E's clock: 5.00 - 3.20 yr = 1.80 yr.

We calculate a time interval by subtracting two clock readings. However, if the readings are from two different clocks, we must make sure the two clocks are synchronized! In the ships' frame, we cannot calculate the time interval in the stations' frame as 5.00 - 0.00 = 5.00 yr, because the two clock readings come from different clocks (5.00 yr on station E's clock at the end of the trip, and 0.00 yr on station A's clock at the beginning of the trip), and the two stations' clocks are not synchronized in the ships' frame.

Note that the two ships' clocks are not synchronized in the stations' frame, either, but this hasn't entered into our calculations because we've been using only ship A's clock.

Last edited: Dec 1, 2008
15. Dec 1, 2008

### JesseM

I mentioned a few times in that post that I was calculating things from the perspective of the station frame. But in relativity all inertial frames always agree about local events like the times on two clocks as they pass each other, so you'd get the same answers if you calculated everything from the perspective of the ship frame.
The diagrams would be mirror images of each other--if the station frame diagram showed the station worldlines as vertical and the ship worldlines as slanting from the lower right to the upper left, then the ship frame diagram would show the ship worldlines as vertical and the station worldlines slanting from the lower left to the upper right. In the station frame the two ships are coming from the right, with the leftmost ship (the earth ship) in front and the other ship behind, while in the ship frame the two stations are coming from the left, with the rightmost station (the alpha station) in front and the other station behind.
It applies equally to their two frames. But again, the spacetime diagrams in each one's rest frame are mirror images, with the alpha ship trailing behind the earth ship in the station frame and therefore analogous to the earth station which is trailing behind the alpha station in the ship frame, and the alpha station being ahead in the ship frame and therefore being analogous to the earth ship which is ahead in the station frame.
With the qualifications I added, I agree with all your statements 1-5.
Each frame of course thinks the other frame's clocks are more dilated. I think maybe where you're confused is that you are imagining that if two clocks pass one another and one reads 5 years while the other reads 3 years, that means the one reading 3 years must have been "more dilated". That's not correct--because of the relativity of simultaneity there is an alternative explanation, namely that clock that ended up reading 5 years had a "head start" on the clock that ended up reading 3 years, the first clock being set to zero and running well before the second clock is set to zero and set running, so that even though the first clock is ticking slower in the frame we're using, it still ends up being ahead by 2 years when it meets the second clock.

If you don't know about the relativity of simultaneity, please read the link above on the subject, it's pretty important to this problem. If you are familiar with the idea, read this section of my example again:
So in the station frame, at the moment that both the earth station and the alpha station clocks reads zero, the earth ship clock reads zero too, but the alpha ship already 3.2 years. Likewise, in the ship frame, at the moment that both the earth ship and the alpha station clocks read zero, the alpha station clock reads zero too, but the earth station clock already reads 3.2 years. Simultaneity is relative--each frame thinks it's the other frame's clocks that are out-of-sync by 3.2 years, while their own clocks are synchronized.

Now, look at how the fact that the ship clocks are initially out-of-sync in the station frame applies to the case of the alpha ship meeting the alpha station:
In this example, when the alpha ship reaches the alpha station, the alpha ship's clock is ahead of the alpha station's when they meet--the alpha ship reads 5 years and the alpha station reads 3. But as I said at the start of the paragraph, this was all calculated from the perspective of the station frame, where the alpha ship's clock is ticking slower than the station ship's clock. So how does the ship's clock end up ahead when they meet? The key is in that sentence in bold--in the station frame, because of the relativity of simultaneity, the alpha ship's clock already read 3.2 years at t=0 when all the other clocks read 0 years. So 3 years later in the station frame, the slowed-down alpha ship clock has only moved forward by 1.8 years, but because it had a head start of 3.2 years, it will read 5 years at this moment.

Last edited: Dec 1, 2008
16. Dec 3, 2008

### Austin0

17. Dec 3, 2008

### Austin0

18. Dec 3, 2008

### Staff: Mentor

Here's the rule I use to calculate "desynchronization."

[added later: I see JesseM did it basically the same way.]

Start with two clocks, at rest with respect to each other, separated by distance $L_0$ in their rest frame, and synchronized in that frame.

In another frame, in which the clocks move with speed v, they are out of sychronization by the amount $vL_0/c^2$. The clock that is "ahead" in terms of position (imagine it being "chased" by the other clock as they both move along a line) is "behind" in terms of clock readings.

In my example, the two stations are $L_0$ = 4.00 ly apart in their rest frame. In the ships' frame (v = 0.8 ly/yr), the two stations' clocks are therefore out of sync by the amount (0.8 ly)(4.00 ly/yr)/(1 ly/yr)^2 = 3.20 yr.

In the ships' frame, the two stations are moving to the left, with station A "in the lead." Station A's clock lags behind station E's clock by 3.20 yr (3.20 yr versus 0.00 yr in the third row of my diagram, or 5.00 versus 1.80 yr in the bottom row of my diagram).

I think of this "desynchronzation" formula (it's usually referred to as "relativity of simultaneity") as part of a three-legged stool of relativity formulas, the other two legs being the formulas for length contraction and time dilation. These three formulas are equally important. You need all three of them, in general, to calculate a complete picture of most relatvity "paradoxes." Omit one and you run into trouble, just like a three-legged stool falls down if you remove one leg.

Last edited: Dec 3, 2008
19. Dec 3, 2008

### JesseM

Yes, I chose to express the relativity of simultaneity formula in terms of the distance d between the ship clocks in the frame where they were at rest. But if you prefer, we can equally well write the formula in terms of the shorter distance d' between them in the frame where they are moving at speed v; in this case the formula would say they will be out-of-sync by $$\frac{vd'}{c^2 \sqrt{1 - v^2/c^2}}$$, or $$\frac{vd'}{\sqrt{c^2 - v^2}}$$. Of course, you can tell this is equivalent to the original formula vd/c^2 since the length contraction formula tells us $$d' = d\sqrt{1 - v^2/c^2}$$. There's nothing that says you can't express the value of a quantity in one frame as a function of the coordinates of a different frame; you just have to make sure you're doing it right and not getting confused.

It's not too hard to show that the formula vd/c^2 follows from the Lorentz transformation which relates different inertial frames. If we have two inertial frames A and A' which we label with primed and unprimed coordinates, and we assume the origins of both coordinate systems coincide (so the event with coordinates x=0, t=0 in the A frame has coordinates x'=0, t'=0 in the A' frame), and in the A frame the origin of the A' frame is moving in the +x direction at speed v, the the Lorentz transformation equations are:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c^2)
with gamma = $$1/\sqrt{1 - v^2/c^2}$$

So imagine we have two clocks at rest in the A frame, with their times synched to the A frame's coordinate time, and the first clock is resting at x=0 while the second clock is resting at x=d, so the distance between them is d. Now let event #1 be the event of the clock at x=0 reading a time of 0, and let event #2 be the event of the clock at x=d reading a time of vd/c^2. So event #1 has coordinates (x=0, t=0) and event #2 has coordinates (x=d, t=vd/c^2). So, use the Lorentz transformation to find the coordinates of event #1 in the primed frame:

x' = gamma*(0 - v*0) = 0
t' = gamma*(0 - v*0/c^2) = 0

So event #1 has coordinates (x'=0, t'=0) in the primed frame. What about event #2?

x' = gamma*(d - (v^2*d/c^2)) = gamma*((c^2*d/c^2) - (v^2*d/c^2)) =
gamma*d*(c^2 - v^2)/c^2 = gamma*d*(1 - v^2/c^2)

...and since gamma = 1/sqrt(1 - v^2/c^2), this means x' = d*sqrt(1 - v^2/c^2). Note that this is the Lorentz-contracted distance between the two clocks A and B.

Now, the t' coordinate of event #2:

t' = gamma*((vd/c^2) - vd/c^2) = gamma*0 = 0.

So, in the primed frame both event #1 and event #2 happen at the same time-coordinate t'=0, meaning they are simultaneous in this frame. So, this shows that at the "same time" clock A reads 0, clock B reads vd/c^2, in the frame where clock B are moving with speed v (and the distance between them is d*sqrt(1 - v^2/c^2) in this frame).

And again, the clocks at different points on the ruler are synchronized using the assumption that light moves at the same speed in all directions relative to me--so, one simple way of synchronizing two clocks in my system would be to set off a flash at the exact midpoint of the line between them, then set each one to read the same time when the light from the flash reaches them. To an observer who sees me and the clocks in motion, though, naturally this will lead them to define the clocks as out-of-sync, because they will see one of the two clocks moving away from the point where the flash was set off while the other is moving towards that point, so naturally if they assume light moves at the same speed in all directions in their frame, the light will catch up with one of the clocks before the other.
Again, the clocks on the A station and the E station are assumed to be pre-synchronized. Since the stations are 4 light years apart, one could set off an explosion at the midpoint 2 light years from each one, and have each station set its clock to some prearranged time when the light from the explosion reached each one. Or, the A station could send a signal when its clock reached some time (say, t=-10 years) and then when the E station received the signal, it could set its own clock to a time 4 years ahead of that time (t=-6 years).
In the E ship's frame, each individual clock would be running slow (dilated time), but each would be out-of-sync with the others, with each one closer to the E station being ahead of the previous one. So for example if each clock is 1 light-day apart in the station frame, in the ship frame each one will be 0.8 days ahead of the previous one at any given moment. Meanwhile they are 0.6 light-days apart in distance in the ship frame, so in the ship frame it takes (0.6 light-days)/(0.8c) = 0.75 days to pass from one to the next. In that time each one only advances forward by (0.75 days)*(0.6) = 0.45 days, but the next one already had a head start on the previous one by 0.8 days, so the next one's time when the ship passes it will be 1.25 days past the time on the previous one when the ship passed it, whereas the ship's own clock has only gone forward by 0.75 days. So in this sense the ship does see the days on the outside clocks passing faster, even though each individual outside clock is running slower.
The convention is that each frame has their +x direction oriented the same way (at least that's what's assumed in the standard form of the Lorentz transformation which I gave earlier), so if the origin of the primed frame is moving in the +x direction in the unprimed frame, that must mean the origin of the unprimed frame is moving in the -x direction of the primed frame.

Anyway, regardless of how the origins are oriented, the A station must be treated as equivalent to the E ship simply because in the station frame the E ship is in front, while in the ship frame the A station is in front.
Not sure exactly what you mean by "the degree of desynchronization was represented by the intersection of this plane, of one world line, with the world line of the other frame". First of all, frames don't have world lines, they're just coordinate systems, only objects with a well-defined position at every time have world lines--were you talking about the world lines of clocks at rest in one frame or another? If in your own frame a horizontal line of simultaneity intersects the worldline of moving clock #1 when it reads 0 and the worldline of moving clock #2 when it reads vd/c^2, then the two clocks are out-of-sync by this amount in your frame. And then if you draw in a slanted plane of simultaneity for the clocks' own frame, and it intersects with the worldline of clock #1 when it reads 0, then this slanted plane will intersect with the worldline of clock #2 when it reads 0 too.
Not sure what you mean here either--maybe you could draw a diagram and post it? What do you mean by "sphere of simultaneity"? A surface of simultaneity looks like a flat plane in a Minkowski diagram, not a sphere. It's true that moving clocks which are closer together will be less out-of-sync then moving clocks which are farther apart, but I don't know if that's what you were talking about.

Last edited: Dec 3, 2008
20. Dec 6, 2008

### Austin0

Hi I understand how you may be suspecting I am a bit dense , this possibility is in my mind also. There is something crucial here I am not getting and that's for sure. But it is not the math or basic concepts. My math is not good but in the past I did the math for various inertial systems in order to understand how the measured invarience of c was possible through clock desynchronization, length contraction and time dilation. I understand the method of light synchronization and the relativity of simultaneity.
I have gone over your and jtbell's work up of this situation and understand the processes i am just unsure of how you applied them in this situation.
I see the reciprocity and symmetry between the two frames. Fine.
But I did an extrapolation of the scenario. Added some more stations on the same vector and continued the journey. SO with each station the ships get 2 years behind the station time. Assuming at some point they turn around and return to Alpha station. There would then be a discrepancy of many years between ship and station time, correct???
So how does this symmetric relation lead to such an asymmetric net result????
And if this is the case, doesn't it mean there would be an overall time dilation aboard the ships???? It is not just clocks, it is also physical processes etc, Correct???Or is this just another erroneous assumption on my part?

.
I was not refering to desynchronization between the stations clocks.
I was talking about the desynchronization at that point between E stations clock and E ships clock which at that point is at A station. In the case of A station with regard to A ship,, you assumed that A ship was running 3.2 yrs ahead of A station time Right??? (t=0) (t'=t +3.2) In the case of E station with regard to E ship you assumed that they were simultaneous. I.e. that (t=0) =(t' =0 ) not t'=t +3.2
The fact that E ship and A station's clocks happened to both read the same at that point makes it confusing.
In the case of A ship's trip of 2.4 light years to A station ,instead of just calculating elapsed ship time as dt' =gamma*(dt=dx/v ) you added in the desynch factor. (+3.2 yrs)
You did not do this with E ship to E station. Does this make any sense???

I sent a drawing. It is just my interpretation of how desynchronization relates to world line diagrams ,,so I may be completely of base. I have seen no sourse for this question so wil be glad to get your take. It was an assumption not a knowledge.

Yes I am aware that frames dont have singular world lines and I was refering to specific clocks A station and A ship.

Isnt that simply because of the limitations of trying to depict 4 dimensions in a 2 or quasi 3d diagram. In the world aren't we talking about light spheres [not cones] and spheres of simultaneity [not planes] that every point in spacetime has attached???
That for any point or any two points, the intersection of their spheres of simultaneity is not confined to a flat plane but is in fact spread throughout spacetime in all directions ???.
Thanks for your help ,,,I hope I am expressing myself somewhat clearly. [and that you dont run out of patience.]

#### Attached Files:

• ###### Picture.jpg
File size:
27.2 KB
Views:
78
Last edited: Dec 6, 2008