# 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.

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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.]

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21. Dec 6, 2008

### JesseM

Are you talking about how things look in the ship's frame, where each station's clock is out-of-sync with the next one's?
Sure, that's just the twin paradox.
Because if the ship turns around it's no longer symmetric--the Alpha station moved inertially throughout the journey but the ship accelerated to turn around. Standard SR relations such as time dilation as a function of velocity only work in inertial frames. Also, note that the definition of simultaneity in the ship's inertial rest frame right before the turnaround would be very different than the definition of simultaneity in the ship's inertial rest frame right after the turnaround, so that a twin on a distant station might be aged 35 simultaneously with the ship just beginning the turnaround in the ship's instantaneous rest frame at that moment, but the same twin might be aged 45 simultaneously with the end of the turnaround in the ship's instantaneous rest frame at that moment (a different inertial frame with a different definition of simultaneity), even if the ship experienced the turnaround period as very brief.
Yes, again, that's the twin paradox. I recommend reading the link I posted above if you're not familiar with it.
If in the station frame it's true that the E ship's clock reading 0 and the A station's clock reading 0 happen at the same time-coordinate (happen simultaneously in that frame), and it's also true that in the station frame the A station's clock reading 0 happens at the same time-coordinate as the E station's clock reading 0, then of course it must be true that in the station frame the event of the E station's clock reading 0 happens at the same time-coordinate as the event of the E ship's clock reading 0--simultaneity in a single frame certainly obeys the transitive property! Of course the E ship's clock will immediately start to go out of sync with the station clocks after this since it's ticking slower, but at that single time coordinate its clock reads the same time as the time on the station clocks, in the station frame.
In the station frame, yes. Of course this is only because we chose to set the E ship's clock to 0 at the moment it passed the A station and the A station's clock read 0. The choice of where to set 0 on any clock is arbitrary, we could also have set things so that the A ship clock read 0 at the same moment the A station clock read 0 in the station frame if we wanted to, but if the A ship and the E ship have synchronized clocks in their own rest frame, this would mean that the E ship clock read -3.2 years when the A ship clock and the A station clock read 0, in the station frame.
Again, simultaneity in a single frame obeys the transitive property. If event A and event B happen at the same time coordinate in my frame, and event B and event C happen at the same time coordinate in my frame, then of course event A and event C must have happened at the same time coordinate in my frame. So if you wanted the E ship to set its clock to 0 at the moment it was next to the A station and the station's clock was also reading 0, then this necessarily means that in the station frame, the E station's clock read 0 at the same moment the E ship's clock read 0. But if you wanted the E ship to zero its clock at some other moment, or you didn't want the A station's clock to read zero at the moment the E ship passed it, then these things would no longer be true (tell me what you want the A station clock and the E ship clock to read at the moment they pass next to each other, or what they each read in the station frame when the E ship is a certain specified distance from the A station, and I can rework the example with these numbers, if you like).
Sure, in the station frame, at time-coordinate t=0 years the E ship's clock read 0 years while the A station's clock read 3.2 years. Those were their starting points at t=0 years in this frame. Then at time t=3 years, each clock has advanced forward by 0.6*3 = 1.8 years in this frame, since each is slowed down by a factor of 0.6. So, at t=3 years in the station frame, the E ship's clock reads 1.8 years while the A ship's clock reads 5 years. In general, if you have an inertial clock moving at speed v and you know its reading T at some initial time-coordinate t0 in your frame, and you want to know what it'll read after a time interval of $$\Delta t$$ in your coordinate system, the answer will be that the clock reads $$T + \Delta t * \sqrt{1 - v^2/c^2}$$ at time coordinate $$t0 + \Delta t$$ in your frame.
In what coordinate system would it be a sphere? It's assumed in spacetime diagrams that whatever coordinate system you're using, you graph it with the spatial axes and the time axis all at right angles to one another, just like you would graph an equal number of spatial dimensions (so if you're drawing two spatial dimensions x,y and a time dimension t, it looks just like a graph of 3D euclidean space with spatial dimensions x,y,z, except that you replace the z-axis with a t-axis). Necessarily a surface of constant t will just look like a horizontal line (if you are using 1 space dimension), a flat plane with zero slope (if you are using 2 space dimensions), or a flat 3D "hyperplane" with zero slope (if you are using 3 space dimensions). Then if you graph the surface of simultaneity for a different inertial frame on this same graph, it will look like a slanted line, a slanted plane, or a slanted hyperplane.

So, I don't know where you're getting the sphere from. What would be a sphere is if I looked at the set of events at some past time-coordinate whose light is just reaching me at this moment--this would be a 3D slice of a 4D light cone, just like if you take a horizontal 2D slice through a 3D cone you get a 2D circle. If we "synchronized" clocks in a way that didn't compensate for light delays--if a reference clock sent out a signal when it read 0, and every distant clock set itself to 0 when it received the signal, instead of setting itself to 0 + (its own distance from the reference clock)/c as in Einstein's synchronization convention--in that case a surface of simultaneity would be a 4D light cone with spherical intersections, at least if you graphed this odd simultaneity convention on a normal spacetime graph whose own definition of t-coordinate is defined by clocks synchronized using the Einstein synchronization convention.
Again, if you want to use all three space dimensions, a surface of simultaneity is a 3D "hyperplane" in a larger 4D spacetime. This is impossible to visualize, but we don't lose anything important if we imagine a universe with only 2 spatial dimensions (like in the story Flatland), in which case we can think of an ordinary 3D graph except with the third dimension standing for the time dimension.

22. Dec 9, 2008

### Austin0

Hi Well I finally figured out and saw the symmetry that was eluding me. My original setup with just two ships made it confusing when the E ship and A station were both t =0.
When I extrapolated in both directions added a beta ship and station, I got it. Each succeding ship agrees with the next station in line and then falls behind 2 yrs at each station after that and from the stations perspective, each succeeding ship is running 2 more years ahead of the station than the last one. And the ships and the stations agree respectively on the elapsed time of a station passing between ships or a ship passing between stations. SO actually my original assumption was correct in that in this paradigm there is no net time dilation in either direction but I didnt factor in desynchronization as far as actual ,current time ,comparison at coincidence.
So I can stop bugging you about that issue. Buuuuuut
Now that I have seen the scnario with actual figures there are new questions.
I have not sat down and worked out the whole problem with jtbell's figures because the way he approached it was not as symmetrical as your method [he used a different rest spacing for the ships] but I am curious and will do so because I dont understand how it worked out that you both arrived at the same times for the station encounters using different perameters to work with.
Secondly I have seen many twins problems including some threads that covered the turn around question. That was actually my motivation for originally posting this question.
It seems to me that acceleration is an area of some ambiguity and lack of concensus as to how it fits in, what are its effects if any etc etc.

SO I see that dilation is symmetric as far as it applies between frames but now I am unsure what is the reality between individual ships and stations. The figures seem to indicate that E ship for instance would accrue a large discrepancy between it and E station if it continued its journey along that path.
So is this simply a matter of asynchronicity?
This does not make real sense because in each locality it is the actual passage of time.
SO is this somehow , not real , unless the ship decides to turn around ,which of course requires acceleration . Does this action of relatively short temporal duration then magically make the apparent clock difference an actual time difference???
From what I know the empirical findings so far indicate that there is no detectable dilation effect purely from acceleration.
So this then is another mystery. The one that did not seem to be resolved in the twins case.
Do you have any thoughts on the Halafe tests which seem to indicate time dilation due to velocity as opposed to gravitation????
I didnt say a coordinate system. I said in the world. I take a very literal view of SR.
I think it is a fundamental description of the actual physical fabric of reality. It seems that there are two prevailing views in this regard. One is that it is a system of coordinates for translation between moving systems and the things described are only relative and cannot even be discussed on a level of being real . Others seem to think that it is relative , in that the effects of motion are not subject to determination within a system or between systems , but it is at the same time real in that ultimately the effects [time dilation etc.] are the result of real physics in action , cause and effect. In that context I personally think that simultaneity is at the root of all. That it is not just a matter of the perception of the temporal relationship of spacially removed events but is the actual physical interconnectedness of matter.
So a local points light sphere is one half of its relationship with the world and the spheres of simultaneity are the worlds relationship with it.
I was not suggesting that there was a better system than Minkowskis I was simply mentioning that ANY coordinate system that tries to cram 4 dimensions into a 2 d representation is neccesarily limited. In this paradigm a light cone is a series of circular slices that represent the propagation of light through time, no? But those 2 d slices actually represent spheres correct???

23. Dec 13, 2008

### JesseM

Sorry I took a while to get back to you on this...
No, there is no lack of consensus about how acceleration works in SR. What aspects of acceleration are you confused about?
It all depends on what frame you use. You could pick an inertial frame whose velocity is exactly midway between that of the E station and the E ship, for example; in this frame the E ship and E station have equal speeds, so there clocks tick at the same rates. Whatever frame you're using, the clock that has the higher speed will be running slower, by the same factor of $$\sqrt{1 - v^2/c^2}$$ (assuming we are using an inertial frame).
What is the actual passage of time? In relativity there is no absolute truth about which of two clocks is running slower at any given moment, if that's what you mean--different frames will disagree on this point, yet all frames always make the same predictions about local events like what two clocks read at the moment they meet at a single location in space. You can see an illustration of this I drew up a while ago on this thread.
No, there are no "actual" facts about the time difference between two clocks until those clocks meet at a single location in space. Only these sorts of local differences have objective, frame-independent truths about them in relativity.

I find it is easiest to think of spacetime in terms of a sort of "geometric analogy". If you draw two dots on a 2D plane, and you draw two paths between them, one of which is a straight path and one of which is a path with a bend in it, then this is enough to guarantee that the straight path will be the shorter one, since a straight line is always the shortest distance between two points. If you draw a cartesian coordinate system on this plane, then you can define the slope at each point on a given path, and you will find that the straight-line path has constant slope while the path with a bend has a slope that changes from one part to another. This is closely analogous to the fact that if you have two paths through spacetime that meet at two different events, the path with constant velocity will always have elapsed less proper time between the events than the path with changing velocity. I expanded on this analogy a bit in this post if you want to look at that.
No, just like if you're driving a car along with a path with a bend in it, your odometer doesn't suddenly increase your mileage suddenly when you come to the bend; nevertheless, if you're driving between point A and point B via a path with a bend, and I'm driving between point A and point B via a straight-line path, my odometer will have accumulated fewer miles when we both reach point B than yours. The bend is enough to guarantee that your path will be longer, but the extra distance is not all accumulated as you drive the bendy part, it's a geometric property of the whole path.
This is not really much different than the twin paradox, in which any inertial frame can calculate the time accumulated on each twin's clock just by knowing their velocity as a function of time v(t) in that frame. Despite the fact that different frames disagree on the exact function v(t) they all agree on the accumulated time when the twins reunite.
How can you talk about the "shape" of anything in spacetime without making a spatial diagram with the time axis represented by a particular direction in the diagram? And to do this, of course you need a coordinate system.
These words don't really seem meaningful to me, they're too vague. If I draw two paths in 2D space and one has a bend in it, would you say that the bend is the "cause" of that path having a greater length? And when you say "the things described are only relative", what "things" specifically? In relativity some facts are frame-independent and therefore objective, like the total time accumulated on a clock between two events on its worldline (like the event of departing from a twin clock and the event of reuniting with the same clock). Other facts are frame-dependent and unless you discard relativity there can be no objective physical truth about them, like which of two objects has a greater velocity at a particular moment, or which of two clocks is ticking slower at a particular moment.
If you're saying you think there's an objective truth about whether two events at different points in space are simultaneous, and you think that there would be some experimental way to determine this truth (as opposed to it being a purely metaphysical truth that we could never know), then you are saying the theory of relativity is wrong, period. And the theory of relativity could only be wrong if some of the fundamental laws of physics were not invariant under the Lorentz transformation, but so far all of the fundamental laws that physicists have found obey equations that are Lorentz-invariant.
I still don't know what you mean by "spheres of simultaneity".
You are free to graph spacetime coordinates in more than 2 spatial dimensions. You could use a 3D euclidean space where the x and y axes represent spatial axes and the z axis represents a time axis, for example. You could even use a 4D euclidean space--this could not be directly visualized, but we can certainly talk about the geometry of 4D space using mathematics. Surfaces of simultaneity graphed in this space would still definitely be 3D hyperplanes, not hyperspheres.
Yes, but light cones are totally different than surfaces of simultaneity. Again, the moment that I receive light from some distant event A is not defined to be "simultaneous" with event A in my rest frame (or in any other inertial frame for that matter).
I don't quite follow--if the spheres are concentric like layers of an onion, and they're expanding outwards with me outside them, then they will not all hit me simultaneously--first the outermost one will hit me, then the next one in will hit me, then the next one in, and so forth. Perhaps you are instead talking about me being at the center of a bunch of spheres, and each sphere emitting light inwards at different times so that all their light hits me simultaneously--the sphere with a radius of 5 light-seconds would have emitted its light 5 seconds ago, the sphere with a radius of 10 light-seconds would have emitted its light 10 seconds ago, and so forth. Is that what you mean? In this case each sphere is just a different cross-section of my past light cone. However, although the light from all events which lie on my past light cone will reach me simultaneously, the events themselves did not happen simultaneously in my frame--the sphere with a radius of 10 light-seconds emitted its light at a time 5 seconds earlier in my frame than the sphere with a radius of 5 light-seconds emitted its own light, for example.
If you're talking about spheres representing different cross-sections of the observer's past light cone as I suggested above, then no matter what frame you use, each sphere will have the same separation from the next largest sphere in all directions, it won't be closer to one side of the next largest sphere than another (this is because no matter what frame you use, light moves at the same speed in all directions in that frame). But maybe this isn't what you mean, I don't know. Perhaps you draw a diagram of what you're talking about and post it as an attachment?
A 4-dimensional graph with 3 axes representing space and the fourth representing time will be directly analogous to a 3-dimensional graph with 2 axes representing space dimensions and the third representing time--you don't lose anything important by dropping the dimensions by 1, you just have circles as cross-sections of 3D light cones instead of spheres as cross-sections of 4D light cones, and so forth.
That diagram seemed to be based on some different assumptions about clock synchronization than the one I was using in my example (I assumed the E ship and the A station both had their clocks reading 0 at the moment they passed next to each other, you seemed to assume something different). Did you have a specific question about how the situation would look when diagrammed?