# I Solve the SR 'paradox'

1. Aug 21, 2016

### Xilor

Hi, I found this 'paradox' which supposedly says that SR is incorrect, I have a hunch why it may be a flawed argument, but can't personally get the numbers right. Perhaps it's a fun exercise for some of you to destroy the argument.

Here it is:

So my hunch why this is incorrect:
This is ignoring length contraction, because of it, d will appear differently in the different reference frames, this changes the calculations for the expected times, making the problem disappear. Unfortunately I couldn't get the numbers right myself, giving me different results for t1 and t2 in the different reference frames regardless. So I'm not entirely sure if that's the correct counter.

2. Aug 21, 2016

### GeorgeDishman

First you should clarify that statement, in which frame is $d$ measured, that of the stations or one of the ships? Since each ship sees the other passing locally, I would say $d=0$ in the frame of each ship. I suspect the intention is that it is measured in the station frame but then ship crossings happen at different times as well as at different locations so you also get the question "when is $d$ measured?". The assumption that d is equal for both ships is only valid in the Newtonian case and you not only have different length contraction but the time of crossings being different also introduces relativity of simultaneity which I suspect is what you have missed.

I would always advise getting the back of an envelope and quickly sketching a spacetime diagram, it often reveals what is being missed.

3. Aug 21, 2016

### Xilor

Yeah, d should be in the station frame as it would be like the newtonian version, with perhaps a d1 and a d2 that refer to the distances seen in the reference frames of ships from s1 and s2. These distances could then be the distance that the other, passing ship would have to cross.

But I still can't work it out in that view.
On ship 1 you would say: Okay, I see some distance that ship 2 has to cross while I stand still. It takes ship 2 some time t to reach me according to my clock. The clock on ship 2 must be dilated, so their time t' must be shorter than mine.
Meanwhile on ship 2 you would say the opposite, that the clock of s1 must be dilated, so that the time noted by s1 is shorter.
On both ships you would seem to think that the others time should be shorter. And of course the station frame would have its own prediction.

The relativity of simultaneity road does seem promising in the sense that disagreement could be expected. But that implies that the predictions made on the ships are in fact going to be different from the messages that will be send and are thus wrong.
What about the stations frame? If the stations predictions are also incorrect, then what are the true values? If the station is correct while the ships aren't, then you would have a difference between reference frames and you could determine which of these frames is truly at rest. There's not supposed to be such a thing right?

Last edited: Aug 21, 2016
4. Aug 21, 2016

### Staff: Mentor

And both would be right. The key here is relativity of simultaneity. If both clocks were initially synchronized at noon, and then S1 finds that at the same time that his clock reads 1300 S2's clock reads 1230, S1 will properly conclude that S2's clock is running slow by a factor of two. However, that conclusion is based on S1's finding that "at the same time that S1's clock reads 1300 S2's clock reads 1230", and because of relativity of simultaneity that does not mean that S2 also finds those two events happened at the same time. On the contrary, S2 finds that the event "S2's clock reads 1230" happened at the same as the event "S1's clock reads 1215" and therefore it is S1's clock that running slow by a factor of two.

If you do all the calculations properly, taking into account time dilation, length contraction, and relativity of simultaneity, you will get the same predictions no matter which frame you use to do the calculation. Any apparently contradictory predictions you make will be the result of mixing distances from one frame with times from another, or (and this is very easy to do - it's the key to most SR paradoxes) overlooking relativity of simultaneity and assuming that two things that happened at the same time in one frame are happening at the same time in another frame.

As an aside, relativity of simultaneity could be considered as more "fundamental" than either length contraction or time dilation. In the previous paragraph I explained how time dilation can be inferred from relativity of simultaneity because comparing the rates at which the clock are running depends on comparing their readings "at the same time". Length contraction can also be inferred from relativity of simultaneity because the length of an object is the distance between its endpoints "at the same time".

Last edited: Aug 21, 2016
5. Aug 21, 2016

### Xilor

So this makes a lot of sense as written in other scenarios. However, in this particular scenario, the moment they are comparing clocks is actually the moment they are together in space and time. They could have a nice chat back and forth while the ships pass each other and must agree that that moment is simultaneous between them. The events of disagreement would be the moments they had their previous ship encounters. This seems problematic to me because of the ships you see at rest. Let me explain:

Suppose that ships from s1 and s2 move at the same velocities in the station frame.
Take a large telescope and look down the line of spaceships from your own station that are ahead of you. They are all at rest compared to you, so you trust the rates their clocks tick at. Every time they meet with a ship going in the opposite direction, they announce to everyone how long it took them to meet their last ship. The signals travel at the speed of light, so you see them the same moment that you see them pass a ship in the opposite direction.
You keep writing down the numbers they announce, and of course, these numbers will be the same as the numbers that you are announcing when you pass other ships. 2 hours, 2 hours, 2 hours they all say. The same time that you keep finding between passes. Now you look at one of the ships coming in your direction, at every pass in front of you it will also announce 2 hours as from its perspective it's facing the exact same scenario. Since you were expecting time dilation, something is off. The observed distance between passes here can be ignored at this point because we only need to look at the times.
If dilation has occurred, it must mean that the time the moving ship announced is too long from your perspective, so you must've seen the moving spaceship take less than 2 hours. But, the spaceship at rest compared to you just announced 2 hours! How are these ships at rest noting 2 hours when you must have been seeing it take less time?
Should the ships at rest have been announcing say 1.5 hours instead? If so, then the identical ship that is coming towards you would have also been announcing 1.5 hours!

Relativity of simultaneity is no fix here, because information about simultaneity is accurately provided thanks to the ships at rest. The moment of passing must seem simultaneous for all involved. You will be able to see it through your telescope. Someone on a ship at rest may track the moving ship with their head, which you could see through that telescope. If they do, you would see an image of the ships passing with someone turning their head as it happens. This must mean that light of the ship traveled to his eyes, so to maintain consistent physics you must accept that the observer did in fact see a ship there (assume the observer does not look outside randomly). You then see him press a button to announce the time. All of these events must appear simultaneous to you, and they must have appeared simultaneous to the observer in both the moving ship and the ship at rest (the moving ship also contains an observer who turns their head).
After another pass, you have two events that appeared simultaneous in all reference frames, at known spacetime coordinates (use your clock for time, use parallax for distance). You see the moving ship cross the space part between those coordinates to move towards the event, while the ship at rest is just moving through time, waiting for the event. The moving ship since it is carrying a clock of the same type should be announcing less time passage because of time dilation, but it doesn't, both say '2 hours'.

6. Aug 21, 2016

### Ebeb

Good to mention this, because some people know how to play with length contraction and/or time dilation but never understood relativity of simultaneity.

7. Aug 21, 2016

### Staff: Mentor

It would be more accurate to say that the moment of passing (two ships in the same place at the same time, with the clock on one ship reading whatever it reads and its pilot pushing a button and announcing the time, and the clock on the other ship reading whatever it reads and its pilot pushing a button and announcinf the time) is a single event. An "event" is everything that happens at a particular place in space time, and all observers agree that either the event happened or it didn't - that's the essential requirement for consistency.

Stop. Right. Now.
Using the frame in which the stations are at rest, write down the coordinates (distance, and reading on the clocks of the at-rest ships) of each ship-meets-ship passing event.

Now use the Lorentz transforms to calculate the corresponding coordinates of these events using the frame in which the outgoing stream of ships is at rest. Use the transformed time coordinates to calculate what what announcements the pilots of those ships make.

Finally, use the Lorentz transforms to calculate the corresponding coordinates of these events using the frame in which the other stream of ships is at rest. Use these transformed time coordinates to calculate what what announcements the pilots of those ships make.

You will find that all the the elapsed times between passings are the same for both, just as you are expecting. You will also find that the passing events that are simultaneous using the frame in which the stations are at rest (all the ships in the outgoing stream nicely aligned with ships in the incoming stream at the same time) are not simultaneous using the frames in which either stream of ships is at rest.

.

8. Aug 21, 2016

### Xilor

Ah right, I seem to be messing events and simultaneity up. Thanks! That helps me understand them better.

So I tried for a bit to attempt this again, but it all fell apart pretty quickly. I think I'm probably messing up which results I'm supposed to use at which points. Might try again tomorrow, but for now I think I'll just believe you on this point. Thanks again for your help!

9. Aug 21, 2016

### Staff: Mentor

Getting the numbers right requires a little bit of careful setup.

Using 2D spacetime (t,x) and in units where c=1 and f=1, in the space stations' rest frame the worldline for the n-th ship from s1 is $(t, 2 v (t-n_1))$, and the worldline for the n-th ship from s2 is $(t,r-v(t-n_2))$ where r is the distance between the space station and the other variables are as you gave above.

A little algebra shows that any two given ships meet at $t=\frac{r+(2 n_1+n_2)v}{3v}$. So, for $n_1=0$ and $n_2=0$, they meet at the event $m_{00}=\left( \frac{r}{3v},\frac{2r}{3} \right)$. For $n_1=0$ and $n_2=1$, they meet at $m_{01}=\left( \frac{r+v}{3v},\frac{2(r+v)}{3} \right)$. For $n_1=1$ and $n_2=0$, they meet at the event $m_{10}=\left( \frac{r+2v}{3v},\frac{2(r-v)}{3} \right)$.

Then the time on a s1 ship's clock between meetings can be calculated as the spacetime interval between m00 and m10 which is $\frac{1}{3}\sqrt{1-4v^2}$

And the time on a s2 ship clock is the interval between m00 and m01 which is $\frac{2}{3}\sqrt{1-v^2}$

10. Aug 22, 2016

### Xilor

Yeah this is approximately the part until which I could still seemingly accurately calculate everything. My calculation problems start appearing when I try to calculate the time gien through announcements of s1 from the restframe of s2 (or s2 from s1). I never managed to get this numbers to be the same numbers as those found in the rest frame. The discussion above made me realize I was probably getting the spacetime coordinates of the other ships previous passing event incorrectly because I hadn't accounted for relativity of simultaneity, I had just been plugging in the coordinates found in the rest frame. How to get these coordinates correctly in a ships frame still eludes me.

11. Aug 22, 2016

### GeorgeDishman

Don't try to take shortcuts using length contraction etc., take the station frame coordinates for an event and put them directly into the Lorentz Transforms for your chosen ship. Apply those formulae to two consecutive events for the ship and you'll get the difference in ship time.

Last edited: Aug 22, 2016
12. Aug 22, 2016

### Staff: Mentor

Just use the Lorentz transform. You can either transform the worldline themselves and re calculate their intersections, or you can simply transform the intersection events directly.

13. Aug 22, 2016

### Staff: Mentor

If you have the coordinates of an event in one frame, the Lorentz transforms will give you the coordinates of that event in any other frame.

Thus, the standard way of attacking these problems is to start with a frame in which its easy to find the coordinates and then transform to the frame in which you want the answer. For this problem, it's easy to figure the coordinates of the meeting events in the frame in which the stations are at rest, so we'll start there. If a ship leaves each station every two hours, the ships travel at .5c, the stations are four light-hours apart, and we choose the moment that the lead ship from one station arrives at the other station to be time zero, then we have following events and their coordinates (measuring time in hours and distances in light-hours):
Event 1 with coordinates (x=0,t=0): Ship s2.0 arrives at station s1 as s1.4 leaves station s1.
Event 2 with coordinates (x=1,t=0): Ship s2.1 inbound passes ship s1.3 outbound.
Event 3 with coordinates (x=2,t=0): Ship s2.2 inbound passes ship s1.2 outbound.
Event 4 with coordinates (x=3,t=0): Ship s2.3 inbound passes ship s1.1 outbound.
Event 5 with coordinates (x=4,t=0): Ship s2.4 leaves station s2 as s1.0 arrives at station s2.
Event 6 with coordinates (x=0,t=1): Ship s2.1 arrives at station s1 as s1.5 leaves station s1.
Event 7 with coordinates (x=1,t=1): Ship s2.2 inbound passes ship s1.4 outbound.
Event 8 with coordinates (x=3,t=1): Ship s2.3 inbound passes ship s1.3 outbound.
......
Now we can use the Lorentz transforms to find the coordinates of each of these events in the frame in which the outbound ships from s1 are at rest. For example, event 3 (ship s2.2 inbound passes s1.3 outbound) has coordinates in that frame:
$$x'=\gamma(x-vt)=1.15(x-.5t)=1.15 \\ t'=\gamma(t-vx)=1.15(t-.5x)=-1.15$$
and likewise for the other events. The same thing works to get the coordinates of these events in the frame in which the ships inbound from s2 are at rest, you just have to remember that $v=-5.c$ for that calculation. If you try a few of these conversions yourself, you'll be able to construct the consistent and paradox-free picture of what's going here as viewed from each frame.

14. Aug 22, 2016

### Xilor

So here's what I've tried:
I took as the variables:
x of spacestation 1 = 0
x of spacestation 2 = 5
v1 = 0.5
v2 = 0.5
c= 1
with new ships released at every t % 0.5 = 0

Using a notation eij where e is some event, and i is the number of the ship from station 1 and j is the number from spacestation 2 I picked some random passing events and after using some algebra found their t and x coordinates in the station frame.

(I used: 0.5t - 0.5i for x after solving for t: 0.5t - 0.5i = 5 - 0.5t + 0.5j)

e47 => t = 10.5 x = 3.25
e57 => t = 11 x = 3
e67 => t = 11.5 x = 2.75
e48 => t = 11 x = 3.5
e58 => t = 11.5 x = 3.25
e68 => t = 12 x = 3

We'll be assumed to be sitting on spaceship 6 from station 1 (and thus experience e67 and e68).

We'd like to set the origin point to where e68 takes place. So let's subtract 12 from all t's and 3 from all x's. We also need to account for the fact that spaceship 5 left the station 0.5 earlier and spaceship 4 left 1 earlier. So we need to add 0.5i back to time so that all rest clocks are synchronized with ours.
We get:

e47 => t = -0.5 x = 0.25
e57 => t = -0.5 x = 0
e67 => t = -0.5 x = -0.25
e48 => t = 0 x = 0.5
e58 => t = 0 x = 0.25
e68 => t = 0 x = 0

Then, using the transforms:
t' = gamma*(t-vx)
x' = gamma*(x-vt)
I get these values:

e47 => t' = -0.72169 x' = 0.57735
e57 => t' = -0.57735 x' = 0.288675
e67 => t' = -0.43301 x' = 0
e48 => t' = -0.28868 x' = 0.57735
e58 => t' = -0.14434 x' = 0.288675
e68 => t' = 0 x' = 0

0.43301 is the same number I had gotten by applying time dillation to the difference in time experienced in the station frame. So that seems to check out. Both station and ship anticipate that the ship says 0.43301. (no wonder since it's basically doing the same thing).

The ship coming towards us that will meet us at e68, has just been seen by us at e58.
Now here's where I think I get a strange result. The t' and x' differences, suggest that we must see the ship approach at v = 0.5, however, that is already its speed in the station rest frame. Because we are moving towards it, we shouldve seen it come faster towards us. By relativistically adding velocities I obtained v = 0.8
So that's confusing to me and it implies to me I did something wrong, but let's continue and try for a bit with both v=0.5 and v=0.8

We see that e58->e68 took 0.14434 in our timeframe. Since the approaching spaceship is moving, its clock must be moving slower compared to us, so we expect a lower value on the clock of the incoming spaceship. That is problematic, because the time we found that the time all ships must be announcing is 0.43301, which is already higher than 0.14434. Again, it seems I must have done something wrong, but I don't know what.

Using v = 0.5 calculating the predicted value the ship moving towards us will call out I get these times since these events:
e47 => t = -0.625
e57 => t = -0.5
e67 => t = -0.375
e48 => t= -0.25
e58 => t= -0.125
e68 => t= 0

At least they're nice round numbers. Using v = 0.8 gives me:
e47 => t = -0.43301
e57 => t = -0.34641
e67 => t = -0.25981
e48 => t= -0.17321
e58 => t= -0.0866
e68 => t= 0

So the time between e58 and e68 seems off from the value called out using both velocities. Interestingly enough, I do get the same 0.43301 between e47 and e68. But those are not the events that should be connected here in the same way. Looking back up, the velocity of v=0.8 also makes sense for those events.

Going back and redoing things, it seems that if I don't adjust for the ships leaving the stations at different times, I do get 0.43301 for the last event while using v=0.8, which also checks out in terms of distance and time travelled. So this would've been the expected answer, so I suppose I shouldn't have been adjusting for that. Why not though? This would just be adjusting for the distance and clock difference between me and the ships at rest. I may be able to see when an event happens in my time, but through adjustment the ships at rest are able to tell me when it really happened according to my clock.

This discrepancy between adjusting and not adjusting appears to imply to me that you could tell the actual velocity of an incoming object and through that your own velocity.
If you don't adjust for the clocks at rest and will judge velocities based on the time you see the passing moment, you'll see the relativistic velocity 0.8 pop out and the incoming ship announces the expected number.
If you do adjust, and judge velocities based on the time the ships at rest tell you the event occurred at, then you'll see the velocity of 0.5 pop out, the one we started with in the station frame. Here, the incoming ship gives an unexpected answer.
With information of both of these velocities, you could tell your own velocity in the station frame.
Now that shouldn't be right as the station frame is 'arbitrarily chosen', so I must still be doing/thinking something wrong. (I checked it with other velocities too just to be sure it wasn't because of the rather nice number 0.5, and got the same results.)

Edit: Weird, it seems to not even require to listen to the clock of those at rest. If you're at eij, then do using your time/space coords of events: (ti(j-1) - t(i-1)j)/x(i-1)j
This somehow always gives you your velocity in the spacestation frame. How does that work?

Last edited: Aug 22, 2016
15. Aug 22, 2016

### GeorgeDishman

Was that a typo, in your first post you said:
Did you mean:
v1 = 0.50
v2 = 0.25

16. Aug 22, 2016

### Xilor

Ah no I switched to both at 0.5 at some point during this thread as it didn't seem to matter so much. Also repeated the same with 0.2 and 0.4 though, giving different numbers obviously but with concenptually similar results. I ended up building an excel sheet out of the calculation so it's easy to change the variables and steps of the calculation.

17. Aug 22, 2016

### GeorgeDishman

Oh, OK. It'll mean the crossing happen at the same time in the station frame which may hide some effects but if you're doing a spreadsheet, you can try a variety of speeds :-)

18. Aug 24, 2016

### Staff: Mentor

I am getting different numbers than you. I get e48 is (t,x)=(8,3). Since they leave every 0.5, then ship 4 will leave station 1 at t=2, and ship 8 will leave station 2 at t=4. By t=8 ship 4 from station 1 will have gone a distance of 3 at 0.5 c, and ship 8 from station 2 will have gone a distance of 2 at 0.5 c, so they will meet.

Similarly e68 should be at (t,x)=(8.5,2.75)

I wouldn't do this. Doing boosts as well as doing translations is just going to cause confusion. I would just leave the coordinates as they are, and calculate proper times from that.

19. Aug 24, 2016

### Xilor

Yeah I ended up finding the same later. I had made an error adjusting properly for the later leaving times. Your version is indeed correct.
Using the corrected method I obtained:

e47 t =7.75, x=2.875
e57 t =8 x= 2.75
e67 t =8.25 x= 2.625
e48 t =8 x= 3
e58 t =8.25 x= 2.875
e68 t =8.5 x= 2.75

Yeah later on in the same post I discovered that this did indeed lead to problems. Using the fixed version the problems disappeared.

No paradoxes appeared after having done the Lorentz transforms correctly as you guys predicted correctly.
All ships that appear at rest call out the same number as you do. And the ships coming towards you will call out that same number under two conditions: their velocity in the station frame is equal to yours and the two stations are outputting ships at the same frequencies.
All the numbers that are called out will be successfully predicted in all the reference frames.
(with v1=0.5,v2=0.5,c=1, outputEveryT=0.5, I got 0.216506 as the value everyone calls out)

The key was indeed in relativity of simultaneity. You will not think that the passing event of the ship at rest in front of you, happens at the same moment as your passing event.
Thank you all for your help, that was a good learning experience.

After messing a little with the variables involved, the earlier mentioned magic formula became less magical as well. I was missing two aspects in it, c and the ratio between output frequencies of the stations. Because the latter is anchored in the station frame, it makes sense why something from the station frame would pop out.

The formula ends up being:
v = ((t1f-t2)*c2)/x

With variables:
v = your speed in the station frame
t1 = time of your last meeting (in your frame)
t2 = time of the last meeting of the ship appearing at rest in front of you (in your frame)
c = speed of light
x = location of the last meeting of the ship appearing at rest in front of you (in your frame)
f = frequency ratio between station outputs with f = s2/s1, with s1=frequency of your station, s2=frequency of the other station (in the station frame)

Apply this with x=0, and t=0 set at the moment and place you pass a ship.
It still works if the approaching object moves at c, but not if you do. It works with all other combinations of velocities wherein passing moments can happen. Stuff like the positions of the stations etc don't matter.

Still seems like an amusing little thing. Not a clue why it works. Especially the interaction between t1f - t2 seems odd.

Last edited: Aug 24, 2016
20. Aug 24, 2016

### Staff: Mentor

Good insight. That is exactly correct. The station frame is not "preferred" in any physical sense, but it happens to be the unique frame where the launches are synchronized. So it will pop out on occasion.

21. Aug 24, 2016

### Xilor

It would seem the link must be somewhere in the Doppler effect. Time dilation would change the frequencies, but not the ratios. It probably accidentally solves for the v in there