1. Jul 21, 2004

### µ³

Suppose I am sitting around floating in the middle of intergalactic space minding my own business, when all of a sudden WOOSH a really weird massive ellipsoid thing starts rushing by me at .99c (where relative gamma is about 7). The thing appears to be 100,000/7 light years across (take the length of our galaxy and divide it by 7 to get the relativistic contraction). Simultaneously, the inhabitants of the galaxy are going "WTF? that flat-person-looking thing is going at .99c." (According to Einstein no inertial reference frame is better than an other so both viewpoints are equally valid as far as each of them can tell). Now let's say that each inhabitant of this galaxy has a watch and all of their watches are synchronized to the exact same time (assume they are all in the same inertial reference with each other). So, as I'm passing the first one I see by (close enough so that time between messages is neglible), I take my flashlight out and morse-code really quickly "sup d00d, what time you got". Immediately, he replies "I got t=0", so I synchronize my watch to t = 0 years. So, after 100,000/7 years (The galaxy is practically travelling at c so it should take 100,000/7 years for it to completely pass by me), I decide to take out my flashlights and talk to one of the inhabitants again. I figure since the galaxy is passing by me at .99c, time in the galaxy is going by 7 times slower than my time, 100,000/49 years should have passed. The inhabitant sees me rushing by not long after hearing the news from the first inhabitant through their intergalactic morse-code network. He figures that since his galaxy is 100,000 light-years across, the message must have been sent at t = 0, and the traveller(me) took a little more than 100,000 years to traverse the galaxy. He decides he wants to be smart and realizes that time must have been going 7x as slow for me as it was for him so he morse-codes me "Hey your clock reads t = 100,000/7 years, right?" I also wanted to be smart and realized that time for him must have been going 7 times slower than it was for me so I morse-code him "Yup, and your clock reads t = 100,000/49, right?". "Wrong, it reads t = 100,000." He replied.

So what was wrong with this?

2. Jul 21, 2004

### VantagePoint72

3. Jul 21, 2004

### Janus

Staff Emeritus
What is wrong is that you would not say that his clock reads t =100,000/49, but t= 100,000(just like he says it does). You've neglected Relativity of simultaneity. While all the clock in the galaxy are synchronized according to the galaxy, they will not be synchronized according to you. So when you pass the first person and he tells you that his clock reads t=0, at that instant, according to you, the second person's clock reads t= 100,000 - 100,000/49.

To understand the reason for this, consider how the clocks in the galaxy are synchronized in the first place. The best way would be for a signal to be sent out from the center and for each clock to set its time depending on its distance form the center.

Assume that your central clock sends a signal thart says "it is now 00:00 , day 1, year 1. A clock sitting 1 lightyear from the center receives this signal, and noting that the signal took 1 year to reach it, determines that the central clock at that moment reads 00:00, day 1, year 2, and sets its clock to that value to sync itself with the central clock. Likewise, the clocks sitting at opposite ends of the galaxy would set their clocks to 00:00, day 1 year 50,000 when they receive the signal so that they will be in sync.

Now let's see how the same sequence appears to to you (assume you are watching in a very powerful telescope.) Remember that c must remain constant relative to you. Notice that since the galaxy is moving at .99c realtive to you, the the relative speed between the signal sent from the central clock and the first person you encounter is .1 c. Meaning that according to your clock it will take 50,000/7/.1 = 71428.571 years for the signal to reach this person. At which point, this person sets his clock to read 00:00, day 1, year 50,000. But the central clock, according to you, at this time reads 19:06, day 29, year 10204(71428.571 years divided by a time factor of 7) so the central clock and the first person's clocks are not sychronized.

Also, the signal traveling to the second person will have a velocity relative to the galaxy of 1.99c, and thus will only take 50,000/7/1.99 = 3589.3754 years by your clock to reach person two and for him to set his clock to 00:00, day 1, year 50,000. At which time,according to you, the central clock would read 07:00, day 280 ,year 512. Person 2 and the central clock are not in sync nor are person 1 and person 2.

If you work things out with the proper velocities and Lorentz transformations, you will find that all the observers will agree as what each other's clocks read when they pass each other (they won't agree however, when each respective clock started)

Last edited: Jul 21, 2004
4. Jul 22, 2004

### µ³

If as the traveler passed the first inhabitant (t=0) a message from the second inhabitant (at the end of the galaxy) arrives giving the time when he sent it: t = -100,000, the traveler would figure that since the galaxy is 100,000/7 light-years across and the second inhabitant is experiencing time dialation by a factor of 7 the current time of the second inhabitant must be t=-100,000 + 100,000/49. However, the galaxy's inhabitant figures that since the galaxy is 100,000 light-years across, the current time of the second inhabitant must be t = 0, which confirms that their clocks are synchronized (in their reference frame). So after the galaxy bypasses him after 100,000/7 years, another 100,000/7 years have gone by in the reference of the traveler and another 100,000/49 years, in reference to the second inhabitant/galaxy, which would put the time of the second inhabitant to be t = -100,000 + 100,000/7, which does not equal 100,000. (How could something have a relative velocity of 1.99c I thought when you relativistically add velocities they can never be greater than c)

Last edited: Jul 22, 2004
5. Jul 22, 2004

### VantagePoint72

The last part of your post is easy to resolve. In Einstein's universe one plus one no longer equals two. The proper way to add velocities is:

w = (u + v)/(1 + uv/c^2)

where w is the total velocity and u and v are velocities in which the positive direction for each velocity point in opposite directions.
If u and v are small than the effect of the denominator is extremely tiny, which is why classical velocity addition works to high precision at common speeds. Add your velocities with that method and see if it resolves your problem with Janus' post.

Last edited: Jul 22, 2004
6. Jul 22, 2004

### Janus

Staff Emeritus
No, remember the traveler measures the speed of the signal from the second inhabitant to the first inhabitant as traveling at c relative to himself and not at c relative to the galaxy. IOW, When the traveler 'backtrack"s the signal, he figures that by his clock it took 1,412429 years to leave inhabitant 1 and reach inhabitant 2.
Look at it this way: The signal leaves #2 at c traveling towards #1, according to the traveler. After 100,000/7 years, it is 100,000/7 lightyears from where it started. But during this time #1 and #2 are moving at .99c in the same direction as the signal. So when it gets there, #1 will have moved 99,000/7 lightyears away from that spot and #2 will have moved 99,000/7 lightyears closer to that same spot. The signal continues to chase after #1 until it finally catches him after 142429(100,000/7/.01) years. This would amount to 198,933 years on #1's clock due to time dilation(using 7 for the factor actually gives you an answer of 204081, but the actual Lorentz factor is closer to 7.09. Using it gives the more accurate answer I gave . Thus the traveler must assume that the present time for #2 (when #1 receives the signal) is t= -100,000+ 198,933 = 98933 years.

Using the more accurate factor, 100,000/7.09 years pass for the traveler as the galaxy passes, so 100,000/50.2681 = 1989 years pass for inhabitant 2, thus inhabitant 2's clock reads 98933+ 1989 = 100,922 years according to the traveler. Now this is more than 100,000, But, that 100,000 figure was gotten by rounding the .99c relative velocity of the traveler to galaxy up to c. If we don't do this, we find that inhabitant 2 measures that it takes 100,000/.99 = 101,010 years for the traveler to cross the galaxy. (the 88 year difference is caused by the fact that even 7.09 isn't the exact factor, 7.08881205 would be even more accurate.)
So yes, the traveler and inhabitant 2 both agree as the the time reading on inhabitant 2's clock as they pass each other.
Two objects cannot have a velocity greater than c relative to each other as measured from either object, But that does not mean that they cannot have a greater than c velocity relative to each other as measured by someone else.

For instance, if you measure an object as moving at .9 c away from you in one direction, and a second object moving away from you in the opposite direction also at .9c, their relative velocity to each other, according to you, will be 1.8c, even though according to them their relative velocity will only be 0.9945c.

Last edited: Jul 22, 2004
7. Jul 22, 2004

### µ³

Ahhh, makes perfect sense now, thanks for that. Last question:
if the Earth and a traveler are sitting in universe with a 4-hypersphere geometry (going in a straight direction for a long enough amount of time you will eventually arrive at the same place you started), or there are wormholes far apart in the universe. Assume the traveling flies past the Earth at some time t, with some speed v (or the Earth flies past the traveler). After circumnavigation of the universe who will experience time dialation?

8. Jul 23, 2004

### VantagePoint72

In the wormhole situation, it depends on how the wormhole is used.

If the observer flying by the galaxy is carrying one end of a wormhole while the other observer holds the other end, neither will experience time dilation. The wormhole will act as bridge through space and time. Picture this: Itchy and Scratchy each have one end of a wormhole. Itchy gets into a rocket, blasts off and takes his end of the wormhole with him, leaving Scratchy behind on the earth. Itchy travels at 0.999999999c (it's important that he had to accelerate to reach this speed), giving a relativistic gamma of about 22,361. He travels for 2 hours by his watch (though due to length contraction/time dilation, he's able to travel significantly far into the depths of the galaxy), meanwhile Scratchy watches him through the wormhole the entire time. Should Itchy turn around and head back to earth at the same speed, he would arrive over 3700 years into the future (remember, he's established himself as being the one who should experience time dilation since he accelerated initially). However, since the wormhole is a single entity and different parts of it can't age at different rates, when Itchy looks back through the wormhole, he sees Scratchy also having aged 2 hours, he's looking into the past. If he steps through the wormhole, Itchy and Scratchy will be in the same reference frame and both will have aged the same: 2 hours.

If Itchy isn't carrying a wormhole with him, but just happens to find the mouth of one after his 2 hours and realizes that the other end is near the earth, he'll still undergo time dilation since he underwent initial acceleration while scratchy didn't and has to undergo negative acceleration to come to rest with Scratchy after passing through the wormhole. Since the mouths of the wormholes were never in motion relative to each other, there is no time gap, so if Itchy steps through the wormhole, the mouth by the earth will be in same time period. So, since due to his inevitable acceleration Itchy caused himself to undergo "true" time dilation while travelling at 0.999999999c, Itchy will have aged 2 hours while Scratchy will have aged about 1800 years. All that Itchy succeeds in doing when stepping through the wormhole is avoiding further time dilation by side stepping another 2 hour journey at 0.999999999c. Due to his original acceleration and final negative acceleration, Itchy still has underwent time dilation when reunited with Scratchy.

The same ideas hold for the 4-hypersphere Universe. One observer will still experience time dilation as he still had to accelerate to reach the velocity he travels at and even if he never has to turn around he still has to undergo negative acceleration in order to stop after ending up at the earth again. Since spacetime is the benchmark for acceleration (Itchy can feel it when he accelerates) he can't say it was really the other observer who was accelerating. Even if the two observers are born into a state of uniform motion relative to each other, someone will still have undergo negative acceleration to stop. Acceleration breaks the symmetry between the two observers, and only if there is something to act as a bridge over time as in the first example, one observer will still be seen to have underwent time dilation when reunited with the other observer.

So, it's been a roundabout way of answering it but as for the question at the end of your last post: the person you call the traveller will have aged less since after circumnavigating the Universe he had to undergo negative acceleration, something he could feel, in order to come to rest with the earth.

9. Jul 23, 2004

### µ³

What if he never comes to rest but he merely observes the Earth as he passes by (or it passes by him) every time? Also, how would he feel acceleration if he is merely traveling in a straight line (along one of the universes' great circles) not changing velocity? Consider the "traveller" synchronizes clock with the Earth as they pass each other what will the times read (i.e. whose will be slow and whose will be fast) on the 2nd, 3rd, etc... pass?

10. Jul 23, 2004

### VantagePoint72

Look back to one of my previous posts. As long as they are in uniform motion, relative to each other each will perceive the other to be experiencing time dilation and there will be know way to see who "right" (because they both are until the come to rest with each other). Whatever information one tries to communicate to the other, it will always agree with the recipients observations. Again, see my first post. Even if someone is traversing the curvature of the universe, the only way that the two observers can re-enter the same reference frame is if one person applies a force in order to undergo the negative acceleration required to bring two observers in motion relative to each other to rest. Otherwise, he'll just pass be the earth and what I said the beginning of this post applies. Until two observers come to rest with each other, they will each have conflicting observations over the rate that each is experiencing the passage of time, and any attempt to communicate their own observations will be affected by such things as the relativistic doppler effect and the recipient will receive information that confirms their own observations. Also, in order for the two observers to come to rest, some form of acceleration must be engaged, regardless of a loop in space or wormholes, since it's required to bring the two observers to rest relative to each other. While in uniform motion, observers are symmetric and will disagree on observations. Once at rest, the symmetry is broken by the required acceleration and both observers will agree on who aged less.

11. Jul 25, 2004

### µ³

This is still confusing, although the scenario is very similar to the galaxy one.

For the sake of simplicity, assume that instead of a 4-hypersphere universe, there is just an infinite number of Earths at regularly spaced intervals. Now, every object in the universe, in motion or not, can send a light signal in any direction and receive it in the opposite one in 100K years so everyone agrees that the universe has a circumference of 100K LY. As Earth1 passes me (the Earth for the first time) I synchronize my watch with it at t=0. Soonly thereafter, I receive a signal from Earth2 sent in the opposite direction I am traveling saying "t=-100,000." Since the Earth is travelling at .99c, the distance between itself should contract by 7.0889. As such, the current time (when t=0 for me) in Earth2 is -100,000 + -100,000/7.08992/.01 (taken into account length contraction, time dialation, and the relative speed between the signal and Earth1). I wait for the Earth2 to pass by me which should take another 100000/7.0899/.99 years (my time) and -100,000/7.08992/.99 Earth-time afterwhich, my clock will read t = 100000/7.0899/.99 = 14,249 and the Earth's will read t= -100,000 + -100,000/7.08992/.01 + -100,000/7.08992/.99 = 101,010 years.
Problems with this:
1) The Earth's time is not synchronized with itself (I can live with this, it just appears that it isn't when measured at relativistic speeds)
2) Suppose I sent out a signal when Earth1 passed then that means I would see Earths(2-7) before I receive the signal again!
3) The faster something goes, the less time it takes to circumnavigate the universe from my perspective, often much less time than it would take light to. (e.g. something travelling at .9999c I would see circumnavigate the universe 70 times for the same time it took light to once)
4) It's not symmetric. Replace "Earth" with "me" and vice versa and you get a conflicting scenario.

Where did I mess up?

Last edited: Jul 25, 2004
12. Jul 25, 2004