A Special Relativity Question

Q

QuarkMoon

I
Registered Senior Member
As we all know, Einstein's theory of Special Relativity asserts that when you travel at a certain fraction of the speed of light, time is warped for each individual observer; however, neither observer's point of view is more valid than the other.

That brings me to a scenario that is repeated in many works dealing with Special Relativity, including the book Fabric of the Cosmos by Brian Greene.

Scientist A and scientist B stand at each opposite end of a boxcar, traveling at a fraction of the speed of light; scientist A stands at the back while scientist B stands at the front. They are attempting to synchronize their watches by lighting a pile of gunpowder in the center of the boxcar and setting their watches as soon as they see the flash of light. Scientist C pours out the gunpowder and gets ready to light it.

Meanwhile, scientists D and E are the stationary observers, standing on a nearby platform; their jobs are to look in the boxcar as it races down the tracks at a fraction of the speed of light and see if scientists A and B properly synchronized their watches.

Now, Special Relativity asserts that when this experiment is carried out, scientists A and B will conclude that their watches are synchronized, and scientist C will confirm their conclusion. They all agree because they were all moving at an equal velocity relative to each other. However, scientists D and E will disagree with those findings. Scientists D and E will conclude that scientist A's watch is faster then scientist B's, because scientist A was traveling toward the light emitted by the gunpowder and scientist B was moving away from; concluding that the light took longer to reach scientist B. Special Relativity also asserts that all of the scientist's observations and conclusions are justified, both the platform observers and the boxcar observers are correct.

And that's usually where the experiment ends. But I kept thinking a little bit further; what about when scientists A, B, C, D, and E all meet up the next night for some drinks, and scientist D (one of the platform observers), who is feeling a little tired and wants to know how late it is, asks scientist A and B for the time? Assuming that scientists A and B never changed their watches, you would conclude that scientist A's watch must still be faster then scientist B's. However, since everyone is now moving at an equal velocity relative to each other, a logical paradox arises. Scientists A and B will claim that their watches both show 9:00PM, with scientist C agreeing with their claims; but when scientists D and E (the platform observers) lean over to see for themselves, they will see that scientist A's watch shows 9:01PM, and scientist B's watch shows 9:00PM!

Now, soak that in and then flip it around. When scientist A, B, and C take a look at the watches of scientist D and E, their watches will show a time that is slower than their own. For example, they will claim their watches show 8:59PM, while the boxcar observers claim their watches show 8:58PM.

This has been killing me, because now all of the observers are moving relative to each other, and yet they are seeing completely different things when comparing their watches. According to scientists D and E; scientists A, B, and C have somehow lost their original places in time. And on the flip side, according to scientists A; B; and C; scientist D and E have lost their places in time. Someone has traveled through time. But who? And I thought time travel (at least time travel to the past) was impossible? What has just happened?

Also, since all observers are now moving at an equal velocity relative to each other, the fact that they are seeing different things kind of violates the idea that the Universe, which includes time, is balanced. If these scientists have all skipped to different moments in time, then something, somewhere must have filled in the gaps and brought balance to the Universe.
 
Here's some of the solutions I tried to come up with:

The "time travel" that has taken place is real, relativity is leads to real consequences. The scientists are now physically aging at a different rate, the cells in their bodies are dividing along a different timeline. Which means in order for them to synchronize their journey through time, they would need to perform the experiment again but in reverse; in essence, they would need to "rewind" time. But that's impossible, right? Time runs along a straight line (for the most part), it's not possible to bend that line and go in reverse.

Another solution is that they run the experiment again just like normal, but this time the scientists switch places. Scientists A; B; and C will remain on the platform, while scientists D and E will perform the experiment on the speeding boxcar. By doing so, the warping of their timelines will be repeated so that they are now synchronized. But, once again, that leads to a paradox. Because now, all of the scientists are traversing through time in a different position relative to everyone else who did not take part in the experiment. Therefore, they have still altered their place in time relative to the rest of the Universe.

So, we're back to square one. How do you resolve this paradox? How do you return the scientists to their original place in time? The two solutions we have just mentioned will simply synchronize them relative to each other, but not the Universe.
 
Hi QuarkMoon,
QuarkMoon said:
Scientists D and E will conclude that scientist A's watch is faster then scientist B's, because scientist A was traveling toward the light emitted by the gunpowder and scientist B was moving away from; concluding that the light took longer to reach scientist B.
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Usually we would say that A's watch is "ahead of" B's, not "faster than". "Faster than" implies that more than 1 second ticks on A's watch for every 1 second that ticks on B's watch. "Ahead of" implies that the number on A's watch is greater than the number on B's.

To determine everyone's watch readings at the end of the day all you have to do is compute the spacetime interval along their respective worldlines. Even though they will not generally be synchronized at the end of the day, all will agree that each person's watch displays the correct time based on their respective travel. There is no mysterious unbalancing of the universe involved.

-Dale
 
Here's one thats always bugged me. Consider two observers, moving relative to eachother some sizable fraction of the speed of light. They both can say the "other" person is moving and will apparently see time moving slower for that person. Now, let us suppose something happens so that the relative velocity between them is zero. What do they see now? Both thought the other was moving is slow motion, and we can suppose this goes on sufficiently long enough for each observer to feel that he is "older" than the other. When they both stop, how is the time descrepancy accounted for? Because their observations of time were just as relative as their velocities, it doesn't seem as though one person could be "right" about being older than the other. So what IS right? I'm sorry if I'm simply restating the above quesiton in a different form, but I think this particular form is gets more to the point of the problem.
 
alyosha said:
Now, let us suppose something happens so that the relative velocity between them is zero. What do they see now?
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That depends critically on what the "something" is. The procedure is the same as always, the proper time (the time reading on a clock) is equal to the spacetime interval along the clock's worldline. This is a frame invariant quantity and so all observers will agree. If two observers performed such an experiment both would agree on who would wind up older.

-Dale
 
It does make sense that it would depend on just what that "something" is, because for their relative velocities to go from nonzero to zero, someone must have accelerated (or both of them may have accelerated in some way). Maybe the answer I'm looking for has to do with this acceleration.
 
DaleSpam said:
Hi QuarkMoon,Usually we would say that A's watch is "ahead of" B's, not "faster than". "Faster than" implies that more than 1 second ticks on A's watch for every 1 second that ticks on B's watch. "Ahead of" implies that the number on A's watch is greater than the number on B's.
Click to expand...

Ah, thanks for the correction. This is only the second time I've actually attempted to articulate my question into words.;)

To determine everyone's watch readings at the end of the day all you have to do is compute the spacetime interval along their respective worldlines. Even though they will not generally be synchronized at the end of the day, all will agree that each person's watch displays the correct time based on their respective travel. There is no mysterious unbalancing of the universe involved.

-Dale
Click to expand...

But doesn't that mean some time travel has occured? Their worldlines (timelines) are no longer synchronized with each other or the rest of the Universe. I guess my question is: Can we reconcile their altered timelines without violating any natural laws?
 
QuarkMoon said:
But doesn't that mean some time travel has occured? Their worldlines (timelines) are no longer synchronized with each other or the rest of the Universe. I guess my question is: Can we reconcile their altered timelines without violating any natural laws?
Click to expand...
Sure, you could call this time travel if you want. But what makes you think that this kind of time travel violates any natural laws?

Suppose that two observers begin a trip from LA to NY with their odometers reading the same. One drives through Denver on the way and the other drives through Dallas. When they arrive in NY they notice that their odometers have different readings. Have any natural laws been violated? Nothing has been violated at all, the two trips had different geometries so they had different lengths.

Similarly, suppose that two observers begin a trip through spacetime from event A to event B beginning with their clocks reading the same. They take different paths. When they arrive at B they notice that their clocks have different readings. Have any natural laws been violated? No, again, the two trips had different geometries so they had different lengths.

No natural laws violated, it is just geometry.

-Dale
 
alyosha said:
It does make sense that it would depend on just what that "something" is, because for their relative velocities to go from nonzero to zero, someone must have accelerated (or both of them may have accelerated in some way). Maybe the answer I'm looking for has to do with this acceleration.
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Precisely. Again, this is just geometry and spacetime intervals. Let's say that initially they are at the same location but inertially moving relative to each other, forming two straight lines in spacetime. One then suddenly accelerates and makes an intercept course. Now we have a triangle in spacetime, one observer traveling along two legs and the other along one. We simply calculate the spacetime intervals on each side of the triangle to calculate the elapsed time.

If both turn, then you have a quadrilateral instead of a triangle, but the concept and the procedure remain the same.

-Dale
 
Quark Moon,

It appears that you have substituted a calculation for an observation made during cocktail hour (your second post). The explanation in your original post tells it all. However, SRT tells us that the A and B clock will register the same times and that the embankment observer’s clocks will register different times as the forward detector is moving away from the light pulse and the rearward detector is closing with the oncoming pulse of light. SRT, however, says that the A and B clocks will register the same time as on the moving frame the observers may consider that the embankment is moving and that they, the moving frame, is actually at rest.

Look at the problem as a restatement of the twin paradox. Using the equivalence of inertial frames postulate each twin sees the other as moving away and therefore each sees the other as the slower aging twin. This embarrassing paradox was resolved in the early 1960s (Feynman et al) by asserting that the twin ion the space ship is the twin that has accelerated and is therefore the twin in motion. Hence, this twin will age slower tyhan the twin remaining on earth. This solution does not claim that the acceleration affects the aging process. Only the inertial motion, the high speed velocity, affects the aging process.

The obvious problem with this resolution is that the equivalence of inertial frames postulate must be scrapped, as each twin will is able to determine the true motion of the other from the get go. It is the motion of the accelerated frame that suffers the so called SR effects.

There is nothing special about your experiment. The model is certainly not unique. Look at the problem using mirrors at the same location as the clocks on the moving frame. When the rearward moving pulse strikes the oncoming detector (after moving a distance ct) the forward moving pulse (also having moved a distance ct) is located a distance 2vt from the forward detector. When the rearward moving pulse reflects from the mirror both pulses are moving in the same direction for the brief period of time that the forward moving pulse must travel to catch the forward detector. This same direction motion is a period of symmetry loss. For instance, in tests where the frame is at rest wrt the embankment, the pulses are always moving in opposite directions.

At the instant both pulses have moved a distance ct, the forward moving pulse must travel a distance ct’ = 2vt + vt. Solving for t’ we arrive at t’ = t(2v)/(c – v). t’ is the distance the frame moves when the pulse travels the 2vt distance. Clearly, t’ is zero if the frame velocity wrt the embankment is zero.

You should have stuck with the findings when everyone compared their watches during cocktail hour.

Geistkiesel​
 
geistkiesel said:
The obvious problem with this resolution is that the equivalence of inertial frames postulate must be scrapped, as each twin will is able to determine the true motion of the other from the get go. It is the motion of the accelerated frame that suffers the so called SR effects.
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Sorry GK, spacetime intervals are frame invariant. No special frame is needed in order to determine the length of a trip through spacetime. Try again next time.

-Dale
 
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