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Assume a period of uniform relative motion (no acceleration) of a spaceship in reference frame, B, going 80% of light speed observed from my reference frame, A, when it passes by Earth (my references frame).

When it passes Planet X which, say, is 20 light years away from Earth, 25 years will have elapsed in my reference frame (A). But also from my reference frame, I will judge 15 years elapsed time on the spaceship's clock which is in reference frame, B.

Two Questions:

1); Since uniform motion is relative, the perspective from the spaceship, i.e., reference frame B, is that my reference frame (A) is moving at 8O% light speed while the spaceship is stationary. So, shouldn't the spaceship's clock (reference frame B) now show the elapsed time of 25 years from when Earth is at the spaceship's location until Planet X arrives at the spaceship's location. And, also from the spaceship's perspective, i.e., reference frame B, shouldn't the elapsed time on Earth (reference frame A) now be judged only 15 years?

2); How can the clocks in each reference frame (A and B) run slower than the other?

In order to "see" what's actually going on here, an explanation in plain English to assist the mathematics would be helpful.

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    $\begingroup$ So many variation on this one question and they all have the same basic answer. physics.stackexchange.com/q/241772 is intended to be a master question for the subject, but I am a huge admirer of this answer and, indeed, eventually adopted Takeuchi as a text for my Modern Physics class. $\endgroup$ – dmckee May 10 '17 at 18:37
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This is a great question--how can each clock run slower than the other? It is not at all intuitive, and the correct answer is that spacetime (Minkowski space) is hyperbolic, and the metric preserves $c^2(\Delta t)^2 - (\Delta x)^2 $, thereby allowing both observes to see the other running slower and length contracted.

Now if by ""see" what's actually going on here" you mean "make intuitive from a Galilean point-of-view", I'm afraid that's not possible. It is a property of the geometry of space-time, and "see"-ing what is going on is analogous to answering: two ships are on a collision course, head-on. Each turns to port and they miss. Both captains see the other captain move to the right, how is that possible?

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  • $\begingroup$ Since mathematics is merely a tool in deciphering more than one set of equally valid abstract relationships, I'm still not giving up on an answer in this forum which is at least reasonably palitable to my intuition. $\endgroup$ – Rob Apr 4 '17 at 19:56
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Well, you must remember that only when the person comes back to earth , will he be able to tell the difference in time elapsed for him and for his counterpart on earth. Then, it follows simply from lorentz formulae that the moving observer has his time dilated and not the one who stayed behind. If else, no such meeting of the two takes place , there is no way of deciding who's clock is running slow - both are right. You can read the " twin paradox" which i presume is present in almost all books on special theory which details this procedure.

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  • $\begingroup$ If the person comes back, that would involve acceleration/deceleration which is absolute (not based on frame of reference). So, if this paradox cannot be answered satisfactorily by orthodox physics (except mathematically), then we will have to conclude that our brains are not capable of grasping the complete ("big") picture, which invokes philosophical explanations perhaps involving the role of consciousness, parallel Universes for each observer, etc. $\endgroup$ – Rob Apr 4 '17 at 19:38
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In this problem there is an apparent symmetry which usually mislead us. The situation is not symmetrical so the conclusions of your point (1) aren't right. I'll try to explain.

In your reference frame, you use two clocks to measure the time elapsed (one on the Earth and one on Planet X) while the "moving" guy in his reference frame uses just one clock to measure its proper time. So what you're really comparing is the proper time interval measured by B, with the time interval measured by two clocks in the reference frame A. You can conclude that if you compare a clock in a reference frame with more than a clock in another reference frame moving with respect to the first one, that will be always late (that is to say that the time interval will be smaller). If you want to compare the clock in B with just one clock in your reference frame A, you have to come back to Earth, and the answer to this case has already been given. Take your time to think about it, I didn't really get it for a while when I first encountered the problem.

You can look at a more complete discussion on time dilation that comprehend the above discussion on Landau-Lifshitz Course of Theoretical Physics Series, Vol 2, The Classical Theory of Fields, chapter one.

I hope it helped.

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  • $\begingroup$ I'm still not satisfied. The ultimate question is whether time dialation is REALLY occurring during the period of relative uniform motion in BOTH reference frames, A and B (which seems illogical), not what we are able or not able to observe. The same ultimate question applies to the idea of relativity of simultaneity in that it is only so due to the fixed max light speed of information in all uniformly moving frames of reference which affects how long it takes for a particular observer to receive it. It doesn't address when the events really happened. $\endgroup$ – Rob Apr 5 '17 at 6:42
  • $\begingroup$ Of course, one might argue that there is no universal "now." But there is a "practical now." For example, let's make it equal to the maximum, un-dialated duration of one half second where all possible dialated times - from the shortest Plank time up to one half second - would be even less significant $\endgroup$ – Rob Apr 5 '17 at 6:48
  • $\begingroup$ @Rob of course it is occurring in both reference frames, you just have to be careful in specifying with respect to what. B proper time is late with respect to the interval measured by two clocks in A, like A's proper time is late with respect to two different clocks in the reference frame B. I don't get what you mean by wanting to know what's really happening. That's what's happening. I'm sorry I'm usually not good enough at explaining concepts. Maybe someone with a better ability will come here and do it. $\endgroup$ – Run like hell Apr 5 '17 at 8:20
  • $\begingroup$ Here's the thing. I understand that in order for one observer to CONFIRM time dilation of the other's clock, he would have to accelerate/decelerate to meet up with him to examine that clock, which would then alter his own clock and cancel the otherwise symmetry presumed if we were to only consider uniform motion and then simply reverse the perspective from the standpoint of other observer's frame of reference. But my question was: Does this symmetry REALLY exist DURING THE PERIOD OF UNIFORM MOTION? How can each clock be running slower than the other during this period? $\endgroup$ – Rob Apr 5 '17 at 16:14
  • $\begingroup$ I incorrectly conflated symmetry with the idea of each clock running slower than the other. Actually, symmetry would be a special case of the latter. So, I'm actually questioning how could each clock be running slower than the other during the period of relative uniform motion (not even to consider by how much each runs slower than the other, i.e., without my even invoking lack of symmetry from the perspectives of both reference frames based on the concept of relativity of simultaneity). $\endgroup$ – Rob Apr 5 '17 at 17:05
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Concerning 1)

Spacetime diagrams might help clarify some ideas.

In the context of your scenario, if two observers in relative motion agree on the elapsed time between two events, then you are back to the absolute time structure of Galilean physics. That structure is not compatible with the principle of relativity and the speed of light principle. For example, if the two observers exchange light-signals "at the same time", they would receive the signals at different times.

So, let's look at the details using a spacetime diagram on rotated graph paper, in accordance with special relativity.

spacetime diagram on rotated graph paper

Let's let each light-clock diamond represent 5 years. I've marked the event when the spaceship (B) meets Planet X. (Implicitly, B continues past Planet X.)

As you've said,...
In the earth frame (along OA), that event occurs 25 years after the separation event O, at a location 20 light-years away from OA. Note these measurements use the earth's light-clock diamonds and uses the event A, which the earth says is simultaneous with event BmeetsX.

In the spaceship frame (along OB), this event BmeetsX occurs 15 years after the separation event O.

However, in the spaceship frame, BmeetsX still occurs 15 years after event O. Note that, in the spaceship frame, event A is not simultaneous with BmeetsX... it's simBmeetsX that is simultaneous. (That occurs at (15 y)/gamma=(15 y)/(5/3)=9 years [1.8 diamonds] according to the earth frame.) In addition, that event is (2.4 diamonds)*(5 yr/diamond)=12 light years away. (Indeed, from the spaceship frame, the earth traveled 12 light years in 15 years... and does so with speed (12 ly)/(15 y)=0.8c.)

All of the proportions in the previous paragraph are correct. But if you want to see a direct comparison, the spaceship should wait until 25 years (5 diamonds) have elapsed on its worldline... the event I labeled B. In that case, the event on the earth that the spaceship says is simultaneous with B is simB. I think you can see that simB is 15 years after event O, and that simB is 20 light years away. So, there is the symmetry displayed.

Underlying these diamonds is the geometric structure of special relativity: Minkowski spacetime geometry, whose "circles" [figures of equal time-elapsed] are hyperbolas. In the diagram below, time runs to the right [rather than upward, as in the earlier diagrams].

https://www.desmos.com/calculator/ti58l2sair hyperbolas in Minkowski space

When an observer's worldline meets that unit hyperbola [marking one tick of that observer's clock], the tangent at the intersection defines the events simultaneous with the intersection event according to that observer. (Those lines are parallel with the spacelike diagonals of the light-clock diamonds.)

Now here is the answer to Question 2.

Note how an observer's line-of-simultaneity cuts the other observer's worldline before that observer reaches the hyperbola... and that fraction is the same for each observer. In fact, that fraction is 1/(time-dilation factor).
In the simulation, if you tune "E" to 0, you get back to the Galilean case.
If you tune "E" to -1, you can see the Euclidean analogue.

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Here's your diary:

2000: Today, my friend Sally took off for Planet X, 20 light years away.

2005: Today marks five years since Sally took off. She's one-fifth of the way there. But her calendar-clock says 2003. It seems to be running slow.

2009: Today marks nine years since Sally took off. She's 9/25 of the way there. But her calendar-clock says 2005.4. Still running slow, I see. Also, I stubbed my toe today.

2015: Today marks fifteen years since Sally took off. She's three-fifths of the way there. That slow calendar-clock of hers says 2009.

2025: Today, Sally arrived at Planet X --- with her calendar-clock reading 2015. It ran slow the whole time.

Here's Sally's diary:

2000: Today, I parted ways with my friend Rob --- and the whole planet earth for that matter. I'm 12 light years from Planet X, and plan to land there in 15 years.

2008.33: It's a third of the way through 2008, and I'm over halfway there. But Rob's slow calendar-clock says 2005. He's currently making a diary entry that says my calendar-clock says 2003.

2015: Ah, I've just arrived! My calendar-clock, which keeps perfect time, says 2015. Meanwhile, Rob's slow calendar-clock says 2009. He's currently making a diary entry that say's I'm 9/25 of the way here, and that my calendar-clock says 2005.4. Also, he stubbed his toe today.

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  • $\begingroup$ Please explain precisely why, in the year 2000, Sally's diary reads only 12 light years between her spaceship and Planet X when, at the same time (in the year 2000), Rob's diary reads 25 light years between Sally's spaceship and Planet X. In other words, please elaborate on the role length contraction and relativity of simultaneity plays in connection with the differing distances between spaceship and Planet X. Thanks. BTW, I commend you on the eloquence with which you attempt to make understandable a topic that is found confusing to many students of physics. $\endgroup$ – Rob Apr 6 '17 at 5:36
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    $\begingroup$ The only answer to your question about why these are the right numbers is that this is what the Lorentz transformation tells you the numbers have to be. As to your use of the phrase "At the same time", keep in mind that Rob and Sally disagree about which events happen at the same time. Rob says that his toe-stub happens at the same time that Sally completes 9/25 of her journey. Sally says that his toe-stub happens at the same time as her arrival at Planet X. $\endgroup$ – WillO Apr 6 '17 at 15:05
  • $\begingroup$ Special Relativity says that the laws of physics (including Special Relativity itself) apply equally in all uniformly moving frames of reference. It seems the reason the effect on clocks is not synchronous in both frames of reference is that special relativity is not applied equally to both frames of reference. $\endgroup$ – Rob Apr 6 '17 at 16:49
  • $\begingroup$ For example, the presumption that there is no preferred state of uniform motion is applied from A's perspective where B's time dilates. But dilated time, instead of normal time, is used to apply the presumption that there is no preferred state of uniform motion from B's perspective. In other words, why is special treatment given to B's reference frame by incorporating dilated time in B's judgment of what A's clock reads if there is no preferred state of uniform motion? $\endgroup$ – Rob Apr 6 '17 at 16:50
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    $\begingroup$ If Jim faces north and Jane faces west, then Jim can say Canada is 1000 miles straight ahead, while Jane says Canada is 1000 miles to the right. ALL of your confusion about Rob applies equally well to Jim and Jane. How can Jim and Jane contradict each other and yet both be right? I suspect you'll not see that as a difficult question. When you understand relativity, you'll see that you're asking exactly the same question about Rob and Sally. $\endgroup$ – WillO Apr 6 '17 at 18:59
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But my question was: Does this symmetry REALLY exist DURING THE PERIOD OF UNIFORM MOTION? How can each clock be running slower than the other during this period?

The measuring procedure has some interesting details.

It is often said, that a single clock dilates relatively to a reference frame of observer, a.k.a rest frame.

Very often observer in special relativity is not a physical person, but the whole frame itself, or team of observers. Each of them possesses a clock and these clocks are Einstein - synchronized by beam of light. The beam of light travels from observer to observer and they adjust their clocks accordingly, since they know velocity of light c and distance the beam travels, so they can calculate how much time it took for the light to go from clock to clock.

https://en.wikipedia.org/wiki/Einstein_synchronisation

You can imagine Einstein - synchronized clock A,B,C,D... X,Y,Z These clock show the same time in reference frame K. All that reference frame K is an observer, or family of observers.

Single clock C1 moves in reference frame K. Then observer of reference frame K compares readings of clock A-Z with clock C1 in immediate vicinity, when clock C1 passes by these clocks.

https://en.wikipedia.org/wiki/Observer_(special_relativity)

If moving clock C1 and clock A showed 12 PM at meeting, moving clock will show 3 PM and clock Z will show 6 PM when they will meet. This is how time dilation works. SINGLE moving clock dilates relatively to a set of synchronized and spatially separated clocks, not vice versa. Set of clock runs faster from the point of view of single clock.

This article emphasizes this detail at page 6 (6)

http://isites.harvard.edu/fs/docs/icb.topic455971.files/l09.pdf

Animation:

https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_dilation02.gif

Related articles:

https://arxiv.org/ftp/physics/papers/0512/0512013.pdf

Good article that gives correct explanation:

http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter039.htm "Two spatially separated clocks, A and B, record a greater time interval between two events than the proper time recorded by a single clock that moves from A to B and is present at both events."

If observer, who was once in motion wants to measure dilation of relatively moving clock, he has to turn himself into one "at rest" and introduce his own rest frame by means of putting two Einstein - synchronized clock at point of departure and arrival of single clock.

In your case your rest frame consist of two Einstein - synchronized clocks. One is on the Earth, another is on Planet X. Spaceship is the Single clock. You compare readings of clocks Earth with clock Spaceship first and clock Planet X with clock Spaceship 20 years later.

Spaceship can measure dilation of single clock Earth or of single clock Planet X. For example, he wants to measure dilation of clock Earth. Spaceship places two Einstein - synchronized clock at point of departure of the Earth and point of arrival of the Earth and compares readings of his synchronized clocks with clock Earth when these clocks are in immediate vicinity.

Good to note, that only observer "at rest" measures dilation of moving clock. To get reciprocal observations, we have to "forget" old frame and replace it with a new one, this way turning that observer who was once in motion into one "at rest".

If Single clock compares it's readings successively with clocks of rest frame, it will measure, that time in reference frame runs faster.

From the point of view of reference frame, single clock dilates. From the point of view of single clock, time in reference frame runs faster, since it's own "time" runs slower.

Mossbauer effect rotor time dilation test demonstrate that effect very well.

Is that what you were looking for?

If we don't change frames, the moving observer measures, that clock of observer "at rest" ticks gamma times faster, not slower that his own. For example, if moving observer measures frequency shift, he will detect blueshift of frequency, but not redshift, i.e. clock at rest runs gamma times faster.

https://en.wikipedia.org/wiki/Relativistic_Doppler_effect

The transverse Doppler effect is the nominal redshift or blueshift predicted by special relativity that occurs when the emitter and receiver are at the point of closest approach. Light emitted at closest approach in the source frame will be blueshifted at the receiver. Light received at closest approach in the receiver frame will be redshifted relative to its source frequency.

Reflection from a transversely moving mirror graphically demonstrates that. https://www.youtube.com/watch?v=FQKp3FU8vR8

P.S. I think there is a mistake in Wikipedia. Light emitted at closest approach is redshifted. Light received at closest approach is blueshifted.

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protected by Qmechanic Apr 13 '17 at 6:31

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