- #1
Artyvr
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I would like to know why it is called time DILATION and not time CONTRACTION?
Why is Time DILATION Called Time DILATION?
- Thread starter Artyvr
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In summary, the phenomenon known as time dilation is called so because it refers to the stretching or lengthening of time intervals observed between two reference frames in relative motion. This is due to the slowing down of clocks in the moving frame compared to the stationary frame, which is observed from either perspective. The terminology can be seen as a matter of perspective, as the moving frame may also observe a "contraction" of time in the stationary frame. This concept is also known as differential aging, where one clock runs slower than another depending on the frame of reference. This can be seen in the Twin Paradox, where one twin ages slower than the other due to a difference in frames of reference. These principles also apply to particles in motion, where
- #2
Doc Al
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The word dilation means stretching out or lengthening, which is just what is observed. Moving clocks run slow (in the observing frame), thus 1 second on the moving clock appears longer (dilated) from the observing frame.Artyvr said:
- #3
Fredster1765
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I think it's just a matter of perspective. When some particle is accelerated at the LHC, the scientists who observe the particle will see time "slowing down" for that particle, whereas the particle (if it had a brain) would see time "contracting" for those scientists.
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Doc Al
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Time dilation is symmetric. Both frames see the other's clocks as slowing down or 'dilating'.Fredster1765 said:
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A specific guy on the train sees the time changing more rapidly on the sequence of clocks he passes along the ground than the time on his own clock (i.e., that he carries along with him). This is why it is called time dilation.
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arindamsinha
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Fredster1765 said:
Yes, there is a bit of terminology issue here. What you are talking about is 'differential aging', as in the Twin paradox. The term 'time dilation' seems to be reserved for slow down only, though it is certainly related to differential aging.
If you look at this from the perspective of the traveling twin (or LHC particle), you could call what is happening to the stationary twin as 'time contraction' from that perspective. The terminology is not in common use however.
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Vaid
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arindamsinha said:
If the first twin (who ventures on a space journey) is moving away from the second one, the second one is also moving away from the first one.
Why don't both age the same relative to each other?
- #8
arindamsinha
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Vaid said:
The twin who is moving away has to accelerate, putting him/her in a non-inertial frame of reference for some time (and breaking the symmetry of the situation). After the acceleration stops, he/she is in another inertial frame, but in this frame time is dilated compared to the twin who never moved.
This is how SR works.
- #9
K^2
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arindamsinha, it sounds like there might be some confusion in your understanding of twin paradox. The moving twin still observes a time dilation. Whether an observer has accelerated in the past or is going to accelerate in the future does not change the current description of physics. And so as a particle drifts along a chamber in accelerator at constant speed, relative to it, it's the scientists who are time-dilated.
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arindamsinha
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K^2 said:
I think you misunderstood me.
I am not talking about the 'observed' time dilation, which would be true based even on classical (Newtonian) Doppler effect.
I am talking about differential aging, i.e. real clock time dilation. The moving twin always has the real clock time dilation, never mind who observes what.
Same applies to the particle in the accelerator. No matter what it sees, it is living longer than expected by the scientists' clocks, and this is aymmetric differential aging (i.e. the scientists do not live longer by the particles clock, they live shorter, leaving aside any observations based on Doppler effect).
- #11
K^2
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There is no "moving" twin. They are both moving relative to each other. There is no absolute frame. You cannot say who is standing still.
The only time it makes sense to distinguish between the two is if one twin left and then came back. After he came back, we can compare the net aging of the two. But while in transit, SR applies to both. The twin that stayed behind is aging slower from perspective of the twin that left.
A particle coasting through accelerator is in an inertial frame. There is nothing special about that frame with respect to the laboratory frame.
The only time it makes sense to distinguish between the two is if one twin left and then came back. After he came back, we can compare the net aging of the two. But while in transit, SR applies to both. The twin that stayed behind is aging slower from perspective of the twin that left.
A particle coasting through accelerator is in an inertial frame. There is nothing special about that frame with respect to the laboratory frame.
- #12
arindamsinha
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K^2 said:
I do not think that is a correct interpretation of SR. If the movement of the "twins" was symmetrical in every way, there could never be a "differential aging" between them.
However, when one of the twins accelerates and then reaches a steady velocity, that makes him the clearly "moving" twin. He has differential slower aging throughout the journey, not at certain arbitrary parts.
There is no necessity for the traveling twin to come back to the origin and compare clocks for differential aging or relative time dilation to happen.
I believe this is well demonstrated by the velocity time dilation of GPS satellites. Even though they never come back, we know clearly that their clocks slow down compared to Earth-based ones (ignoring gravitational time dilation). You may contend that they are never in an inertial frame, but I believe they are in a good enough approximation of an inertial frame for us to apply SR for the velocity part.
- #13
K^2
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Ok, that's a fairly jumbled mess you've got there. Don't take it as criticism, but try to bear with me, and we'll sort it out.arindamsinha said:
Lets forget twins for a moment. Let's focus on particles. Now, you might think that asking about the age of particle is silly, I mean, it's not like it celebrates birthdays, or anything, but there is a special case. Radioactive particles. They are ideal time keepers. We know that half-life of radioactive particles in accelerator increases as a direct confirmation of special relativity. Of course, that's perfectly consistent with your point of view. We accelerated a particle, so it ages slower, and we should expect a longer half-life. Good so far.
Picture the following thought experiment. We have a big chunk of radioactive material with a long half-life. We need long half-life so that the number of radioactive atoms remains roughly constant. That gives us a constant rate of radioactive decay events. So we can set up a detector nearby, connect it to a light, and the light will flash, on average, a constant number of flashes per unit time. Say, N times per second. Now rather than accelerate all of this mess, which we agree on predictions for, let's say a ship with scientists accelerates towards it. They speed up for a while, reach constant velocity v ~ c and then drift towards the sample with the flashing light.
Suppose, this drifting ship recorded flashes for some time t in frame of the device. Naturally, during that time the crew aged t/γ. Again, simple time dilation you agree with. How many flashes did the team record? Well, there are the Nt flashes generated in that time, plus the flashes that were emitted earlier and were located in that space the ship covered. So the total is Nt + N(vt/c) = Nt(1+v/c). Of course, the researchers only aged t/γ, so the rate they record is Nγ(1+v/c). But these researchers aren't stupid. They know the source of flashes is traveling at them at v. They know they'll see flashes more frequently than they are actually emitted. So, you have an object coming at you at v, and you receive flashes from it at the rate of Nγ(1+v/c). What is the actual rate according to you? Well, if it emits a signal every T seconds, the second pulse has vT less to travel than the first one. So you get it with a vT/c shorter delay. In other words, the time between pulses you see becomes T(v/c). So to get the actual rate, you need to divide what you measure by v/c. So according to the researchers, the light signals are emitted at a rate of Nγ(1+v/c)/(v/c) = Nγ/γ² = N/γ. (Check all the algebra as an exercise.)
So the rate of radioactive decay appears lower! And by the same factor γ. Despite the fact that it's the researchers that were accelerated and not the particles. The half-life of the particles is longer in either scenario!
This is a very important result in Special Relativity and is the crux of the entire matter. Two observers traveling with respect to each other would observe the other being time-dilated, regardless of which one had to accelerate to get them to the current state.
This also makes sense physically. How should the particle know which of the two accelerated? It has no memory. Theory that suggests otherwise would be very suspicious.
So in order for the twin paradox to manifest, one of the twins has to leave and come back. Only afterwards can we talk about which of the two really aged.
First of all, GPS satellites do have to take gravitational time dilation into consideration. It is a significant enough factor. Second point is more philosophical. A lot of SR results with time dilation follow from the fact that you cannot be moving with respect to a source and still remain at the same distance. If you are moving, you will experience Doppler effect, and you have to correct for it. As you can see from above, it makes all the difference.arindamsinha said:
Not so with circular motion. An object traveling around the source can have high velocity and experience no Doppler effect. This let's you synchronize the clocks. This means that in flat space-time, an object traveling in a circle around you must not experience any time dilation at all. So the acceleration effect is canceling the effect due to the velocity. So you can never claim this effect negligible.
Of course, in case of a satellite, you are also looking at time dilation due to gravity, and things get a lot more interesting. But that's General Relativity.
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PAllen
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K^2 said:
I'm not sure I get your meaning here. If you have a radioactive source traveling rapidly in a circle around another radioactive source in flat spacetime (gravity ignorable), then the central one detects the circling source decaying slower; and the circling source detects the central source decaying faster. (For simplicity, assume each source sends a spherical light pulse every N of its decays, so there is no concern about direction particles are emitted).
- #15
arindamsinha
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PAllen said:
PAllen, you have crisply summarized the point I am trying to make.
K^2 said:
Yes K^2, I am aware. I meant we ignore that gravitational time dilation part for the sake of isolating the velocity time dilation, which is the part under discussion.
K^2 said:
This has been one of the key issues with SR interpretations. Contrary to your initial premise, you have changed frames and now you are saying the other body is coming towards you at a certain velocity v instead of being at rest, and therefore need to divide/compute etc. etc. If you looked at things this way, there would never be any relative time dilation this way. All is 'observation', none of it is real 'differential aging'
But differential aging does happen. In a setup like this, only one of the participants is moving (the one who accelerated) and the other is at rest. Otherwise you would never have measurable differential aging. Over the years this has been tacitly recognized based on experiments.
K^2 said:
Again, very good point from the original SR derivations, but experiments suggest otherwise. Whether a particle has memory or otherwise, it is the moving one if it has "felt" acceleration.
A traveling body doesn't have to come back to the origin for establishing this. Where do you draw the line anyway? If your twin has traveled and come back to within 100m, you still can't be sure who aged more? Or is it 10m? Or is it when your clocks are touching each other? In other words, at what stage does the magic of "differentially aging" suddenly materialize?
I think it is more reasonable to accept what SR is saying - the actual differential aging happens throughout the journey at a predictable rate (depending on velocity), and what the twins compare when they meet is the "cumulative differential aging".
If you can answer the following simple twin-based thought experiment, it will help establish what we are discussing more clearly:
Thought Experiment:
Two twins on Earth syncronize clocks. One twin stays on Earth. Another twin travels at almost light speed for a certain distance (say 10 seconds on the Earth clock, reaching a distance of 10 light seconds) and then stops. (Assume the acceleration and deceleration periods are negligible).
At exactly half the distance (i.e. 5 light seconds from Earth), we have previously placed a device that can send a light signal simultaneously to both twins. This signal is activated when the twin who left Earth stops, and sends a signal to the device, saying "I've stopped". Once the twins receive the device light signal, each twin immediately sends their "current clock readings" to their other twin. (They can subsequently communicate and establish whether there was a difference in their clocks, and if so, how much. Since they are now mutally at rest, there's no complex hanky panky about this.)
Now, tell me which of the following possibilities is correct:
1) The twin who left Earth has his clock behind the Earth twin's (about 10s or so)
2) The twin who left Earth has his clock ahead of the Earth twin's (about 10s or so)
3) There is no difference between the clock readings of the two twins
4) We cannot predict the outcome without actually doing an experiment
Once you have answered this, you will probably understand what I am saying, or we can discuss further.
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- #16
K^2
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Sorry, yes. That's right. I got myself confused there for a bit. The moving one is still time-dilated. But not the static one. Got that mixed up. I should have written down the metric tensor right away. Though, it's really obvious in the retrospect.PAllen said:
I really chose a horrible example. This problem behaves exactly as arindamsinha expects, since the acceleration of the frame of the circling source results in accelerated time for the central source.
That's still the point. The acceleration of the second source is responsible for the effect, but it's a bad example to use with somebody who is confused on inertial case.
RELATIVITY. You can't say A is moving and B is static. The whole POINT is shifting frames. The whole POINT is that we can look at it from A's rest frame or the B's rest frame. In both frames the OTHER is dilated.arindamsinha said:
You can read Einstein's original works, and he derives time dilation exactly the same way, via exchange of clock pulses using light beams. In every frame, the other's clock appears to run slow.
Precisely. STOPS. They now have the same rest frame. We are talking about a particle that's still moving. Situation where two observers have different rest frames.arindamsinha said:
- #17
arindamsinha
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K^2 said:
Can you square the above statement with your post #11 where you say 'There is no "moving" twin'. They are both moving relative to each other.'?
K^2 said:
Huh?! Can you please explain this part further?
K^2 said:
You mean the acceleration of the GPS satellite is responsible for its velocity time dilation? Can you please explain a little better?
BTW, who is this 'somebody' who is confused on inertial case? Me?
It would be nice not to have such patronizing statements in a civilized discussion. While I may not have your expertise in relativity, I may possibly understand it just a wee little bit better than you think.
K^2 said:
In other words, everything is symmetrical, and therefore, there is no question of 'relative time dilation' or 'differential aging'?
K^2 said:
Let me refer to one of Einstein's original works here: http://www.fourmilab.ch/etexts/einstein/specrel/www/.
(a) In Section 4, on what basis does he conclude that 'the clock moved from A to B lags behind the other'? By your logic, the clock not moved also has a symmetrically equal motion w.r.t. the moving clock, so in the end there is no 'relative time dilation'. Why do experiments show relative time dilation?
(b) Note that he also takes an example of a clock traveling in 'a closed curve with constant velocity until it returns to [the origin]', and says that such a moving clock will be slower. This is exactly what the GPS satellite is showing. Einstein doesn't bring into picture any acceleration for this. It is all about the velocity here.
K^2 said:
No, we are not splitting hairs about whether a moving object has 'stopped' or is 'still moving'. We are talking about 'when' and 'where' the relative time dilation or differential aging occurs between two entities, whatever the stage when we inspect them (stopped or still moving).
Why don't you answer the thought experiment I mentioned, and see where we go from there? It is a very simple question after all.
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K^2
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I'm not trying to say this is too complex for you to understand, or anything. And I am explaining some of these things less than well, which is entirely my fault. But there is a definite flaw in your understanding of SR. If I try to explain it to you, a certain degree of patronization is necessary. If I'm overstepping the necessity, and am more patronizing than absolutely necessary, I apologize. It is not my intention.arindamsinha said:
Absolutely. A bit of it is sloppy language on my part, but it is fairly common usage. While the motion is still relative, the acceleration in GR, and to the best of our knowledge in general, is absolute. So there is no ambiguity in deciding which object is going around in circles around which.arindamsinha said:
Perhaps, a slightly better illustration is choice of rotating coordinate system. In a rotating coordinate system there is a definite center. Points away from center experience a centrifugal force which arises due to the choice of an accelerated reference frame. (Ref for clarity, if needed.)
So unlike linear, unaccelerated motion, where we cannot say which is moving and which is still, in case of rotation, we can clearly state which one is rotating. That breaks the symmetry of SR and introduces a difference.
Indeed, in the rotating frame, since both objects are static, time dilation is computed from the tt term of the metric tensor. This term unambiguously tells us that the object undergoing centripetal acceleration is the one that will age slower by the factor 1/γ. Which, of course, agrees with prediction from the inertial frame.
Hopefully, the above already helps. If you aren't afraid of metric tensors, I can write out the GR treatment of this in a rotating frame. Because in the rotating frame nothing is moving, it is extremely simple, and you really don't need to understand anything about curvature of manifolds or differential geometry, or any of the other stuff that usually scares people away from GR.arindamsinha said:
Depends on choice of coordinate system. If you look at it from perspective of the satellite itself, yes! I mean, from perspective of satellite itself, it's not moving. Yet (ignoring gravity) the Earth's clocks run fast. The only explanation to that is the fact that the satellite is accelerating.arindamsinha said:
With gravity things get slightly more complicated. Satellite isn't actually accelerating. It follows a geodesic. So now to understand the clock differences we have to consider the curvature that actually causes it to go around in circles. Schwarzschild metric describes it, and it does give you a time dilation effect which does depend on your altitude. Of course, in Scwarzschild metric, the satellite is also moving, so you have both the gravitational time dilation and the time dilation due to satellite's velocity.
Again, none of this is terribly complex, so long as you take Scwarzschild metric on faith and accept the satellite velocity at given radius as given. If you want to actually verify the former and compute the later, it is going to involve some tensor calculus.
Everything is symmetrical, so the time dilation is too. Observer A claims that B's clock runs slow, and observer B claims that A's clock runs slow. They are in a disagreement, but they cannot find a contradiction by simply exchanging light signals.arindamsinha said:
a) On the contrary. I've taken that from radioactive source's perspective the ship is time-dilated as a given. Since it accelerated, and you seem to accept that fact. I then derived the identical time dilation of the source from perspective of the ship. It's in perfect agreement with the section.arindamsinha said:
b) Yes. Einstein, being quite a bit smarter than myself, didn't make the same mistake of confusing oneself with accelerated systems. If you have a clock that travels along an accelerated curve, you can still describe it from an inertial frame. In inertial frame, you can use SR and consider time dilation of an accelerating object. There is no problem with that. Lorentz factor will simply be a function of time rather than a constant. No big deal. It's only when you decide to try and describe that inertial observer from perspective of an accelerating clock that you run into a headache.
All inertial frames, on the other hand, are equivalent under SR. So time dilation formula applies exactly the same way, regardless of whether you are looking at source flying towards the rocket or if you are looking at a rocket flying towards the source. So long as both of these are traveling at uniform velocity, both can use time dilation formula under assumption of self being static.
It's not splitting hairs. It's precisely where all of the Special Relativity is. What happens to the twin that accelerated and then decelerated by the same amount is absolutely clear to both of us. We are in agreement on that. The question is, who's clock runs slow relative to whom in between. We aren't talking about the fact that the clocks themselves are already in disagreement by this point. We are talking about the rates at which each clock runs at this point. The twins can communicate while in relative motion. And they can correct for Doppler Effect quite easily, because they both know that speed of light is always c. And after they correct for Doppler effect and compare their clocks, they still find that the other twin's clock is running slow. Not in terms of absolute difference. But in terms of the rate at which the clocks advance.arindamsinha said:
- #19
PAllen
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K^2 said:
There is a nuance here that I am not sure has been addressed adequately (I admit I haven't read the whole thread). Imagine the turnaround twin deriving the rate of the distant clock as described. Assume, for simplicity, instant turnaround. Then, throughout the trip they consider the stay at home clock running slow. For example, from 1 pm (when they separate) to 2 pm on their clock the see a redshifted clock going from e.g. 1 pm to 1:15 pm, and they figure it is slow (but by less than visual after correction for 'pure doppler'). Then, from 2 pm to 3 pm (at which point they re-unite), they see the stay at home clock advance uniformly from 1:15 pm to e.g. 3:15 pm. Correcting for doppler, it is considered to run slow during this whole time - yet advances more than their own clock.
Obviously, the resolution, is that to use this standard approach for removing Doppler, they must also accept the standard approach to simultaneity, which says that much of the blueshifted history occurred before the turnaround, even though seen after and indistinguishable from the period they consider the distant clock running slow (this is unsurprising, due to finite light speed). With a non-instant turnaround, you would consider this to be delayed reception of the remote clock running fast during the turnaround.
The key point is that for a significant period after turnaround (even for non-instant turnaround), after the turnaround twin is inertial, they must interpret the signals they receive in a way that is cognizant of the fact of their turnaround - if they want to avoid a logical contradiction, while still using the standard removal of doppler convention.
Personally, I prefer to give much less emphasis the uniqueness, let alone, objective reality of this interpretation. I consider that time dilation is a coordinate dependent, non-observable quantity whose character is a matter of convention - the choice of coordinates.
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arindamsinha
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K^2 said:
OK, I feel there has been a disconnect in what you and I are saying. As per my posts in threads #9 and #11 where this discussion started, I was specifically referring to the disagreement or difference in age between the two clocks at any point in the journey, should that be periodically measured.
My contention is that a periodic measurement will prove that the traveling clock is ticking at a slower rate throughout the journey, and that there is no need for it to come back to the other clock for such 'relative time dilation' or 'differential aging' to be proven. It happens throughout the journey at an uniform rate given an uniform velocity.
You are talking about how they would 'appear' to each other from a distance. Let me know if that is a correct interpretation of what you are saying.
PAllen said:
I understand the standard approach you mentioned.
However, my preference is to consider that this is objective reality - the traveling clock always ticks slower at a specific predictable rate, compared to the stationary one, throughout the journey.
Moreover, the materialization of such relative time dilation does not depend on the traveling clock returning to the stationary clock's location for a face-to-face comparison. (@PAllen - I realize you have not referred to this, but I have seen that in many posts)
Let me explain why I am partial to this way of thinking.
Refering back to Einstein's paper http://www.fourmilab.ch/etexts/einstein/specrel/www/, I see in Section 4:
- He writes "A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2 tv2/c2"
- He does not state that the clock stationary at B could similarly be considered slower by the moving clock A in its own rest frame (which is strange since he does say that about length contraction earlier). I am not saying he meant it would not happen, just that he does not stress that part
- Nevertheless, he then goes on to talk about one clock at the equator and another at a pole of Earth, and concludes that the equator one "must go more slowly, by a very small amount".
This last part to me implies a clear objective reality. He seems to tacitly state that in any real situation the stationary and moving clocks would become clear, the situation will be aymmetric, and real relative time dilation will show up between the clocks (unless the conditions of both clocks are really completely symmetrical). Moreover, such difference between the clocks is an ongoing and predictable amount at any point of the journey of the moving clock.
Also, the equator/pole relative time dilation happens even though the two clocks never get together at a location.
Would you say my thinking is correct, or is there something wrong with it?
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- #21
Drakkith
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What is "real relative time dilation"?
- #22
arindamsinha
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Drakkith said:
Experimentally proven and measurable differential aging between clocks.
What I meant:
"Experimentally proven and measurable" = "real"
"Relative time dilation" = "differential aging"
- #23
Drakkith
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arindamsinha said:
So the first would be the difference between two clocks after one has accelerated and then returned to an inertial frame at rest with the original observer? And the second would be the observed slowing of clocks between observers moving relative to each other?
- #24
arindamsinha
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Drakkith said:
Actually, both refer to real clock time difference as seen in experiments. I am not referring to any 'as observed from different frames' phenomena dependent on Doppler effect etc.
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Drakkith
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arindamsinha said:
What does doppler effect have to do with this? Am I misunderstanding what you mean by "doppler effect"? I only know the shift in wavelength type of doppler effect.
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Mentz114
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Yes, and the difference in clock rates depends only on the relative velocity at that moment. Nothing more needs to be said.arindamsinha said:
If there is no face-to-face ( ie co-located) comparison, you are talking about something which cannot be observed which is a waste of time. If the worldlines of the clocks involved are known, then the elapsed time on the clocks are invariants whose values are easily calculated.
You won't find any new physics by looking at time dilation - it is a coordinate dependent effect and not physical. Differential ageing is physical.
PS
This is nonsense. Are you going to reject all of quantum mechanics because it is counter to (your) intuition ?
- #27
PAllen
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arindamsinha said:
Doppler is invariant, not frame dependent. Any measurement by a detector is a frame invariant quantity - it's result will be computed to be the same no matter what frame or coordinates are used. The only direct observables applicable to a twin scenario (any variant) are:
- clock time along each world line (each observer's own clock) (differential aging is thus invariant because it is just record of each world line's clocks at a pair of coincidence events)
- Doppler observed by each observer for the other. Everything else you might want to say is computed per some set of conventions. Note that Doppler as used here reflects both frequency shift and visually observed rate of a distant clock - these are always the same, because the frequency of emitted light from a standard physical process functions as clock. Thus, if you see blue shift factor of 1.5, you also would visually see a clock comoving with the emitter going at 1.5 times the rate of a similarly constructed clock of your own.
- #28
K^2
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The problem here is with your assumption of instantaneous turnaround. It implies infinite acceleration, and that does cause a singularity in the metric from perspective of the turning ship, which causes clocks to jump instantly, as well as whole lot of other problems.PAllen said:
In a more realistic case of finite turning time, when the traveling twin steps on the throttle to decelerate the rocket and accelerate for the return trip, that's where he'll see his Earth twin's clock starting to run rather fast.
You can also solve this without GR. If you do the same treatment with signals being sent and received, with Doppler correction, you can see the accelerating twin concluding that his brother's clocks run fast.
If you do take this on as an exercise, please, keep in mind that you need to decide what you are calling constant acceleration. Best way, I think, is to assume that rocket delivers constant thrust, so acceleration is constant from the perspective of the twin in the rocket. That will mean that dv/dt is a function of v. (You can burn a rocket to deliver experience of 1g on board constantly, but you'll never reach the speed of light relative to any inertial observer, so dv/dt from perspective of that observer must decrease after a while.) So be careful with that. But if you do this right, you'll be able to derive the formula for time correction for an accelerated frame of reference.
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PAllen
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K^2 said:
Instant turnaround is not critical. As long as the turnaround occurs in a small interval of proper time for the turnaround twin, with long inertial periods, the turnaround twin will observe blue shift/fast clock , throughout the inertial return phase. During this time, the observed passage of time on the distant clock will exceed that on their clock, yet, by Doppler removal reasoning, they conclude the distant clock is running slow this whole time. The only way out of this is to reject that they can use the Doppler removal assumption as soon as they are inertial. Instead, they must not use it until receiving images from the distant clock that they consider simultaneous to the beginning of their inertial phase. This will be quite some time after they become inertial. Otherwise, they are led to the nonsense conclusion that a clock running always slow during the inertial return phase ends up reading more time than theirs accumulates during this phase.
[Let me add: What is unambiguous is what the turnaround twin sees, for any inertial+acceleration+inertial round trip that is symmetric:
- they see the distant clock running slow for half of their time (the amount of visual slowdown decreases during first half of smoothed turnaround)
- they see the distant clock running fast for half of their time (the amount of speed up is seen to increase during during second half of smoothed turnaround)
There are many consistent approaches to interpreting this visual experience (= direct measurement). One that does not work is to say that the during the whole inertial return leg, the distant clock is running slow[edit2: well, you can if you completely decouple this statement from what is seen; as I propose above, you say that almost all of what is seen on the inertial return leg corresponds to delayed reception of events that occurred before the beginning of the inertial return leg] . Even with smoothed turnaround, this leg can be made to include almost all the time seen to elapse on the distant clock. Just draw a spacetime diagram with slightly smoothed turnaround, and figure the reception points of times on the stay at home clock. You can make the inertial return leg receive arbitrarily close to 100% the time of the home world line by making initial speed sufficiently close to c (fly by, if you will), and the turnaround sufficiently sharp.]
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ghwellsjr
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1) Since the traveling twin's clock is essentially stopped during the trip, it will end up 10 seconds behind the Earth twin's clock when they do the synchronization verification test you described.arindamsinha said:
But having answered this, it doesn't help me understand what you are saying.
- #31
ghwellsjr
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He does not say that "B could similarly be considered slower by the moving clock A" because A is not at rest in an Inertial Reference Frame (IRF).arindamsinha said:
There is something wrong with your thinking.arindamsinha said:
All IRF's will agree that the total amount of time difference per rotation of the Earth between the clock on the equator and the clock at the pole will be the same but they will not agree on the time dilations of the two clocks.
In the IRF in which the pole clock is at rest, it is not time dilated and the equator clock is time dilated by a constant amount, the same as the ratio of the accumulated times after one day.
But in other IRF's, the pole clock can have a constant time dilation while the equator clock has a fluctuating time dilation.
Observers cannot observe time dilation because it is a function of the chosen IRF.
- #32
ghwellsjr
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Time Dilation is not the same as Differential Aging.arindamsinha said:
The reason Differential Aging requires the two clocks to have a face-to-face meeting at the beginning and ending of the measurement interval is to remove any assumptions about clock synchronization. You may have found some specific examples where the Time Dilation factor for a particular Inertial Reference Frame (IRF) is equal to the Differential Aging factor, you cannot extrapolate to all other general examples. Even for the same example where it does work, picking a different IRF will make it not work.
- #33
ghwellsjr
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I have made some diagrams to illustrate this example. I'm assuming that each twin either sends a light signal to the other one every 15 minutes (quarter hour) or that we just pay attention to the image of each twin at 15-minute intervals.PAllen said:
To summarize your scenario, the traveling twin leaves at a high speed such that his observation of the Earth twin's clock is Doppler shifted to the point that he sees it running at 1/4 of his own on the outbound portion of his trip which takes one hour. (This fixes the speed of the traveling twin to be 0.882353c.) That means the traveling twin will send out four signals during this portion of the trip (not counting the initial one when he starts) and he will just be receiving the first signal from the Earth twin when he gets ready to turn around. Since the inbound portion of the trip also takes one hour, the inbound speed will be the same and the Doppler shift will be the reciprocal of the outbound Doppler shift. I realize that you were just using an example when you said that the traveling twin will see two hours pass on the Earth twin's clock but it is actually going to be four hours.
We can easily calculate the differential aging of the twins by simply taking the average of the two Doppler shifts since the time for the traveling twin is the same for the outbound and inbound portions of the trip. The average of 1/4 and 4 is 4.25 divided by 2 or 2.125. This is also the gamma at the speed of 0.882353c.
I show the Earth twin's progress as a heavy blue line with dots every 15-minutes indicating when he sends out a signal depicted as a yellow line traveling at c towards the traveling twin. The traveling twin is shown as a heavy black line with dots every 15-minutes indicating when he sends out a signal depicted as a thin black line traveling at c towards the Earth twin.
The first diagram is for the Earth's Inertial Reference Frame (IRF) in which the Earth twin is at rest and in which his clock is not time dilated so the blue dots are spaced identically to the coordinate grid lines at 15-minute intervals. However, since the traveling twin's speed is 0.882c, his clock is time dilated by gamma or 2.125 and the black dots are spaced at every 2.125 of the coordinate time. This is easy to see precisely if you consider the entire trip takes 2 hours or 8 quarter hours and if we multiply 8 by 2.125 we get 17 which is the coordinate time at the end of the trip.
The next diagram is for the IRF in which the traveling twin is at rest during the outbound portion of the trip. This diagram was generated by taking the coordinates of each event depicted in the first diagram and using the Lorentz Transformation equations to generate a new set of coordinates for an IRF moving at 0.882353c with respect to the first IRF. It depicts exactly the same information that was contained in the first diagram.
Note that now the black traveling twin's clock is not time dilated during the outbound portion of the trip and so the first five dots line up with the coordinate grid lines (look down in the lower right corner). However the blue Earth twin's clock is time dilated by the same amount that applied for the traveling twin in the first diagram. Also note that the Doppler signals are sent out and received in exactly the same way as in the first diagram even though they travel at c in this IRF and may have different distances to go.
The third diagram is for the IRF in which the traveling twin is at rest during his inbound portion of the trip. Again, this was generated simply by taking the coordinates for all the events from the first IRF but transformed to a speed of -0.882353c. Similar comments could be made about the how the clocks are time dilated differently than in the other two IRF's and yet all the Doppler signals are sent and received identically.
Great comments. Hopefully, this will enable arindamsinha to see the difference between Time Dilation and Differential Aging.PAllen said:
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Mentz114 said:
I explained this with the thought experiment in post #15. Co-location is not necessary for establishing differential aging. Signals exchanged at the speed of light between distant locations can establish the same.
What exactly is 'co-location' anyway? Can you define it in terms of 'observations' not using signals at the speed of light?
Mentz114 said:
I am trying to have a discussion on existing physics - terminologies and interpretations. Not looking for 'new physics' here, as you so sweetly call it.
If time dilation is just the 'apparent, or coordinate-dependent' effect, what use is it anyway, since it must be symmetrical between two bodies? As you mention, it is not 'physical', meaning it is an 'apparent' effect depending on 'where you observe it from' - analogous to 'parallax error'. Both are very readily understandable and correctable using Newtonian mechanics and common sense. It would appear using Doppler effect even in Newtonian mechanics.
"Differential aging" or "relative time dilation" is the real essense of relativity theory, in my opinion. I don't understand why people make such a big fuss about separating "time dilation" and "differential aging". When Einstein talked about "time dilation" he was clearly talking about "differential aging". If it was just a coordinate-dependent observational phenomenon, it would have been within the purview of Newtonian mechanics anyway, using Doppler effects.
ghwellsjr said:
Thanks for taking this up and answering the question.
I was trying to establish that velocity-based differential aging between two bodies (and agreement on the same by both), is not dependent on the moving twin coming back to origin to be 'co-located' with the stationary twin to 'compare clocks'. It can be done at any point during the traveling twin's journey using light-signals.
ghwellsjr said:
You misunderstood me. I said A in 'its own rest frame' (which of course is an IRF as well, and A is at rest in that IRF, in the situation considered). As I said, he didn't deny it, but just that he didn't stress it, and then went on to an example where there is a clearly established stationary and moving frame - which I found very interesting.
ghwellsjr said:
Subjective. Can you point out exactly what?
ghwellsjr said:
This is a SR situation we are discussing. So, without bringing in GR or acceleration, what prevents us from seeing the polar clock as 'rotating' w.r.t. an IRF fixed to a point on the equator?
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arindamsinha said:
By "fixed to a point on the equator", you mean that there is a point on the equator that is at rest in the frame, right? No such frame can be inertial because it is accelerating, and I see no way to discuss it without considering the acceleration.
(We need not bring in GR, of course)