Twin Paradox: Explaining the Relativity of Aging

[JesseM]

JesseM' Hello again. Re your post 100. I think there is great difficulty separating the conditions observed in the twins journey from the conditions used in the analysis.
You said '... you can calculate the time elapsed on each segment using the time dilation equation and then just add the two times..."
If you use this procedure aren't you excluding any external forces or accelarations from your analysis?
The time dilation equation was derived for two coordinate systems in constant relative velocity v. If you use this equation as you said aren't you treating the two clocks as inertial?

I'm saying one clock has a polygonal path consisting of different inertial segments--for example, the clock might have moved at constant velocity 0.6c in one direction for 20 years, then instantaneously changed velocity so it was moving at velocity 0.8c in the opposite direction, and continued at this constant velocity for another 15 years. Einstein talked about this sort of polygonal path in section 4 of the 1905 paper. Are you disagreeing that in SR we should be able to calculate the total time elapsed on this sort of polygonal path just by using the time dilation equation to calculate the time elapsed on each segment, i.e. 20*sqrt(1 - 0.6^2) + 15*sqrt(1 - 0.8^2), which gives a sum of 25 years? The acceleration phase between these two inertial phases is assumed to be instantaneous, so it shouldn't cause any sudden change in the clock's reading. The only way this sort of sum wouldn't give the right answer is if the clock had some kind of "memory" that it had been accelerated which affected its rate of ticking, so even if on the second segment it was traveling right next to a clock that had been moving inertially at 0.8c for all time, somehow its past history would make it so that its rate of ticking didn't match that of the inertial clock it was traveling alongside. Perhaps this wouldn't explicitly contradict either of the 2 basic postulates of SR, but the fact that a clock's time dilation is predicted purely by its velocity is something that can be tested experimentally, I think to a pretty high degree of accuracy.

So the analysis you propose ignores turn around effects and treats both clocks as inertial, just as the 1905 paper does. So what is the purpose of insisting that one clock is not inertial, again?

The 1905 paper doesn't ignore acceleration, rather in section 4 Einstein is effectively proposing the postulate that any accelerated path can be taken as the limit of a polygonal path as the length of the segments goes to zero, along with the postulate that for a polygonal path with instantaneous accelerations we can assume the instantaneous accelerations don't change the reading so the time elapsed is just the sum of the time elapsed on each segment. It's true that these postulates might not follow directly from the two basic postulates of SR, but they certainly seem pretty plausible as additional postulates if you accept the basic SR postulates, and ultimately the real test of both the basic SR postulates and these additional postulates is experimental evidence, which supports all of them quite well.

 

[JM]

To whom...
This has been a long thread. It may be useful to gather the features of SR that apply to the twin paradox as descrebed in par. 4 of the 1905 paper " on the Electrodynamics of Moving Bodies'.
A.The path of the moving clock is given as a constant speed motion along a closed path of connected straight line segments beginning at, and endding at, the same point. The path is describes as 'polygonal' which suggestrs many more than two segments.
B. The time of the moving clock is calculated by entering the expression x = vt, representing the position of the clock at the origin of the moving coordinates, into the Lorentz equation relating 'moving' time to 'stationary' time. These transfforms were derived under the condition that both coorddinates are inertial, i.e. moving in a straight line at constant relative speed v. The use of this transform implies the following assumptions:
1.The behavior of the moving clock when moving along the polygonal patthis the same as when moving along a straight inertial line,
2.The forces required to constrain the clock to the polygonal line and any accelerations are excluded from the analysis. This is consistent with the kinematic procedure named in the introduction of the paper.
3. Both coordinates/clocks are represented as inertial for the polygonal path, as they are for the straight line.
C. This last condition, along with the idea that there is no location of absolute rest, allows each twin/clock equal entitlement to consider himself to be at rest.
D. The original question for this thread was whether it is possible to determine who went for the trip and who didn't. The answer, according to the 1905paper, is no. Each twin caan equally consider himself to be at rest and the other one to be traverling.

 

[Ich]

The answer, according to the 1905paper, is no.

The answer, according to the 1905 paper, is yes. It's explicitly given.
The answer, according to your understanding, may be no. But then, that's wrong.

 

[ZikZak]

D. The original question for this thread was whether it is possible to determine who went for the trip and who didn't. The answer, according to the 1905paper, is no.

Let's see if that is, in fact, true.

...if the clock at 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...

if one of two synchronous clocks at A is moved in a closed curve with constant velocity until it returns to A, the journey lasting t seconds, then by the clock which has remained at rest the traveled clock on its arrival at A will be 1/2 tv^2/c^2 second slow.

No, it looks like you are reading the Bizarro Version of the 1905 paper, because in the real 1905 paper, Einstein explicitly states which clock arrives reading behind the other. Please stop putting (wrong) statements into Einstein's mouth. He is not here to defend himself.

 

[JesseM]

To whom...
This has been a long thread. It may be useful to gather the features of SR that apply to the twin paradox as descrebed in par. 4 of the 1905 paper " on the Electrodynamics of Moving Bodies'.
A.The path of the moving clock is given as a constant speed motion along a closed path of connected straight line segments beginning at, and endding at, the same point. The path is describes as 'polygonal' which suggestrs many more than two segments.

Sure it does, I was just picking an example of a polygonal path with two segments to make writing out the sum easier. I'm sure if Einstein postulated that we can just calculate the elapsed time on each segment using time dilation and add them together to get total time elapsed in the case of a polygonal path with many segments, he'd say the same about a polygonal path with just two segments--do you disagree?

As a matter of fact, in this paragraph below from section 4 he seems to be talking about a clock A that's originally at rest relative to B and then is accelerated once, moving inertially at constant velocity v towards B after the acceleration--that would in fact be a polygonal path with two segments.

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at 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)*tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

B. The time of the moving clock is calculated by entering the expression x = vt, representing the position of the clock at the origin of the moving coordinates, into the Lorentz equation relating 'moving' time to 'stationary' time.

You could use the full Lorentz transformation for the event (t, vt) to find the time t' of that event in the moving frame (assuming the clock was initially at (0,0)), but you can also just use the time dilation equation t_{your frame} = t_{moving clock} / sqrt(1 - v^2/c^2), which you can rearrange to solve for time on the moving clock if you know the time t in your frame: t_{moving clock} = t_{your frame} * sqrt(1 - v^2/c^2). Of course the time dilation equation is derived directly from the Lorentz transformation, so this is just a shortcut.

These transfforms were derived under the condition that both coorddinates are inertial, i.e. moving in a straight line at constant relative speed v. The use of this transform implies the following assumptions:
1.The behavior of the moving clock when moving along the polygonal patthis the same as when moving along a straight inertial line,

Yes, more specifically, the assumption is that during each segment of the polygonal path, the clock would be ticking at the same rate as a clock at rest next to it that had been moving inertially for all time.

2.The forces required to constrain the clock to the polygonal line and any accelerations are excluded from the analysis. This is consistent with the kinematic procedure named in the introduction of the paper.

Yes, given that the accelerations are instantaneously brief in the case of a polygonal path, it seems reasonable to postulate that the accelerations won't cause any sudden jump in the clock reading. And in any case, the postulate that the rate of ticking at each moment depends solely on the velocity at each moment is a testable one, and experimentally it seems to be true.

3. Both coordinates/clocks are represented as inertial for the polygonal path, as they are for the straight line.

What do you mean? Each segment is inertial, but the entire polygonal path is clearly not inertial, since the clock does not remain at rest throughout the journey in any inertial frame.

C. This last condition, along with the idea that there is no location of absolute rest, allows each twin/clock equal entitlement to consider himself to be at rest.

No! Where are you getting that idea? Einstein clearly defines the meaning of an inertial frame in his paper, and a clock moving on a polygonal path is clearly not the sort of clock he was talking about when he imagined constructing an inertial frame using a network of inertial clocks. In particular, if you tried to construct a coordinate system using a network of clocks moving on polygonal paths and then write the equations for the laws of physics in such a coordinate system, it would not be "a system of co-ordinates in which the equations of Newtonian mechanics hold good" as he says at the beginning of section 1.

 

[JM]

To Ich and ZikZak: OK, OK, here's the missing piece. Refer to post 65 and the reference cited there, Relativity 1952 by Einstein. He says that all inertial frames are equal, or in my paraphrase, the carriage and the embankment are both equally entitled to consider themselves at rest and the other to be in motion. The 1905 paper gives results for only one being at rest. Remember the saying ' moving clocks run slow'? If A is at rest he sees B's clock to be slow, and if B is at rest he sees A's clock to be slow. Apply that to each twin (considered to be inertial as described in my recent post ) and you get the same result for each.
To JesseM: I see many points of agreement. But let's focus on the segments. You agree that each segment can be considered to be inertial,and the time dilation calculated using the formula for two inertial clocks, yes? If you do this for each segment and add the results for the total trip, isn't that the same as calculating the time dilation for a single segment of the same length as the round trip? Thats what I'm thinking.

 

[JesseM]

To JesseM: I see many points of agreement. But let's focus on the segments. You agree that each segment can be considered to be inertial,and the time dilation calculated using the formula for two inertial clocks, yes? If you do this for each segment and add the results for the total trip, isn't that the same as calculating the time dilation for a single segment of the same length as the round trip? Thats what I'm thinking.

The answer is the same only if the speed is constant on each segment and the direction is the only thing that changes, from the perspective of whatever frame you're using. If the speed isn't constant, then the time dilation factor is obviously different in different segments.

Even if the speed is constant--if the traveling twin travels away at 0.6c for 10 years and then travels back at 0.6c for another 10 years, from the perspective of the Earth twin--the situation is still fundamentally different from a purely inertial trip at 0.6c for 20 years to a destination 12 light years away (as opposed to the polygonal trip where the traveling twin ends up back at Earth). In the case of a purely inertial trip the Earth twin could calcuate that the traveling twin has only aged 20 years * sqrt(1 - 0.6^2) = 20*0.8 = 16 years when he reaches the destination 12 light years away, but the situation is symmetrical, the traveling twin can also correctly conclude that in his own inertial rest frame, after 16 years he reaches the destination but the Earth twin has only aged 16*0.8 = 12.8 years. This has to do with the relativity of simultaneity and the fact that different inertial frames disagree about whether two events that are separated in space are "simultaneous" or not--in the Earth-twin's inertial rest frame, the event of the traveling twin's clock showing 16 years is simultaneous with the event of the Earth-twin's clock reading 20 years, but in the traveling twin's inertial rest frame, the event of the traveling twin's clock showing 16 years is simultaneous with the event of the Earth-twin's clock reading 12.8 years.

It's because of this relativity of simultaneity that the situation is not symmetric if the traveling twin was taking a polygonal path, so the traveling twin can't calculate how much the Earth twin has aged just by adding the amount the Earth twin ages during the first segment in the traveling twin's inertial rest frame #1 during the first segment + the amount the Earth twin ages during the second segment in the traveling twin's inertial rest frame #2 during the second segment. For example, suppose the traveling twin is observed in the Earth twin's frame to move away at 0.6c for 10 years, aging 8 years during this time while the Earth twin ages 10 years, then return at 0.6c for another 10 years, aging another 8 years while the Earth twin ages another 10 years. In the traveling twin's inertial rest frame #1 where the traveling twin is at rest during the first half of the journey, the event of the traveling twin's clock reading 8 years (when the traveling twin turns around) is simultaneous with the event of the Earth twin's clock reading 6.4 years. But then in the traveling twin's inertial rest frame #2 where the traveling twin is at rest during the second half of the journey, the same event of the traveling twin's clock reading 8 years (and the traveling twin turning around) is simultaneous with a totally different event on the Earth twin's worldline, when the Earth twin's clock reads not 6.4 years but 13.6 years. So, during the 8 years it takes in this frame for the traveling twin to return to Earth, the Earth twin will age an additional 6.4 years, meaning the Earth twin's clock will show 13.6 + 6.4 = 20 years when the traveling twin returns. So the traveling twin cannot just say "well, in my inertial rest frame #1 before I turned around, the Earth twin aged 6.4 years from the moment I left Earth until the moment of the turnaround, and in my inertial rest frame #2 after I turned around, the Earth twin aged another 6.4 years from the moment I turned around to the moment I returned to Earth, therefore the Earth twin should have aged 6.4 + 6.4 = 12.8 years when I return". This calculation would ignore the difference in simultaneity which means the two frames have very different ideas about how old the Earth twin is at the moment the traveling twin turns around, and thus this calculation would fail to take into account the "missing" 7.2 years between the Earth twin's clock reading 6.4 and the Earth twin's clock reading 13.6.

 

[phyti]

Dr Gregg;

"No. Neither is the train having a force applied to it (1st law)"

It requires a force to balance the frictional forces. If the engine stopped delivering power to the wheels, the train would come to a stop.

 

[DrGreg]

Dr Gregg;

"No. Neither is the train having a force applied to it (1st law)"

It requires a force to balance the frictional forces. If the engine stopped delivering power to the wheels, the train would come to a stop.

I agree. I was referring to the total force, what you get when you add all the forces acting on the train; it is zero. And the total of all the forces acting on the embankment is zero too.

By the way, it helps if you use the QUOTE button when quoting what I've said (you can then delete irrelevant parts of the quote); then I can click on the arrow to find the original post in which I said it.

 

[phyti]

I agree. I was referring to the total force, what you get when you add all the forces acting on the train; it is zero. And the total of all the forces acting on the embankment is zero too.

By the way, it helps if you use the QUOTE button when quoting what I've said (you can then delete irrelevant parts of the quote); then I can click on the arrow to find the original post in which I said it.

It was one of those word things!
Normally use the quote button, except for 1 or 2 liners, but you make good point for back tracking.