Twin Paradox Problem: Do Twins Age Differently?

  • #251


kamenjar said:
I don't think that's it's a valid example. I thought someone "devised" a formula that you integrate velocity times the traveled distance over the duration of time to deduce reduction in aging over a period of time...

Wouldn't you say that is exactly what the Lorentz transformation or it's alternate form the line element does?
 
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  • #252


kamenjar said:
I don't think that's it's a valid example. I thought someone "devised" a formula that you integrate velocity times the traveled distance over the duration of time to deduce reduction in aging over a period of time...

It is a valid example. Apply the formula, and you will see that the example is valid. Specifically, if you call the longer, west moving distance w, and the shorter east moving distance e, then for long T (elapsed time between separation and meeting), the total time dilation for the slow twin is O(w^2/T); for the near lightspeed short trip twin, it is O(e). For any e, however, small, and any w, however large, you can make time dilation for the long distance traveler less by making T sufficiently large. Then, the twin that travels the longer distance ages more.
 
  • #253


kamenjar said:
I thought that OP was looking for an answer. Not what's possible or impossible. It doesn't mean that we can't use a third reference frame, which for all purposes would be the Earth twin's reference frame because CMB is quite isotropic in his.

So we're in the market for impossible answers? Not on this forum.
kamenjar said:
I said, anywhere where CMB is isotropic. It's definitely not in inertial observer's frame. Or is there something else I didn't specify.

CMB is an artifact of an expanding universe. In an expanding universe, distant clocks stationary with respect to CMB perceive each others clocks to be slow. No one clock can be picked out as the fastest in the universe (your words).
kamenjar said:
What the observer sees/measures is different what it actually is. We see all kinds of things that aren't. We see/measure some galaxies that appear to move at speeds faster than speed of light, but it's not what it is.

We don't measure any galaxies to be going faster than light. A recession velocity is not a relative speed. It is analogous to the growth in distance between to bodies traveling near c away from each other in some frame. In that frame, the growth in proper distance between them is 2c. Recession velocity is the same sort of thing. As for relative velocity, GR says it cannot be defined at all for distant objects.

Let me ask again: please specify in some clear way what clock is the fastest clock in the universe. So far, you have not proposed anything that begins to make sense, and SR and GR say there cannot be such a thing.
 
  • #254


kamenjar said:
I actually don't. Convincing anyone anything with exception of having them only understand your point of view, is in my personal opinion wrong. And that task in itself is quite difficult. We all have different beliefs and are not equally open to other's views.

I thought that it would be helpful to break the symmetry. I don't know yet how exactly, but I'm thinking along the lines of:
1) Each observer looks at CMB, compensates for clock rate (symmetry is broken here) to establish their clock rate.
2) Each observer looks at angular size of the object they are observing. "Sizes" are known at rest.
3) Each observer looks at the light signal's doppler shift and clock reading.
4) Given the above three values, the observer projects their location and the other's path in space WRT the CMB frame.
5) Given above value, the observer can compensate for the delay of the signal received form the other twin and only then make a conclusion about the other's clock rate.

There is only one way, sort of, to do this (ignoring GR/gravity). Pick any arbitrary inertial clock, and its rest frame and call it universal (say, pick the clock tower at Emerald City). Declare, by fiat, that all other clocks times are not 'real', and the time assigned by this chosen clock's frame to any event is the 'real' time. You can go ahead and do this, but you have not explained anything by doing so.

Try to define a universal time in any less arbitrary way, and you will fail.
 
  • #255


kamenjar said:
Which is what I said - a frame with the "absolute clock rate". No?
No, not even close. You said such a clock was the fastest ticking clock in the universe, which is not true in general.
 
  • #256


um hello everyone as far as i think there can't be something called fastest clock in the universe, i have read this thread and was curious to join.
As i think if both twins see each other at same speed then the fact in aging maybe that whoever enters other frame of reference ie it depends on whether the Earth twin approaches the moving twin or vice versa will cause the difference in aging it's just a rough idea though.
 
  • #257


PAllen said:
...SR and GR say there cannot be such a thing.
I know that exactly.
DaleSpam said:
No, not even close. You said such a clock was the fastest ticking clock in the universe, which is not true in general.
I may be wrong, but give me an example of a faster clock.
The moving twin ages less - clock is slower. CMB is distorted.
Gravity field makes you age less - clock slower. CMB is distorted.
What are you suggesting? White holes and antigravity may exist only in theory. We haven't seen it yet.
 
  • #258


PAllen said:
CMB is an artifact of an expanding universe. In an expanding universe, distant clocks stationary with respect to CMB perceive each others clocks to be slow. No one clock can be picked out as the fastest in the universe (your words).
...
My words are also that each clock sees another to be slow. If I was anywhere in the galaxy where CMB is isotropic right now (and "right now" is a big term here), the rate of my clock would be equal to the one I'm looking at right now (+/- gravity galaxy movement etc.). The rate of such stationary clock is the fastest rate of a clock possible.

Edit: In other words - There is no possible way to make me age faster - hence it is the fastest clock.
 
  • #259


kamenjar said:
My words are also that each clock sees another to be slow. If I was anywhere in the galaxy where CMB is isotropic right now (and "right now" is a big term here), the rate of my clock would be equal to the one I'm looking at right now (+/- gravity galaxy movement etc.). The rate of such stationary clock is the fastest rate of a clock possible.

Edit: In other words - There is no possible way to make me age faster - hence it is the fastest clock.

Sure there is. Any distant clock isotropic with CMB sees you age slower. Who is right? You say seeing doesn't count. Then what does? Any measurement each does to compare their rate with another, shows the other is slower.
 
  • #260


seiginotekken said:
um hello everyone as far as i think there can't be something called fastest clock in the universe, i have read this thread and was curious to join.
As i think if both twins see each other at same speed then the fact in aging maybe that whoever enters other frame of reference ie it depends on whether the Earth twin approaches the moving twin or vice versa will cause the difference in aging it's just a rough idea though.
It should be symmetrical in that way. They have to take one frame of reference and use that to find out who is actually the one moving.
PAllen said:
There is only one way, sort of, to do this (ignoring GR/gravity). Pick any arbitrary inertial clock, and its rest frame and call it universal (say, pick the clock tower at Emerald City). Declare, by fiat...

I don't have to do any of that, I can just use common sense. If I make a trip to Centauri that lasts 1 year:
- When I get there, i'll look at your clock in Emerad City and see t0+1y. My clock shows t0+1y. I am not stupid, so I know that the picture of your clock took 4y to arrive, so I conclude that if you didn't move from your couch in 4 years, it must be that your clock that's running faster.
- When you see me arrive at Centauri, your clock will show t0+9y and you'll see that the picture of my clock shows t0+1y. You are not stupid either, so you conclude the same way that my clock is running slower.
- In fact we can do this for any leg of the trip, without needing Mikowski, or Einstain, or his wife -- which was also Serbian by the way - like myself and Tesla :) -- to conclude whose clock is running slower.

Why is such a simple thing made so complicated? I don't understand.
 
  • #261


kamenjar said:
I don't have to do any of that, I can just use common sense. If I make a trip to Centauri that lasts 1 year:
- When I get there, i'll look at your clock in Emerad City and see t0+1y. My clock shows t0+1y. I am not stupid, so I know that the picture of your clock took 4y to arrive, so I conclude that if you didn't move from your couch in 4 years, it must be that your clock that's running faster.
- When you see me arrive at Centauri, your clock will show t0+9y and you'll see that the picture of my clock shows t0+1y. You are not stupid either, so you conclude the same way that my clock is running slower.
- In fact we can do this for any leg of the trip, without needing Mikowski, or Einstain, or his wife -- which was also Serbian by the way - like myself and Tesla :) -- to conclude whose clock is running slower.

Why is such a simple thing made so complicated? I don't understand.

Ok, let's say you pass right by Earth at high speed and synchronize clocks. Then reach Centauri, who has previously synchronized clocks with earth. You see these clocks as totally non-synchronized. If you factor out non-relativistic Doppler and travel time, you consider both Earth and Centauri clock going slow. When you pass by Centauri, you see its clock ahead of yours, but that is only because it started way ahead of yours due to 'erroneous' synchronization done by earth/Centauri. So which clock is faster? No experimental procedure can say.

Your argument shows absence of any understanding of relativity.

(Note: you create an asymmetry by having one clock accelerate and decelerate. If they both keep moving the way the always were, they each think the other is slower, and you can't pick out which is faster. If you try to bring in CMB, then you have my unanswered question of which CMB-isotropic clock is faster, since they each one thinks it is faster than all others.)
 
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  • #262


PAllen said:
Sure there is. Any distant clock isotropic with CMB sees you age slower. Who is right?
If they assume that our clocks ARE slower and we conclude the same, then nobody is right. They should use common sense and account for space expansion. After you do that, you conclude that clocks are the same.
PAllen said:
You say seeing doesn't count. Then what does? Any measurement each does to compare their rate with another, shows the other is slower.
First of all, we can not know anything for sure about them. We can only compensate for space expansion and draw conclusions based on that. We only see a delayed picture of them. We can not conclude that their clock is slower than ours now. I use common sense and assume that universe is homogeneous, so I can assume that right now their clock ticks at the same rate unless they are moving or near a larger gravity well.

If you were blind, could you say anything about my clock "right now" by just hearing its ticks? If you thought yes - then what if I was 300m form you? You only hear a 1-second delayed sound, so you can't say anything about "right now". If you even saw me, it wouldn't be "right now" because you are looking at everything around you with a delay.
 
  • #263


PAllen said:
Ok, let's say you pass right by Earth at high speed and synchronize clocks. Then reach Centauri, who has previously synchronized clocks with earth. You see these clocks as totally non-synchronized. If you factor out non-relativistic Doppler and travel time, you consider both Earth and Centauri clock going slow. When you pass by Centauri, you see its clock ahead of yours, but that is only because it started way ahead of yours due to 'erroneous' synchronization done by earth/Centauri. So which clock is faster? No experimental procedure can say.

Your argument shows absence of any understanding of relativity.

(Note: you create an asymmetry by having one clock accelerate and decelerate. If they both keep moving the way the always were, they each think the other is slower, and you can't pick out which is faster. If you try to bring in CMB, then you have my unanswered question of which CMB-isotropic clock is faster, since they each one thinks it is faster than all others.)
I just wrote a lengthy response and the site logged me out, so I lost it...

I'll try again but it will be shorter and summarized...

Mehod of Synchronization:
Define "real" time to be such that when remote clock is viewed from each system to be showing "real"-4years. We can say that "real" clock is set in such way that when local time on Earth is 10/19/2012, the clock seen at remote system Centauri. is seen as 10/19/2008. The same way, when local time on Centauri is 2012, the time observed on on Earth form that system is 2008. However they know (even though they can't see) with some degree of certainty that time "now" on Earth is the same as on their system.

Your scenario:
I just pass Earth and see Centauri clock read 2008, but is real 2012, Earth 2012, my watch 2012. Nothing odd. All times in sync.

I arrive at Centauri. Local time is 2017, Earth clock reads 2013, but is 2013+ 4= 2017, my (slow) watch is 2013 -I aged a year, it took me 5 "real" years to get there. Nothing odd. All times correct and equally faster than my slow watch.

I arrive at Earth. Local time is 2022, Centauri clock reads 2018, but is 2018+ 4= 2022, my (slow) watch is 2014 -I aged a year, it took me another 5 "real" years to get back from Centauri. Nothing odd. All times correct and equally faster than my slow watch.

Your statement about my knowledge of relativity:
I don't need to know more than this to be leading this argument and be correct in my view of the universe.

Your CMB comment about accelerating, moving, and decelerating a clock to make it offset:
If you are saying that you just moved a clock arond to make it slower then you returned it to the same place where the standing clock is? You just made it offset. Nothing else. It was running slower while you were moving it around and that's why it's offset. Nothing strange there and no paradoxes.

Edit: One possible reason for argument: non-english language barrier.
In English, someone may say that a clock is running "fast" if it is showing 12:05 and correct time is 12:00. However no statement should be made about the rate of the clock by only observing it once. Maybe it "was" running wast previously, or someone set it wrong. The clock is just wrong/offset at the given time. We can only say that it's running fast (at a faster rate) of it is off by even more the next morning. So let's please be consistent about wording if anyone is interpreting "fast" as "single positive offset reading".
 
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  • #264
PAllen said:
I was referring to SR. You can have accelerated observers in SR. You can compute what they observe in an inertial frame. You can also construct various 'non-inertial' coordinates. Mainly, I was arguing in favor of terminology with invariant meaning: ( inertial vs non-inertial; path of shorter/longer proper time) for example. [..]
Of course you can have accelerated observers in SR; and in SR, acceleration is not rest. :wink:

Here's how Einstein clarified, in the context of the twin paradox, the use of SR (which is a label that he invented):

"according to the special theory of relativity the coordinate systems K and K' are by no means equivalent systems. Indeed this theory asserts only the equivalence of all Galilean (unaccelerated) coordinate systems, that is, coordinate systems relative to which sufficiently isolated, material points move in straight lines and uniformly. K is such a coordinate system, but not the system K', that is accelerated from time to time."
- https://en.wikisource.org/wiki/Dialog_about_Objections_against_the_Theory_of_Relativity

(note to Dalespam: another definition of "Galilean coordinate systems").

My point was that just as in classical mechanics, in SR an accelerated observer is not "in rest" - that is wrong terminology for both. The traveler can certainly say to be in rest in his capsule, but he can not "prefer to think to have traveled zero distance", for that equals "being in rest all the time", and thus being "unaccelerated all the time".

As this subtle point is at the heart of the "twin paradox", it is not something to overlook.
 
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  • #265


Hi All
I have a somewhat different analogy to offer:

The twins Bill and Bob are driving in separate cars down a long straight highway from point A to b
Bob drives the whole way at a constant 50mph. while bill is erratic sometimes driving at 20 , others at 90 etc.etc.

They are both traveling the exact same path ,in the same direction. The only difference between them is their relative motions over the course.

It is obvious that upon meeting they must both have the same average speed for the trip.

As Bob's speed was constant this means that the overall portion of the journey that Bill was traveling faster than Bob must be exactly balanced by an equivalent portion where he was traveling slower in order for him to arrive simultaneously with Bob at B

Assume that the clocks in the cars are linked to the speedometers in such a way that the clock rates decrease as the speedometer readings increase. From a base rate at 0 mph

It is clear that if the correlation is linear that the overall elapsed times would be the same.

The difference in accumulated time where Bill was going faster (with a slower clock) would likewise be exactly countered by an equivalent cumulative difference when his clock was ticking faster.

But of course the physics of time and velocity as expressed by the gamma function is not linear it is exponential.
The amount of retardation for a given increase in velocity getting proportionately larger with higher velocities

SO the slowing of Bill's clock when traveling faster than Bob must be greater than the increase in rate during the portion of the trip when he is traveling slower.This necessarily results in a larger cumulative difference in the fast phase than the reciprocal cumulative difference resulting from the phase moving slower.than Bob.

The end result being less total elapsed time on Bill;s clock, even though the paths and relative average velocities were equivalent and perfectly symmetrical.

Of course the highway is any inertial frame and the above relationships hold for any constant velocity for Bob and any accelerations or velocities for Bill. Acceleration per se and relative simultaneity are completely irrelevant.
It is simply an inevitable consequence of the condition of initial and final co-location,relative motion and the intrinsic asymmetry of the gamma function.
It appears to me that it is the non-linearity of this function which is the root of the asymmetric difference in aging (elapsed time) in any separation and rejoining of worldlines.

Of course this scenario is not the general case and was chosen for maximum symmetry and equivalence. Many other situations would result in unequal path lengths and average speeds but the principles still apply in all cases.
 
  • #266


seiginotekken said:
um hello everyone as far as i think there can't be something called fastest clock in the universe, i have read this thread and was curious to join.
As i think if both twins see each other at same speed then the fact in aging maybe that whoever enters other frame of reference ie it depends on whether the Earth twin approaches the moving twin or vice versa will cause the difference in aging it's just a rough idea though.
Hi seigino, Welcome to physicsforums :smile:

What you perhaps meant is that, with reference to "Galilean frames" (see my post #264 here above), the twin who undergoes a change of velocity is the one who will be found to have aged less. That is correct because no matter which Galilean frame you use as reference, that twin undergoes more time dilation over the course of travel.

However, if you meant that it has to do with distance, then that is wrong. In special relativity, distance plays no role in aging. A Galilean frame has infinite distance (it is a construction of the mind, an infinite extension of a material inertial body), so that everyone is always in all reference frames.

How that exactly works follows directly from the time dilation equation, and some people find it helpful to picture this in space-time diagrams.
 
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  • #267


Rishavutkarsh said:
ghwellsjr said:
Yes, it is, if you are referring to your comment, "the important thing was that the traveling one was noninertial and the other observer was inertial". It is neither important, relevant, significant, nor is it a requirement for the Twin Paradox.

In the simplest presentation of the Twin Paradox, we talk about the Earth twin as if the Earth had no gravity and no acceleration, which are of course not true and so the Earth and the Earth twin are considered to be inertial for the purpose of discussing the scenario. In this simplest presentation the other twin is non-inertial and so without knowing anything else, we can always say that the traveling twin is the one who ages less because he experienced acceleration whereas the Earth twin did not. This leads many people to falsely jump to the conclusion that it is the acceleration that causes the difference in aging between the twins and they look for explanations (a non-inertial frame or jumping between two frames) that support that incorrect idea. These people tend to reject the simple explanation that I offered in post #42 and quoted in post #52.

But we can complicate the Twin Paradox by having both twins follow exactly the same accelerations. They can both take off in identical spaceships and achieve 90%c. Then one of them immediately turns around and lands back on Earth for the rest of the time while the other twin continues on for a long time before repeating what the first twin did. So now both twins are non-inertial in exactly the same way so we can't jump to the false conclusion that the acceleration is what caused the difference in aging. But we can still analyze this more complicated Twin Paradox using the same inertial Earth reference frame as before. It's the simplest one to use, mainly because it is the one that is used to describe their motions.

In fact, it doesn't matter how the two twins accelerate or what speeds they achieve or what directions they travel in (polygons or circles or some of each) or where they end up together. We can always analyze their individual aging during the entire process from the standpoint of the inertial frame that we use to describe their activities.

Furthermore, with a little more work, we can determine what each twin sees of the other ones clock during the entire scenario, not just at the beginning and the end, and any analysis we do (whatever frame(s) we use) will all provide the same answers. We can make this as complicated as we want. But the more complicated analyses will not provide any more insight or information into what is happening.
so you mean that in fact acceleration has nothing to do with this and it's just used to simplify this. so would you be kind enough to tell us what exactly is seen by the both twins and why? and is acceleration absolute or relative?
Acceleration is not like velocity which is relative. Velocity is something that you cannot sense except by reference to something outside you that is not comoving with you. But you can tell if you are accelerating or not. When you step on the accelerator in your car, you can feel a push into the back of your seat.

Now I didn't say acceleration has nothing to do with this, I said acceleration is not what causes the difference in aging, but it can cause a difference in the speed of an observer as defined by the frame of reference. So if one twin accelerates, only his speed can change (and not the other one), so his time dilation can change (and not the other one). Time dilation means that time for him runs at a slower rate than the coordinate time of the frame of reference. In order for this to result in an age difference, the twin has to maintain this new speed for an extended period of time.

It's kind of like an alarm clock that I have that plugs into the wall which provides perfect time keeping. But if I unplug it to take on a trip, it reverts to a battery operated internal electronic circuit that happens to run slow and when I get where I'm going and plug it back in, it is off by several minutes. The longer it takes for me to get there, the more it is off. If I unplug the clock and move it to another room and plug it right back in, it has not lost enough time to notice.

So do you see the point? In order for a difference in aging to occur, the clock has to be time dilated for an extended period of time, the longer the time and the greater the time dilation, the more the age difference will be.

Now you also asked about what would both twins see of the other ones clock during the trip. I only answered this question completely for the traveling twin in post #5:
ghwellsjr said:
When the traveling twin departs at 90%c, they both do see each others clock equally slow--a factor of 0.2294 times their own. But this is true only for the outbound portion of the trip. Things are different for the inbound portion of the trip. As soon as the traveling twin turns around, he immediately sees the Earth twin's clock going fast--4.359 times his own. Since he spends an equal amount of time going out as coming in, you can easily calculate how much of a difference there will be in the amount the two twins aged by simply taking an average of the two factors. The average of 0.2294 and 4.359 is 2.2942 so however much the traveling twin aged during the trip, his Earth twin will age 2.2942 times as much. Simple, isn't it?
So as the traveling twin looks back at the Earth twin's clock prior to turning around, he sees it ticking at a rate that is 0.2294 times his own. After he turns around, he sees it ticking at a rate that is 4.359 times his own. The traveling twin can keep a list of what times he sees on his own clock and what times he sees on the Earth twins clock. (Every frame will agree with this list.)

But what does the Earth twin see? He also sees the traveling twins clock ticking at 0.2294 times his own but he continues to see this slowed-down tick rate for more than half of the entire trip. In fact, it isn't until near the end that he finally sees his twin turn around and start heading for home and it's at this point that he sees his twin's clock running faster than his own by a factor of 4.359, the same as his twin sees of his clock. When his twin gets back home a short time later, he will see that his clock has advanced by 2.2942 times what the traveling twin's clock has advanced by. The two twins will, of course agree. Now the Earth twin can keep a list of what times he sees on his own clock and what times he sees on the Earth twins clock and every frame will agree with this list.

So what is it that causes a difference in the aging of the twins even though they both see the other ones clock experiencing exactly the same Doppler factors? It's simply the fact that one of them sees the switch from slow rate to fast rate at the 50% mark of the trip while the other one doesn't see the same thing happen until much later in the trip.

With the information about how each twin observes the other ones clock compared to his own, can you figure out at what percentage of the trip will the Earth twin see his brother turn around?
 
  • #268


kamenjar said:
...If I make a trip to Centauri that lasts 1 year:
If it takes 1 year by your clock, then your speed was 97%c.
kamenjar said:
- When I get there, i'll look at your clock in Emerad City and see t0+1y.
No, you won't see that one year has passed, you will only see about a month and a half has passed. At 97%c, the Relativistic Doppler Factor is about 8.1 so you divide your 1 year by 8.1 to see how much time has advanced on the Earth clock.
kamenjar said:
My clock shows t0+1y. I am not stupid, so I know that the picture of your clock took 4y to arrive, so I conclude that if you didn't move from your couch in 4 years, it must be that your clock that's running faster.
You may not be stupid but you are ignorant of how to do these calculations.
kamenjar said:
- When you see me arrive at Centauri, your clock will show t0+9y and you'll see that the picture of my clock shows t0+1y. You are not stupid either, so you conclude the same way that my clock is running slower.
Wrong again, the Earth clock will be at just over 8 years. Remember, you are traveling at almost the speed of light so it takes you just over 4 years to get there and 4 years for the image of you to get back to earth.
kamenjar said:
- In fact we can do this for any leg of the trip, without needing Mikowski, or Einstain, or his wife -- which was also Serbian by the way - like myself and Tesla :) -- to conclude whose clock is running slower.

Why is such a simple thing made so complicated? I don't understand.
I think we all agree that you don't understand.
 
  • #269


kamenjar said:
...
Mehod of Synchronization:
Define "real" time to be such that when remote clock is viewed from each system to be showing "real"-4years. We can say that "real" clock is set in such way that when local time on Earth is 10/19/2012, the clock seen at remote system Centauri. is seen as 10/19/2008. The same way, when local time on Centauri is 2012, the time observed on on Earth form that system is 2008. However they know (even though they can't see) with some degree of certainty that time "now" on Earth is the same as on their system.

Your scenario:
I just pass Earth and see Centauri clock read 2008, but is real 2012, Earth 2012, my watch 2012. Nothing odd. All times in sync.

I arrive at Centauri. Local time is 2017, Earth clock reads 2013, but is 2013+ 4= 2017, my (slow) watch is 2013 -I aged a year, it took me 5 "real" years to get there. Nothing odd. All times correct and equally faster than my slow watch.
No, it only took you just over 4 "real" years to get there, 4 years plus a month and a half. So when you get there, the local time is early December 2016, not mid October 2017. And the Earth clock reads early December 2012, not mid October 2013.
kamenjar said:
I arrive at Earth. Local time is 2022, Centauri clock reads 2018, but is 2018+ 4= 2022, my (slow) watch is 2014 -I aged a year, it took me another 5 "real" years to get back from Centauri. Nothing odd. All times correct and equally faster than my slow watch.
Your trip back also takes just a little over 4 years, 4 years plus a month and a half. So when you get back it will be mid January 2021, not mid October 2022 and the Centauri clock reads mid January 2017, not mid October 2018.

But you did do the 1+1=2 calculation correctly, congratulations!
kamenjar said:
Your statement about my knowledge of relativity:
I don't need to know more than this to be leading this argument and be correct in my view of the universe.
Maybe you should rethink your assessment of yourself and think about the rules of this forum before disaster hits you.
 
  • #270


kamenjar said:
I may be wrong, but give me an example of a faster clock.
Any other clock is faster in its own frame. For any clock, at rest wrt the CMB or not, it is always possible to find an infinite number of frames where it is slower than another clock.
 
  • #271
harrylin said:
"according to the special theory of relativity the coordinate systems K and K' are by no means equivalent systems. Indeed this theory asserts only the equivalence of all Galilean (unaccelerated) coordinate systems, that is, coordinate systems relative to which sufficiently isolated, material points move in straight lines and uniformly. K is such a coordinate system, but not the system K', that is accelerated from time to time."
- https://en.wikisource.org/wiki/Dialog_about_Objections_against_the_Theory_of_Relativity

(note to Dalespam: another definition of "Galilean coordinate systems").
Yes, this is the definition I am familiar with. In the absence of gravity the phrase "sufficiently isolated, material points" means that the material point does not feel any acceleration.

Btw, don't try to tell me that you don't want to discuss something here and then continue to discuss it here. If you really think it is off topic then that is fine, but as it is it just seems to be a tactic to avoid answering a difficult question.
 
  • #272


PAllen wrote:
There is only one way, sort of, to do this (ignoring GR/gravity). Pick any arbitrary inertial clock, and its rest frame and call it universal (say, pick the clock tower at Emerald City). Declare, by fiat...

kamenjar said:
I don't have to do any of that, I can just use common sense. If I make a trip to Centauri that lasts 1 year:
- When I get there, i'll look at your clock in Emerad City and see t0+1y. My clock shows t0+1y. I am not stupid, so I know that the picture of your clock took 4y to arrive, so I conclude that if you didn't move from your couch in 4 years, it must be that your clock that's running faster.
- When you see me arrive at Centauri, your clock will show t0+9y and you'll see that the picture of my clock shows t0+1y. You are not stupid either, so you conclude the same way that my clock is running slower.
- In fact we can do this for any leg of the trip, without needing Mikowski, or Einstain, or his wife -- which was also Serbian by the way - like myself and Tesla :) -- to conclude whose clock is running slower.

Why is such a simple thing made so complicated? I don't understand.

It's not complicated. It's just that, as PAllen says, you can pick ANY reference frame to serve as your basis for your absolute time. The "common sense" works out exactly the same.

You say that "I know that the picture of your clock took 4y to arrive". HOW do you know that? That's true in some frames of reference, but not in other frames of reference. So your reasoning is circular--you picked a frame of reference to use for your measurements, and then concluded that that frame of reference was the one whose clocks are running the fastest. If you had picked a different frame of reference, you would have gotten different measurements, and you would have gotten a different answer to the question of which clock is running fastest.
 
  • #273
harrylin said:
Of course you can have accelerated observers in SR; and in SR, acceleration is not rest. :wink:

Here's how Einstein clarified, in the context of the twin paradox, the use of SR (which is a label that he invented):

"according to the special theory of relativity the coordinate systems K and K' are by no means equivalent systems. Indeed this theory asserts only the equivalence of all Galilean (unaccelerated) coordinate systems, that is, coordinate systems relative to which sufficiently isolated, material points move in straight lines and uniformly. K is such a coordinate system, but not the system K', that is accelerated from time to time."
- https://en.wikisource.org/wiki/Dialog_about_Objections_against_the_Theory_of_Relativity

(note to Dalespam: another definition of "Galilean coordinate systems").

My point was that just as in classical mechanics, in SR an accelerated observer is not "in rest" - that is wrong terminology for both. The traveler can certainly say to be in rest in his capsule, but he can not "prefer to think to have traveled zero distance", for that equals "being in rest all the time", and thus being "unaccelerated all the time".

As this subtle point is at the heart of the "twin paradox", it is not something to overlook.

Saying that an accelerated observer is objectively different from an inertial observer does not mean that an accelerated observer is somehow not allowed to do distance measurements relative to themself. They cannot use Lorenz transforms, or Minkowski metric, but they can choose to set up a coordinate system at which they are the origin.

Remember, this was all in relation to whether 'distance traveled' has anything to do with the twin paradox. I have shown multiple ways that this is not true (distance traveled does not correlate with aging differential in general, and is therefore not a valid way to understand it). [see esp. #245, #252]

We've also been over in what way acceleration is responsible for differential aging. The most you can say is:

- in flat spacetime with normal topology, at least one path having acceleration is a necessary but not sufficient condition for differential aging. You cannot localize the differential aging to the acceleration, and it is up for debate whether you consider the acceleration per se causative.

(I say normal topology, because in a closed universe with locally flat geometry everywhere, you can have differential aging with no path experiencing any acceleration. Obviously, we don't normally assume this applies to our universe, but we don't know for sure. We only know our universe is very big.)
 
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  • #274


PAllen said:
Saying that an accelerated observer is objectively different from an inertial observer does not mean that an accelerated observer is somehow not allowed to do distance measurements relative to themself. They cannot use Lorenz transforms, or Minkowski metric, but they can choose to set up a coordinate system at which they are the origin.
As long as they say to be accelerated, they still can use Lorenz transformations - be it one, two, or an infinite set of them. That is still SR. However, if they say to be not accelerated but "in rest", then they are talking GR - exactly as Einstein explains in his paper.
Remember, this was all in relation to whether 'distance traveled' has anything to do with the twin paradox. I have shown multiple ways that this is not true (distance traveled does not correlate with aging differential in general, and is therefore not a valid way to understand it). [..] We've also been over in what way acceleration is responsible for differential aging. [..]
Sure - I merely spotted a significant phrasing issue. In sofar as I followed it, I have no issues overall with what you say (except philosophical ones perhaps; but that is out of place here).
 
  • #275


harrylin said:
As long as they say to be accelerated, they still can use Lorenz transformations - be it one, two, or an infinite set of them. That is still SR. However, if they say to be not accelerated but "in rest", then they are talking GR - exactly as Einstein explains in his paper.

This is our core terminological difference. Because it is terminology, there is ultimately no right answer. As I understand it;

rest - your definition: inertial; my definition: stationary in some coordinate system, equally for SR as GR. I don't see any contradiction between this and the statement that SR says only that all inertial frames are equivalent. To distinguish simple form for laws versus complex, I distinguish inertial from non-inertial. This is based on a local measurement with an accelerometer. Distance is a non-local measurement, based on conventions. A passenger in a rocket can measure distances from themself, and call themself at rest in their coordinate system.

SR vs. GR You say if you use generalized coordinates and metric, it is GR even if no curvature. I say it is GR only if the EFE is used, and there is curvature.
 
  • #276


ghwellsjr said:
I never said any of that stuff you are supposedly agreeing with me about. I don't even know what you're talking about.

What's so hard about saying that the time that a clock displays and measures is legitimate and not compromised just because it doesn't agree with what another clock with a different history displays? Einstein says they're all correct. You never have to make an excuse for one of them saying it lost some time and the other one is correct. They're all correct. Furthermore, all the objects and observers that are local to a clock, experience objective time (and subjective, if we can identify that) the same as the clock. This doesn't have anything to do with the number of events in the universe, which is infinite, by the way.

I won't refer to anything you post again after this.

Robinpike is only asking for an explanation of why the two reunited clocks show different times.

Einstein referred to the observers time as subjective. Who am I to argue with him.
 
  • #277


PAllen said:
I am interested in avoiding so called explanations that are tied to a specific formulation of differential aging. A good understanding will apply equally to any formulation.

What is 'longer path'? The only invariant along timelike paths through spacetime is proper time. If you mean 'distance traveled' each twin may prefer to think they have traveled zero distance and the other has done the traveling. Travel distance has really nothing to do with differential aging, and is strictly frame or coordinate dependent quantity.

Maybe you mean longer as drawn on a conventional spacetime diagram? If so, I missed where you stated this. It is true that for different (timelike) paths between two given events drawn on conventional SR spacetime diagram, the longer one has less proper time.

There should be someone here who can give robinpike a SIMPLE explanation for why the reunited clocks read differently. If the clocks read the same before parting, and read differently when rejoined, it doesn't require a government study to conclude one ticked less than the other. Labelling the time is not an answer. It comes down to the physics of objects in motion. Who will answer?
 
  • #278


phyti said:
Einstein referred to the observers time as subjective. Who am I to argue with him.
For the record, here is what you said:
phyti said:
I agree with you, the same amount of time (number of events in the universe) does not change for any observers.
If we say a clock measures the rate of activity for each observer, it would correspond more to reality, i.e. subjective time.
I don't believe Einstein ever equated reality with an observer's subjective time. I'm sure he pointed out that a person's perception of time is quite fickle and rarely corresponds to any objective measure of time on a clock.

When we are discussing the passage of time in the context of Relativity, and we use a phrase like the "observer's time", we always mean the time on a clock colocated with him (even if there isn't really one there), never his subjective perception of time. Subjective time is irrelevant in discussions of relativity, except to point out that we are not talking about an observer's perception of time.
 
  • #279


phyti said:
There should be someone here who can give robinpike a SIMPLE explanation for why the reunited clocks read differently. If the clocks read the same before parting, and read differently when rejoined, it doesn't require a government study to conclude one ticked less than the other. Labelling the time is not an answer. It comes down to the physics of objects in motion. Who will answer?
He's been given many answers and he rejects every one because he wants an answer that fits his preconceived notion that the traveling twin's clock must lose time at a particular point in the trip:
robinpike said:
I have suggested that the reason for the loss in time is because his rate of time slowed down at some point in his journey. If that is the reason, then how does Relativity explain this (and please note I am not asking for the calculation).

If the deduction that his rate of time slowed down at some point in his joureny is incorrect, then all I ask is that the correct explantion for the loss in time is stated for me to see.
The problem is that nature does not reveal to us the answer in a way that makes him feel comfortable. He wants an absolute answer. "It happened right here or it happened in this particular way." He's not going to get that answer. The best we can do is explain how the Theory of Special Relativity deals with it which is that time is relative. Every clock keeps track of time legitimately. It's just what happens.

It's really no different than the concept of relative speed. After the traveling twin departs, he can consider himself to be at rest and the Earth twin to be moving at some relative speed or the Earth twin can consider the traveling twin to be the one to be moving at that same speed but in the opposite direction. What would you tell someone that rejects that kind of ambiguity in assigning speed and insists that there must be only one correct answer?

If robinpike can ever feel comfortable with the notion that speed is relative to an arbitrary frame of reference and that the two twins will have different speeds in a different frame of reference, then he should have no trouble understanding that the rate of ticking on each twin's clock is related to his speed in the arbitrarily chosen frame of reference, the faster the speed, the slower the clock ticks. Keep track of the ticks of each clock throughout the scenario and you have the answer. Transform the scenario into any other frame of reference where the twins have different speeds and therefore their clocks tick at different rates, keep track of the ticks of each clock throughout the scenario and you end up with the same answer, even though the tick rates of the clocks are totally different in the different frames.

And it's not just the final answer at the end that comes out the same in different frames of reference, it's also the observations of each twin looking at the other ones clock compared to their own throughout the entire scenario from start to finish that comes out the same no matter which frame of reference is used.

That is the simplest way I know to answer robinpike's question but it doesn't nail down where along the trip the slow down occurred because that comes out different in each frame of reference and nature won't disclose to us anything more than the explanation we already have with Special Relativity.
 
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  • #280


ghwellsjr said:
If it takes 1 year by your clock, then your speed was 97%c.

No, you won't see that one year has passed, you will only see about a month and a half has passed. At 97%c, the Relativistic Doppler Factor is about 8.1 so you divide your 1 year by 8.1 to see how much time has advanced on the Earth clock.

You may not be stupid but you are ignorant of how to do these calculations.
Ok I admit, I missed that part and I added 1+1=2. I don't do relativity every day.

We can just re-do the scenario where numbers are made so that return trip takes 10 years. In that case only my understanding of what I would read on my watch is incorrect. My watch would show more than 2 years when I return to Earth and 1/2 that when I arrive at Centauri. The clocks on centauri and Earth are still synched (and have same timestamps stated in the thought experiment), my watch is still slow and that's what matters.

ghwellsjr said:
Wrong again, the Earth clock will be at just over 8 years. Remember, you are traveling at almost the speed of light so it takes you just over 4 years to get there and 4 years for the image of you to get back to earth.

I think we all agree that you don't understand.
I absolutely do not disagree here. First of all, you have a wrong conception of simultaneity in the common sense. Drop your 100% relativity thinking relativity for 1 second and use it to apply common sense. Maybe too much relativity gets someone thinking too much in one direction that it clouds their judgement. The part where you say and 4 years for the image of you to get back to earth is where your basis of thinking confuses you. You consider two synchronized timestamps to be equal when one receives the timestamp of another. In static frames, if any of the receivers receive clock timestamp of another receiver that is equal or later than their own, their clocks are not synchronized. Clock rates maybe, but not timestamps.

And here is an example: If we synchronize our watches that our times are in synch to UTC time, if we would be looking at Greenwitch's clock and my watch at the same time, it would be off by a fraction of a second. It doesn't mean that my clock is off, it means that it appears to me that Grenwitch clock is off by an amount that makes it correct. That fraction of a second corresponds to the time the light takes form Greenwitch to us. Same thing would be true if we were on Centauri, when looking at Greenwitch (UTC -

You are saying at what time "on earth" is the time when Earth receives my image and that is why your doppler based explanation is asymmetric (though correct) where my analysis is 100% symmetric. You didn't read how I synchronized clocks and what is and what is seen. You can consider that the real time on Earth and centauri you forgot that Earth sees 2008 on centauri's clock at t=0. Saying that time on Earth is Centauri + 8 is not correct when applying common sense thinking.

The only way to make clock timestamps symmetric is to is to make each clock off by 4 years when compared to the observed clock. Remember - if you are looking (both at rest) at someone's clock and it it shows the same exact time as yours, your clock is wrong!

Edit: To clarify -- Two rest clock timestamps are truly synched if only and only if an observer placed exactly in the middle between clocks is reading the same time on both clocks.
 
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  • #281
stevendaryl said:
It's not complicated. It's just that, as PAllen says, you can pick ANY reference frame to serve as your basis for your absolute time. The "common sense" works out exactly the same.

You say that "I know that the picture of your clock took 4y to arrive". HOW do you know that? That's true in some frames of reference, but not in other frames of reference. So your reasoning is circular--you picked a frame of reference to use for your measurements, and then concluded that that frame of reference was the one whose clocks are running the fastest. If you had picked a different frame of reference, you would have gotten different measurements, and you would have gotten a different answer to the question of which clock is running fastest.

I guess the flaw in kamenjar's view is that one cannot mix GR with SR like that and pretend that it still works.
The CMB frame cannot be introduced in the twin paradox SR set up as an absolute frame.
The inertial frames are always preferred in SR, and any of them can be used as reference.
 
  • #282


phyti said:
There should be someone here who can give robinpike a SIMPLE explanation for why the reunited clocks read differently. If the clocks read the same before parting, and read differently when rejoined, it doesn't require a government study to conclude one ticked less than the other. Labelling the time is not an answer. It comes down to the physics of objects in motion. Who will answer?

People have already tried to give the simple explanations, they mostly seem to get ignored as not being explanations.

The problem with robinpike's request is that there are an infinite number of ways one could compare the clocks - basically one for every inertial observer. This is what we mean by "the relativity of simutaneity". This result isn't going to even give one explanation - it can't.

As a consequence, one will have multiple, DIFFERENT "explanations" of why the time is less on one clock than the other. You won't have an explanation, you'll have many. The two major ones have (inertial observers comoving with each of the twins) have already been discussed, I'm sure. Discussing all the infinite numbers of different possible viewpoints isn't going to make anyone any happier, I'm sure, because you can't make a "simple" explnaation out of an infinite number of different ones.

The simple explanation is really simple - just deal with it! The clocks read differently - and there really wasn't ever really any good reason to ever think that they SHOULD be the same when they reunite. Thinking that they should be the same is ultimately a consequence of some latent belief in absolute time.

To go back to the simple explanations again (the ones that seem to be ignored as not-explanations), if you drive two cars over different courses and they later reunite, you're not surprised when their odometers do not read the same. In fact, you don't particularly expect them to be the same. But you are surprised when the clocks don't read the same. The reason for this is that you are thinking of the clocks as being different from the odometers. But if you take the block universe approach, for example you can see that measuring time is measuring the "length" of a wordline on a space-time diagram, just like an odometer measures distances on a space diagram. And it's not paradoxical at all for odometers to read differently when cars re-unite. So why the hangup when clocks re-unite? It's not because of logic, it's because of habit. You're used to absolute time, where the clocks read the same after they take different paths. If you use pure logic, untainted by habit, you can realize that there isn't any particular reason to demand that the clocks read the same when they reunite.
 
  • #283


kamenjar said:
I don't do relativity every day.
You don't do relativity any day.
kamenjar said:
I absolutely do not disagree here.
I'm glad you do not disagree. I like it when people agree with me. (Maybe you should preview your posts and proof-read them before you hit the Submit button.)
kamenjar said:
IFirst of all, you have a wrong conception of simultaneity in the common sense. Drop your 100% relativity thinking relativity for 1 second and use it to apply common sense. Maybe too much relativity gets someone thinking too much in one direction that it clouds their judgement.
I'm sure someone's judgment is clouded. I wonder if we can figure out whose.
kamenjar said:
The part where you say and 4 years for the image of you to get back to earth is where your basis of thinking confuses you. You consider two synchronized timestamps to be equal when one receives the timestamp of another. In static frames, if any of the receivers receive clock timestamp of another receiver that is equal or later than their own, their clocks are not synchronized. Clock rates maybe, but not timestamps.

And here is an example: If we synchronize our watches that our times are at any point in time are showing atomic UTC time, if we would be looking at Greenwitch's clock and my watch at the same time, it would be off by a fraction of a second. It doesn't mean that my clock is off, it means that it appears to me that Grenwitch clock is off by an amount that makes it correct. That fraction of a second corresponds to the time the light takes form Greenwitch to us. Same thing would be true if we were on Centauri, when looking at Greenwitch (UTC -

You are saying at what time "on earth" is the time when Earth receives my image and that is why your doppler based explanation is asymmetric (though correct) where my analysis is 100% symmetric. You didn't read how I synchronized clocks and what is and what is seen. You can consider that the real time on Earth and centauri you forgot that Earth sees 2008 on centauri's clock at t=0. Saying that time on Earth is Centauri + 8 is not correct when applying common sense thinking.

The only way to make clock timestamps symmetric is to is to make each clock off by 4 years. Remember - if you are looking (both at rest) at someone's clock and it it shows the same exact time as yours, your clock is wrong!
I was just following your method:
kamenjar said:
Mehod of Synchronization:
Define "real" time to be such that when remote clock is viewed from each system to be showing "real"-4years.
Your mistake has been in thinking that if 1 year progressed on your clock as you traveled from Earth to Centauri, then that would take 1 extra year beyond what it takes for light to make the trip, from the Earth's frame and it doesn't. It only takes an extra month and a half. So the total time it takes for you to get from Earth to Centauri is four years plus 1.5 months. But it takes another four years (your definition) of how long it takes for the image of your arrival at Centauri to propagate back to earth. That's a total of 8 years plus 1.5 months. You said it would be 9 years which is incorrect.

This issue, by the way, has absolutely nothing to do with any synchronized clocks because we're not talking about what the Earth observer sees or measures or calculates of the clock that was originally on Centauri, we're talking about the clock that was transported a distance of 4 light-years in such a way that 1 year transpired on the clock during the trip and what time is on the Earth clock when it sees the traveling clock arrive at Centauri.

Kamenjar, it's time for you to back off and learn Special Relativity instead of accusing those of us who are trying to help you of not knowing what we're talking about.
 
  • #284


kamenjar said:
Edit: To clarify -- Two rest clock timestamps are truly synched if only and only if an observer placed exactly in the middle between clocks is reading the same time on both clocks.
That's one way but it's not the only way. And it's not a way that works in your scenario because there is no one half way between Earth and Centauri.

It's also not the original way that Einstein proposed in his 1905 paper. There he said that two clocks are synchronized if a light pulse is sent from one and reflected off the other and arrives back at the first, and the average time between the start and finish on the first one equals the time when the reflection occurs on the second.

And another way is the one you proposed which is to say if both clocks see the same difference between their local time and the remote time, then they are synchronized.
 
  • #285


ghwellsjr said:
Your mistake has been in thinking that if 1 year progressed on your clock as you traveled from Earth to Centauri, then that would take 1 extra year beyond what it takes for light to make the trip
You are correct. I remodeled the trip so that total trip takes 10 years, so please use new numbers.

I tried to tell you that the only thing that was wrong is that I made a wrong call that traveler would age 1 year during legs of the trips. I used to be making those kinds of mistakes before, but forgot and it's a very dumb mistake. I apologize about the error, but I believe that nothing is invalidated when I make the correction. I will rephrase to correct it, and maybe there truly is a fatal flaw here:

Given: Return trip time = 10 years. Distance to Centauri=4ly. Velocity = 0.8c. Lorentz factor reciprocial=0.6. Traveler ageing at arrival on earth=6y

At take off: Earth 2012, Viewed Centauri Clock 2008, "real" time in both systems = 2012. Traveler watch = 2012

At Centauri landing: Viewed Earth Clock 2013, Centauri 2017, "real" time in both systems = 2017. Traveler watch = 2012+3=2015. Traveler concludes that his watch runs slow compared to ("real" time).

At Earth landing: Earth Clock 2022, Viewed Centauri Clock 2018, "real" time in both systems = 2022. Traveler watch = 2012 + 6=2018, where 2018 is less than "real" time (2022). Traveler concludes that his watch runs slow compared to ("real" time).
ghwellsjr said:
... accusing those of us who are trying to help you of not knowing what we're talking about.

I never said that what you said was not correct, nor that you don't know what you are talking about. I was suggesting that you (or someone) tries to introduce previously synchronized clocks and event times (wrt timestamps received) by method that I described into their solution, it should at least help explain the solution.
 
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  • #286


kamenjar said:
Given: Return trip time = 10 years. Distance to Centauri=4ly. Velocity = 0.8c. Lorentz factor reciprocial=0.6. Traveler ageing at arrival on earth=6y

At take off: Earth 2012, Viewed Centauri Clock 2008, "real" time in both systems = 2012. Traveler watch = 2012

At Centauri landing: Viewed Earth Clock 2013, Centauri 2017, "real" time in both systems = 2017. Traveler watch = 2012+3=2015. Traveler concludes that his watch runs slow compared to ("real" time).

At Earth landing: Earth Clock 2022, Viewed Centauri Clock 2018, "real" time in both systems = 2022. Traveler watch = 2012 + 6=2018, where 2018 is less than "real" time (2022). Traveler concludes that his watch runs slow compared to ("real" time).
Perfect. And thanks for the excellent job of proof-reading. It really helps.

Let me just add in the Doppler Analysis. At 0.8c, the Doppler Factor is 3. So since it takes 3 years for the traveler to get to Centauri, he will see the Earth clock progress through 3/3 or 1 year as you pointed it. The Earth clock goes from 2012 to 2013 as viewed by the traveler.

For the return trip, which also takes 3 years, the traveling sees the Earth clock going 3*3 of his rate so it progresses through 9 years bringing it to 2022 when they reunite.

Excellent post. Thanks for getting it right.
 
  • #287


wikipedia said:
Starting with Paul Langevin in 1911, there have been numerous explanations of this paradox, many based upon there being no contradiction because there is no symmetry—only one twin has undergone acceleration and deceleration, thus differentiating the two cases. Max von Laue argued in 1913 that since the traveling twin must be in two separate inertial frames, one on the way out and another on the way back, this frame switch is the reason for the aging difference, not the acceleration per se.[1] Explanations put forth by Albert Einstein and Max Born invoked gravitational time dilation to explain the aging as a direct effect of acceleration.[2]
Explanation put forth by (lol infamous noob) kamenjar invoked clock rate difference and virtual simultaneity to explain aging as a direct effect of the clock rate being reduced by travel under relativistic speeds in any direction wrt rest frame.

Seriously, I am still not sure if there's a consensus on cause of difference in aging, or is this opening a new can of worms again?
 
  • #288


Ever hear of the expression, "there's more than one way to skin a cat?" :smile:
 
  • #289


kamenjar said:
Explanation put forth by (lol infamous noob) kamenjar invoked clock rate difference and virtual simultaneity to explain aging as a direct effect of the clock rate being reduced by travel under relativistic speeds in any direction wrt rest frame.

Seriously, I am still not sure if there's a consensus on cause of difference in aging, or is this opening a new can of worms again?

If you state this as: for any arbitrarily chosen reference rest frame, this is fine. If you insist there is one unique choice for 'the rest frame', you have to specify it, and this attempt will fail. Let's say there are only two solar systems in the universe, with high relative velocity. How do you pick 'the unique rest frame'?

Further, even if you come up with your rule for picking, SR guarantees I can pick a different one, or none, and have identical results. Thus your explanation will satisfy no one but you.
 
  • #290


And that is a good place to lock this thread.
 

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