0
   

Twin paradox chancy

 
 
Reply Wed 1 Feb, 2017 12:24 pm
Seve forgive this new OP but I'd like to get inputs from many who won't respond to yours

http://able2know.org/topic/366028-1

Quote:
I might have a followup question. In the traditional description of the twins paradox, only "Betty" (the travelling twin) accelerates.
That's right Steve; with you and I watching her performance sitting back here throughout while she does at least two accelerations and one deceleration. Reason I say 'at least': At the end of our experiment she isn't required to stop

https://www.scientificamerican.com/article/how-does-relativity-theor/

To simplify the discussion however, let's recapitulate, her very powerful rocket able in an instant to achieve nearly c. So ten minutes later (to us) as she reaches Mars, her clock is still reading zero we see her fire her retros for her return home (supposing this is what you've called the 'arrival event'). At this same instant, however, she is indeed back home, her clock still reading zero, ours of course 10 min


Now as I understand your variation, Steve, owning a similar rocket you take off five minutes after Betty has left us, that is, presumably at the moment she's firing her retros (which I don't see of course til 5 min later) so you meet her halfway between, your clock still reading 5 min. and hers zero.

Since that was the end of your variation, we don't care whether either of you Forgive me Stevedecides to stop

Returning now to your original query, re
Quote:
STR:......I have searched online ...but none of the basic site seem to directly answer it. The...linked articles seem to get close and then sort of gloss over and omit the answer.


Sims, that has constituted precisely my own objection to the classical 'explanation'. In short, how can a short burst of acceleration stop your clock for the entire remainder of your journey out or your return

In previous discussions a year or two ago I attempted to resolve that apparent discrepancy in 'resolution' of the Twin Paradox but didn't get much support, owing probably to how we're so 'dug in' to the Einstein version. Not sure we can resurrect it after so long but if you'd like to look for it and get back to us....

I always like to chat about relativity
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Question • Score: 0 • Views: 3,019 • Replies: 119
No top replies

 
stevesims
 
  1  
Reply Thu 2 Feb, 2017 01:36 pm
@dalehileman,
hi dalehileman,

Let me start out with what I have done so far in achieving an understanding of SR.

In the so-called "traditional story", Alice and Betty are twins of the exact same age. Betty takes off in a high-powered spaceship to a star 8LY distant at a speed of 80%c. Both metrics are specified from the Earth-star frame. (i.e. in the example, the Earth and the star are 8LY distant, but have no relative motion.) Both Alice and Betty have high-powered telescopes and clocks, and each observe the others clock throughout the trip. The rest is a description of what they see.

At time zero, Betty has accelerated, but has not moved yet. At time=0 they both see each others clocks reading zero.

If we assume newtonian mechanics and universal time, with the speed of light = c, then at the arrival event (yes, the event of Betty reaching the star), Betty will read 10 years on her own clock, Alice will read 10 years on her own clock, however, Alice will not see the arrival event because of the 8LY of distance. However, when she look in her 18th year, she will see Betty arrive and observe Betty's clock reading 10 years. As Alice monitored this continuously, she will have seen Betty's clock move from zero to 10 in 18 years of her own time, or move, therefore at a rate of 10/18ths or 5/9ths normal speed.

When Betty arrives and looks at Alice's clock, she, likewise won't read 10 years, because she is observing light that is 8 years out of date, and therefore will see Alice's clock read 10-8 = 2 years. Since she has been monitoring the clock continuously she will have seen it pass from 0 to 2 years in 10 years of her own time, or move, therefore at a rate of 2/10ths or 1/5th normal speed.

However, experimentation indicates that in reality, both of them will see the clocks moving at the same rate. In the paradox, the only thing we really don't know by definition, or given, is what Betty will read on her own clock at the arrival event. If we rewrite these fractions using 'x' for that value, and force them to be equal, then we get the equation:

x / 18 = 2 / x
Why is the 18 still valid? Because the speed of the ship was given in the Earth frame, and the speed of light is a constant.
Why is the 2 still valid? Because 10 is the value that light is carrying off the Earth-bound clock face at the expiration of the trip from it's speed in the Earth-frame, and the distance to the star in the Earth-frame, and the speed of light is a constant.

And therefore x=6. Betty only experiences 6 years of time passage.

In a little bit, I will further generalize this to show the derivation of the Lorentz factor.
dalehileman
 
  0  
Reply Thu 2 Feb, 2017 03:45 pm
@stevesims,
Steve thanks most kindly for that expl which I might have to read 18 times...
...min


No offense Steve but what might require my first 10 readings is, 'If we assume newtonian mechanics and universal time...,' where my immediate reaction is, 'Why would we do that anyway, introducing still another factor to complicate things'

So continuing to read two or three additional times it's yet not clear to your Average Clod (me) when [or even if] relativity is introduced

I wonder however if anything critical would be lost if someone might simplify the problem even furtrher by assuming v very nearly c; and acceleration to/from instantaneous as I've done


Edited with embarrassment to note at the saving of the above that you've continued the expl with another posting (the next one). Alas, another 18 readings....
stevesims
 
  1  
Reply Thu 2 Feb, 2017 03:46 pm
@stevesims,
The equation shown in the previous post was

x / 18 = 2 / x

Where the 18 is

10 + 8

Where the 10 is the 10 years that it takes for an object to move 8 LY at 80%c.
And 8 is the time it takes light to move 8 LY.

The 2 in the above equation is likewise 10 - 8

Where the 10 is the 10 years that it takes for an object to move 8 LY at 80%c.
And 8 is the time it takes light to move 8 LY.

The relevant formula is v = d/t, or rather t = d/v. Or, i.e. time = distance / velocity.

The 10, in generalized form is 8 LY (distance) / 80%c (velocity)
The 8, in generalized form is 8 LY (distance) / c (velocity)

If we now let d represent the 8 LY, and v represent 80%c, we can substitute:

x / 18 = 2 / x --- becomes

x / (10 + 8) = (10 - 8) / x --- becomes

x / (d/v + d/c) = (d/v - d/c) / x --- using words that would becomes

BettysTime / (distancebetweenEarthandStar/VelocityOfBettysShip + distancebetweenEarthandStar/VelocityOfLight) =
(distancebetweenEarthandStar/VelocityOfBettysShip - distancebetweenEarthandStar/VelocityOfLight) / BettysTime

Lets use t' (i.e. tPrime) instead of saying 'x' or 'BettysTime', and keep in mind that t would be AlicesTime (i.e. t without the Prime) and that t = d/v so long as we are using values from a single frame of reference.

t' / (d/v + d/c) = (d/v - d/c) / t'

If we cross multiply those fractions we get:

t'^2 = (d/v + d/c) (d/v - d/c)

Now on the right we get a common denominator by multiplying each fraction by 1, in the form c/c, or v/v.

t'^2 = (dc/vc + dv/vc) (dc/vc - dv/vc)

Each binomial factor on the right now has a common denominator, so we simplify:

t'^2 = ((dc+dv)/vc) ((dc-dv)/vc)

Now we multiply the binomials on the right:

t'^2 = (dc+dv) (dc-dv) / (vc)^2

t'^2 = ((dc)^2 +dvdc -dvdc -(dv)^2) / (vc)^2

t'^2 = ((dc)^2 - (dv)^2) / (vc)^2

t'^2 = (d^2c^2 - d^2v^2) / (v^2c^2)

Factor out the common d^2 in the numerator:

t'^2 = (d^2 (c^2 - v^2)) / v^2c^2

t'^2 = (d^2/v^2) ((c^2 - v^2)/c^2)

Now recall that t = d/v:

t'^2 = t^2 ((c^2 - v^2)/c^2)

Normalize the fraction:

t'^2 = t^2 ((c^2/c^2) - (v^2/c^2))

t'^2 = t^2 (1 - (v^2/c^2))

Divide by t^2:

t'^2 / t^2 = 1 - (v^2/c^2)

Take the square root of both sides:

t'/t = sqrt(1 - (v^2/c^2))

Take the reciprocal of both sides:

t/t' = 1 / sqrt(1-(v^2/c^2))

This last equation is the Lorentz factor (gamma), which can also be seen here: https://en.wikipedia.org/wiki/Lorentz_factor in the "Definition" section on that page.
0 Replies
 
stevesims
 
  1  
Reply Thu 2 Feb, 2017 03:50 pm
@dalehileman,
oops...well, I apologize. What I was doing was taking you through Einstein's thinking that is published in his book "Special Relativity". I have a copy that is translated to English.

He started out assuming Newtonian mechanics, got to that same contradiction with experimentation of his day, and proceeded to let Betty's clock reading at the arrival event be 'x', and made the two fraction be equal. That is how he arrived postulating SR to begin with.
dalehileman
 
  0  
Reply Thu 2 Feb, 2017 03:53 pm
@stevesims,
Wow again and thanks again Steve

Incidentally, there's an assertion hereabout, something I've detected in the literature on and off for some decades now, that the Twin Paradox hasn't actually been resolved. In simple sentences of common words in the usual order and little or no arithmetic, I wonder if you could clarify my notion

By the way it's just dawning that your two expl postings aren't discussing the Paradox at all but merely showing how and where the Lorenz factor came from

Or did I get that wrong too...


Edited to remark that months or yeears ago here on a2k I had attempted to explain my own weird theory, which not only explains the Twin Paradox in the simplest of terms but also explains to the satisfaction of one's intuition all those apparent changes taking place in the moving object

If interested, I'd do a search and see if one or two of the best ones (best to me) are still on file. But thanks so far for your patience
stevesims
 
  1  
Reply Thu 2 Feb, 2017 04:42 pm
@dalehileman,
I don't know, to be honest. When I first read "Special Relativity" some years ago, I had a buddy that I went to lunch with who was interested and in talking to him about that text, he said some things that lead me to doing the mathematics of it.

There could conceivably be limitations to what you can get out of the "Twins Paradox", but until I somehow gain an understanding where all of my mental experiments work out and make sense, I just don't know. I was hoping to get some insight that might lead me to a vague understanding of GR. And it is possible that some of the concepts I am getting are helpful.

When you are accelerating, for the purposes of relativity, you are transisting between different frames of reference even though the ground is holding you in place. The key concept is that when you leave one frame of reference, and enter another you have accelerated, at least briefly, and time flows differently.

But there is something odd about that. You can accelerate in any direction, and time always seems to do the same thing according to SR, but that definitely isn't the whole story, because you can always accelerate back in the opposite direction and reenter a frame of reference that you were in before and left....and it only stands to reason that you would be back to the same original flow of time that you were in before you left it to begin with.

If I think I am sure of something I will definitely mention it though.
0 Replies
 
stevesims
 
  1  
Reply Thu 2 Feb, 2017 05:18 pm
@dalehileman,
I would like to see your theory.
stevesims
 
  1  
Reply Thu 2 Feb, 2017 05:24 pm
You know what? Lets consider that Betty goes all the way to the star, but doesn't stop. If she had stopped, then we do know that SR says she would be exactly 4 years younger than Alice. However, lets consider that she keeps going. Also consider that Alice takes off in her own spaceship at the point Betty arrives, which is when Betty's clock reads 6 and Alice's clock reads 10.

Alice should definitely be older at that point even so.
0 Replies
 
dalehileman
 
  0  
Reply Thu 2 Feb, 2017 06:56 pm
@stevesims,
Quote:
I would like to see your theory
I shall conduct a search Steve first thing tomorrow
0 Replies
 
stevesims
 
  1  
Reply Fri 3 Feb, 2017 08:09 am
I think I may be starting to get just a glimmer of understanding on this question. My problem has been in a way of thinking that is too "caught up with" concepts of before and after, older and younger....and thinking of time in a temporal manner rather than a spatial one.

At the beginning of the experiment, Betty and Alice are coincident in space-time, although they are in different inertial frames.

Thereafter, they are heading in different directions in 4 dimensional space-time. Significant to the question, the time components of their vectors through space-time are not the same.

If we were to ask which of two physical objects was spatially before the other in space, we would ask the question "From which direction?" That's because they both can be said to be "before" or "after" the other depending on which side of the two that you are on when you are looking and judging before, and after.

Once we consider the same notion from a temporal point of view, that question kind of gets lost. I think that in order to answer my own question that I must consider a complete journey through space-time for each with some well-defined starting and ending points, and then ask the question: "With regards to the time-domain of each journey, which one followed the longer path?"
dalehileman
 
  0  
Reply Fri 3 Feb, 2017 12:01 pm
@stevesims,
Quote:
time components of their vectors through space-time are not the same.

Steve, part of the problem I think, is that all such approaches I've noted depend upon a v significantly less than c, introducing an unnecessary factor into the equations that complicate them severely. Of course we can't achieve c but we can suppose a super rocket that accelerates to 'almost' that value, and a superlady rocketeer Betty who can stand such acceleration, with only a slight diff in the results...I think

...or can we....
stevesims
 
  1  
Reply Fri 3 Feb, 2017 02:45 pm
@dalehileman,
hi Dalehileman,

What I am starting to see (I think) is that I have been thinking of time as being a uni-directional concept. And it surely is, in a single inertial frame. But once you consider 2 different inertial frames, time itself heads in different directions. That's why both seem slower to each other.

Imagine:
Take a sheet of paper and draw two non-parallel lines on it. Mark out a length of 6 inches on one, and then mark out 6 inches on the other. If you consider the length that you have marked out on one, superimposed on the other it will be less than 6 inches. But if you do the same thing with the other, it will also be less than 6 inches on the first.

And that's a fair analogy, I think to why observers in two different inertial frames both see time on the other's clock running slower.

And to extend this analogy, it may be that acceleration / deceleration is a process that tilts the angle between the lines, either increasing it, or decreasing it.

Bear with me, I am still thinking on all this.
dalehileman
 
  2  
Reply Sat 4 Feb, 2017 04:16 pm
@stevesims,
Quote:
Bear with me, I am still thinking on all this.
With ya Steve. Do however try my system and see if it doesn't avoid a lot of the difficulties. Describe your travelers' situation in the simplest possible terms and I will attempt to convert it where v is in effect infinite


Alas, it does however complicate 'time at a distance,' which is then just as difficult to explain as the Twin Paradox
0 Replies
 
stevesims
 
  1  
Reply Thu 9 Feb, 2017 02:46 pm
I've been searching and searching on this puzzle. And I ran across a graphic that seemed like it was going to straighten out my basic thinking. You can see it here:
https://www.researchgate.net/figure/269104759_fig1_Figure-1-Light-cone-of-future-present-and-past

However, as I began thinking about that diagram, the more confused I got. The main element of it that is confusing me now is that it shows a time axis for space-time itself that does not appear to be relative. In other words, it is uni-directional for all objects in space-time.

So are there 2 kinds of time? One that varies with a change of motion and a different sort of time that is uni-directional for all objects in space-time as shown on the graphic?
Krumple
 
  1  
Reply Thu 9 Feb, 2017 02:53 pm
@stevesims,
stevesims wrote:

I've been searching and searching on this puzzle. And I ran across a graphic that seemed like it was going to straighten out my basic thinking. You can see it here:
https://www.researchgate.net/figure/269104759_fig1_Figure-1-Light-cone-of-future-present-and-past

However, as I began thinking about that diagram, the more confused I got. The main element of it that is confusing me now is that it shows a time axis for space-time itself that does not appear to be relative. In other words, it is uni-directional for all objects in space-time.

So are there 2 kinds of time? One that varies with a change of motion and a different sort of time that is uni-directional for all objects in space-time as shown on the graphic?



I personally say there has to be two, but if you ask most physicists, they would say only time relative to motion.

I wondered if we could cease all motion of an atom would its relative time speed up. Same for escaping the galaxies influence field and had a zero sum velocity relative to space. Time would speed up for this reference frame.

If I'm right, physical space "ages" faster than objects in motion, atoms, galaxies, photons, etc.
stevesims
 
  1  
Reply Fri 10 Feb, 2017 05:23 pm
@Krumple,
Quote:
I wondered if we could cease all motion of an atom would its relative time speed up.

I think you are onto a substantial idea with this thought. Since there _is_ motion, there should be an element of time dilation.....and perhaps there is a relationship between that time dilation and gravity.

I wonder it is possible to test to see if there is a gravitational difference of a Bose-Einstein condensate.
0 Replies
 
Kevin Sorbello
 
  1  
Reply Wed 22 Feb, 2017 09:49 am
@dalehileman,
Each post seems to imply that the relativistic effect of time dilation is apparent, rather than actual. I agree that using a v equal to c makes examples much easier to understand. For example, if person A instantly accelerates to c, their clock would appear to stop to person B who stayed in the initial frame of reference. However, it doesn't just "appear" to stop...it "does" stop...and if that person arrived at a destination 1 light year away and could instantly stop, their clock would have experienced "no time." It is meaningless to ask how the clock at rest would have appeared to person A because person A experienced no time. If Person A were able to return to Person B at the speed of c (instant acceleration and deceleration) as soon as they arrived at their initial destination, the clock of Person A would still be at zero and the clock at Person B would show 2 years had elapsed...and "that" is the twin paradox. How a clock "appears" is therefore meaningless. Why? because experiments show the clock actually slows...not just appears to slow, and the effect is at the basic structure of matter, since atomic clocks show the same change in time based on speed or energy. Which raises another issue...since clocks on top a mountain run at a different speed than those at sea level...and at a difference not connected to angular velocity...but rather potential energy...which, according to mc2 means the one running slow actually has more mass.
layman
 
  0  
Reply Wed 22 Feb, 2017 10:43 am
I have read this thread, but not carefully, and I find what I have read to be confusing.

Most people misconstrue, deliberately or not, the nature of the paradox to begin with.

Feynman gave the obvious answer: According to the LT, the MOVING clock runs slow.

Not BOTH clocks, as SR posits.

The accelerated clock is the one moving (even though SR tries to deny this).

Does that help?

The reason people find SR confusing is because it is confusing. Or, more precisely, just wrong in it's fundamental postulates. The theory incorporates logical contradictions, then says you're being illogical if you dispute it.
layman
 
  0  
Reply Wed 22 Feb, 2017 11:07 am
Some other misconceptions that may be creating confusion here.

1. It has been proven in the lab that acceleration has absolutely NO independent effect on time dilation. In fact this assumption has been called the "third postulate" of special relativity.

2. Delays in light travel have absolutely NOTHING to do with time dilation due to speed differential. Any consideration of the doppler effect is therefore a totally irrelevant red herring (non sequitur) and should be ignored.
 

Related Topics

Relativistic mechanics - Discussion by Granpa
Tesla's take on relativity - Discussion by gungasnake
Cesium clocks??? - Question by gungasnake
Why c, revisited still again - Question by dalehileman
Is there a relativist in the crowd - Question by dalehileman
relativity - Question by alexjlaonnae
Does light have Mass? - Question by peter jeffrey cobb
simple relativity question - Question by ralphiep
 
  1. Forums
  2. » Twin paradox chancy
Copyright © 2021 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 03/01/2021 at 11:54:37