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Twin Paradox again 
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#1
Feb1512, 12:55 AM

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Hello, All
I read the article: http://math.ucr.edu/home/baez/physic...n_paradox.html But, I cannot understand how just asymmetry (without acceleration) can cause less age of moving observer. The article says: Please, explain me this. But, no maths please. 


#2
Feb1512, 01:54 AM

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#3
Feb1512, 03:14 AM

P: 215

Thanks PAllen.
But, I need some detail to understand it. So I am going to describe what is my understanding about the paradox. A is stationary and B is moving with constant speed. There is no acceleration involved (Just assume because we trying to solve it only with asymmetry). (The x and t is in A's reference frame) Suppose A at (x=0, t=0) and B (x=0, t=0). A flashes a beam each per second. A knows the speed of B. So A can calculate at which (x=?, t=?) beam will meet to B. At t=1 A at (x=0, t=1) and B at (x=1, t=1). At t=1 A calculate that beam and B will meet at (x=a, t=b). where a > 1, b > 1. At t=2 A at (x=0, t=2) and B at (x=2, t=2). At t=2 A calculate that beam and B will meet at (x=c, t=d). where c > 2, d > 2 ,(c2) > (a1), (d2) > (b1). As time elapse, B sees A's clock slowing down more. Because B gets beam more late. (The frequency of receiving beam increases until B is in motion relative to A. Whenever B stops it receives quick signals for some time because of signals coming behind it. But after some time B starts receiving signals in time which is flashed after stopping B.) In inward journey At t=10 A at (x=0, t=10) and B at (x=10, t=10). At t=10 A calculate that beam and B will meet at (x=p, t=q). where p < 10, q > 10. At t=11 A at (x=0, t=11) and B at (x=9, t=11). At t=11 A calculate that beam and B will meet at (x=r, t=s). where r < 9, s > 11, (r9) < (p10), (s11) < (q10). So, as time elapsed, B sees A's clock running more fast. Because B gets beam more quickly. (If B doesn't stay motionless for some time described in above round braces, then B get much more signals in starting phase of returning journey which is flashed before returning of B. when B starts receiving signals flashed after returning, we can describe above scenario) Is my understanding of the paradox is right? If no then please give me some detail. 


#4
Feb1512, 08:51 AM

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Twin Paradox again
Did you look at this picture:
http://math.ucr.edu/home/baez/physic...e.html#doppler Which part of this picture is confusing you (you requested no math, so here is just a a picture). 


#5
Feb1512, 08:58 AM

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#6
Feb1512, 11:58 PM

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After he does this, B will see A's clock ticking more slowly as he is moving away from A, AND as he is moving towards A. 


#7
Feb1612, 02:44 AM

P: 215

I am very sorry.
I don't need to consider this. I need to consider this. a=(ca) and b=(db). And from this B receives signals at same frequency in outward journey. @PAllen The first diagram shows doppler shift analysis. But, there is same length of time axis for both, then why stella can send only 16 signals while terence can 32? The sedond diagram shows lines of simultaneity. As I understand from the 6th diagram of the link:http://en.wikipedia.org/wiki/Relativity_of_simultaneity that the diagram shows lines of simultaneos events according to stella in FoR of terence. The same diagram we can draw for simultaneos events for terence in FoR of stella. what is the asymmetry? @ghwellsjr No, x is just a unit of length. I am just trying to say that at t=1 B should be at x=1. @Rap sorry, I cannot get you. 


#8
Feb1612, 03:12 AM

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Are you trying to solve this problem in a general sense, that is, for all values of speed? 


#9
Feb1612, 09:41 AM

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I seem to be repeating myself. Can you try to explicitly say what you don't understand about the following: Stella and Terence each send out one pulse per second. Each pulse is an expanding spherical wave front. While they are moving away from each other (thus each one's pulse has to 'catch up' to the other), each receives (for example) one pulse every two seconds. As soon as Stella turns around, Stella starts 'running into' Terence's already emitted pulses faster. So now Stella is getting (for example) two pulses per second. So, let's say, Stella experiences 10 seconds before turnaround and 10 seconds after turnaround. Then Stella receives 5 pulses on her outward leg and 20 pulses on her inward leg. So she concludes Terence ages 25 seconds while she aged 20 seconds. However, when Stella turns around, Terence has not received many of the pulses from Stella's outward leg. Terence will continue to receive one pulse every two seconds for 20 of Terrence's seconds  that is, until receiving all of the 10 pulses Stella sent on her outward leg. Then, for 5 of Terrence's seconds, Terrence will receive two pulses per second from Stella  getting the 10 from Stella's inward leg. Thus Terrence will get 20 pulses from Stella in 25 of Terrence's seconds. Again, they both agree that Stella aged 20 seconds to Terrence's 25. 


#10
Feb1712, 07:03 AM

P: 215

Ok,
So, stella experience 20 second in whole journey, 10 second in outward leg, and 10 second in inward leg. Stella receives 5 pulse in outward (1 per 2 sec) and 20 in inward (2 per 1 sec). So she concludes terence age as 25. And terence receives 20 pulse which stella has sent, so stella's age is 20. Actually we guess stella's age and conclude terence's age, if we guess terence's age and try to get stella's age then what? Suppose, we guess terence experience 25 second during whole journey. Stella gets total 25 pulse in 20 second, where terence gets 20 pulse in 25 second. Now, stella gets 5 pulse in 10 second in outward leg. (0.5 per sec) and terence gets x pulse in t second in outward leg. (x/t=r(out) per sec) Now stella gets 20 pulse in 10 second in inward leg. (2 per sec) and terence gets 20x pulse in 25t second in inward leg. ((20x)/(25t)=r(in) per sec) (where x can be 2 to 19, and t can be 2 to 24) So, for any values of x and t there is r(out) != 0.5 and r(in) != 2. But, it should be. because at least if we consider only outward or inward both traveling related to each other (without considering frame change). How can we explain the different rate? Different rate is the result of guessing different time elapsed. But, that is the thing which to be proved, not to be guessed. I am very sorry but I cannot understand from where terence gets 25 second while stella gets only 20. 


#11
Feb1712, 09:02 AM

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Do you also understand and agree that if Stella started out far away from Terrence and was approaching Terrence, they would each see the other ones clock ticking at 2 times the rate of their own? These are both symmetrical situations and so they both must see the same thing in the other one, correct? Do you agree with these statements? 


#12
Feb1712, 11:53 PM

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#13
Feb1812, 01:36 AM

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Good.
Now while Stella and Terrence have a great distance between them, does it make sense to you that if something happened to one of them, the other one wouldn't see it until some time later because it would take time for the image of the change to propagate across that great distance? 


#14
Feb1812, 01:44 AM

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#15
Feb1812, 02:09 AM

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Fine.
Now does it make sense to you that while Stella and Terrence have a great distance between them, if one of them changes their speed and/or their direction toward or away from the other one, that first one will instantly see a change in the tick rate of the other ones clock? 


#16
Feb1812, 03:14 AM

P: 215

Ok, I am describing again what I understand: 1. In outward leg stella receives 5 pulse in 10 second, as per stella terence's age is 5 second. 2. In outward leg terence receives 5 pulse in 10 second, as per terence stella's age is 5 second. 3. In inward leg stella receives 20 pulse in 10 second, so stella say terences's age is 25. 4. But, after turn around terence continue receives 5 pulse in 10 second which is sent by stella during outward leg. And this is the thing which cause age difference. 5. After finishing that 5 pulse terence also starts receiving pulse at rate of 2 pulse/second, and he gets 10 in 5 second. So he concludes her age as 20. But, my concern is other one. The above one is says that terence is stationary and stella is moving. So pulse sent by moving body or stationary body from a single point to terence always reach at same time because moving body cannot affect light speed. Problem is "first some time in inward leg", during that terence receives at same rate stella at different. Terence cannot know quickly about turn around because he receive pulse at same rate. while stella has quick knowledge that terence is comming toward her, and she starts receiving pulse quickly!!! Actully, if we imagine that stella has power to measure one way speed of pulse in inward leg which is fired in outward leg. she will measure the speed < c. because that pulse is reaching to terence as same rate (We can imagine this). But, if terence has power to measure one way speed of pulse he sent during whole journey, he concludes the speed = c. The situation seems asymmetrical only when we accept that stella measure one way speed of pulse < c in inward leg fired in outward leg. Light has constant velocity relative to everything, then who determines that stella is coming near so she can receive quickly, and terence is stationary, so he should receive as per same rate (point 4)? My point is if there are only the two in universe, and stella doesn't know that she is actually moving. She thinks that terence is moving. And if both meets stella is less aged than terence, then who has determined that stella was moving and stella had chaged her frame? 


#17
Feb1812, 04:11 AM

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But if you are going to incorporate Special Relativity to help understand the Twin Paradox using Relativistic Doppler, you must use a single Frame of Reference, any one you choose. You can't have Stella "changing her frame". Both Stella and Terrence should be discussed in terms of the same frame. It can be one in which Terrence is always stationary, or it can be one in which Stella is stationary for no more than half the time, but there is no inertial frame in which Stella is stationary for the entire trip. Please note that the questions I asked you in the preceding posts were not related to any particular frame, nor were they concerned with who was moving and who was stationary. I specifically worded my questions to you in the context of symmetry. 


#18
Feb1812, 05:04 AM

P: 215

Suppose, during inward leg, from stella's FoR, terence is moving, and stella measures one way speed of light as c, then terence should get light pulse quickly, because terence is moving towards stella and stella measures one way speed as c. If we think that stella measures one ways speed of light < c, then only terence doesn't receive pulse quickly. From terence's FoR, he always measures c and stella receives pulse quickly. this is ideal situation. Both fires 10 pulse during outward leg, but after turn around stella receives quickly and terence not. Both have to receive remaining 5 pulse. Both is moving relative to each other. Then why only stella receives quickly and terence not. If we say that stella measures speed < c, then only we can say that stella is moving and stella has changed her frame. If we say that stella always measure speed as c, then how can we determine that who is moving, and if we cannot determine that who is moving then how can we determine who is less aged. And yet, one of both is less aged then who has determine that "actually" who was moving? We say that motion is relative and speed of light is c relative to everything, then how can we determine that stella is moving and stella has changed her frame? This is also the question that how terence know that he didn't to anything, where how stella know that she had changed her frame? If this seems silly discussion and I am not understanding well then please excuse me. Thanks 


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