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Twin Paradox again

by mananvpanchal
Tags: paradox, twin
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mananvpanchal
#1
Feb15-12, 12:55 AM
P: 215
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:

All well and good, but this discussion at first just seems to sharpen the paradox! Stella sees what Terence sees: a slow clock on the Outbound Leg, a fast clock on the Inbound Leg. Whence comes the asymmetry between Stella and Terence?

Answer: in the duration of the Inbound and Outbound Legs, as seen. For Stella, each Leg takes about a year. Terence maintains that Stella's turnaround takes place at year 7 at a distance of nearly 7 light-years, so he won't see it until nearly year 14. Terence sees an Outbound Leg of long duration, and an Inbound Leg of very short duration.
I cannot understand why do Terence see an Outbound Leg of long duration and an Inbound Leg of very short duration?

Please, explain me this. But, no maths please.
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PAllen
#2
Feb15-12, 01:54 AM
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Quote Quote by mananvpanchal View Post
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:



I cannot understand why do Terence see an Outbound Leg of long duration and an Inbound Leg of very short duration?

Please, explain me this. But, no maths please.
Because if Stella turns around at 7 light years, it takes 7 years for the light from this turnaround to reach Terrence. Thus, for 14 years, Terence sees Stella's clock run slow.
mananvpanchal
#3
Feb15-12, 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 ,(c-2) > (a-1), (d-2) > (b-1).
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, (r-9) < (p-10), (s-11) < (q-10).
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.

PAllen
#4
Feb15-12, 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).
ghwellsjr
#5
Feb15-12, 08:58 AM
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Quote Quote by mananvpanchal View Post
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 ,(c-2) > (a-1), (d-2) > (b-1).
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, (r-9) < (p-10), (s-11) < (q-10).
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.
It looks like you are saying that B is traveling at the speed of light but I don't think you are so I don't understand what you are trying to communicate in the bolded values.
Rap
#6
Feb15-12, 11:58 PM
P: 789
Quote Quote by mananvpanchal View Post
Thanks PAllen.

But, I need some detail to understand it. So I am going to describe what is my understanding about the paradox.

outbound:

As time elapse, B sees A's clock slowing down more. Because B gets beam more late.

In inward journey

So, as time elapsed, B sees A's clock running more fast.
This is not correct. B knows how far away A is and knows that A is moving away or towards B, so B will have to take into account the speed of light and the fact that A is moving before he calculates how fast or slow A's clock is ticking.

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.
mananvpanchal
#7
Feb16-12, 02:44 AM
P: 215
I am very sorry.

I don't need to consider this.
where c > 2, d > 2 ,(c-2) > (a-1), (d-2) > (b-1).
where r < 9, s > 11, (r-9) < (p-10), (s-11) < (q-10)
So, my conclusion
(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.)
is wrong.

I need to consider this. a=(c-a) and b=(d-b).
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.
ghwellsjr
#8
Feb16-12, 03:12 AM
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Quote Quote by mananvpanchal View Post
@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.
Is the unit of time one second? In other words, are all the integer values of t when a new flash is emitted?

Are you trying to solve this problem in a general sense, that is, for all values of speed?
PAllen
#9
Feb16-12, 09:41 AM
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Quote Quote by mananvpanchal View Post
@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.
Ignore the second diagram. You originally asked about the Doppler explanation. Let's focus on that because it is the most physical explanation.

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.
mananvpanchal
#10
Feb17-12, 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 20-x pulse in 25-t second in inward leg. ((20-x)/(25-t)=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.
ghwellsjr
#11
Feb17-12, 09:02 AM
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Quote Quote by mananvpanchal View Post
...
But, I cannot understand how just asymmetry (without acceleration) can cause less age of moving observer.
...
I cannot understand why do Terence see an Outbound Leg of long duration and an Inbound Leg of very short duration?

Please, explain me this. But, no maths please.
Do you understand and agree that if Stella never turned around then they would each see the other ones clock ticking at 1/2 the rate of their own?

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?
mananvpanchal
#12
Feb17-12, 11:53 PM
P: 215
Quote Quote by ghwellsjr View Post
Do you understand and agree that if Stella never turned around then they would each see the other ones clock ticking at 1/2 the rate of their own?

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?
Yes, I am agree.
ghwellsjr
#13
Feb18-12, 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?
mananvpanchal
#14
Feb18-12, 01:44 AM
P: 215
Quote Quote by ghwellsjr View Post
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?
Yes.
ghwellsjr
#15
Feb18-12, 02:09 AM
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P: 4,745
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?
mananvpanchal
#16
Feb18-12, 03:14 AM
P: 215
Quote Quote by ghwellsjr View Post
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?
I am sorry, but this seems that our discussion is accelerated too fast in last post.

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?
ghwellsjr
#17
Feb18-12, 04:11 AM
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Quote Quote by mananvpanchal View Post
I am sorry, but this seems that our discussion is accelerated too fast in last post.

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.
You've explained this perfectly.
Quote Quote by mananvpanchal View Post
But, my concern is other one.

The above one is says that terence is stationary and stella is moving.
The Relativistic Doppler analysis, which you just did, is not concerned with who is stationary or who is moving. The non-relativistic Doppler Effect is concerned with who is stationary and who is moving because it depends on each observer's motion through the medium. Not so with Relativistic Doppler. It doesn't say anything about a Frame of Reference or any particular theory or anything about the one-way speed of light. It doesn't offer any explanation for why the effect is happening but merely describes what each observer sees with their own eyes. Now of course, Special Relativity can help us figure out what each observer sees, but it is not defining what each observer sees.
Quote Quote by mananvpanchal View Post
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!!!
That's correct.
Quote Quote by mananvpanchal View Post
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.
Even if they had special knowledge of how light propagates, it wouldn't change what they actually see, would it? Of course not. This has no bearing on the Relativistic Doppler analysis that you just did.
Quote Quote by mananvpanchal View Post
The situation seems asymmetrical only when we accept that stella measure one way speed of pulse < c in inward leg fired in outward leg.
No, the situation is actually asymmetrical because Stella immediately sees a change in Terrence's tick rate coinciding with her turn around and Terrence doesn't have that experience. When he sees a change in Stella's tick rate, it is not related to anything he did.
Quote Quote by mananvpanchal View Post
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)?
Stella determined that she is coming near to Terrance when she turned around and started coming near to him. Terrence didn't do that.
Quote Quote by mananvpanchal View Post
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?
As I said, the Relativistic Doppler is not related to any establishment of any frame, that is, it doesn't happen because of Einstein's particular definition of a Frame of Reference or Einstein's postulate that the one-way speed of light is c for any inertial observer. It works the same way under Lorentz Ether Theory in which the one-way speed of light is a constant only in an absolute fixed "ether" frame.

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.
mananvpanchal
#18
Feb18-12, 05:04 AM
P: 215
The Relativistic Doppler analysis, which you just did, is not concerned with who is stationary or who is moving. The non-relativistic Doppler Effect is concerned with who is stationary and who is moving because it depends on each observer's motion through the medium
Ok, I am agree with this. No one can claim that I am moving with relativistic doppler analysis.

Even if they had special knowledge of how light propagates, it wouldn't change what they actually see, would it? Of course not. This has no bearing on the Relativistic Doppler analysis that you just did.
Sorry, but I am not understanding this.
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?

No, the situation is actually asymmetrical because Stella immediately sees a change in Terrence's tick rate coinciding with her turn around and Terrence doesn't have that experience. When he sees a change in Stella's tick rate, it is not related to anything he did.
That is the question, why stella sees immediate change while terence not?
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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