Explaining the twin paradox with special relativity and doppler effect.

[xyzt]

Still stand by that. Ok no hard feelings then, you don't have to respond if you don't want to.

I already corrected your multiple, oft-repeated, misconceptions. Repeatedly. The fact that you do not understand what they are is your problem.

Edited March 27, 2015 by xyzt

[Janus]

I gave you negative feedback because I out of 4 posts, not a single one added any value and you were getting rude and personal. I stand by it.

 

I do not see where is special relative the sign makes any difference in the calculations. I am open to that idea, please just show me your logic.

You need to take the Relativity of Simultaneity into account. To illustrate what this means, imagine that your traveling twin is traveling past a string of clocks that are at rest with respect to his brother, and synchronized to each other according to both themselves and his brother.

As our traveler flies past these clocks, he will note three things. These clocks will run slow compared to his own identically constructed clock, the clocks and the distances between them will be length contracted (less than that as measured by the clocks themselves), and the clocks will not be synchronized to each other ( the clocks will not all read the same time at any given instant).

 

The first two have no dependence on which direction he is traveling along the string of clocks. (they will be the same on the both the outbound and return legs).

The third does. Because of the third effect, clocks that are ahead of him will read ahead of the clock he is passing at any given instant, and clocks behind him will read behind. The greater the separation between the clocks, the greater the difference in their readings. For example, if he is passing a clock at the instant it reads 12:00, a clock 1000 km ahead of him might read 12:01 and one 1000 km behind him would read 11:59.

 

Now if he were to reverse direction, clocks that were ahead of him will now be behind him and vice-versa; these clocks will also reverse roles as to which clocks read ahead of which.

 

So let's say that he starts at rest with respect to clock 1 while both his clock and clock 1 read the same time. He then instantly accelerates up to some fraction of c. The time on clock 1 remains unchanged (though it will run slow from this moment on), but the time of all of the other clocks ahead him have jumped forward in time as compared to clock 1.

 

As he travels along the string of clocks, they all, including clock 1, run slow compared to his own. At some point, he instantly turns around and accelerates back towards clock 1.

Just like above, the time reading on the clock he is adjacent to is not changed by this. However, since Clock 1 is now in the direction he is traveling in, it reads ahead instead of behind this clock now. In fact, it will now read ahead of his his clock. In effect, when our traveler turns around, by his accounting, clock 1 gains enough time to more than compensate for the time it lost by running slow.

On the return leg, clock 1 still runs slow according to our traveler, but this is still not enough to make up for the gain caused by the traveler's reversal of direction. He will find that clock 1 will have accumulated, in total, more time than his own clock has.

 

The above description deals what out traveler would determine to be happening once he factored out the propagation delay for light.

[CasualKilla]

You need to take the Relativity of Simultaneity into account. To illustrate what this means, imagine that your traveling twin is traveling past a string of clocks that are at rest with respect to his brother, and synchronized to each other according to both themselves and his brother.

As our traveler flies past these clocks, he will note three things. These clocks will run slow compared to his own identically constructed clock, the clocks and the distances between them will be length contracted (less than that as measured by the clocks themselves), and the clocks will not be synchronized to each other ( the clocks will not all read the same time at any given instant).

 

The first two have no dependence on which direction he is traveling along the string of clocks. (they will be the same on the both the outbound and return legs).

The third does. Because of the third effect, clocks that are ahead of him will read ahead of the clock he is passing at any given instant, and clocks behind him will read behind. The greater the separation between the clocks, the greater the difference in their readings. For example, if he is passing a clock at the instant it reads 12:00, a clock 1000 km ahead of him might read 12:01 and one 1000 km behind him would read 11:59.

 

Now if he were to reverse direction, clocks that were ahead of him will now be behind him and vice-versa; these clocks will also reverse roles as to which clocks read ahead of which.

 

So let's say that he starts at rest with respect to clock 1 while both his clock and clock 1 read the same time. He then instantly accelerates up to some fraction of c. The time on clock 1 remains unchanged (though it will run slow from this moment on), but the time of all of the other clocks ahead him have jumped forward in time as compared to clock 1.

 

As he travels along the string of clocks, they all, including clock 1, run slow compared to his own. At some point, he instantly turns around and accelerates back towards clock 1.

Just like above, the time reading on the clock he is adjacent to is not changed by this. However, since Clock 1 is now in the direction he is traveling in, it reads ahead instead of behind this clock now. In fact, it will now read ahead of his his clock. In effect, when our traveler turns around, by his accounting, clock 1 gains enough time to more than compensate for the time it lost by running slow.

On the return leg, clock 1 still runs slow according to our traveler, but this is still not enough to make up for the gain caused by the traveler's reversal of direction. He will find that clock 1 will have accumulated, in total, more time than his own clock has.

 

The above description deals what out traveler would determine to be happening once he factored out the propagation delay for light.

+1 Janus, that is exactly the explanation I needed to get around the confusion caused by the propagation of light signals, Doppler effect and was just very well rounded in general, thankyou.

 

Ok I can get down with the say 12:00 "event" being observed at different times for each clock from the spaceship, but what about the clock periods? (since TD factor is only depended on relative speed)? If we observe clock 1 from the spaceship, will we see a constant period, or will it change depending on our distance from the earth and whether we move away or towards it?

Edited March 27, 2015 by CasualKilla

[xyzt]

+1 Janus, that is exactly the explanation I needed to get around the confusion caused by the propagation of light signals, Doppler effect and was just very well rounded in general, thankyou.

 

Ok I can get down with the say 12:00 "event" being observed at different times for each clock from the spaceship, but what about the clock periods? (since TD factor is only depended on relative speed)? If we observe clock 1 from the spaceship, will we see a constant period, or will it change depending on our distance from the earth and whether we move away or towards it?

Read here. I wrote a large part of it.

[Janus]

+1 Janus, that is exactly the explanation I needed to get around the confusion caused by the propagation of light signals, Doppler effect and was just very well rounded in general, thankyou.

 

Ok I can get down with the say 12:00 "event" being observed at different times for each clock from the spaceship, but what about the clock periods? (since TD factor is only depended on relative speed)? If we observe clock 1 from the spaceship, will we see a constant period, or will it change depending on our distance from the earth and whether we move away or towards it?

That depends what you mean by "see". If you mean as what you physically see, then moving away or towards does effect what you see. If you mean what is happening to the clock after you factor out the light delay, then clock 1 runs slow no matter what direction your are moving. The Doppler shift formula for light I gave earlier includes both the light delay effect and time dilation. I'd go into this deeper, but I'm about to head out of town for the weekend, and won't be back until late Sunday

[swansont]

 

I do not see where is special relative the sign makes any difference in the calculations. I am open to that idea, please just show me your logic.

 

They don't show up in the calculations, because the calculations don't directly allow for an acceleration. They are derived assuming that you have two objects moving relative to each other in inertial frames. That assumption is violated. Also, I was correcting your terminology. The variable in the equation is speed, not velocity.

[CasualKilla]

That depends what you mean by "see". If you mean as what you physically see, then moving away or towards does effect what you see. If you mean what is happening to the clock after you factor out the light delay, then clock 1 runs slow no matter what direction your are moving. The Doppler shift formula for light I gave earlier includes both the light delay effect and time dilation. I'd go into this deeper, but I'm about to head out of town for the weekend, and won't be back until late Sunday

Yes I mean after you factor out signal delay, trying to avoid doppler and light propagation since it is complicating the issue.

 

 

They don't show up in the calculations, because the calculations don't directly allow for an acceleration. They are derived assuming that you have two objects moving relative to each other in inertial frames. That assumption is violated. Also, I was correcting your terminology. The variable in the equation is speed, not velocity.

Ah ok, it finally clicked, thanks for clearing that up.

Read here. I wrote a large part of it.

Well written, thanks.