According to SR...

This is pretty typical of how any discussion with MD goes.

Since he doesn't pay any attention to anything he doesn't like, it's difficult to carry on a rational discussion.

 

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No, you're wrong. I asked him how much time elapsed from being reunited until the light returns. I even gave them each stop watches so they could see for themselves while they are standing side by side, waiting on the light to return. Where do you think he came up with the figure, 1 minute and 9 seconds?

 

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You have been told several times, 120s=2min. IF you reset the clocks to zero when the traveling twin has returned. IF you don't reset the clocks, then you get 1m and 1m9s respectively. Time for you to stop trolling.

 

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Why did you tell me 1 minute and 9 seconds elapses from the time they are reunited until the time the light arrives?

Why did you say the stay at home twin says one minute elapses from the reunion until the light returns?

Why are you telling me 2 minutes, when that is not the question?

 

Because, as I explained several times, the elapsed time depends on how and when you start/stop the clocks. Until you understand this basic fact, you have no business trolling the forum.

You don't understand your own question(s). You specified that the twins are next to each other, starting and stopping their clocks simultaneously. Light travels 1m out and 1m return, this is 2m total. Go troll elsewhere.

 

So I'll ask you once again.

When the twins are reunited standing side by side they each start a stop watch simultaneously. They each stop their stop watch simultaneously when the light returns. How much time does each stop watch indicate has elapsed since reunion?

 

Few ideas but fixed : 2m.

 

Wrong, try again.

In order for the stop watches to read 2 minutes the light would have had to be emitted when they started their stop watches. The light was emitted at the same time the traveling twin started his journey. His journey took time to travel, so it is impossible for their stop watches to read 2 minutes, since they weren't started until AFTER the completion of the journey when the twins were reunited standing side by side once again.

 

Good grief.
Since both twins are in the same frame, and since the laws of physics are the same everywhere, they will both record the same intervals of time. This has nothing to do with one of the twins traveling somewhere, because they're not traveling, they're standing next to their twin.

So the time the light was emitted might not be the same for both twins (unless it was emitted when they were in the same frame), but the time it arrives back is the same, for that simple reason--they're both in the same frame of reference.

Why would anyone (except perhaps you) think this scenario is some kind of SR "test"?

 

Then why did Tach tell me the traveling twin says 1 minute and 9 seconds?

You seem to be confused. The light was emitted much earlier than when the stop watches are started. The stop watches are timing the elapsed time from when the twins were reunited until the light finally returns.

 

The light is traveling 2 light-minutes, and the stop watches are started just as the light is hitting the mirror and stopped when the light returns. That's one minute.

The fact that one twin travelled has nothing to do with the answer, and this question has nothing to do with the twin paradox.

Tach was assuming that both watches were started when the travelling twin started travelling.

 

Light is being sent when the twin travel starts.
Total travel time for light trip is 2m.
The traveling twin returns in 1m.
The twins reset their clocks.
Light comes back at t1=2m but according to your new scenario the clocks were reset at t2=1m , so both clocks show elapsed time t3=1m. How much longer do you plan to troll?

 

IF you do NOT reset the clocks, dishonest troll. This has been explained to you several times.

 

Because without resetting the clocks the reunion times AND the light returning times will BOTH be offset by the amount accumulated during the one twin's travel. Tach seems to have been telling you what both would read when the light returned, but the reunion time readings would be equally offset.

 

So, this is the root of your mad attempt at "disproving" relativity.
The "stay at home" twin measures the distance \(D\) being covered in the time \(\tau\). In your example \(D=20lys\) and \(\tau=60s\)

The "traveling" measures the distance \(D'=D \sqrt{1-(v/c)^2}\) being covered in the time \(\tau'=\tau \sqrt{1-(v/c)^2}\).

Likewise, the "stay at home" twin measures the ray of light emitted at \(\tau=0\) to have traveled \(R=c \tau\) while the "traveling" twin, while in motion will measure \(R'=c \tau'=R \sqrt{1-(v/c)^2}\). All distances are measured as being contracted from the frame of reference of the traveling twin. When the traveling twin STOPS, at the end of his journey and he is in the SAME FRAME as the "at home" twin, they both AGREE that light traveled \(R=c \tau\) (60lys, in this case).
You have been told this many times before, you will not manage to "disprove" relativity, despite your deluded efforts.

 

I feel the same way. I am not familiar with the members of this forum but I can see posting in this thread is not constructive. I am, however, still curious as to resolution of the issue as I defined it.

Would you know how the speed of light behaves with respect to an accelerating observer? Would I be correct in believing that, under certain circumstances, an accelerating observer can measure the speed of light to be faster than c?

 

All observers will measure the velocity of light to be c. There may seem to be some conflict in this when considering an observer who is also moving near the speed of light. Something to keep in mind is that when an observer is moving, especially at relativistic velocities, their clocks are time dilated and their measuring rods are length contracted. These are the tools they use to measure the velocity of light. It turns out that the time dilation and length contraction are such that when measured in the moving frame of reference, light continues to have the same velocity, c as it does in all other frames.

 

So you are saying the stop watches are started as the light hits the mirror. The traveling twin has just returned from his trip and the light is just hitting the mirror as the stop watches are started. Are you suggesting that the traveling twin acknowledges that the light is just hitting the mirror when he returns? So the traveling twin thinks the light traveled to the mirror in ~51 seconds?

No he was not, otherwise he never would have posted the one minute and 9 second answer.

 

What are you talking about reset the clocks? I am asking you what each twin thinks the elapsed time is from the time they are reunited to the time the light returns. Stop trolling, troll. You know damn well what I am asking, and you already gave the incorrect answer. Where did you get the one minute and 9 second answer???????

 

There's really no point to continuing this. This is pretty much the way any discussion with MD ends up.

Pointless and confused.