Twin Paradox Problem?

Question

At 12:00 noon, Twin A with clock stays at rest while Twin B with telescope jumps to √(4/5) speed of light. Twin B, using telescope, observes Twin A's clock going more slowly because of relativistic doppler. After some time, Twin B suddenly reverses course and heads home at √(4/5) speed of light. Still using the telescope, Twin B observes Twin A's clock going faster, again because of relativistic doppler. At 1:00 pm, he notices that Twin A's clock now also shows 1:00 pm. When Twin B arrives back home with Twin A, how much older is Twin A than Twin B?

Bonus question: Why does Twin A, using telescope, NEVER sees Twin B's clock showing the same time as Twin A's clock?

Twin B's path is always on the same straight line from and to Twin A's position.

Prof Zikzak, you do not see another's relativistic proper time by viewing his clock with your telescope. Hence, it's quite easily possible to "view" another clock going faster than your own because of doppler shift.

Prof Zikzak, whenever one uses a telescope, I would assume that an observation is being made. But we quibble. Use the term "view" then, we know that we're not talking about relativistic proper time.

IMPORTANT NOTICE: I meant to say at 4/5ths of the speed of light, it makes the math easier to work out.

At 4/5ths light speed, the doppler factor is redshifted 1/3 and blueshifted 3. Really, this is basic algebra.

Remo Aviron writes: I believe the correct answer is time dilation of 5/3. Twin B travels out 55 mins and come back 5. The light from Twin A's clock starts as 60 minutes and travels 40 minutes out to Twin B. (Note speed 4/5c means 50 minutes out equals 40 minutes of time per twin A) Twin B then spends another 50 minutes his time coming home for a 110 minute journey. Twin A, meanwhile has waited around for 183.33 minutes. The difference being 73 min and 20 seconds.

Answer 1

They give you v, so you have the time dilation factor:
beta = sqrt (1 / (1 - (v/c)^2))

The apparent time dilation viewed by A has to take doppler shifts into account in addition to time dilation
redshifted time seen by A on B's watch through A's scope
= B's time / (1 - v/c)
= A's time / (beta * (1 - v/c))
= A's time * sqrt ( (1 + v/c) / (1 - v/c) )
blueshifted time seen by A on B's watch through A's scope
= B's time / (1 + v/c)
= A's time * sqrt ( (1 - v/c) / (1 + v/c) )

The time it takes B to get to the turnaround point at a distance d is:
t = d / (velocity * beta)
The observed (redshifted time) is:
t' = d / (velocity * beta * (1 - v/c))
The time it takes B to get a distance d2 partway back is:
t2 = d2 / (velocity * beta)
The observed (blueshifted time) is:
t2' = d2 / (velocity * beta * (1 + v/c))
So we know that
t1+t2 = t1'+t2' = a given time (1 hour)
That's two equations and two unknowns, d and d2. It's just a matter of algebra (albeit some nasty algebra I'll leave to you) to find them.

Once you know d, it's a piece of cake to find the total times.
For A, total time = 2d / v
For B, total time = 2d / (beta v)

Have fun!

BONUS: Traveling twin never sees the stationary twin's clock read the same as his. It always reads later. Why? He sees it run slow at first because of the redshift (during this phase there's a symmetry between A and B). The problem is that there isn't enough of a redshift time to overcome the time dilation. On travelling twin's clock, the trip takes an equal time both ways. But to the stationary twin, it looks like the trip there took a lot longer than the trip back.

EDIT2: That doesn't make any sense. Either you edited the question or I read it wrong. The traveling twin sees the times cross over because his blueshift time is equal. The stationary twin does not see crossover because he sees blueshift for a much shorter period of time. If you do the algebra the way I set it up (for when the stationary twin sees clock crossover), you shouldn't get a solution. Only the travelling twin sees crossover, so that at the end of the day, he is younger than stationary twin. So if you want to get the answer you want, you'll have to modify the equations slightly.

EDIT3: Re prof Zikzak. Toward the end of the journey, when the traveling twin looks through the telescope, he sees the stationary twins watch, and he sees that that watch has a later time than his own. He sees time passing faster than on his own watch because of the blueshift, which is stronger than time dilation. So I guess you want to call that viewing, because observation would take into account the travel time of light, hence killing the doppler shift. Since the problem is entitled doppler shift, I think you can assume that you are meant NOT to correct away the doppler factor from your observation.

Answer 2

You've copied the question wrong, or you're teacher doesn't know what s/he's talking about. Neither twin ever observes the other's clock to move faster than his own. There's a difference in Relativity between "viewing" and "observing." Is your teacher talking about "viewing?" That gets quite complicated.

I see. Your teacher doesn't know what s/he is talking about then. The word "observed" has a very specific meaning in Relativity, and it has nothing to do with Doppler shifting. You are talking about viewing through a telescope. That is not "observing" by the Relativistic definition, which requires viewing, and then accounting for light-travel times. Tell your teacher not to use the word "observe" in this context. It is incorrect. You can *view* another's clock running fast, but in SR you can never, ever *observe* it.

Answer 3

♥ aaah Scyth my regards! The famous twin problem, and so nicely twisted! Here’s precise timing (12:00 noon, 1:00 pm), telescope, relativistic Doppler, etc. Be merciful to us answerers! Look, everybody is red-hot of excitement to prove their truth and honour of science. Mr. Zikzak in particular recommends you to change your physics teacher. I love it! So bambino Scyth do it just to please us, LOL!
♠ meanwhile who was exerted to acceleration? ---Twin B! so his clock will be late!
Why? --- coz every molecule, atom, every electron becomes more massive due to relativism;
So what? ---increasing of inertia slows down the processes – mechanical, chemical, and any others where change of speed is involved, including Twin B’s digestion and his clock;
How much? ---integrating should show the total delay of Twin B’s clock;
Do it kaksi! ---I can’t, but if you’d give me a graph v(t), t being Twin A’s time, I’d venture to (numerically, I’d prefer trapezoid-like graph);
♣ how much am I right my yea-who compatriots?
My regards to Remo Aviron, too;

Answer 4

No difference. The relativistic effects of the turn-around's deceleration and re-acceleration cancel out the effects of the two main legs of the trip.