special relativity - What does the accelerated twin see in the twin paradox?

I am referring to the usual twin "paradox" where one twin remains on Earth while the other one takes a round trip journey on a rocket. I understand how the experience of the twins is not symmetrical, because one is accelerating, and that the solution involves changing reference frames. What I haven't seen in common answers, and what I am curious about, is what would one twin (on the rocket) see if she could constantly look at the other twin's clock. Imagine that the travelling twin had a super telescope and watched the Earth-bound twin's clock constantly during her trip. This scenario includes the following stages:

1. Positive acceleration, +a, away from Earth


2. Floating freely for a while at velocity +v


3. Negative acceleration, -a, to slow to a stop relative to Earth


4. Briefly motionless relative to Earth (v=0)


5. Accelerating towards Earth at -a until reaching velocity -v


6. Floating freely towards earth at velocity -v



7. Final positive acceleration, +a, to land on Earth (v=0).

What would the twin on the rocket see on Earth, looking from afar:

1. While accelerating away?


2. While floating freely before starting the deceleration?


3. While decelerating but still moving away?


4. During transition (turn around)?


5. While decelerating but now moving back towards Earth?


6. While floating freely back towards Earth?



7. While accelerating to a stop on Earth?

Imagine the traveling twin constantly looking through the telescope and comparing the Earth clock to her rocket clock. I'm looking for an answer like "...now the Earth-clock appears to be ticking slower than her own clock on the rocket". And "...at this point the clocks are momentarily showing the same time." You can keep the math to the bare minimum, in fact no math is best! Thank you in advance.

Answer

I made some videos about this a few years back, on YouTube. It shows what the traveler sees on a variety of clocks in the scene, for the entire duration of the journey.

I would suggest "Earth gravity front view" to start with ("Earth gravity" represents the value of acceleration in this context!)

The terse explanatory text is reproduced here:

The twin "paradox" (in quotes because it is NOT a real paradox!) is a basic teaching scenario for Special Relativity: http://math.ucr.edu/home/baez/physics...

This is a series of visualizations of the journey from the point of view of the traveling twin, who flies 20 light years away from his home station then returns. The journey consists of four parts, joined together. The first quarter is an acceleration away from the station. During the second and third quarters the ship accelerates towards the station (so that at the half way point the ship is stationary 20 light years away). In the fourth quarter the ship accelerates away from the station in order to come to rest there.

The total coordinate travel time (shown as a red dot in the top left HUD clock) is 43.711/58.918 years for acceleration at earth/moon gravity levels, whilst proper time (green dot) is 12.101/38.694 years. The yellow dot represents the time the traveler would see on the station clock face through a very powerful telescope! The octahedral stations (spaced one light year apart and one light year to the left of the flight path) are all synchronized to coordinate time and rotate once over the course of the whole journey. There is a 2x2 light year wall one light year beyond the far end of the journey, and large rectangular frames every 5 light years. Where a floor is shown it is 1 light year per stripe, and there are small 1 ly milestone spheres along the way, with a larger one every five light years.

The flights are rendered without relativistic effects and then for two values of acceleration (currently earth gravity and moon gravity). The distortion artifacts are due to aberration of light, the Doppler effect and the headlight effect. The earth gravity videos exhibit some nice penrose-terrel "rotation" effects. Magenta markers in a circle show where you really are in the scene (in the sense that things outside the circle are behind you, whilst those inside the circle are in front of you. Doppler shift = gamma for this circle), and grey markers show the circle where the doppler shift is 1 (gamma is in theory directly observable on this circle).