This is the scenario:

A spaceship is launched from Earth at a relativistic speed towards a distant planet. One observer is placed on the spaceship and one is placed on the planet.

At the time of launch, two stopwatch clocks are started simultaneously -- one on the spaceship and one on the distant planet.

When the planet-based observer sees 20 years on his stopwatch, the spaceship flies by the planet. Here are my questions:

1. What will the planet-based observer see on his own stopwatch? answer: 20 years
2. What will the planet-based observer see on the ship stopwatch?
3. What will the ship-based observer see on the ship stopwatch?
4. What will the ship-based observer see on the planet-based stopwatch?

Assume that there is zero space between the planet and the ship at the time of its passing so the ship-based observer will see the planet stopwatch in real time. Also assume ideal conditions (the planet gives off no gravity etc.)

I've been looking for an answer to this for some time now yet I still don't have any. Perhaps you could help me?

There is an equation for time dilation. I believe the gist of the answer is that time dilation is a factor of the relative speed between two locations or 2 locations plus some significant gravity producing thing, not based off of the speed of an individual object relative to everything else.

http://en.wikipedia.org/wiki/Time_dilation

http://en.wikipedia.org/wiki/Twin_paradox#Resolution_of_the_paradox_in_special_relativity

The times they will see on oneanothers stopwatch will be identical when they indeed pass at negligible distance.

For the rest, your question is ill posed.

If the spaceship makes a roundtrip, then the time passed on the spaceship will be < 20 years, or zero years, if it has been going at lightspeed all the time. The precise outcome depends on its trajectory of acceleration. It cannot make a roundtrip without acceleration; thats the assymetry that explains the twin 'paradox'; both observers are unequal in that regard.

However, there is a more interesting case which the typical textbooks unfortunately omits: if the spaceship comes back without ever reversing its direction or otherwise accelerating, by flying around a spherical universe, say, all bets are off. In that case, it is possible to triangulate a special rest frame (yes, really), and it all depends; has the planet shot around the universe while the spaceship, while seeming to move away at lightspeed from the planet, actually has stood still relative to the rest frame? If so, your question breaks down; time would be a crawl on the planet, relative to the rest frame, and the spaceship would not be back in 20 years, but instantly, with its clock all worn out waiting for you and your planet to fly around the universe. Or vice versa, depending who is really moving and who isnt.

The times they will see on oneanothers stopwatch will be identical when they indeed pass at negligible distance.

I don't believe that this is correct.

[...] Here are my questions:

1. What will the planet-based observer see on his own stopwatch? answer: 20 years
2. What will the planet-based observer see on the ship stopwatch?
3. What will the ship-based observer see on the ship stopwatch?
4. What will the ship-based observer see on the planet-based stopwatch?

Answers are:

2. A time longer than 20 years, depending on the ship's velocity. For example, that watch would read approximately 40 years if the ship was traveling at 0.866c.
3. 20 years.
4. A time longer than 20 years, depending on the ship's velocity. For example, that watch would read approximately 40 years if the ship was traveling at 0.866c.

That is, they both see each other's clock as being at a time which is later than their own clock's time. This is a counter intuitive effect of special relativity.

Another observation is this:

1. What will the planet-based observer see his own stopwatch looks like? answer: it will look like a regular watch
2. What will the planet-based observer see the ship stopwatch looks like? answer: the watch will be contracted along the direction of the ship's trajectory
3. What will the ship-based observer see on the ship stopwatch looks like? answer: it will look like a regular watch
4. What will the ship-based observer see on the planet-based stopwatch? answer: the watch will be contracted along the direction of the ship's trajectory

This is the scenario:

A spaceship is launched from Earth at a relativistic speed towards a distant planet. One observer is placed on the spaceship and one is placed on the planet.

At the time of launch, two stopwatch clocks are started simultaneously -- one on the spaceship and one on the distant planet.

This alone is meaningless statement in STR.

Simultaneity depends on reference frame for any spatially separated events (in this case spaceship and planet). I.e you have to specify in which reference frame - "Earth/Planet" or "moving spaceship" these stopwatches were started simultaneously.

This issue is muddled by the acceleration spaceship has to undertake in above example. But you can make it cleaner so, that spaceship flies with constant speed and synchronizes its clock at the moment, it is passing by Earth in its way towards the planet.

When the planet-based observer sees 20 years on his stopwatch, the spaceship flies by the planet. Here are my questions:
1. What will the planet-based observer see on his own stopwatch? answer: 20 years

Correct by definition.

2. What will the planet-based observer see on the ship stopwatch?

If clocks were started simultaneously in Earth/planet reference frame, then something << 20 years
If clocks were synchronised in moving spaceship reference frame, then something >> 20 years

3. What will the ship-based observer see on the ship stopwatch?

Exactly the same value that planet-based observer is seeing (if they are close enough to each-other you can ignore all relativistic effects)

4. What will the ship-based observer see on the planet-based stopwatch?

Also the same value that planet-based observer is seeing (i.e. 20 years)

If the spaceship makes a roundtrip, then the time passed on the spaceship will be < 20 years, or zero years, if it has been going at lightspeed all the time. The precise outcome depends on its trajectory of acceleration. It cannot make a roundtrip without acceleration; thats the assymetry that explains the twin 'paradox'; both observers are unequal in that regard.

You can formulate twin paradox without acceleration:

Spaceship A passes Earth in way to Alpha Centauri. Precisely at the passing moment, it synchronises its clock (B) with Earth-based observer (A).

Precisely at the moment spaceship A is passing by Alpha Centauri, another spaceship C comes from opposite direction flying towards Earth. At the moment they pass each other, it synchronizes its clock with the first ship - i.e. clock (C ) with (B)

Now when spaceship C arrives Earth. its clock (C ) shows less, than Earth-based clock (A).
The inequality does not come only from acceleration, but from the shift between two reference frames - frame of ship B and ship C.

At the time of launch, two stopwatch clocks are started simultaneously -- one on the spaceship and one on the distant planet.

How on earth is this synchronisation achieved (no pun intended)?

Okay, I think I understand your confusion. For simplicity, let's assume that the ship is going at speed 0.661c relative to the two planets, hence the time and space dilation factor is gamma = 3/4.

From the reference frame of the two planets, the ship is moving, hence the ship's clock is ticking slower. With a speed of 0.661c the clock on the ship will show approximately 15 years when it arrives at the destination planet due to time dilation. As others have pointed out, an observer on the planet will read the ship's clock as indicating 15 years too and an observer on the ship will also read the clock on the ship as indicating 15 years.

Now, from the reference frame of the ship, however, it is the destination planet that is moving. Hence, the clock on the planet is ticking slower. So at a speed of 0.661c, it is correct to say that the clock on the planet will only have ticked for 15 years throughout the journey and in the reference frame of the spaceship.

I think your mistake is to assume that the clock on the planet will read 15 years from the observer on the spaceship. That is wrong, the clock will read 20 years from the observer on the spaceship, as others have pointed out.

The problem lies with the notion of simultaneity of events. In relativity, there is no such thing as absolute simultaneity of events. So you cannot say without ambiguity that the two clocks were started at the same time. You need to specify in which frame of reference the clocks are being synchronized. So if the clocks were started at the same time in the reference frame of the planets, then in the reference frame of a spaceship that has just been launched from Earth, the two clocks were not started at the same time, in fact the clock on the destination planet was started earlier.

This explains why the clock on the destination planet reads 20 years from the ship-based observer, even though he knows the clock must have only been ticking for 15 years since the ship left Earth. It's because from the ship's reference frame, the two clocks were not started at the same time and the clock on the destination planet has been started 5 years before the ship's clock was started. This part can be calculated using a Lorentz transformation to compare the space-time coordinates of an event in different inertial reference frames.

[quote name='Lesan' timestamp='1320151999' post='4879222']At the time of launch, two stopwatch clocks are started simultaneously -- one on the spaceship and one on the distant planet.

How on earth is this synchronisation achieved (no pun intended)?
[/quote]

If you know the distance between the planets, you can easily achieve that (in theory). Suppose the distance is 1 light-year. Send a beam of light from planet A to planet B and wait 1 year. After exactly 1 year, immediately start the clock on the spaceship and launch the spaceship (assuming the spaceship can be given a high acceleration burst such that the desired velocity is almost instantaneously achieved). On planet B, start the clock exactly when the light beam is received from planet A.

This only synchronizes the watches in the reference frame of the planets, however.

How on earth is this synchronisation achieved (no pun intended)?

By the way, I am having serious difficulty understanding why time is connected to speed in the first place... and don't even get me started on quantum mechanics. That is serious nonsense to me... and I've been selling quantum-based generators for a while! That nonsense actually works!