I haven't done much mathematics with SR, so please bear with me.
Let's say we have two points in space (could be Earth and a star) that are at rest with respect to each other and they are separated by 20 light-years.
Somebody is going 0.8c from one point to the other. From their point of view the two points in space are 12 light-years apart from each other:
12 = 20*squareroot(1-0.8^2) So it will take 12 years for light to travel from one point to the other point. After 12 years the person travelling will have gone 0.8c*12 years = 9.6 light-years. So the light will have travelled only (12-9.6) = 2.4 light-years in 12 years for a speed of only 0.2c for the person moving at 0.8c. So how does the person moving at 0.8c measure the light going at c and not 0.2c?
I'm kind of new to this so it's a little difficult. Maybe somebody can clarify.

A constant medium...

Oh, doppler shift, wave length changes, but c is always c in any frame of refernence, be it .2c, .8c, or .99999c.

there is a formula used in SR for adding velocities - a part of lorentz transformations. try adding 0.2c to 0.8c with that formula. it will yield less than 1c. infact any two numbers less than 1c when added will yield less than 1c, but closer to 1c.
in your case, you, if your person fires a light beam while traveling to the star you are adding 1c to your persons 0.8c which will yield 1c. if you fire the light beam from earth, you are adding 1c to 0c which will again yield 1c.

You have to be careful to specify the reference frame you're measuring things in.

For an observer who is stationary, light takes 20 years to travel from Earth to the star, and the traveller takes 25 years.

For the traveller, the star is 12 light years from Earth. Light takes 12 years to make the trip, and the traveller takes 15 years.

If the light leaves the Earth at the same time as the traveller, then when light reaches the star the traveller will measure himself to be 9.6 light years from Earth, with 2.4 light years still to travel.

I haven't done much mathematics with SR, so please bear with me.
Let's say we have two points in space (could be Earth and a star) that are at rest with respect to each other and they are separated by 20 light-years.
Somebody is going 0.8c from one point to the other. From their point of view the two points in space are 12 light-years apart from each other:
12 = 20*squareroot(1-0.8^2) So it will take 12 years for light to travel from one point to the other point. After 12 years the person travelling will have gone 0.8c*12 years = 9.6 light-years. So the light will have travelled only (12-9.6) = 2.4 light-years in 12 years for a speed of only 0.2c for the person moving at 0.8c. So how does the person moving at 0.8c measure the light going at c and not 0.2c?
I'm kind of new to this so it's a little difficult. Maybe somebody can clarify.
Hi Zeno,
From the point of view of the traveller, the points are not points in space - they are moving. Earth is falling behind at 0.8c while the star is approaching at 0.8c. The time for light to get from Earth to the star is 12/1.8 = 6.67 years. The time for light to get from the star to Earth is 12/0.2 = 60 years.

It is difficult to see how that could be:
If a bit of light left Earth at the same time as a bit of light left the star, then they'll obviously both arrive at their destination at the same time as well, right? So how can one arrive in 6 years and the other in 60 years?

This is where many people get stuck, because getting your head around the notion that "at the same time" depends on how you are moving is difficult.

But, it must be true if the speed of light is constant:
Two things that happen at the same time in different places according to one point of view don't happen at the same time in other points of view.

Essentially, a point in time is relative, just like a point in space is relative.

Thanks for the response Pete. So somebody who is at rest with respect to the Earth and star will measure 20 years for light to travel from Earth to star and also star to Earth. So that means that 20 years for this person will be equal to both 6.67 years and 60 years for the person who is travelling from Earth to star?

Put as simply as possible, energy and or force exchanges from mass to mass takes time.

In a solid where there are many compressed fields “Mass” the span of each field is not as large as those fields that are near unity or considered near vacuum where the spans of each field of mass is massive.

If we allowed two exchanges of energy in a solid compared to two exchanges in near vacuum it would seem like there was hardly any velocity gained in the solid because of the very little distance a given charge was exchanged because each field of mass doesn’t take up a large area.

Now if we consider an area or field around hundreds of thousands of kilometers of span to it then, it would seem a massive velocity had been gained when the fact is each field passed on the charge using the same time for each exchange, and what gives us the reasoning of a fast velocity is that we have to consider all the fields distance that is covered in two exchanges.

Obviously if these fields distances are allowed to be greater “c” would be even faster!
But what limits this is “Unity” and if we mess around with E/M=c^2 you should get an idea how “c” has been derived..

Thanks for the response Pete. So somebody who is at rest with respect to the Earth and star will measure 20 years for light to travel from Earth to star and also star to Earth. So that means that 20 years for this person will be equal to both 6.67 years and 60 years for the person who is travelling from Earth to star?
That's correct. Relativity says that the time between two widely separated events could be pretty much anything, depending on both the motion of the chosen frame of reference (what we choose to be "stationary), and on the distance and direction between the events.

The maths looks like this:
If Earth and the traveller set their clocks to zero just as they pass each other, and if we know the time (Te) and distance from Earth in the direction of trael (Xe) of some event according to Earth observers , then...

...We can calculate the time of the event according to the traveller:
Tt = (Te - v.Xe/c²) / √(1 - v²/c²)

...And the distance of the event from the traveller according to the traveller:
Xt = (Xe - v.Tt) / √(1 - v²/c²)


If distance is in light years, time is in years, and v = 0.8c, then these equations simplify to:

Tt = 5(Te - 0.8Xe)/3
Xt = 5(Xe - 0.8Te)/3

The reverse equation is the same, except that v = -0.8c:
Tx = 5(Te + 0.8Xe)/3
Xx = 5(Xe + 0.8Te)/3

(You should check that these equations are equivalent).


Let's do an example.
If some light leaves the star at the moment that the traveller and Earth pass each other, then:
Te = 0
Xe = 20
...so...
Tt = 5(0 - 0.8 x 20)/3 = -26.7
Xe = 5(20 - 0.8 x 0)/3 = 33.3

This means that from the traveller's point of view, the light left the star when the star was 33.3 light years from the traveller, and 26.7 years before Earth passed the traveller.

The light will arrive at Earth when:
Te = 20
Xe = 0
...so...
Tt = 5(20 - 0.8 x 0)/3 = 33.3
Xt = r(0 - 0.8 x 20)/3 = -26.7

This means that from the traveller's point of view, the light catches up with Earth when it is 26.7 light years behind the traveller, and 33.3 years after Earth passed the traveller.

Putting these together, from the traveller's point of view the light travelled 60 light years in 60 years - as we expected!


Here's an exercise for you:

According to the traveller, the light left the star when the star was 33.3 light years away, and 26.7 years before Earth passed the traveller.
According to the traveller, when does the light reach the traveller? (Tt = ?, Xt = ?)
Use the equations to find Te and Xe (transform the event coordinates to Earth's reference frame).
Is the result what the Earth observer would expect?