# I Relativistic velocity scenario

1. Apr 23, 2017

### calinvass

Lets suppose there is an observer at a certain distance from a planet that is moving at 0.9c as seen by the observer. We take as a reference, a frame where the observer is at rest. A spaceship leaves the planet and begins accelerating relative to the planet in the oposite direction. The observer should see the space ship decelerating and at some point having zero velocity, then the ship will apear to acceletate further in the opposite direction. Relative to the observer I suppose it is possible in principle can reach 0.9c. But when the observer checks the relative velocity between the planet and the spaceship it will conclude the spaceship managed to change its velocity bu 1.8c. Is this correct ?

2. Apr 23, 2017

### Staff: Mentor

Moderator's note: moved to relativity forum.

3. Apr 23, 2017

### Staff: Mentor

You have described this incorrectly. A correct description is: when the observer checks the difference in velocity, in his reference frame, between the planet and the spaceship, he will find it to be 1.8c. This is because the ship and the planet are moving in opposite directions, each at 0.9c, in his frame.

However, this 1.8c is not "the relative velocity between the planet and the spaceship". The latter term means either the velocity of the planet in the spaceship's frame, or the velocity of the spaceship in the planet's frame. (Numerically these are the same, they only differ in sign.) This can be found by the relativistic velocity addition formula: $(0.9 + 0.9) / (1 + 0.9^2)$, which gives about 0.995c.

4. Apr 23, 2017

### calinvass

Thank you for the correction. In other words the observer sees the spaceship and the planet are moving away from each other faster than the speed of light. But that is acceptable since the observer basically sees two objects traveling at 0.9c that happen to be in opposite directions.
But it is still funny because the ship will see the planet going away at 0.995c and the observer can talk to the crew and tell them he sees them separating at 1.9c.

Another scenario can be with two spaceship. They both are initially in a frame at rest then continuously accelerate (constant proper acceleration) in opposite directions. The relative velocity between them can in principle will approach c, but there must exist a reference frame where the spacecraft are separating away from each other at more than 1.9 c after a certain amount of time has passed.

5. Apr 23, 2017

### Staff: Mentor

Yes. The frame in which they were originally rest in one of these.

6. Apr 23, 2017

### Mister T

No. He sees the distance between them increasing at a rate that's faster than the speed of light. He knows, as a result of a calculation, how fast they move relative to each other. Do the calculation incorrectly and you get 1.8 c. Do it correctly and you get 0.95 c.

7. Apr 23, 2017

### Arkalius

The important takeaway here is that it is still the case that nothing with mass can reach or exceed the speed of light relative to you, but you can observe two different things with mass approach or move apart from each other at a rate faster than c (but always slower than 2c).

From your perspective, light emitted from the ship will barely outpace the ship at 0.1c, and slowly catch up to earth at 0.1c, so nothing is actually moving faster than light relative to you.

8. Apr 24, 2017

### calinvass

Supposing only one ship starts to accelerate continuously , and the other remains still in the observer's frame. Again at some point I suppose there should be a frame where the difference between them is 1.9c (or whatever value greater than c we choose, depending on how much we wait), although in the observer's frame, this time the difference will never reach c.

9. Apr 24, 2017

### Staff: Mentor

If by "whatever value greater than c we choose" you mean "whatever value greater than c and less than 2c we choose", and by "the observer's frame" you mean "the frame in which both ships are initially at rest", then yes.

10. Apr 25, 2017

### calinvass

Thank you. Of course, less than 2c. But in this case, I think, if the observer remains in the frame where both spacecraft were at rest, it will only see a maximum just below c. (In this second example one of the spacecraft remains in the initial position) In order to see a difference of just below 2c it needs to follow the first spacecraft that leaves and to accelerate until it sees the spacecraft that remained at the initial position, just below c, then to stop accelerating and wait until the first one reaches near c.

Last edited: Apr 25, 2017
11. Apr 25, 2017

### FactChecker

Not sure exactly what this all means, but the situation is fairly simple:

1) Nothing can move in any inertial reference frame faster than c.
2) Two objects can move apart in a "stationary" reference frame faster than c.
a) Suppose two objects are moving apart faster than c in a "stationary" reference frame. In either of their moving reference frames, the other is not moving faster than c.
b) A person in the "stationary" frame can use SR to understand why a person in one moving reference frame would say that the other moving object was not moving away faster than c.​

Last edited: Apr 25, 2017
12. Apr 25, 2017

### calinvass

Thank you, I agree with the above statements. My example uses them but if what I said is correct (to me it looks clear) it may not be so obvious to say how it the thought experiment works.