Relativistic velocity scenario

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    Relativistic Velocity
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Discussion Overview

The discussion revolves around relativistic velocities, particularly focusing on scenarios involving observers, planets, and spaceships moving at significant fractions of the speed of light. Participants explore the implications of relativistic velocity addition and how different observers perceive the motion of objects in various frames of reference.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a scenario where an observer sees a spaceship accelerating away from a planet moving at 0.9c, questioning the implications of relative velocities and suggesting that the spaceship could appear to reach 0.9c relative to the observer.
  • Another participant corrects the initial claim, explaining that the observer would calculate the relative velocity between the planet and the spaceship using the relativistic velocity addition formula, resulting in approximately 0.995c, not 1.8c.
  • Some participants discuss the concept that while two objects can appear to move apart at a rate faster than the speed of light in a stationary reference frame, no object with mass can exceed the speed of light relative to any observer.
  • A scenario is introduced where two spaceships accelerate in opposite directions, with participants debating the maximum relative velocities that can be observed and calculated from different frames of reference.
  • There is a discussion about the conditions under which an observer can perceive the separation of two objects at rates approaching but never exceeding the speed of light, with some participants suggesting that the observer must follow one of the objects to see a difference approaching 2c.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of relative velocities and the implications of relativistic effects. While some agree on the principles of special relativity, there remains no consensus on the specifics of how these principles apply in the scenarios discussed.

Contextual Notes

Participants highlight the importance of correctly applying the relativistic velocity addition formula and the distinction between relative velocity in different frames of reference. There are unresolved assumptions regarding the conditions under which certain velocities are observed and calculated.

Who May Find This Useful

This discussion may be of interest to those studying special relativity, particularly in understanding the nuances of relativistic velocity addition and the implications of different reference frames in high-speed scenarios.

calinvass
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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 spaceship 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 ?
 
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calinvass said:
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.

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.
 
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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.
 
calinvass said:
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.
Yes. The frame in which they were originally rest in one of these.
 
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calinvass said:
In other words the observer sees the spaceship and the planet are moving away from each other faster than the speed of light.

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

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.
 
calinvass said:
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.
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.
 
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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.
 
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  • #11
calinvass said:
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.
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.​
 
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  • #12
FactChecker said:
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.​

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.
 

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