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Harmony
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Let says there's a starting point A. Two observer, B and C travel from point A in opposite direction with 90%c. How can we calculate the speed of observer C as observed by observer B?
At the same time according to who? Realize that observers in relative motion will disagree as to whether two events happen at the same time or not.Harmony said:Let say both of them stop at the same time after some distance?
What do you mean?What will happen?
Harmony said:I apologize for not making my question clearer.
1.Stop at the same time according to an observer at the starting point A, which doesn't move from the point.
2. Observer A notice that observer B remain young while he himself becomes old. I believe that observer B will thinks the same way as observer B. If they both stop, will both observer be young?
Janus said:The end result will be that each will note a period (while they themselves are coming to a "stop".) where the other ages more rapidly then they do and that just cancels out the period where the other aged more slowly.
Wizardsblade said:Just a curious question, if B and C where in identical ships and where each told to stop after 30 seconds would they both stop at the same time to A?
Is this because of the loss of simultaneity or the propagation time of the light?
Yes - observations of space - length, time - duration, inertial (relativistic) mass and simultaneity are relative to the observer's world-line through space-time. (Her 4-velocity). That is why it is called the Theory of Special or General Relativity.Harmony said:Is it correct to say,that every observer in different velocity and position observes events differently?
Why do you think the observer would see no time pass, as opposed to the photon? Are you saying that since the relative speed between the observer and the photon is c, we can look at a frame where the observer is moving at c and the photon is at rest? The problem is that photons don't actually have their own rest frame in relativity--it would violate the rule that the laws of physics must work the same way in all inertial reference frames. When people talk about what things would look like from a photon's point of view, they're either misusing the concept of different reference frames or they're talking about the limit of what you would see as your speed approached c (relative to some outside landmarks like the Earth and the galaxy).tbc said:Because the speed is relative, no time would pass for an observer of the photon. This would lead to the idea of the observer seeing the photon being created AND being destroyed at the same time!
This does not happen, so i know I've gone wrong somewhere...I just don't know where!
The speed of Observer C can be calculated using the following formula: Speed of C = (Speed of A - Speed of B) / (1 - (Speed of A x Speed of B / (Speed of light)^2)), where the speed of light is approximately 299,792,458 meters per second.
The necessary measurements needed to calculate Observer C's speed include the speed of Observer A and Observer B, as well as the speed of light. These values can be obtained through experiments or measurements.
No, according to Einstein's theory of relativity, the speed of light is the maximum speed at which anything can travel in the universe. Therefore, it is not possible for Observer C's speed to be greater than the speed of light.
The relative motion of Observer A and Observer B plays a crucial role in the calculation of Observer C's speed. The formula takes into account the difference in their speeds and adjusts for the effects of relativity.
No, the speed of Observer C cannot be calculated without knowing the speeds of Observer A and B. These values are necessary to determine the relative motion and apply the formula for calculating Observer C's speed.