Relativistic Velocity Transformations

[jayjay713]

A quasar is moving away from the Earth with a speed of 0.850C. It emits a proton that eventually reaches earth, and is traveling at a speed of 0.519C relative to the earth. How fast is the proton moving relative to the quasar?

Is this answer as simple as it seems?

is the answer simply 0.850 + 0.519 = 1.369C?
i am confused
thanks for any help

 

[BetoG93]

No, you can't have a speed greater than c.
In relativity, you must use Lorentz's velocity transformation equations.

 

[jayjay713]

but if I used that formula, wouldn't it give me a speed relative to the earth? Or would it be relative to the quasar? :O

 

[Dovahkiin]

but if I used that formula, wouldn't it give me a speed relative to the earth? Or would it be relative to the quasar? :O

Well you know the quasars velocity relative to the Earth and you know the protons velocity relative to the earth. So you know the protons velocity in the Earths rest frame, and you can lorentz transform it into the quasar rest frame using the quasar velocity. Don't forget the signs of the velocities! Post your answer when you have worked it out :)

 

[asteriatic]

I can confirm that the answer is not as simple as adding the two velocities together. Relativistic velocity transformations involve using the Lorentz transformation equations, which take into account the effects of time and space dilation at high speeds. In this case, we would need to use the equation v' = (v + u)/(1 + vu/c^2), where v is the velocity of the quasar (0.850C) and u is the velocity of the proton (0.519C). This would give us a relative velocity of approximately 0.913C. It is important to note that this is not a simple addition of velocities and is only valid for velocities much smaller than the speed of light. Relativistic effects become more significant as speeds approach the speed of light. I hope this helps clarify any confusion.