How Do Relative Velocities Transform in Special Relativity?

[wansui]

If A is moving to the left at a speed of 0.9c relative to E, B is moving to the right at a speed of 0.7c relative to E.

In the reference frame of A, what is the speed of B?
In the reference frame of A, what is the speed of E?
In the reference frame of B, what is the speed of A?
In the reference frame of B, what is the speed of E?

 

[JesseM]

Have a look at this page:

http://www.math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

Try to see if you can answer your own questions using the equations there, and then post again if you're still having trouble. Also, note that speed in special relativity is symmetrical between any two reference frames--if in my frame you are moving at speed v relative to me, then in your frame I will also be moving at speed v relative to you.

 

[kyle8921]

I would like to clarify that "special velocity" is not a commonly used term in physics. However, based on the given information, we can use the principles of special relativity to determine the speeds in different reference frames.

In the reference frame of A, the speed of B can be calculated using the relativistic velocity addition formula: VBA = (VBE + VAE)/(1 + (VBE*VAE)/c^2), where VBA is the velocity of B as seen by A, VBE is the velocity of B relative to E, and VAE is the velocity of A relative to E. Plugging in the values, we get VBA = (-0.7c + 0.9c)/(1 + (-0.7c*0.9c)/c^2) = 0.98c. This means that in A's reference frame, B appears to be moving to the left at a speed of 0.98c.

In the reference frame of A, the speed of E can be calculated using the same formula as above, but with the values of VBE and VAE swapped. This gives us VAE = (VAB + VBE)/(1 + (VAB*VBE)/c^2) = (-0.9c + 0.7c)/(1 + (-0.9c*0.7c)/c^2) = -0.92c. This means that in A's reference frame, E appears to be moving to the right at a speed of 0.92c.

In the reference frame of B, the speed of A can be calculated using the same formula as above, but with the values of VAB and VBE swapped. This gives us VAB = (VAE + VBE)/(1 + (VAE*VBE)/c^2) = (-0.92c + 0.7c)/(1 + (-0.92c*0.7c)/c^2) = -0.98c. This means that in B's reference frame, A appears to be moving to the left at a speed of 0.98c.

In the reference frame of B, the speed of E can be calculated using the relativistic velocity addition formula: VEB = (VAE + VAB)/(1 + (VAE*VAB)/c^2) = (-0.