(7) Show that if a particle moves at an angle θ with respect to the x-axis with speed u in frame S, then it moves at an angle θ′ with the x′-axis in frame S′ with

Question

(7) Show that if a particle moves at an angle θ with respect to the x-axis with speed u in frame S, then it moves at an angle θ′ with the x′-axis in frame S′ with

tanθ′ = sinθ /( γ(cos θ − v/u))

Assume that S′ moves in the +x direction with velocity v with respect to frame S, and that the two x-axes are parallel.

Answer 1

Well, I must say, this expression looks quite "tan"tastic! But let's dive into it and see what we can do.

First, let's consider the Lorentz transformation for velocity in the x-axis direction. We have:

v' = (v - u)/(1 - v*u/c^2)

Since we're only interested in the x-axis, we can disregard the y and z components. Now, let's break down the Lorentz transformation for velocity into its components:

v' = v*cosθ - u*sinθ / (1 - (v*u/c^2))

Now, let's multiply numerator and denominator by γ = 1/√(1 - (v^2/c^2)) to simplify things:

v' = (v*cosθ - u*sinθ) * γ / (1 - (v*u/c^2)) * γ

Now, since we're dealing with angles here, let's play around with some trigonometric identities. We know that sin * γ = sinh, cos * γ = cosh, and sinh/cosh = tanh. So, let's use these identities:

v' = (v*cosh(θ) - u*sinh(θ)) / (1 - (v*u/c^2)) * γ

Now, let's separate the numerator into two terms:

v' = v*cosh(θ)/γ - u*sinh(θ)/γ / (1 - (v*u/c^2)) * γ

Now, let's remember that γ = 1/√(1 - (v^2/c^2)). So, we can substitute that in:

v' = v*cosh(θ)/√(1 - (v^2/c^2)) - u*sinh(θ)/√(1 - (v^2/c^2)) / (1 - (v*u/c^2)) * √(1 - (v^2/c^2))

Now, let's combine the two terms in the numerator:

v' = (v*cosh(θ) - u*sinh(θ))/√(1 - (v^2/c^2)) / (1 - (v*u/c^2)) * √(1 - (v^2/c^2))

Now, let's remember that sinh/√(1 - (v^2/c^2)) is equal to tanθ:

v' = (v*cosh(θ) - u*tan(θ))/√(1 - (v^2/c^2)) / (1 - (v*u/c^2)) * √(1 - (v^2/c^2))

Finally, let's simplify the expression a bit more:

v' = (v*cos(θ) - (v*u*sin(θ)/c^2))/√(1 - (v^2/c^2)) * √(1 - (v^2/c^2))/(1 - (v*u/c^2))

We can simplify further by canceling out some of the terms:

v' = (v*cosθ - v*u*sinθ/c^2)/(1 - (v^2/c^2))

And voila! We have our expression:

v' = (v*cosθ - v*u*sinθ/c^2)/(1 - (v^2/c^2))

So, I'm not exactly a math wiz, but I hope you found my attempt at humor and explanation helpful. Remember, if all else fails, "cos you can't tan properly without a good chuckle!"