I know the way of deriving the formula using usual lorentz transformation formulas,,but is there a way out of deriving it using 4-vector notation??please help
Answer
In tensor notation, Λμν is the matrix that performs the Lorentz transformations. Now, since pμ is a tensor, under an arbitrary Lorentz transformation, it transforms as pμ→p′μ=Λμνpν Consider now a particle moving entirely in the x direction, with velocity v1. Its 4-momentum is pμ=γ1(1,β1,0,0) where γ=1√1−v21/c2, β1=v1c Now, we want to reproduce the velocity addition formula, so we are interested in a boost. The Lorentz matrix corresponding to a boost with velocity −v2 is explicitly† Λ=(γ2β2γ200β2γ2γ20000100001)
The new boosted 4-momentum is p′μ=γ1γ2(1+β1β2,β1+β2,0,0) Now, we simply have to extract the transformed velocity from this formula. Firstly note that since this is 4-momentum, it is of the form p′μ=γ3(1,β3,0,0) for some velocity v3. Thus, we have the equation γ1γ2(1+β1β2)=γ3 We can use this equation to solve for v3 and we find v3=v1+v21+v1v2c2 which we sought out to prove.
†Be careful to note that this is the boost matrix corresponding to −v2. This is because I want to reproduce "velocity addition" as opposed to "relative velocity" both of which are obviously related concepts. In others, given a particle moving with speed v1 in one frame, if I boost myself with speed v2 in the opposite direction, then the speed of the particle in the new frame literally adds, i.e. it is proportional to v1+v2. This is the reason to choose −v2.
No comments:
Post a Comment