1. The problem statement, all variables and given/known data During class today, I was told that: x' = γ(x-vt) y' = y z' = z t' = γ[t-(v/c2)x], where γ=(1-v2/c2)-1/2 (This is just the standard Lorentz transformation.) Then I was told that we could find dx/dt, the inertial velocity, by finding dx' and dt' and dividing them. 2. Relevant equations dx' = dγ(x-vt)+γ(dx-v.dt-t.dv) dt' = dγ[t-(v/c2)x]+γ[dt-(v/c2)x-(1/c2)x.dv)] dγ = (-1/2)(1-v2/c2)-3/2(-2v.dv) = vγ3.dv Hence, dx' = [vγ3(x-vt)-γt]dv + γ(dx-v.dt) dt' = [vγ3(t-v/c2x)-γt]dv + γ(dt-v/c2.dx) We set dv=0, and we end up with dx'/dt' = (dx-v.dt)/(dt-v/c2.dx) = (dx/dt)[(1-v(dt/dx))/(1-v/c2(dx/dt))] = [(dx/dt)-v]/[1-v/c2(dx/dt)] and consequently: dx/dt = [dx'/dt'+v]/(1+v/c2(dx'/dt')], which is what one would expect. 3. The attempt at a solution But then I inquired about what happens when dv is non-zero, since the entire equation looks like: dx' = [vγ3(x-vt)-γt]dv + γ(dx-v.dt) dt' = [vγ3(t-v/c2x)-γt]dv + γ(dt-v/c2.dx) My TA told me that the equations don't hold, and that people have actually proven that the equation doesn't hold unless dv=0. My question is: why? I feel that my TA was just trying to escape answering the question.