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Homework Statement
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.
Homework 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.
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.