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Harrisonized

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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/c

^{2})x],

where γ=(1-v

^{2}/c

^{2})

^{-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/c

^{2})x]+γ[dt-(v/c

^{2})x-(1/c

^{2})x.dv)]

dγ = (-1/2)(1-v

^{2}/c

^{2})

^{-3/2}(-2v.dv) = vγ

^{3}.dv

Hence,

dx' = [vγ

^{3}(x-vt)-γt]dv + γ(dx-v.dt)

dt' = [vγ

^{3}(t-v/c

^{2}x)-γt]dv + γ(dt-v/c

^{2}.dx)

We set dv=0, and we end up with

dx'/dt' = (dx-v.dt)/(dt-v/c

^{2}.dx)

= (dx/dt)[(1-v(dt/dx))/(1-v/c

^{2}(dx/dt))]

= [(dx/dt)-v]/[1-v/c

^{2}(dx/dt)]

and consequently:

dx/dt = [dx'/dt'+v]/(1+v/c

^{2}(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' = [vγ

^{3}(x-vt)-γt]dv + γ(dx-v.dt)

dt' = [vγ

^{3}(t-v/c

^{2}x)-γt]dv + γ(dt-v/c

^{2}.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.