# Derivation of Lorentz Transformation for Acceleration

1. Jan 28, 2014

### MostlyHarmless

1. The problem statement, all variables and given/known data
Starting with the Lorentz transformation for the components of the velocity, derive the transformation for the components of acceleration.

2. Relevant equations
Lorentz Transformation for position and time :
$x'={\gamma}(x - vt)$
$t'={\gamma}(t - {\frac{vx}{c^2}})$
Resulting transformation by taking ${\frac{dx'}{dt'}}$
$u'_x={\frac{u_x-v}{1-{\frac{vu_x}{c^2}}}}$

3. The attempt at a solution
Before I go through the trouble of typing out my attempt in LaTex (I'm on my phone), let me make sure I'm going at this right because I think I'm over thinking this and now its all twisted in my head.
Based on what they did to get the velocity transformation, I should be able to just take $du'_x$ and divide by the same $dt'$ used in the velocity transformation? Because it should be CHANGE of velocity over change in time. So ${\frac{du'_x}{dt'}}$ should give me what I want right?

EDIT: I've refined my confusion.

Its about how they are getting dx' and dt'.
They are given as:
$dx'= {\gamma}(dx - vdt)$
$dt'={\gamma}(dt - {\frac{vdx}{c^2}})$

I'm not sure what they are differentiating with respect to so I don't know how to treat $u'_x$ as far as differentiation goes. I'm assuming(hoping) that my confusion is a result of the my notions of classical mechanics not being consistent here and not a result of losing my grip on Cal 1 stuff.

Last edited: Jan 28, 2014
2. Jan 29, 2014

### voko

It does seem that you are losing your grip on calculus. $x'$ and $t'$ are functions of $x$ and $t$, so their total differentials will be linear forms of total differentials $dx$ and $dt$.