# 4 Force (Homework 9-323)

1. Transformation of force.

a. Show that by transforming the numerator and denominator between different frames, you get Griffith’s equations 12.65 & 12.66. (Basically, just go through Griffith’s derivation). And thereby demonstrate you recover the special case of Griffith’s 12.67 when the particle is at rest in the initial frame.

The 4 Force is defined as

$F=d\textbf{p}/d t$

In S’ (moving with speed v on the x direction) we have

$F_x’=(d p_x’)/(dt’ )$    and $F_y’=(d p_y’)/(dt’)$
We need to know first how 4 momentum $\textbf{p’}$  transforms:

$\textbf{p’}=γ’ m\textbf{u’}$    which means $p_x’=γ’ m v_x’$      and $p_y’=γ’ m v_y’$

We write
$u’=(v-u)/(1+v u/c^2)$      and we obtain (this is EXTREMELY COMPLICATED and would take 2 pages):
$\begin{pmatrix} p_x’\\ p_y’\\ p_z’\\ E’/c \end{pmatrix}=\begin{pmatrix} \gamma & 0 & 0 &-\beta\gamma \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\beta\gamma & 0 & 0 & \gamma \end{pmatrix}\begin{pmatrix} p_x\\ p_y\\ p_z\\ E/c \end{pmatrix}$

(Lorentz transformation of momentum-energy vector)

We can write therefore

$p_x’=γ(p_x-β E/c)$   and $p_y’=p_y$

Hence the differentials of momentum transforms as:

$d p_x’=γ(d p_x-βd(E/c) )$    and $d p_y’=d p_y$
For time we know time is slower in the moving reference:

$t=γt’$

$d t=γdt’+t’ dγ=γdt’+(t/γ) dγ$

with

$dγ=d(1/\sqrt{1-v^2/c^2})=-1/[2\sqrt{(1-v^2/c^2 )^3}]*(-2v/c^2 )d v=γ^3*(β/c)*d v$

$d t=γdt’+γ^2*(β/c) (t d v)$   with $d x=d(v t)=t d v+v d t$

$d t=γdt’+γ^2*(β/c) (d x-v d t)$   or

$d t(1+γ^2*v^2/c^2 )=γdt’+γ^2*(β/c) d x$

$1+γ^2*v^2/c^2 =1+(v^2/c^2 )/(1-v^2/c^2 )=γ^2$

$γ^2 d t=γdt’+γ^2*β/c*d x$

$dt’=γdt-γ*β/c*d x$

Therefore we have

$F_y’=(d p_y’)/dt’=(d p_y)/(γdt-γβ/c*d x)=$

$=((d p_y)/d t)/γ(1-β/c*d x/d t) =F_y/γ(1-β/c*v_x )$         (12.65)

$F_x’=(d p_x’)/(dt’ )=γ(d p_x-β*d(E/c) )/(γdt-γ*β/c*d x)=$
$=((d p_x)/d t-β/c*d E/d t)/(1-β/c*(d x/d t) )=(F_x-β/c*d E/d t)/(1-β/c*v_x)$

$d E/d t=d/d t (γm_0 c^2 )=m_0 c^2*dγ/d t=$
$=m_0 c^2*γ^3*β/c*d v/d t=m_0 γ^3 v*d v/d t=$

$=v*d/d t ((m_0 v)/√(1-v^2/c^2 ))=v*d p/d t=v*F$

$F_x’=(F_x-β/c(v F))/(1-β/c*v_x)$                    (12.66)

For $v=0$  (speed of charge) we have
$F_y’=F_y/γ$      and $F_x’=F_x$       (12.67)