We outline
a short derivation the formula for Lorentz transform.
Please pay attention to Question 1 (below).
I. By $O$ , $O'$ denote coordinate origin of $K$ and $K'$
Suppose that for $t=t'=0$, $O'=O$.
If the clock in rest in $K$ shows time $t$ at a point $A$ then
the coordinate of $A$ is $vt$ ?
In a moment $t$ from point $M'=(0,0, d')$ (denote by $M=(x,0,
d')$ in K) we reflect light ray to $O'$ (denote by $A=x$ in K),
which arrives for $\Delta t'$ at $O'$ (denote by $B$ in K)
$MB= c \Delta t$, $d'= c \Delta t'$, $AB= v \Delta t$
Hence, by The Pythagorean theorem, $\Delta t= \gamma \Delta t'$
II.
Imagine stick $p=O'x'$ in $K'$ of length $l'=x'$
Let $l$ be length of the stick $p=O'x'$ in $K$; then (i) $x=vt +
l$.
We introduce the concept of length (measure) of sticks in $K$ and $K'$.
Question 1. It seems to me that we need to prove more rigorously (i).
Further let $O'$ arrives for time $ t_1'$ (for $t_1$ wrt $K$)
from $vt$ to $x$). Then, since the stick is in rest in $K'$,
$t_1'=\gamma t_1$ and therefore
$l=t_1 v= v t_1'/\gamma =l'/\gamma= x'/\gamma$,so $l=l'/\gamma$.
Hence we find
$x=vt + l= (vt +x')/\gamma $
and therefore
(1) $x'= \gamma(x-vt)$.
As in the Galilean transformation, the transport velocity has to
be changed from v to -v when passing from one frame to the other.
(1) $x'= \gamma(x-vt)$ since K moves with speed $-v$ with respect
to $K'$,
(2) $x= \gamma (x'+ vt)$
If we substitute (1) in (2), using
$1-\gamma^2=- \frac{v^2}{c^2}\gamma^2$, we have
$t'= \gamma (t-vx/c^2)$.
@Forrest Anderson,
A temporarily version is enclosed for RG-purposes. As soon as possible we we will polish it.
We try to make a mathematical model which is motivated by STR and try to give a rigorous proof from mathematical point of view of LT-formula.
We outline a few short derivation the formula for Lorentz transform. Our derivations are based on the concept of length of moving rods and simple Principle of synchronization of clocks.
This paper can help you Mr. Mateljevic
http://www.phys.ufl.edu/courses/phy3221/spring11/lecture4.pdf
Regards,
Syahrial Shaddiq
We try to prove (rigorously from mathematical point of view) that Lorentz transformation are isometry in the corresponding Minkowski spaces.
If $x_1,y_1,z_1,t_1$ and $x_2,y_2,z_2,t_2$ coordinates of two events, then the quantity
\begin{equation}
s_{12}= [c^2 (t_2-t_1)^2 - (x_2-x_1)^2 - (y_2-y_1)^2 - (z_2-z_1)^2] ^{1/2}
\end{equation}
is called the space-time interval between two events.
The corresponding length element $ds$ ( the scalar product) is given by
\begin{equation}
ds^2= c^2 dt^2-( dx^2 + dy^2 +dz^2).
\end{equation}
Suppose that we have two inertial frames $S$ and $S'$.
If $ds=0$ in $S$,
then $ds'=0$ in $S'$.
Hence $ds= e ds'$.
By symmetry $ds'= e ds$. Hence $ds=ds'$ (Landay- Lefsic).
We will show (rigorously from mathematical point of view) that by homogeneity of space and time it follows that $e$ does not depend of events.
Here, we outline an explanation for this !
We denote Lorentz transformation by $L$: $(x',y',z',t')= L (x,y,z,t)$.
Let $ M=(x_0,t_0)$, $T_M$ tangent space at $M$ and $A=dL$ differential of $L$
defined on $T_M$. If $M'=L(M)$, then $ A : T_M \rightarrow T_{M'}$.
Note that we do not know that a priory that $L$ is a linear operator.
We will first prove that $ds'(Ah,Ah)= e s(h,h)$, for every $h \in T_M$, where $e=e(M)$ is independent of \, $h$.
Let
$E_1'=E_1=(1,1/c)$, $E_2'=E_2=(1,-1/c)$ be vectors such that $E_1,E_2 \in T_M$ and $E_1',E_2' \in T_{M'}$, $h= \lambda E_1 + \mu E_2 \in T_M$.
Since, if $ds=0$ in $S$, then $ds'=0$ in $S'$,
we conclude that $ A(E_1) =a E_1'$ and $ A(E_2) =b E_2'$, where $a,b > 0$.
Since $A$ is linear $A h= \lambda A E_1 + \mu A E_2$ $h= (\lambda + \mu, (\lambda - \mu)/c )$ and $A(h)= (a \lambda + b \mu, (a \lambda - b \mu)/c )$, we find
$(h,h)_1= (\lambda - \mu)^2- (\lambda + \mu)^2=-4 \lambda \mu$ and
$(Ah,Ah)_1= -4 \lambda \mu a b $. Hence $(Ah,Ah)_1 = d\cdot (h,h)_1$, where $d=ab$; therefore $ds'= e ds$. More precisely set $L_0(x,t)= (x+x_0,t+t_0)$,
$L^1(x',t')= (x'-x'_0,t'-t'_0)$, $L_0^0= L^1\circ L\circ L_0$ and $A_0= dL_0^0(0,0)$. Then by the above procedure $ds'(0,0)=e_0 ds(0,0)$ and since $L_0$ and $L^1$
are Minkowski isometry, we get $ds'(M')=e_0 ds(M)$.