Classical Mechanics - 2: Lorentz Transformation

Lorentz Transformations are a set of linear transformations from one co-ordinate in space-time to another frame that moves at a constant velocity (the parameter) relative to the former Read more.

Doing justice to the title of the blog, I will straightaway dive into the mathematics of the concept,

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Imagine a basic cartesian co-ordinate system in three dimensions, we this this $\text{S- Frame}$, refer Figure-1 below,

Now, the image below, denoted as Figure-2 shows another frame $\text{S'-Frame}$.

Note that the origins of both frames, i.e., $O$ and $O'$ coincide at $t=0$.

Now imagine a ray of light being emitted from the origins of both the origins.

As per three-dimensional co-ordinate system, the distance covered by light will the distance of the point where light currently is at that particular instant of time.

So, if light is a point $P(a,b,c)$ at $t=t_1$, then the distance from origin is,
$d = \sqrt{a^2+b^2+c^2}$
And, since light is moving at speed $\mathrm{c}$ and the time is $t_1$, we can write above equation as,
$\mathrm{c}t_1=\sqrt{a^2+b^2+c^2}$

Now that the basic is clear, let us formulate the Progress of light equations for fram $S$ and $S'$ when light is emitted from both its origins,

Progress of light described in relation to frame $S$ is,
$ct = \sqrt{x^2+y^2+z^2} \\ c^2t^2 =x^2+y^2+z^2 \\ x^2+y^2+z^2-c^2t^2 =0\tag*{(1)}$

Similarly, progress of light described in relation to frame $S$ is,
$ct' = \sqrt{x'^2+y'^2+z'^2} \\ c^2t'^2 =x'^2+y'^2+z'^2 \\ x'^2+y'^2+z'^2-c^2t'^2 =0 \tag*{(2)}$

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Now, it is important that you pay attention to Figure-2, there are 4 cases we consider to formulate the next equations leading us to Lorentz Transformations,

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Case 1:
You (the observer) is standing at $\text{S-Frame}$ and loking at $O'$.
You will see that $O'$ is moving at velocity$v$ (not using $c$ for generalization’s sake) along $x$-axis.
[time = $t$, distance = $x$ and velocity = $v$]
$v = \frac{x}{t} \\ x - vt = 0 \tag*{(3)}$

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Case 2:
You (the observer) is standing at $\text{S'-Frame}$ and loking at $O'$.
You will see that $O'$ is not moving with respect to $\text{S'-Frame}$.
So, $x'=0$,
which can be written as,
$x' = a( x - vt) \tag*{(4)}$
Since, $x-vt=0$.
$a$ is a constant that is used to show that both the equations are in different frames of references.

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Case 3:
You (the observer) is standing at $\text{S'-Frame}$ and loking at $O$.
PAY ATTENTION
Note that $v$ is the velocity and hence is dependent of direction (a vector quantity).
You will see that $O$ is moving at velocity$-v$ (not using $c$ for generalization’s sake) along $x$-axis.
$-v = \frac{x'}{t'}\\ x'+vt'=0 \tag*{(5)}$

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Case 4:
You (the observer) is standing at $\text{S-Frame}$ and loking at $O$.
You will see that $O$ is not moving with respect to $\text{S-Frame}$.
So, $x=0$,
which can be written as,
$x = a'( x' + vt') \tag*{(6)}$
Since, $x'+vt'=0$.
$a'$ is a constant that is used to show that both the equations are in different frames of references.

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Please note that the velocity is same in all frames, which is in line with the special theory of relativity.

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See, eqn $(4)$,
$x' = a( x - vt)$
See eqn $(6)$,
$x = a'( x' + vt')$
Now, use $(4)$ in $(6)$, to find $t'$
$x = a'x'+a'vt' \\ \implies x = a' (a(x-vt))+a'vt' \\ \implies x = a'ax - a'avt + a'vt' \\ \implies -a'vt' = a'ax - a'avt-x \\ \implies a'vt' = x + a'avt - a'ax\\ \implies t' = \frac{1}{a'}\frac{1}{v}x + at - a \frac{x}{v} \\ \implies \boxed{t' = a [ t + \frac{x}{v}[\frac{1}{aa'}-1]]} \tag*{(7)}$

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We will have to now solve for $a$ and $a'$, which will help us in expressing $x'$ and $t'$ in terms of $x$, $v$ and $c$.

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In the $\text{S'-Frame}$, there is no motion along the $y'$ and $z'$ axis, thus,
(We do this step to bring motion in $S'$ frame in terms of parameters of $S$ frame)
$y'=y \\ z' = z$
And from $(4)$,
$x' = a(x-vt)$
Also, from $(7)$,
$t' = a [ t + \frac{x}{v}[\frac{1}{aa'}-1]]$
Substitute the above values in eqn $(2)$, we get,
$[a(x-vt)]^2+y^2+z^2 - c^2a^2[t + \frac{x}{v}[\frac{1}{aa'}-1]]^2=0$
After some algebraic rigor (which I have gone through, here is the proof :p), you will arrive at,
$[a^2 + \frac{c^2a^2}{v^2}(\frac{1}{aa'}-1)^2]x^2+y^2+z^2-t^2(-a^2v^2+c^2a^2)\\-2xt[a^2v+\frac{c^2a^2}{v}(\frac{1}{aa'}-1)]=0 \tag*{(8)}$
Comparing coefficients of eqn $(8)$ with that of eqn $(1)$,

Comparing $x^2$,
$a^2 + \frac{c^2a^2}{v^2}(\frac{1}{aa'}-1)^2 = 1 \tag*{(9)}$


Comparing $y^2$ and $z^2$ will yield no significant information as the coefficients are same.


Comparing $t^2$
$-a^2v^2+c^2a^2=c^2 \tag*{(10)}$


Comparing $-2xt$,
$a^2v+\frac{c^2a^2}{v}(\frac{1}{aa'}-1)=0 \tag*{(11)}$

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Solve $(10)$ for $a$,
$a^2 = \frac{c^2}{c^2-v^2}\\ \boxed{a = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(12)}$


Solve $(11)$ for $a'$,
$a^2v+\frac{c^2a^2}{v}(\frac{1}{aa'}-1)=0 \\ v+\frac{c^2}{v}(\frac{1}{aa'}-1)=0 \\ \frac{c^2}{v}(\frac{1}{aa'}-1)=-v \\ (\frac{1}{aa'}-1)=-\frac{v^2}{c^2} \\ a' = \frac{1}{(1-\frac{v^2}{c^2}).a} \\ a' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \tag*{(13)}$

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$\therefore \boxed{a = a' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(14)}$
$\text{This is important.}$

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Now, use derived $a$ and $a'$ in equations describing $x'$ and $t'$ to find the space and time Lorentz Transformations,

Eqn $(4)$,
$x' = a(x-vt) \\ \implies \boxed{x' = \frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(15)}$


Eqn $(7)$
$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[t + \frac{x}{v}[\frac{1}{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}-1]] \\ \implies t'=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[t+\frac{x}{v}[1-\frac{v^2}{c^2}-1]\\ \implies \boxed{t' = \frac{t-\frac{xv}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}} \tag*{(16)}$
Equation $(15)$ and $(16)$ are the Lorentz Space and Time Transformation.


Cheers and a very Happy Diwali 2019 $\smile$ !!