To Frame My Frame…(Frame II : Einstein) -4. The end of the argument

December 8, 2005

In fact, we are almost at the end of our arguments. To complete the argument, consider now a third observer C moving with a velocity $v _1$ with respect to A.(Remember that A is moving with a velocity $v$ with respect to B already.) So, the vectors $\hat{e}_t''$ and $\hat{e}_x''$ of the third observer are related to $\hat{e}_t'$ and $\hat{e}_x'$ of the observer A by the equations

\begin{displaymath}\hat{e}_t'' = \gamma_1 \hat{e}_t' + \gamma_1 v_1 \hat{e}_x' \end{displaymath}
\begin{displaymath}\hat{e}_x'' = \gamma_1 \hat{e}_x' + \frac{\gamma_1}{v_1}\left(1-\frac{1}{\gamma_1^2}\right) \hat{e}_t' \end{displaymath}

We can substitute for $\hat{e}_t'$ and $\hat{e}_x'$ in terms of $\hat{e}_t$ and $\hat{e}_x$ to get

\begin{displaymath}\hat{e}_t'' = \gamma\gamma_1\left(1+\frac{v_1}{v}\left(1-\frac{1}{\gamma^2}\right)\right) \hat{e}_t + (\ldots)\ \hat{e}_x \end{displaymath}
\begin{displaymath}\hat{e}_x'' = \gamma\gamma_1\left(1+\frac{v}{v_1}\left(1-\frac{1}{\gamma_1^2}\right)\right) \hat{e}_x + (\ldots)\ \hat{e}_t \end{displaymath}

where we have lazily omitted the terms which we don’t need for further argument😉

There is another way we can find how $\hat{e}_t''$ and $\hat{e}_x''$ of the third observer C are related to $\hat{e}_t$ and $\hat{e}_x$ of the observer B at rest. If the third observer C is moving with a velocity $v_2$ with respect to B, then we can directly write

\begin{displaymath}\hat{e}_t'' = \gamma_2 \hat{e}_t + \gamma_2 v_2 \hat{e}_x \end{displaymath}
\begin{displaymath}\hat{e}_x' = \gamma_2 \hat{e}_x + \frac{\gamma_2}{v_2}\left(1-\frac{1}{\gamma_2^2}\right) \hat{e}_t \end{displaymath}

The only feature of interest to us is the fact that the coefficient of $\hat{e}_t$ in the first equation is same as the coefficient of $\hat{e}_x$ in the second equation. Looking back at the equations in the last paragraph, we conclude that this can be true only if

\begin{displaymath}\frac{v_1}{v}\left(1-\frac{1}{\gamma^2}\right)=\frac{v}{v_1}\left(1-\frac{1}{\gamma_1^2}\right)\end{displaymath}

or

\begin{displaymath}\frac{1}{v^2}\left(1-\frac{1}{\gamma^2}\right)=\frac{1}{v_1^2}\left(1-\frac{1}{\gamma_1^2}\right)\end{displaymath}

Now, since we can choose $v$ and $v _1$ to be anything among the velocities that are physically possible,we conclude that for any velocity the following expression should hold.

\begin{displaymath}\frac{1}{v^2}\left(1-\frac{1}{\gamma^2}\right)= \ some\ constant\ =\lambda (say)\end{displaymath}

This implies that $\gamma$ should be related to $v$ as

\begin{displaymath}\gamma=\frac{1}{\sqrt{1-\lambda v^2}} \end{displaymath}

which is the condition on $\gamma$ if principle of relativity and our other “sensible? assumptions are true ! Now, to conform with notation used by everybody else, we will call $\frac{1}{\sqrt{\lambda}}$ as $c$ or we define $c$ such that $\lambda\equiv\frac{1}{c^2}$(Note that $c$ has the same units as speed).Until now, with all we know now, since $\lambda$ can be any real number, $c$ can be any real or any purely imaginary number.Now of course we will be very much ineinterested knowing its value !

One way is to directly measure the behaviour of moving clocks, and it can be done. But, there are other ways which are widely used to determine it. And we will return to this thing later.Anyway, we can now write down with all due glory,

\begin{displaymath}\hat{e}_t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(\hat{e}_t + v \hat{e}_x \right) \end{displaymath}
\begin{displaymath}\hat{e}_x' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(\hat{e}_x + \frac{v}{c^2} \hat{e}_t \right) \end{displaymath}

And in these equations lies the essence of relativity !I’ll just conclude for now with an interesting observation by Minkowski. Say, we take the notation of a time-like vector seriously. AtlAt leastnoenoughriously to imagine dot product of time-like vectors with the other vectors. Then, what Minkowski said was basically this- If we assume $\hat{e}_t.\hat{e}_t=-c^2$, $\hat{e}_x.\hat{e}_t =0 $ then the above equations imply that $\hat{e}_t'.\hat{e}_t'=-c^2$ and $\hat{e}_x'.\hat{e}_t' =0 $ ! which essentially means that time-like vectors are more like vectors than we imagined and the above transformations are a “kind? of rotation in which the “angle? between vectors are preserved ! This is among the most valuable insights that relativity offers- time is very much like one another direction in space, only slightly different !

“The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.? HERMANN MINKOWSKI

(To be continued)

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