Monday, August 4, 2014

EMII Notes: Special Relativity and Gamma via Initial Conditions

I'm working on a new writing project as I prepare for Electricity and Magnetism II this semester.  I'll be reviewing the class lecture notes each day leading up to the start of the semester.  After each half hour review, I intend to write up a summary of the notes.  (Yes, I took a reflective writing workshop this summer, and I'm trying it out.)

Pages covered today:  7 & 8

We consider light as a spherical wavefornt

$x^2 + y^2 + z^2 - c^2t^2 = 0$

$x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - c^2t^{\prime 2} = 0$

We also look at the usual situation with two frames, $S$ and $S^{\prime}$ that are moving parallel to each other where $S^{\prime}$ has positive velocity $v$ in the x direction with respect to $S$.

We have the usual postulate that c, the speed of light is constan in all frames.

We need a way to transform the distance and time coordinates in the two frames.  Because we want to preserve invariance with respect to translation, (conservation of momentum), and invariance with respectc to translations in space-time, (conservation of energy), we're stuck with a linear transformation.  so

$x^{\prime} = Ax + Bt$

$t^{\prime} = Cx + Dt$

Question:  Is the above an example of a bilinear transform?

We won't really bother with C and D I don't think.

At time t = 0, we set the origins of both frames at the same position and get

$x^{\prime} = 0 => x = vt$

so

$0 = Avt + Bt$
$B = -Av$

Now, since we know where we're headed, go ahead and rename $A$ to be $\gamma$.

This all gives:

$x^{\prime} = \gamma x - \gamma v t$
$x^{\prime} =  \gamma\left( x - v t\right)$

For the $S$ frame at $x = 0$, we have $x^{\prime} = -vt^\{prime}$

We can use this reversed velocity to do a nice little trick giving

$x =  \gamma\left( x^{\prime} - v t^{\prime}\right)$


without doing any math to solve for $x$ in terms of $x^{\prime}$

Since the speed of light is constant in every frame.  That's our postulate.  we get

$x = ct <==> x^{\prime} = ct^{\prime}$

Plugging this into the $AB$ transform equation above we get

$ct^{\prime} = \gamma x - \gamma v t$
$ct^{\prime} = \gamma\left(ct - vt\right)$

and the following string of algebra from solving the two equations above for gamma and then multiplying them together.  

$\gamma = \dfrac{ct}{t^{\prime}\left(c+v\right)}$
$\gamma = \dfrac{ct^{\prime}}{t\left(c - v\right)}$

multiply the two equations for gamma to get.  (Why?  Because I know the answer I'm headed for and this seems simplest).

$\gamma^2 = \dfrac{c^2  t t^{\prime}}{t t^{\prime} \left(c + v\right)\left(c - v\right)}$
$ = \dfrac{c^2}{c^2 - v^2}$
$ = \dfrac{1}{1-\dfrac{v^2}{c^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Question:
Since we never used the $BD$ transform equation, is there more information that can be gleaned from it?



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