The Second postulate of Relativity; what are the implications?

The Second postulate of Relativity states that the speed of light has the same value in any inertial system of reference. The critical question is what is an inertial system of reference? We consider a simple device constructed by a measuring rod L, a pulsing light source and a sensor at one end of the measuring rod L and a mirror on the other end of the measuring rod L. IF this measuring rod is an inertial system of reference, then the light will move from the source to the mirror with velocity c and is reflected and moves from the mirror to the sensor with a velocity c. Then it is clear that the time-difference between emitting the light by the source and receiving the light by the sensor is given by \displaystyle \Delta T = 2L/c. We may represent the device by a vector \vec{L} which represents both the length of the measuring rod as well as the direction.
If so we may consider that this device is moving with respect to another system of reference, such that the velocity is uniform. The time required for the light ray to reach the mirror is given by \Delta T'_+. The time required for the reflected light ray to reach the sensor is given by \Delta T'_-. The path of the light ray to reach the mirror is given by \vec{v} \Delta T'_+ + \vec{L}' and the path of the light ray to reach the sensor is given by \vec{v} \Delta T'_- - \vec{L}'. We can generaly say that \vec{P}_\pm = \vec{v} \Delta T'_\pm \pm \vec{L}'. But we also have that |\vec{P}_\pm| = c \Delta T'_\pm. Here + is associated with the light ray moving toward the mirror and - is associated with the reflected light ray moving to the sensor. We have

(1) |\vec{P}_\pm| = c \Delta T'_\pm thus |\vec{v} \Delta T'_\pm \pm \vec{L'}| = c \Delta T'_\pm,

Then it is clear that

(2a) v^2 \Delta T'^2_\pm + L'^2 \pm 2 \Delta T'_\pm \vec{v} \cdot \vec{L}' = c^2 \Delta (T'_\pm)^2

Then we find that

(2b) \displaystyle \left( \Delta T'_\pm \mp \cos(\phi) \frac{vL'}{c^2-v^2} \right)^2 = \left( \frac{L'}{c^2-v^2} \right)^2 \left( c^2 - v^2 \sin^2(\phi) \right)

So

(2c) \displaystyle \Delta T'_\pm = \frac{L'}{c^2-v^2} \left( \pm v \cos(\phi) + \sqrt{c^2 - v^2 \sin^2(\phi)} \right)

Then

(3) \displaystyle \Delta T' = \Delta T'_+ + \Delta T'_- = \frac{2L'}{c^2-v^2} \sqrt{c^2-v^2\sin^2(\phi)}

This can be written as

(4a) \displaystyle \frac{2L'}{\Delta T'} = c \frac{1-\beta^2}{\sqrt{1-\beta^2\sin^2(\phi)}},

where \beta = v/c. When we argue that \displaystyle c = \frac{2L}{\Delta T} we obtain

(5) \displaystyle \frac{2L'}{\Delta T'} = \frac{2L}{\Delta T} \frac{1-\beta^2}{\sqrt{1-\beta^2\sin^2(\phi)}}.

Therfore

(6) \displaystyle \frac{L'}{L} = \frac{\Delta T'}{\Delta T} \frac{1-\beta^2}{\sqrt{1-\beta^2\sin^2(\phi)}}.

Now let \displaystyle \tau = \frac{\Delta T'}{\Delta T} and \displaystyle \lambda_\phi = \frac{L'}{L}, then

(7) \displaystyle \lambda_\phi = \tau \frac{1-\beta^2}{\sqrt{1-\beta^2\sin^2(\phi)}}.

Now let \displaystyle \gamma(v) = \frac{1}{\sqrt{1-\beta^2}}, \lambda_{\phi=0} = \lambda_x and \lambda_{\phi=\pi/2} = \lambda_{\perp x} then we obtain

(8a) \displaystyle \lambda_x = \tau / \gamma^2(v) and
(8b) \displaystyle \lambda_{\perp x} = \tau / \gamma(v) thus
(8c) \displaystyle \lambda_{\perp x} = \lambda_x \gamma(v).

Two special cases are:

(9) \begin{array}{r|ccc} & \tau & \lambda_x & \lambda_{\perp x} \\ \hline \textrm{Lorentz} & \gamma(v) & 1/\gamma(v) & 1 \\ \textrm{Voigt} & \gamma^2(v) & 1 & \gamma(v) \end{array}

But in general, length-contraction depends on the direction; this is a direct conclusion from the Second postulate of Relativity. Any claim that distance-contraction does not depend on the direction therefore violates the Second principle of Relativity.

Advertisements
This entry was posted in Relativity, Transformations. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s