Consequences:

There are two consequences of Lorentz's Transformation

Length Contraction (Lorentz-Fitzgerald Contraction):
Lorentz- Fitzgerald, first time, proposed that When a body moves comparable to the velocity of light relative to a stationary observer, then the
length of the body decreases along the direction of velocity. This decrease in length in the direction of motion is called '

Expression for Length Contraction:

Let us consider two frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ relative to frame $S$ along the positive x-axis direction. Let a rod is associated with frame $S'$. The rod is at rest in frame $S'$ so the actual length $l_{0}$ is measured by frame $S'$. So

$ l_{0}=x'_{2}-x'_{1}\quad\quad (1)$

Where $x'_{2}$ and $x'_{1}$ are the x-coordinate of the ends of the rod in frame $S'$.

According to Lorentz's Transformation

$ x'_{1}=\alpha (x_{1}-vt)\quad\quad (2)$

$ x'_{2}=\alpha (x_{2}-vt)\quad\quad (3)$

Now put the value of $x'_{1}$ and $x'_{2}$ in equation $(1)$, then

$ l_{0}=\alpha (x_{2}-vt)-\alpha (x_{1}-vt)$

$ l_{0}=\alpha \left (x_{2}-vt-x_{1}+vt \right )$
$ l_{0}=\alpha \left (x_{2}- x_{1}\right )$

$ l_{0}=\alpha l $ <
div>

$ l_{0}= \frac{l}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\quad\quad\left \{ \because \alpha =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \right \}$

Here $l$ is the length of the rod measured in frame $S$.

Here The factor $\sqrt{1-\frac{v^{2}}{c^{2}}}$ is less than unity. It means
that

$ \sqrt{1-\frac{v^{2}}{c^{2}}}< 1 $

so

So the length of the rod in frame $S$ will be less than the proper length or
actual length which is measured in frame $S'$.

Case:
If $v=c$, Then $l=0$, i.e. a rod moving with the velocity of light will
appear as a point to a stationary observer. So from the above discussion, we
can conclude that in relativity there is no absolute length'.

The length of the rod is measured by a stationary observer relative to the length of the rod in the frame.

Time Dilation (Apparent Retardation of Clocks):

Let us consider two frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ relative to frame $S$ along the positive x-axis direction.

Let two events occur in frame $S$ which is at rest at time $t_{1}$ and $t_{2}$. These two event measured in frame $S'$ at time $t'_{1}$ and
$t'_{2}$. So time interval between these two events in frame $S'$ is
$ t=t'_{2}-t'_{1}\qquad (1)$

According to Lorentz's Transformation

$ t'_{1}=\alpha '\left ( t_{1}-\frac{xv}{c^{2}} \right )\qquad(2)$

$ t'_{2}=\alpha '\left ( t_{2}-\frac{xv}{c^{2}} \right )\qquad(3)$

Now put the value of $t'_{1}$ and $t'_{2}$ in equation $(1)$.so

$ t=\alpha '\left ( t_{2}-\frac{xv}{c^{2}} \right )-\alpha '\left (t_{1}-\frac{xv}{c^{2}} \right )$

$ t=\alpha '\left (t_{2}-\frac{xv}{c^{2}}- t_{1}+\frac{xv}{c^{2}} \right )$

$ t=\alpha '\left (t_{2}- t_{1} \right )$

$ t=\alpha 't_{0}$

Here the factor $\sqrt{1-\frac{v^{2}}{c^{2}}}$ is less then unity. i.e

$ \sqrt{1-\frac{v^{2}}{c^{2}}}< 1$

Then

So the time interval between two events in frame $S'$ will be longer than the time interval taken in frame $S$.

The time dilation is a real effect. All clocks will appear running slow for
an observer in relative motion. It is incorrect to say that the clock in
moving frame $S'$ is slow as compared to the clock in stationary frame $S$. The
correct statement would be that All clocks will run slow for an observer in
relative motion.

Case:

If $v=c$, then $t=∞ $ i.e. When a clock moving with the speed of light appears to be completely stopped to an observer in a stationary frame of reference.

The time interval between two events that occur at the same position recorded by a clock in the frame in which the events occur (or frame at rest) is called 'proper time'.

The time interval between the same two events recorded by an observer in a frame that is moving with respect to the clock is known as 'Non -proper time or relativistic time'.

Experimental Verification of Time Dilation:

The direct experimental confirmation of time dilation is found in an experiment on cosmic ray particles called mesons. Î¼-mesons are created at
high altitudes in the earth's atmosphere (at the height of about 10 km) by the interaction of fast cosmic-ray photons and are projected towards the earth's surface with a very high speed of about $2.994\times10^{8}$ m/s which is $0.998$ of the speed of light $c$. Î¼-mesons are unstable and decay into electrons or positrons with an average lifetime of about $2.0\times10^{-6}$sec. Therefore, in its lifetime a Î¼-mesons can travel a
distance.

$ d=vt$

$ d=vt=(2.99\times10^{8})\: (2\times10^{-6})\simeq 600\:m$

$ d=0.6\:km$

Now the question arises how Î¼-mesons travel a distance of 10 km to reach the earth's surface. This is possible because of the time dilation effect. In fact, Î¼-mesons have an average lifetime $t_{0}=2.0\times10^{-6}$ sec in their own frame of reference. In the observer's frame of reference on the earth's surface, the lifetime of the Î¼-mesons is lengthened due to relativity effects to the value $t$ given as,

$t=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{2.0\times10^{-6}}{\sqrt{1-(0.998)^{2}}}=\frac{2.0\times10^{-6}}{0.063}$

$ t=3.17\times10^{-5} sec$

In this dilated lifetime, Î¼-mesons can travel a distance

$ d_{0}=(2.994\times10^{8})(3.17\times10^{-5})$

$ d_{0}=9500\: meter=9.5\:km$

This explains the presence of Î¼-mesons on the earth's surface despite their brief lifetime.

- Length Contraction (Lorentz-Fitzgerald Contraction)
- Time Dilation (Apparent Retardation of Clocks)

*Length Contraction*'.$ l_{0}= \frac{l}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\quad\quad\left \{ \because \alpha =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \right \}$

$l=l_{0}\sqrt{1-\frac{v^{2}}{c^{2}}}\quad$ |

$l< l_{0}$ |

***What is the proper length?*$t=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ |

$t> t_{0}$ |

*** Proper and Non-Proper Time:*
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