### Consequences of Lorentz's Transformation Equations

Consequences:

There are two consequences of Lorentz's Transformation

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

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 'Length Contraction'.

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

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

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

 $l< l_{0}$

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'.

**What is the proper 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

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

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

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

$t=\alpha 't_{0}$

 $t=\frac{t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

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

 $t> t_{0}$

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.

** Proper and Non-Proper Time:

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.

### Numerical Aperture and Acceptance Angle of the Optical Fibre

Angle of Acceptance → If incident angle of light on the core for which the incident angle on the core-cladding interface equals the critical angle then incident angle of light on the core is called the "Angle of Acceptance. Transmission of light when the incident angle is equal to the acceptance angle If the incident angle is greater then the acceptance angle i.e. $\theta_{i}>\theta_{0}$ then the angle of incidence on the core-cladding interface will be less than the critical angle due to which part of incident light is transmitted into cladding as shown in the figure below Transmission of light when the incident angle is greater than the acceptance angle If the incident angle is less then the acceptance angle i.e. $\theta_{i}<\theta_{0}$ then the angle of incidence on the core-cladding interface will be greater than the critical angle for which total internal reflection takes place inside the core. As shown in the figure below Transmission of light w

### Fraunhofer diffraction due to a single slit

Let $S$ be a point monochromatic source of light of wavelength $\lambda$ placed at the focus of collimating lens $L_{1}$. The light beam is incident normally from $S$ on a narrow slit $AB$ of width $e$ and is diffracted from it. The diffracted beam is focused at the screen $XY$ by another converging lens $L_{2}$. The diffraction pattern having a central bright band followed by an alternative dark and bright band of decreasing intensity on both sides is obtained. Analytical Explanation: The light from the source $S$ is incident as a plane wavefront on the slit $AB$. According to Huygens's wave theory, every point in $AB$ sends out secondary waves in all directions. The undeviated ray from $AB$ is focused at $C$ on the screen by the lens $L_{2}$ while the rays diffracted through an angle $\theta$ are focussed at point $p$ on the screen. The rays from the ends $A$ and $B$ reach $C$ in the same phase and hence the intensity is maximum. Fraunhofer diffraction due to

### Particle in one dimensional box (Infinite Potential Well)

Let us consider a particle of mass $m$ that is confined to one-dimensional region $0 \leq x \leq L$ or the particle is restricted to move along the $x$-axis between $x=0$ and $x=L$. Let the particle can move freely in either direction, between $x=0$ and $x=L$. The endpoints of the region behave as ideally reflecting barriers so that the particle can not leave the region. A potential energy function $V(x)$ for this situation is shown in the figure below. Particle in One-Dimensional Box(Infinite Potential Well) The potential energy inside the one -dimensional box can be represented as $\begin{Bmatrix} V(x)=0 &for \: 0\leq x \leq L \\ V(x)=\infty & for \: 0> x > L \\ \end{Bmatrix}$ $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2m}{\hbar^{2}}(E-V)\psi(x)=0 \qquad(1)$ If the particle is free in a one-dimensional box, Schrodinger's wave equation can be written as: $\frac{d^{2} \psi(x)}{d x^{2}}+\frac{2mE}{\hbar^{2}}\psi(x)=0$ \$\frac{d^{2} \psi(x)}{d x