Galilean Transformation Equations and Failure of Galilean Relativity

What is Transformation Equation?
A point or a particle at any instant, in space has different cartesian coordinates in the different reference systems. The equation which provide the relationship between the cartesian coordinates of two reference system are called Transformation equations.

Galilean Transformation Equation:

Let us consider, two frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ relative to an inertial frame $S$. Let

  1. The origin of the two frames coincide at $t=0$

  2. The coordinate axes of frame $S'$ are parallel to that of the frame $S$ as shown in the figure below

  3. The velocity of the frame $S'$ relative to the frame $S$ is $v$ along x-axis;

Relative Motion of Frames
The position vector of a particle at any instant $t$ is related by the equation

$ \overrightarrow{r'}=\overrightarrow{r}-\overrightarrow{v}t\qquad (1)$

In component form, the coordinate are related by the equations

$\overrightarrow{x'}=\overrightarrow{x}-\overrightarrow{v}t;\quad y'=y; \quad z'=z\qquad (2)$

The equation $(1)$ and equation $(2)$ express the transformation of coordinates from one inertial frame to another. Hence they are referred to as Galilean transformation.

The equation $(1)$ and equation $(2)$ depending on the relative motion of two frames of reference, but it also depends upon certain assumptions regarding the nature of time and space. It is assumed that the time t is independent of any particular frame of reference. i.e. If $t$ and $t'$ be the times recorded by observers $O$ and $O'$ of an event occurring at $P$ then

$ t=t'\qquad (3)$

Now add the above assumption with transformation equation $(3)$ so the Galilean transformation equations are

$ \begin{Bmatrix} x'=x-vt\\ y'=y\\ z'=z\\ t'=t \end{Bmatrix}\qquad (4)$

The other assumption, regarding the nature of space, is that the distance between two points (or two particles) is independent of any particular frame of reference. For example if a rod has length $L$ in the frame $S$ with the end coordinates $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ then

Length or Distance is Invariant under the Galilean Transformation
$L=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\quad (5)$

At the same time, the end coordinate of the rod in frame $S'$ are $(x'_{1}, y'_{1}, z'_{1})$ and $(x'_{2}, y'_{2}, z'_{2})$ then

$ L'=\sqrt{(x'_{2}-x'_{1})^{2}+(y'_{2}-y'_{1})^{2}+(z'_{2}-z'_{1})^{2}}\qquad (6)$

But for any time $t$ from equation $(4)$

$ \begin{Bmatrix} x'_{2}-x'_{1}=x_{2}-x_{1}\\ y'_{2}-y'_{1}=y_{2}-y_{1}\\ z'_{2}-z'_{1}=z_{2}-z_{1} \end{Bmatrix}\quad\quad\quad (7)$

So from equation $(5)$, equation $(6)$ and equation $(7)$, we can write as:


$L=L'$

Thus, the length or distance between two points is invariant under Galilean Transformation.

The hypothesis of Galilean Invariance:(Principle of Relativity)

The hypothesis of Galilean invariance is based on experimental observation and is stated as follows:

The basic laws of physics are identical in all reference system which move with uniform velocity with respect to one another.

OR in other words
The basics laws of physics are invariant in inertial frame.

Modify the hypothesis of Galilean Invariance by giving the following statement-

The basic law of physics are invariant in form in two reference system which are connected by Galilean Transformation

Failure of Galilean Relativity OR Galilean Transformation:

There are the following points that could not explain by Galilean transformation:

  1. Galilean Transformation failed to explain the actual result of the Michelson-Morley experiment.

  2. It violates the postulates of the Special theory of relativity.
According to Maxwell's electromagnetic theory, the speed of light in a vacuum is $c$ $(3\times10^{8} m/sec)$ in all directions. Let us consider a frame of reference relative to which the speed of light is $c$ in all directions, According to Galilean transformation the speed of light in any other inertial system, which is in relative motion with respect to the former, will be different in a different direction. For example- If an observer is moving with speed $v$ opposite or along with the propagation of light, The speed of light $c_{0}$ in the frame of the observer is given by
$ c_{0}=c\:\pm \: v$

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 )$
Diagram for Length Contraction
$ 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
Diagram for Time Dilation
$ 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. μ-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 $c$. μ-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 μ-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 μ-mesons travel a distance of 10 km to reach the earth's surface. This is possible because of the time dilation effect. In fact, μ-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 μ-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, μ-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 μ-mesons on the earth's surface despite their brief lifetime.

Derivation of Lorentz Transformation's Equations

Derivation:

Let us consider two inertial frames $S$ and $S'$ in which frame $S'$ is moving with constant velocity $v$ along the positive x-axis direction relative to the frame $S$. Let $t$ and $t'$ be the time recorded in two frames. Let the origin $O$ and $O'$ of the two reference systems coincide at $t=t'=0$.

Now suppose, a source of light is situated at the origin $O$ in the frame $S$, from which a wavefront of light is emitted at time $t=0$. When the light reaches point $P$, the time required by a light signal in travelling the distance OP in the Frame $S$ is
Diagram of Inertial Motion with Respect to Speed of Light
$ t=\frac{OP}{c}$

$ t=\frac{\left (x^{2}+y^{2}+z^{2} \right )}{c}$

$ x^{2}+y^{2}+z^{2}=c^{2}t^{2}\qquad (1)$

The equation $(1)$ represents the equation of wavefront in frame $S$. According to the special theory of relativity, the velocity of light will be $c$ in the second frame $S'$. Hence in frame $S'$ the time required by the light signal in travelling the distance $O'P$ is given by

$ t'=\frac{O'P}{c}$

$ x'^{2}+y'^{2}+z'^{2}=c^{2}t^{2}\qquad (2)$

According to the Galilean transformation equation:

$ \begin{Bmatrix} x'=x-vt\\ y'=y\\ z'=z\\ t'=t \end{Bmatrix}$

Now substitute these values in equation $(2)$ then we get

$ (x-vt)^{2}+y^{2}+z^{2}=c^{2}t^{2}$

$ x^{2}+v^{2}t^{2}-2xvt+y^{2}+z^{2}-c^{2}t^{2}=0$

The above equation is certainly not same as the equation $(1)$ because it contains an extra term $(-2xvt+v^{2}t^{2})$. Thus the Galilean transformation fails.

Further $t=t'$ because $\left( t=\frac{OP}{c} \: and \: t'=\frac{O'P}{c} \right)$ which does not agree with Galilean transformation equations.

The extra term $(-2xvt+v^{2}t^{2})$ indicates that transformations in $x$ and $t$ should be modified so that this extra term is cancelled. So modification in transformation

$ x'=\alpha (x-vt) \quad for \: x'=0,\: x=vt$

$ t'=\alpha (t+fx)$

Where $α$, $α'$ and $f$ are constant to be determined for Galilean Transformations $α= α'=1$ and $f=0$. Now substituting these modified values in equation $(2)$ so

$ \alpha ^{2}(x-vt)^{2}+y^{2}+z^{2}=c^{2}\alpha'^{2}(t+fx)^{2}$

$ \alpha ^{2}(x^{2}+v^{2}t^{2}-2xvt)+y^{2}+ z^{2} =c^{2}\alpha{'}^{2}(t^{2}+f^{2}x^{2}+2fxt)$

$\alpha^{2}x^{2}+\alpha^{2}v^{2}t^{2}-2xvt\alpha^{2}+y^{2}+ z^{2} =c^{2} \alpha'^{2}t^{2}+c^{2} \alpha'^{2}f^{2}x^{2}+c^{2} \alpha'^{2}(2fxt)$

$ x^{2}(\alpha^{2}-c^{2} \alpha'^{2}f^{2})- 2xt(v\alpha^{2}+c^{2}\alpha '^{2}f)+y^{2}+ z^{2} =c^{2}t^{2}\left ( \alpha'^{2}-\frac{\alpha ^{2}v^{2}}{c^{2}} \right )\qquad(3)$

This result obtained from applying transformations from frame $S'$ to $S$ must be identical to equation $(1)$. Therefore

$ \alpha ^{2}-c^{2}\alpha '^{2}f^{2}=1\qquad (i)$

$ v\alpha ^{2}+c^{2}\alpha '^{2}f=0\qquad (ii)$

$ \alpha '^{2}-\frac{\alpha ^{2}v^{2}}{c^{2}}=1\qquad (iii)$

After solving equation $(i)$ equation $(ii)$ and equation $(iii)$, we get

$ \alpha =\alpha '=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ $ f=-\frac{v}{c^{2}}$

Thus, The new transformation equation is invariant at the velocity of light $c$. These are:

$ \begin{Bmatrix} x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\ y'=y\\ z'=z\\ t'=\frac{(t-\frac{vx}{c^{2}})}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{Bmatrix}$

These equations are called Lorentz Transformations because they were first obtained by Dutch Physicist H. Lorentz.

The above transformation equation shows that frame $S'$ is moving in positive x-direction with velocity $v$ relative to the frame $S$. But if we say that frame $S$ is moving with $v$ velocity relative to frame $S'$ along negative x-direction then the transformation is:

$ \begin{Bmatrix} x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\ y=y'\\ z=z'\\ t=\frac{(t'+\frac{vx'}{c^{2}})}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \end{Bmatrix}$

These are known as inverse Lorentz Transformations equations.


Force between multiple charges (Superposition principle of electrostatic forces)

Principle of Superposition for Electric force:

If a system contains a number of interacting charges, then the net force on anyone charge equals the vector sum of all the forces exerted on it by all the other charges. This is the principle of Superposition for electric force.

If a system contains n point charges $ q_{1},q_{2},q_{3}........q_{n}$. Then according to the principle of superposition, the force acting on the charge $q_{1}$ due to all the other charges

$\overrightarrow{F_{1}}=\overrightarrow{F_{12}}+\overrightarrow{F_{13}}+\overrightarrow{F_{14}}+...+\overrightarrow{F_{1n}} \qquad (1)$

Where $\overrightarrow{F_{12}}$ is the force on charge $q_{1}$ due to charge $q_{2}$, $\overrightarrow{F_{13}}$ that is due to $q_{3}$ and $\overrightarrow{F_{1n}}$ that due to $q_{n}$.

If the distance between the charges $q_{1}$ and $q_{2}$ is $\widehat{r}_{12}$ (magnitude only) and $\widehat{r}_{21}$ is unit vector from charge $q_{2}$ to $q_{1}$, then

$\overrightarrow{F_{12}}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r_{12}^{2}}\:\hat{r_{21}} \qquad (2)$

Similarly, the forces on charge $q_{1}$ due to other charges are given by

$ \overrightarrow{F_{13}}=\frac{1}{4\pi \epsilon _{0}}\:\frac{q_{1}q_{3}}{r_{13}^{2}}\:\hat{r_{31}}\qquad (3)$

$.............................$

$.............................$

$ \overrightarrow{F_{1n}}=\frac{1}{4\pi \epsilon _{0}}\:\frac{q_{1}q_{n}}{r_{1n}^{2}}\:\hat{r_{n1}}\qquad (n)$

Hence, putting the value of $\overrightarrow{F_{12}},\overrightarrow{F_{13}},\overrightarrow{F_{14}}......\overrightarrow{F_{1n}}$, in equation $(1)$, the total force on charge $q_{1}$ due to all other charges is given by

$ \overrightarrow{F_{1}}=\frac{1}{4\pi \epsilon _{0}}[\:\frac{q_{1}q_{2}}{r_{12}^{2}}\:\hat{r_{21}}+\frac{q_{1}q_{3}}{r_{13}^{2}}\:\hat{r_{31}}+....\\ \quad\quad\quad ..+\frac{q_{1}q_{n}}{r_{1n}^{2}}\:\hat{r_{n1}}]$

The same procedure can be applied to finding the force on any other charge due to all the remaining charges. For example, the force on $q_{2}$ due to all the other charges is given by

$ \overrightarrow{F_{2}}=\frac{1}{4\pi \epsilon _{0}}[\:\frac{q_{2}q_{1}}{r_{21}^{2}}\:\hat{r_{12}}+\frac{q_{2}q_{3}}{r_{23}^{2}}\:\hat{r_{32}}+... \\ \quad\quad\quad ..+\frac{q_{2}q_{n}}{r_{2n}^{2}}\:\hat{r_{n2}}\:]$

Some Observations Points of Coulomb's Law:

There are the following point has been observed in Coulomb's law, that are
  1. Coulomb's force between the two charges is directly proportional to the product of the magnitude of the charge.

    $ F\propto q_{1}q_{2}$

  2. Coulomb's force between the two charges is inversely proportional to the square of the distance between the two charges.

    $F\propto \frac{1}{r^{2}}$

  3. The electrostatic force acts between the line joining the charges. In two charges, one charge is assumed to be at rest for the calculation of the force on the second charge. So It is also known as a central force.

  4. The magnitude of the electrostatic force is equal and the direction of force is opposite. So the electrostatic force is also known as the action and reaction pair.

  5. The electrostatic force between two charge does not affect by the presence and absence of any other charges but the net force increase on the source charge.

Definition and derivation of the phase velocity and group velocity of wave

Wave:
A wave is defined as a disturbance in a medium from an equilibrium condition that propagates from one region of the medium to other regions.

When such type of wave propagates in the medium a progressive change in phase takes place from one particle to the next particle.

Propagation of Wave: Wave propagation in the medium occurs with two different kinds of velocity. i.e. phase velocity and group velocity.

1. Phase Velocity:
The velocity with which plane of constant the phase of a wave propagates through the medium at a certain frequency is called the phase velocity or wave velocity.

A plane wave traveling in the positive x-direction is represented by

$y=A \: sin \omega (t-\frac{x}{v})$

Where $ \omega $ – angular frequency

$y=A \: sin(\omega t-\frac{\omega x}{v})$

$y=A \: sin(\omega t-kt) \qquad(1)$

For plane-wave $(\omega t-kx)$ is the phase of wave motion. For the plane of constant phase (wavefront). We have

$( \omega t-kx) = constant (\phi) \qquad(2)$

Differentiate with respect to time $(t)$ of the above equation

$ \omega -k.\frac{dx}{dt}=0$

$\frac{dx}{dt}=\frac{ \omega}{k}\qquad(3)$

$V_{p}=\frac{ \omega}{k} \qquad \left [ \because V_{p}= \frac{dx}{dt} \right ]$

Where $V_{p}$ - Phase Velocity of a wave

Question- Show that the phase velocity of matter-wave always exceeds the velocity of light.

Answer- Method-I

$V_{p}=\frac{ \omega }{k}$

$V_{p}=\frac{2\pi \nu }{\frac{2\pi }{\lambda }}$

$V_{p} = \frac{2 \pi h \nu}{\frac{2 \pi h}{\lambda }}$

$V_{P} = \frac{E}{P}$

$V_{P} =\frac{mc^{2}}{mv}$

$V_{P} =\frac{c^{2}}{v}$

Method-II:

$V_{P}=\nu \lambda$

$V_{P}=\frac{mc^{2}}{h} \frac{h}{mv}$

$V_{P}=\frac{c^{2}}{v}$

Where $v$ is the velocity of matter particles.

Wave packet→
A wave packet is an envelope or packet which contains the number of plane waves having different wavelengths or wavenumbers. These numbers of waves superimpose on each other and form constructive and destructive interference over a small region of space and a resultant wave obtain. The spread of amplitude of the resultant wave with distance determines the size of the wave packet. A wave packet is also called a wave group.
Diagram of a wave packet

Group Velocity→
The velocity of propagation of wave packet through space is known as group velocity.

Group Velocity also represents the velocity of energy flow or transmission of information in a traveling wave or wave packet. It is represented by $V_{g}$. So

$V_{g} = \frac{d\omega }{dk}$


Expression for group velocity $(V_{g})$→

Let two plane simple harmonic waves of the same amplitude and slightly different wavelength traveling simultaneously in the positive x-direction in a dispersive medium be represented by –

$y_{1} = A\: sin{(\omega t-kx)} \qquad (1)$

$y_{2}= A\:sin[(\omega +\delta \omega )t-(k+\delta k)x] \qquad (2)$

From the superposition principle the resultant displacement of waves

$y=y_{1}+y_{2}\qquad (3)$

$y=A \: sin(\omega t-kx)+ A \: sin[(\omega +\delta \omega)t-(k+\delta k)x]$

$y=A[sin(\omega t-kx)+ sin{(\omega +\delta \omega )t-(k+\delta k )x}]$

$y=2Asin(\omega t-kx)cos\frac{1}{2}(t\delta \omega –x\delta k) \qquad (4)$

This is the analytical equation for the group of waves. This equation represents the following points –
  1. The sine factors represent a carrier wave that travels with the phase velocity.

  2. The amplitude of the resultant wave is


$R=2A cos \frac{1}{2}(t \delta\omega-x \delta k) \qquad (5)$

(A) For maximum amplitude

$ cos\frac{1}{2}(\delta\omega t-x\delta k)=1 \qquad (6) $

Then the resultant amplitude of the wave packet will be

$R_{m}=2A \qquad $ {from equation $(5)$ }

Where $R_{m}$ - Resultant maximum amplitude

From equation$(6)$-

$cos\frac{1}{2}(t\delta\omega –x\delta k)=1$

$cos\frac{1}{2}(t\delta\omega –x\delta k)=cos0$

$\frac{1}{2}(t\delta\omega –x\delta k)=0$

$\frac{x}{t}=\frac{\delta\omega }{\delta k}$

$V_{g}=\frac{\delta\omega }{\delta k}$

OR

$V_{g}=\frac{d\omega}{dk}$

Thus maximum amplitude i.e. the center of the wave packet moves with velocity $\frac {d\omega}{dk}$ or group velocity.

(B) For minimum amplitude

Let a wave packet that has minimum (zero) amplitude on two successive points $x_{1}and x_{2}$. So for minimum amplitude at point $x_{1}$ –

$cos\frac{1}{2}(t.\delta \omega –x_{1}.\delta k)=0$

$cos\frac{1}{2}(t.\delta \omega –x_{1}.\delta k)=cos(2n+1)\frac{\pi }{2}$

$\frac{1}{2}(t.\delta \omega –x_{1}.\delta k)=(2n+1)\frac{\pi }{2}$

$t.\delta \omega – x_{1}.\delta k=(2n+1)\pi \qquad(1)$

Similarly minimum amplitude at the second successive point $x_{2}$-

$t\delta \omega –x_{2}.\delta k =(2n-1)\pi \qquad(2)$

Now subtract the equation $(2)$ in equation $(1)$

$(t.\delta \omega – x_{1}.\delta k) – (t.\delta \omega - x_{2}.\delta k)= (2n+1)\pi–(2n-1)\pi $

$(x_{2}-x_{1}).\delta k=2\pi$

$x_{2}-x_{1}=\frac{2\pi}{\delta k}$

$x_{2}-x_{1}=\frac{2\pi}{d(\frac{2\pi}{\lambda})} $

$\delta x=\frac{-\lambda ^{2}}{\delta  \lambda }$

Here $\delta x$ -Length of the wave packet

The above equation also can be written as-

$ dx=\frac{-\lambda^{2}}{d \lambda }$

Prove that

$V_{g}= -\lambda^{2} \frac{d\nu }{d\lambda}$

Where $\nu $ is the frequency and $\lambda $ is the wavelength.

Proof:

We know that

$V_{g}=\frac{d\omega }{dk}$

$V_{g}=\frac{d(2\pi\nu )}{d(\frac{2\pi }{\lambda })} \qquad \left( \because \omega= 2\pi \nu \: and \: k=\frac{2\pi}{\lambda } \right)$

$V_{g}=\frac{2\pi d\nu }{2\pi d( \frac{1}{ \lambda })}$

$V_{g}=-\lambda ^{2}\frac{d\nu }{d\lambda }$

Note: A moving particle cannot be equal to a single wave train. The speed of a single wave train is called the phase velocity so moving particles are equivalent to a group of waves or wave packets.

Product of phase velocity and group velocity is equal to square of speed of light

Prove that $\rightarrow$
The Product of phase velocity and group velocity is equal to the square of the speed of light i.e. $\left( V_{p}.V_{g}=c^{2} \right)$

Proof → We know that

$V_{p}=\nu \lambda \qquad(1)$

And de Broglie wavelength-

$\lambda =\frac{h }{mv}\qquad(2)$

According to Einstein's mass-energy relation-

$E=mc^{2}$

$h\nu=mc^{2}$

$\nu=\frac{mc^{2}}{h}\qquad(3)$

Now put the value of $\lambda $ and $\nu $ in equation$(1)$

$V_{p}= [\frac{mc^{2}}{h}] [\frac{h}{mv}]$

$V_{p}=\frac{C^{2}}{v}$

Since group velocity is equal to particle velocity i.e. $V_{g}=v$. So above equation can be written as

$V_{p}=\frac{C^{2}}{V_{g}}$

$V_{p}.V_{g}=C^{2}$

Note →

➢ $V_{g}=V_{p}$ for a non-dispersive medium ( in a non-dispersive medium all the waves travel with phase velocity).

➢ $V_{g}< V_{p}$ for normal dispersive medium

➢ $V_{g}> V_{p}$ for anomalous dispersive media.

Dispersive medium → The medium in which the phase velocity varies with wavelength or frequency is called a dispersive medium. In such a medium, waves of different wavelengths travel with different phase velocities.

Non-dispersive medium → The medium in which the phase velocity does not vary with wavelength or frequency is called a Non-dispersive medium.

Dispersive waves → Those waves in the medium for which phase velocity varies with wavelength or frequency are called dispersive waves.

Non-dispersive waves → Those waves in which phase velocity does not vary with wavelength are called non-dispersive waves. So phase velocity independent of wavelength.

Energy distribution spectrum of black body radiation

Description→ The energy distribution among the different wavelengths in the spectrum of black body radiation was studied by Lummer and Pringsheim in 1899. There are the following important observations of the study.
  1. The energy distribution in the radiation spectrum of the black body is not uniform. As the temperature of the body rises the intensity of radiation for each wavelength increases.

  2. At a given temperature, the intensity of radiation increases with increases in wavelength and becomes maximum at a particular wavelength with further in increases wavelength the intensity of radiation decreases.

  3. Energy distribution in the spectrum of black body
    Energy distribution in the spectrum of black body radiation
  4. The points of maximum energy shift towards the shorter wavelength as the temperature increases i.e. $\lambda _{m} \times T=constant$. It is also known as Wein’s displacement law of energy distribution.

  5. For a given temperature the total energy of radiation is represented by the area between the curve and the horizontal axis and the area increases with increases of temperature, being directly proportional to the fourth power of absolute temperatures.
    The total amount of heat radiated by a perfectly black body per unit area per unit time is directly proportional to the fourth power of its absolute temperature $(T)$.

    $E = \sigma T^{4}$

    Where $\sigma$ = Stefan constant having value $\left ({5}\cdot{67}\times{10}^{-8}Wm^{-2}K^{4}\right )$

    This is called Stefan-Boltzmann's law of energy distribution.

Derivation of torque on an electric dipole in an uniform and a non-uniform electric field

Torque on an electric dipole in a uniform electric field: Let us consider, An electric dipole AB, made up of two charges $+q$ and $-q$, is placed at a very small distance $2l$ in a uniform electric field $\overrightarrow{E}$. If $\theta$ is the angle between electric field intensity $\overrightarrow{E}$ and electric dipole moment $\overrightarrow{p}$ then the magnitude of electric dipole moment →

$\overrightarrow{p}=q\times\overrightarrow{2l}\qquad(1)$

Force exerted on charge $+q$ by electric field $\overrightarrow{E}$ →

$\overrightarrow{F_{+q}}=q\overrightarrow{E}\qquad(2)$

Here in the above equation(2), the direction of $\overrightarrow{F_{+q}}$ is along the direction of $\overrightarrow{E}$

Force exerted on charge $-q$ by electric field $\overrightarrow{E}$ →

$\overrightarrow{F_{-q}}= q\overrightarrow{E}\qquad(3)$

Here in the above equation(3) the direction of $\overrightarrow{F_{-q}}$ is in the opposite direction of $\overrightarrow{E}$

So, the net force of on an electric dipole→

$\overrightarrow{F}=\overrightarrow{F_{+q}} - \overrightarrow{F_{-q}}\qquad (4)$

Now substitute the value of equation $(2)$ and equation $(3)$ in above equation $(4)$. So net force→

$\overrightarrow{F}=0\qquad (5)$
Torque on an electric Dipole
Torque on electric dipole
Hence, the net translating force on an electric dipole in a uniform electric field is zero. But these two force is equal in magnitude, opposite in direction, and act at different point of the dipole so these force form a coupling force that exerts a torque on an electric dipole→

Torque = force x Perpendicular distance between the two forces

$\overrightarrow{\tau}=(qE).2l\:sin\theta\quad\quad\quad\quad(6)$

$ \overrightarrow{\tau}=pE\:sin\theta$

$\overrightarrow{\tau}=\overrightarrow{p}\times\overrightarrow{E}$

Case(I)→ If the dipole is placed perpendicular to the electric field i.e. $\theta=90^{\circ}$, the torque acting on it will be maximum. i.e.

$ \tau_{max}=pE $

Case(II)→ If the dipole is placed parallel to the electric field i.e. $\theta=0^{\circ}$ or $\theta=180^{\circ}$, the torque acting on it will be minimum. i.e.

$ \tau_{min}=0$
Direction of torque
Direction of torque
Electric Dipole Moment:

We know that the torque

$ \overrightarrow{\tau}=pE\:sin\theta$

If $E=1$ and $\theta=90^{\circ}$ Then

$ \tau_{max} =p$

Hence Electric dipole moment is the torque acting on the dipole placed perpendicular to the direction of uniform electric field intensity.

Torque on an electric dipole in a non-uniform electric field:

In a non-uniform electric field, the $+q$ and $-q$ charges of a dipole experience different forces (not equal in magnitude and opposite in direction) at a slightly different position in the electric field, and hence a net force $\overrightarrow{F}$ act on the dipole in a non-uniform field. A net torque acts on the dipole which depends on the location of the dipole in the non-uniform field.

$\overrightarrow{\tau}=\overrightarrow{p}\times\overrightarrow{E}(\overrightarrow{r})$

Where $\overrightarrow{r}$ is the position vector of the center of the dipole.
When p  and E  is Parallel
When p and E are Parallel
In the non-uniform field, If the direction of the dipole moment $\overrightarrow{p}$ is parallel to electric field intensity $\overrightarrow{E}$ or antiparallel to electric field intensity $\overrightarrow{E}$ the net torque on the dipole is zero because the force on charges becomes linear.
When p  and E  is antiparallel
When p and E are antiparallel
However, If $\overrightarrow{p}$ is parallel to $\overrightarrow{E}$, a net force on the dipole in the direction of increasing $\overrightarrow{E}$. When $\overrightarrow{p}$ is antiparallel to $\overrightarrow{E}$, a net force on the dipole in the direction of decreasing $\overrightarrow{E}$. As shown in the figure above.

Electric Dipole and Derivation of Electric field intensity at different points of an electric dipole

Electric Dipole:

An electric dipole is a system in which two equal magnitude and opposite point charged particles are placed at a very short distance apart.

Electric Dipole Moment:

The product of magnitude of one point charged particle and the distance between the charges is called the 'electric dipole moment'. It is vector quantity and the direction of electric dipole moment is along the axis of the dipole pointing from negative charge to positive charge.
Electric Dipole
Electric Dipole
Let us consider, the two charged particle, which has equal magnitude $+q$ coulomb and $-q$ coulomb is placed at a distance of $2l$ in a dipole so the electric dipole moment is →

$\overrightarrow{p}=q\times \overrightarrow{2l}$

Unit: $C-m$ Or $Ampere-metre-sec$

Dimension: $[ALT]$

Electric field intensity due to an Electric Dipole:

The electric field intensity due to an electric dipole can be measured at three different points:

  1. Electric field intensity at any point on the axis of an electric dipole

  2. Electric field intensity at any point on the equatorial line of an electric dipole

  3. Electric field intensity at any point on an electric dipole


1. Electric field intensity at any point on the axis of an electric dipole:

Let us consider, An electric dipole $AB$ made up of two charges of $-q$ and $+q$ coulomb are placed in a vacuum or air at a very small distance of $2l$. Let a point $P$ be on the axis of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of electric field intensity due to dipole's charge.
Electric field intensity at any point of the axis of an electric dipole
Electric field intensity at any point on the axis of an electric dipole
So, Electric field intensity (magnitude only) at point $P$ due $+q$ charge of electric dipole

$ E_{+q}=\frac{1}{4\pi\epsilon}\frac{q}{(r-l)^{2}} \qquad(1)$

Electric field intensity (magnitude only) at point $P$ due $-q$ charge of electric dipole

$ E_{-q}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{(r+l)^{2}} \qquad(2)$

The net electric field at point $P$ due to an electric dipole

$ E=E_{+q}-E_{-q}\qquad(3)$

Subtitute the value of $E_{+q}$ and $E_{-q}$ in equation $(3)$, Then the above equation $(3)$ can also be written as follows

$E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{1}{(r-l)^{2}}-\frac{1}{(r+l)^{2}} \right ]$

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r+l)^{2}-(r-l)^{2}}{(r+l)^{2}(r-l)^{2}} \right ]$

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r^{2}+l^{2}+2lr-r^{2}-l^{2}+2lr)}{(r+l)^{2}(r-l)^{2}} \right ]$

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{4rl}{(r^{2}-l^{2})^{2}} \right ]\qquad(4)$

Here $l \lt r $ so $l^{2} \lt \lt r^{2}$ therefore neglect the term $l^{2}$ in above equation $(4)$ so we can write above equation

$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{4rl}{(r^{2})^{2}} \right ]$

$E=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2p}{r^{3}} \right ]\qquad (5)$

This is the equation of electric field intensity at a point on the axis of an electric dipole.

The vector form of the above equation $(5)$ is

$\overrightarrow{E}=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2\overrightarrow{p}}{r^{3}} \right ]$

2. Electric field intensity at any point on the equatorial line of an electric dipole:

Let us consider, An electric dipole $AB$ made up of two charges of $+q$ and $-q$ coulomb are placed in vacuum or air at a very small distance $2l$. Let a point $P$ be on the equatorial line of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of electric field intensity due to dipole's charge.

So, Electric field intensity (magnitude only) at point $P$ due $+q$ charge of electric dipole
Electric field intensity at any point of the equatorial line of an electric dipole
Electric field intensity at a point of the equatorial line of an electric dipole
$E_{+q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{(r^{2}+l^{2})} \right ] \qquad(1)$

Electric field intensity (magnitude only) at point $P$ due to $-q$ charge of electric dipole

$E_{-q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{(r^{2}+l^{2})} \right ] \qquad(2)$

The net electric field at point $P$ due to an electric dipole

$E=E_{+q}\:cos\theta + E_{-q}\: cos\theta \qquad(3)$

Put the value of $E_{+q}$ and $E_{-q}$ in the above equation $(3)$, So equation $(3)$ can also be written as follows

$E=2\left [ \frac{q}{4\pi\epsilon_{0} }\frac{1}{(r^{2}+l^{2})} \right ]cos\theta \qquad (4)$

From figure, In $\Delta \: POB$,

$cos\:\theta=\frac{l}{\sqrt{(r^{2}+l^{2})}}$

Put the value of $cos\theta$ in equation $(4)$, so equation $(4)$ can also be written as follows

$E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times2l}{(r^{2}+l^{2})^{3/2}} \right ] \qquad (5)$

Here $l \lt r$ so $l^{2} \lt \lt r^{2}$ so neglect the term $l^{2}$ in above equation $(5)$ so we can write above equation

$ E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times2l}{r^{3}} \right ] \qquad (6)$

$E=\frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{3}} \qquad (7)$

This is the equation of electric field intensity at a point on the equatorial line of an electric dipole.

The vector form of the above equation $(7)$ is

$\overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}} \frac{\overrightarrow{p}}{r^{3}}$

3. Electric field intensity at any point of an electric dipole:

Let us consider, An electric dipole $AB$ of length $2l$ consisting of the charge $+q$ and $-q$. Let's take a point $P$ in general and its position vector is $\overrightarrow{r}$ from the center point $O$ of the electric dipole AB.

Electric field intensity at any point of an electric dipole
Electric field intensity at any point of an electric dipole

The electric dipole moment is a vector quantity that has a direction from $-q$ charge to $+q$ charge. So the electric dipole moment's direction is resolved in two-component one is along the vector position $\overrightarrow{r}$ i.e pcosθ and the other is normal to vector position $\overrightarrow{r}$ i.e $psinθ$. So

Electric field intensity due to dipole moment of component $p\:cos\theta$ {Electric field along the Axial Line }

$ \overrightarrow{E_{\parallel }}=\frac{1}{4\pi\epsilon_{0}} \frac{2pcos\theta}{r^{3}} \qquad(1)$

Electric field intensity due to dipole moment of component $p\:sin\theta$ {Electric field along the Equatorial Line}

$ \overrightarrow{E_{\perp}}=\frac{1}{4\pi\epsilon_{0}} \frac{psin\theta}{r^{3}} \qquad(2)$

The resultant electric field vector $E$ at point $P$

$ \overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2}+2 E_{\perp}E_{\parallel}cos90^{\circ}}$

From figure, The angle between $E_{\perp}$ and $E_{\parallel}$ is $90^{\circ}$. So

$\overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2} }\qquad (3)$

Now substitute the value of equation $(1)$ and equation $(2)$ in equation $(3)$. Then

$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(sin^{2}\theta+4cos^{2}\theta)}$

$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(1+3cos^{2}\theta)}$

This is the equation of electric field intensity at any point due to an electric dipole.

The direction of the resultant electric field intensity vector $\overrightarrow{E}$ from the axial line is

$tan\alpha =\frac{\overrightarrow{E}_{\perp }}{\overrightarrow{E_{\parallel }}} \qquad(4)$

Put the value of $\overrightarrow{E_{\perp}}$ and $\overrightarrow{E_{\parallel}}$ in equation (4). we get

$tan\alpha =\frac{sin\theta}{2cos\theta}$

$tan\alpha =\frac{1}{2}tan\theta$

Here $\alpha$ is the angle between the resultant electric field intensity $\overrightarrow{E}$ and the axial line.

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