Bernoulli's Theorem and Derivation of Bernoulli's Equation

Statement of Bernoulli's Theorem:

When an ideal fluid (i.e incompressible and non-viscous Liquid or Gas) flows in streamlined motion from one place to another, then the total energy per unit volume (i.e Pressure energy + Kinetic Energy + Potential Energy) at each and every of its path is constant.

$P+\frac{1}{2}\rho v^{2} + \rho gh= constant$

Derivation of Bernoulli's Theorem Equation:

Let us consider that an incompressible and non-viscous liquid is flowing in streamlined motion through a tube $XY$ of the non-uniform cross-section.

Now Consider:

The Area of cross-section $X$ = $A_{1}$
The Area of cross-section $Y$ = $A_{2}$

The velocity per second (i.e. equal to distance) of fluid at cross-section $X$ = $v_{1}$
The velocity per second (i.e. equal to distance) of fluid at cross-section $Y$ = $v_{2}$

The Pressure of fluid at cross-section $X$ = $P_{1}$
The Pressure of fluid at cross-section $Y$ = $P_{2}$

The height of cross-section $X$ from surface = $h_{1}$
The height of cross-section $Y$ from surface = $h_{2}$

The work done per second by force on the liquid Entering the tube at $X$:

$W_{1}$ = Force $ \times $ Distance covered in one second

$W_{1}= P_{1} \times A_{1} \times v_{1} \quad \left( Force =Pressure \times Area \right)$

Similarly
The work done per second by force on the liquid leaving the tube at $Y$:

$W_{2}= P_{2} \times A_{2} \times v_{2}$

The net work done on the liquid:

$\Delta W=W_{1}-W_{2}$
$\Delta W= P_{1} \times A_{1} \times v_{1} - P_{2} \times A_{2} \times v_{2} \qquad(1)$

Now according to the principle of continuity:

$A_{1} v_{1} = A_{2} v_{2} = \frac{m}{\rho} \qquad(2)$

Now from equation $(1)$ and equation $(2)$

$\Delta W=\left( P_{1} -P_{2} \right) \frac {m}{\rho} \qquad(3)$

The kinetic energy of the fluid entering at $X$ in 1 second

$K_{1}=\frac{1}{2}mv_{1}^{2}$

The kinetic energy of the fluid leaving at $Y$ in 1 second

$K_{2}=\frac{1}{2}mv_{2}^{2}$

Therefore, The increase in kinetic energy

$\Delta K = K_{2}-K_{1}$

Now substitute the value of $K_{1}$ and $K_{2}$ in above equation then

$\Delta K = \frac{1}{2}mv_{2}^{2} - \frac{1}{2}mv_{1}^{2}$
$\Delta K = \frac{1}{2}m \left( v_{2}^{2} - v_{1}^{2} \right) \qquad(4)$

The potential energy of fluid at $X$

$U_{1}= mgh_{1}$

The potential energy of fluid at $Y$

$U_{2}= mgh_{2}$

Therefore, The decrease in potential energy

$\Delta U = U_{1}-U_{2}$

Now substitute the value of $U_{1}$ and $U_{2}$ in above equation then

$\Delta U = mgh_{1} - mgh_{2}$
$\Delta U = mg \left( h_{1} - h_{2} \right) \qquad(5)$

This increase in energy is due to the net work done on the fluid, i.e.

Net Work done = Net increase in energy

Net Work done $(\Delta W)$ = Net increase in Kinetic Energy $(\Delta K)$ - Net decrease in Potential Energy $(\Delta U)$

$\left( P_{1} -P_{2} \right) \frac {m}{\rho} = \frac{1}{2}m \left( v_{2}^{2} - v_{1}^{2} \right) - mg \left( h_{1} - h_{2} \right) $

$\left( P_{1} -P_{2} \right) = \frac{1}{2}\rho \left( v_{2}^{2} - v_{1}^{2} \right) -\rho g \left( h_{1} - h_{2} \right) $

$P_{1} + \frac{1}{2} \rho v_{1}^{2} + \rho g h_{1} = P_{2} + \frac{1}{2}\rho v_{2}^{2} + \rho g h_{2} $

$P + \frac{1}{2} \rho v^{2} + \rho g h = Constant $

This is Bernoulli's theorem equation.

Pressure Head, Velocity Head, and Gravitational Head of a Flowing Fluid:

According to Bernoulli's Theorem equation

$P + \frac{1}{2} \rho v^{2} + \rho g h = Constant $

Now dividing the above equation by $\rho g$, then we get

$\frac{P}{\rho g}+\frac{v^{2}}{2g}+ h = Constant$

Where

$\frac{P}{\rho g}$ = Pressure Head

$\frac{v^{2}}{2g}$ = Velocity Head

$h$ = Gravitational Head

The dimension of each of these three is the dimension of height. The sum of these heads is called the 'Total Head'

Therefore, Bernoulli's theorem may also state as follows:

In streamlined motion of an ideal fluid, the sum of pressure head, velocity head and gravitational head at any point is always constant.

When the fluid flows in a horizontal plane $(h_{1}=h_{2})$, then Bernoulli's equation

$P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} + \frac{1}{2}\rho v_{2}^{2} $

$\frac{P}{\rho g}+\frac{v^{2}}{2g} = Constant$

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Numerical Problems and Solutions of Power

Formulas and Solutions to Numerical Problems of Power

Important Formula of Power:

1.) $P=\frac{Work(W)}{Time(t)}$

2.) $P=\frac{F.s}{t}$

3.) $P=\frac{F \: s \: cos \theta}{t}$

4.) $P=\frac{F \: s }{t} \qquad \left( \because \theta =0^{\circ} \right)$

5.) $P=F \: v \qquad \left( \because v = \frac{s}{t} \right)$

Numerical Problems and Solutions

Q.1 A woman pulls a bucket of water of mass $5 Kg$ from a well which is $10 m$ deep in 10 sec. Calculate the power used by her $(g=10 \: m/sec^{2})$.

Solution:
Given that:
The mass of bucket of water $(m)=5 \: Kg$ Depth of well $(h)=10 \: m$

The time taken to pull a bucket from the well $t= 10 \: sec $

The value of gravitational acceleration $g=10 \: m/sec^{2}$

The power used $(P)=?$


Now the power used by her:


$P=\frac{W}{T}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now substitute the given values in the equation of power:

$P=\frac{5 \times 10 \times 10}{10}$

$P=50 \: Joule/sec=50 W$

Q.2 A man whose mass is $50 \: Kg$ climbs up $30$ steps of the stairs in $30 \: Sec$. If each step is $20 \: cm$ high, Calculate the power used in climbing the stairs $(g=10 \: m/sec^{2})$

Solution:
Given that:
Mass of a man $(m) = 50 \: Kg$

Climbs up the number of Steps $(N) = 30$

The time is taken to climb up the 30 steps $(t)= 30 \: sec$

The length or height of one step $(l)=20 cm \: = \: 0.20 \: m$

The total length or height of 30 steps $h=30 \times 0.20 \: m$

The power used in climbing the stairs=?

Now, from the equation of power:

$P=\frac{W}{t}$

$P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$

Now substitute the given values in the above equation:

$P=\frac{50 \times 10 \times 30 \times 0.20 }{30}$

$P=100 \: Joule/sec = 100 \: W$

Q.3 A horse exerts a pull of $300 N$ on a cart so that the horse-cart system moves with a uniform speed of $18 Km/h$ on a level road. Calculate the power in watts developed by the horse and also find its equivalent in horsepower.

Solution:
Given that:
The horse exerts the pull, i.e, force $(F)=300 N$

The unifrm speed of the horse-cart $(v)=18 Km/h$

So the distance moved in $(s)=18 Km=18000m$

The time $(t)=1h= 60 \times 60 sec$

The power developed by the horse$(P)=?$

From the equation of the power, the power developed by a horse in one hour:

$P=\frac{W}{t}$ $P=\frac{F.s}{t}$

Now substitute the given values in the above equation:

$P=\frac{3000 \times 18000}{60 \times 60}$

$P=1500 W$

$P=\frac{1500}{746}$

$P=2 \: hp$

Q.4 A man weighing $60Kg$ climbs up a staircase carrying a load of $20 Kg$ on his head. The staircase has 20 steps, each of height $0.2m$. If he takes $10 sec$ to climb, find his power.

Solution:
Given that:
The weight of man $(m_{1})=60Kg$

The weight of load $(m_{2})=20Kg$

The number of steps in the staircase $(N)=20$

The height of each step $(H)=10s$

The total mass of the man and load $(m)=m_{1}+m_{2}= 60+20=80 Kg$

The total heght of stair case $(h)=N \times H = 20 \times .2= 4 m$



The power of man: $P=\frac{W}{t}$

$P=\frac{mgh}{t}$

$P=\frac{80 \times 9.8 \times 4}{10}$

$P=313.6 W$

Q.5 A car of mass $2000 Kg$ and it is lifted up a distance of $30m$ by a crane in $1 \: min$. A second crane does the same job as the first crane in $2 \: min$. Do both cranes consume the same or different amounts of fuel? Find the power supplied by each crane? Neglecting power dissipation against friction.

Solution:
Given that:
The mass of car $(m)=2000 Kg $

The lifted up distance $(h)=30m$

The time taken by first crane $(t_{1}= 1\: min)$

The time taken by second crane $(t_{2}= 2\: min)$

The work done by each crane:

$W=mgh$

$W=2000 \times 9.8 \times 30$

$W=5.88 \times 10^{5} J$

As both the cranes do the same amount of work, both consume the same amount of fuel.

The power supplied by the first crane:

$P_{1}=\frac{mgh}{t_{1}}$

$P_{1}=\frac{2000 \times 9.8 \times 30}{60}$

$P_{1}=9800 W$

The power supplied by the second crane:

$P_{2}=\frac{2000 \times 9.8 \times 30}{1.2}$

$P_{2}=4900 W$

Q.6 The human heart discharges $75 \: mL$ of blood at every beat against a pressure of $0.1 m$ of Hg. Calculate the power of the heart assuming that the pulse frequency is $80$ beats per minute. Density of $Hg=13.6 \times 10^{3} Kg/m^{3}$.

Solution:
Given that:
The volume of blood discharge per beat $(V)=75 \: mL = 75 \times 10^{-6} m^{-3} $

The pressure of blood $(P)=0.1 m \: of \: Hg$ i.e.

$P= 0.1 \times 13.6 \times 10^{3} N-m^{-2} \quad (\because P=\rho g h)$

The work done per beat $=PV$

The workdone in 80 beats$W =80 \times PV$

The power of heart:

$P=\frac{80 \times PV}{60}$

$P=\frac{80 \times 0.1 \times 13.6 \times 10^{3} \times 75 \times 10^{-6} \times 9.8}{60}$

$P=1.33 W$

Q.7 A machine gun fires $60$ bullets per minute with a velocity of $700 m/sec$. If the mass of each bullet is $50 g$, then find the power developed by the gun.

Solution:
Given that:
The mass of one bullet is $M=50 g$

The number of bullets $N=60$

The mass of $60$ bullets $m=M \times N= 50\times 60 = 300g= 3Kg$

The velocity of the bullet $v=700 m/sec$

The time take to fire $60$ bullets $t=1 \: min = 60 sec$

The power developed by gun:

$P=\frac{W}{t}$

According to the work-energy theorem:

$W=\Delta K$

$P=\frac{\Delta K}{t} $

$P=\frac{m v^{2}}{2t} \quad( \because K=\frac{mv^{2}}{2})$

$P=\frac{3 \times (700)^{2}}{2 \times 60} $

$P=12250 W $

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Variation of Mass with Velocity in Relativity

Derivation of variation of mass with velocity: Consider two systems of reference (frame of reference) $S$ and $S’$. The frame $S’$ is moving with constant velocity $v$ relative to frame $S$.

Let two bodies of masses $m_{1}$ and $m_{2}$ be traveling with velocities $u’$ and $-u’$ parallel to the x-axis in the system $S’$. Suppose the two bodies collide and after collision coalesce into one body.

The principles of conservation of mass and of momentum also hold good in relativity same as in classical mechanics. So now apply the principle of conservation of momentum.

$m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v\qquad(1)$

Apply the law of addition of velocities, the velocities $u_{1}$ and $u_{1}$ in the system $S$ corresponding to $u’$ and $-u’$ in frame $S’$ are given by $\rightarrow$

$u_{1}= \frac{u'+v}{1+\frac{u'v}{c^{2}}}\quad or \quad u_{2}= \frac{-u'+v}{1-\frac{u'v}{c^{2}}}\qquad(2)$

Now substitute the value of $u_{1}$ and $u_{1}$ in equation $(1)$

$ m_{1}\frac{u'+v}{\left ( 1+\frac{u'v}{c^{2}} \right )}+m_{2}\frac{-u'+v}{\left ( 1-\frac{u'v}{c^{2}} \right )}=\left ( m_{1}+m_{2} \right )v $

$ m_{1}\left [ \frac{u'+v}{1+\frac{u'v}{c^{2}}}-v \right ]=m_{2}\left [ v- \frac{-u'+v}{1-\frac{u'v}{c^{2}}} \right ]$

$ m_{1}\left [ \frac{u'+v-v-u'\frac{v^{2}}{c^{2}}}{1+u'\frac{v}{c^{2}}} \right ]=m_{2}\left [ \frac{v-u'\frac{v^{2}}{c^{2}}+u'-v}{1- u'\frac{v}{c^{2}}} \right ]$

$ \frac{m_{1}}{m_{2}}= \frac{\left ( 1+u'\frac{v}{c^{2}} \right )}{\left ( 1-u'\frac{v}{c^{2}} \right )}\qquad(3)$

From equation$(2)$

$u_{1}^{2} =\left ( \frac{u'+v}{1+\frac{u'v}{c^{2}}} \right )^{2}$

$ 1-\frac{u_{1}^{2}}{c^{2}}=1- \frac{1}{c^{2}} \left ( \frac{u'+v}{1+\frac{u'v}{c^{2}}} \right )^{2}$

$ 1-\frac{u_{1}^{2}}{c^{2}}= \frac{\left ( 1+\frac{u'v}{c^{2}} \right )^{2}-\left ( \frac{u'+v}{c} \right )^{2}}{\left ( 1+\frac{u'v}{c^{2}} \right )^{2}}$

$ 1-\frac{u_{1}^{2}}{c^{2}}= \frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{\left ( 1+\frac{u'v}{c{2}} \right )}$

$ \left ( 1+\frac{u'v}{c^{2}} \right )^{2}=\frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{1-\frac{u_{1}^{2}}{c^{2}}}$

$ \left ( 1+\frac{u'v}{c^{2}} \right ) =\left [ \frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{1-\frac{u_{1}^{2}}{c^{2}}} \right ]^\frac{1}{2}\qquad(4)$

Similarly, square the velocity of $u_{2}$ and solve the as above, so

$\left ( 1-\frac{u'v}{c^{2}} \right ) =\left [ \frac{\left ( 1-\frac{u'^{2}}{c^{2}} \right )\left ( 1-\frac{v^{2}}{c^{2}} \right )}{1-\frac{u_{2}^{2}}{c^{2}}} \right ]\qquad(5)$

Now substituting the values from equation $(4)$ and equation $(5)$ in equation $(3)$

$\frac{m_{1}}{m_{2}}=\frac{\left ( 1-\frac{u_{2}^{2}}{c^{2}} \right )^{\frac{1}{2}}}{\left ( 1-\frac{u_{1}^{2}}{c^{2}} \right )^{\frac{1}{2}}}$

If the body of mass $m_{2}$ is at rest i.e. $m_{2}=m_{0}$ so velocity of the body in frame-S will be zero. i.e. $u_{2}=0$

Where $m_{0}$ is the rest mass.

$m_{1}=\frac{m_{0}}{\left ( 1-\frac{u_{1}^{2}}{c_{2}} \right )^{\frac{1}{2}}}$

Let $ m_{1}=m$ and $u_{1}=v$ so above equation

$m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

This is the generalized formula of variation of mass with velocity.

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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;

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

$L=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\quad (5)$

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$

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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. 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.

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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

$ 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.

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