Question

prove that e = mc^2
please answer in detail
​

Answer 1

E=mc^2 is one of the most famous equation of all

time.

To understand this equation first we need to know

about special relativity. One of the consequence of

special relativity is that mass appears to increase

with speed. The faster an object goes, the heavier it

seems to get.

there is an equation connecting our current mass

speed and mass at that speed.

it is given by,

M =M1/(sqrt(1-v^2/c^2))

where M is the relativistic mass,ie the mass at the

sped we are travelling.

M1 is the rest mass(at stationary)

v is the speed at we are travelling

c, the speed of light.

This is a table showing the relativistic increase in

mass in relative to the speed of light. Here the mass

of an object is kept as unity it cannot be below the unity.

we see that as speed increases the mass increases,

so more energy is required to move that object.

we have Kinetic Energy, KE=1/2 my^2

So this is the condition for which the increase in

kinetic energy occurs.

From our mass -velocity equation M =M 1/(sqrt(1

v^2/c^2)),M and c are relative in some way.consider

that the V is very low ,then we can express the mass

increase in mc^2 term .so this incremental value will

be add on with total KE.

So E=1/2 m^2+mc^2

or

E-mc^2=1/2 mv 2

this result is fine for low speed.

We know that at low speed Kinetic energy is

approximately equals to E-mc^2

ie, Relativistic kinetic energy=E-mc^2

or

E=Relativistic kinetic energy +mc^2.

so the Relativistic energy has two parts a)Relativistic

kinetic energy which depend on the speed of the

object, b)mass increase which does not depends

on

the speed of the object.Both these parts are a form

of energy.we can simplify the equation by

considering a zero velocity for the object.

then, Relativistic kinetic energy =0

then E=0+mc^2

or

E=mc2

Answer 2

✦ MASS ENERGY EQUIVALENCE ✦

❏ PROviNG E=mc² :-

The rate of doing work is given by under force F and Velocity V.

$\sf\tt\therefore\large \frac{dW}{dt}=\vec{F}.\vec{v}$

$\sf\tt\implies\frac{dW}{dt}=\frac{d\vec{P}}{dt}.\vec{v}$

$\sf\tt\implies \frac{dW}{dt}=\frac{d}{dt}(m\vec{v}).\vec{v}$

$\sf\tt\implies \frac{dW}{dt}=\frac{d}{dt}[\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}\vec{v}].\vec{v}$

$\sf\tt\implies \frac{dW}{dt}=m_o \vec{v}[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\frac{d\vec{v}}{dt}-\frac{\vec{v}(-\frac{2\vec{v}}{c^2})}{2(1-\frac{v^2}{c^2})^{\frac{3}{2}}}\frac{d\vec{v}}{dt}]$

$\sf\tt\implies \frac{dW}{dt}=m_o \vec{v}[\frac{1}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}+\frac{\vec{v}(\frac{\vec{v}}{c^2})}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}]\frac{d\vec{v}}{dt}$

$\sf\tt\implies \frac{dW}{dt}=m_o \vec{v}[\frac{1-\cancel{\frac{v^2}{c^2}}+\cancel{\frac{v^2}{c^2}}}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}]\frac{d\vec{v}}{dt}$

$\sf\tt\implies \frac{dW}{dt}=\frac{m_o \vec{v}.\frac{d\vec{v}}{dt}}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}$

Now,

$\sf\tt \frac{d}{dt}[\frac{m_o c^2}{\sqrt{1-\frac{v^2}{c^2}}}]=m_o c^2[-\frac{1(-\frac{2\vec{v}}{c^2})\frac{d\vec{v}}{dt}}{2(1-\frac{v^2}{c^2})^{\frac{3}{2}}}]$

$\sf\tt \frac{d}{dt}[\frac{m_o c^2}{\sqrt{1-\frac{v^2}{c^2}}}]=\frac{m_o \vec{v}.\frac{d\vec{v}}{dt}}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}$

Therefore;

$\sf\tt \implies\frac{dW}{dt}=\frac{d}{dt}[\frac{m_o c^2}{\sqrt{1-\frac{v^2}{c^2}}}]$

➾ In "theory of special relativity" body moves with high velocity and approaches towards the speed of light. So, energy of the body is totally kinetic and potential energy term is neglected.

So,

$\sf\tt\frac{dT}{dt} =\frac{dW}{dt}$

$\sf\tt\implies\frac{dT}{dt} =\frac{d}{dt}[\frac{m_o c^2}{\sqrt{1-\frac{v^2}{c^2}}}]$

∴ $\sf\tt\int\limits_{0}^{T} \, dT=\int\limits_{0}^{v} \, d(\frac{m_o c^2}{\sqrt{1-\frac{v^2}{c^2}}})$

$\sf\tt\implies\boxed{ \red{T=(\frac{m_o c^2}{\sqrt{1-\frac{v^2}{c^2}}})-m_o c^2}}$

Now, $\sf\tt\longrightarrow m=(\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}})$

$\bf\tt\sf\therefore\boxed{\large{\red{ T=mc^2-m_o c^2}}}$

The term $\bf m_o c^2$ is called rest mass energy.

$\bf\tt\sf\therefore T=mc^2-m_o c^2$

$\bf\tt\sf\implies T+m_o c^2=mc^2$

$\bf\tt\sf\implies \boxed{\large{\red{E=mc^2}}}$ [proved].

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$