Question

Mathematically derive the expression :-

E = mc²

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Answer 1

Hello friend, Harish here.

Here is your answer:

To Prove :

The equation E = mc².

First Approach:

Let us assume a object with a velocity which is equal to velocity of light (approximately), and an uniform force acts on it.

NOTE: That force does not affect the object's motion due to its huge velocity.

So, we know that,

E = F × d.

Here the object moves a distance of c per unit time. 

Then, 

E = F × c.  - (i)

And we know that;

⇒ Momentum (p) = Force(F) × Time during which the force acts.

During the time at which the force acts the mass increases and the velocity stays constant at very close to c.

And we also know that momentum (p) = mass(m) x velocity(c), the momentum gained is m x c.

Therefore by these we can conclude that : F = m × c. - (ii)

From (i) & (ii) by equating the forces we get:

$\frac{E}{c} = m \times c$

⇒ $E = mc^2$  -(Derived)
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Second Approach:

We know that:

→ Force is the rate of change of momentum.

⇒ $F = \frac{d(mv)}{dt}$  - (i)

And,  

$Kinetic \ Energy\ (K.E) = \int\limits^s_0 {F} \, ds$  - (ii)

Here, s - displacement, F - Force. , m - relativistic mass, v - velocity.

 Substituting (i) in (ii),

$KE = \int\limits^s_0 { \frac{d(mv)}{dt}} \, ds = \int\limits^{mv}_0 {v} \, dmv$

We know that, 

$m = \frac{m_0}{ \sqrt{1- \frac{v^2}{c^2} }}$  

m₀ - Rest mass. C - velocity of light, v - velocity of the object.

Substituting the value of m in the above equation we get.

⇒ $KE = \int\limits^v_0 {v} \, d (\frac{m_0v}{ \sqrt{1- \frac{v^2}{c^2} }})$

By use integration by parts we get,

⇒ $KE = \frac{m_0v^2 }{ \sqrt{1- \frac{v^2}{c^2} }} +\left[(m_0 c^2) \frac{m_0v^2}{ \sqrt{1- \frac{v^2}{c^2} }} \right ] ^v _{0}$

⇒ $KE = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} }} - m_0c^2$

⇒ $KE  = mc^2 - m_0 c^2$

We also know that,

$KE = E - E_0$

Now by comparing these equation we can confirm that,

$E = mc^2 \ \ \ \& \ \ \ \ E_0 = m_0 c^2$

Hence derived.

There are still more approaches to derive the formula of Einstein's special relativity. 
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Hope my answer is helpful to you. 

Answer 2

HEY

HERE IS ANSWER

it refers to pic

here

E is energy.

M is mass.

and

c^2 is speed of light

or velocity of light

in the theory of relativity a postulate is given that.
nothing can travel faster than the speed of light.

light is the only thing which can travel at speed..

299792458m/s

and universe is limited at that speed and it is used in this.

HOPE IT HELPS

THANKS