# Einstein mass energy relation discussion and derivation

**Einstein’s mass energy relation:**

You must have known the relation E = mc^{2}. Let us derive it:

**Derivation:**

Consider an object of rest mass m’. If force is applied to the object such that it starts moving with relativistic velocity (that is comparable with the speed of light), then its mass will also vary with variation of mass with velocity relation

m = m’/(1-v^{2}/c^{2})^{1/2} (1)

Now suppose that work dw will be done due to this force. If the object is displaced along x axis, then work will be:

dw = Fdx

or dw = (dp/dt)dx (because from Newton’s 2^{nd} law F = dp/dt)

or dw = [d(mv)/dt]dx (because p =mv)

Differentiate R.H.S.

dw = (mdv/dt + vdm/dt)dx (here m is also a variable quantity, thus m is also differentiated)

or dw = mdvdx/dt + vdmdx/dt

or dw = mvdv + v^{2}dm (2)

Now square equation (1) and cross-multiply

m^{2} (1-v^{2}/c^{2}) = m’

or m^{2} [(c^{2}-v^{2})/c^{2}] = m’

or m^{2}c^{2} – m^{2}v^{2} = m’^{2}c^{2}

Differentiating, we get

c^{2}(2mdm) – m^{2}(2vdv) – v^{2}(2mdm)

or v^{2}dm + mvdv = c^{2}dm (3)

Comparing equations (2) and (3), we get

dw = c^{2}dm (4)

The total amount of work done by the applied force in order to change its velocity from 0 to v (or mass from m’ to m) is achieved by integrating the L.H.S of the following equation with limits 0 to W and R.H.S. from m’ to m (because when work is 0 then body has rest mass m’ and when work W is done then body has variable mass m).

∫dw = c^{2}∫dm

Or W = c^{2}(m – m’) (5)

As this work W is done to give motion to the object. Therefore, W will appear in the form of kinetic energy acquired by the body, Thus relativistic kinetic energy will be

K = = c^{2}(m – m’) (6)

By definition of potential energy or the rest mass energy, it is equal to the internal energy of the body. It is also equal to the work done to bring all the particles which make the object of rest mass m’. Thus the rest mass energy of the body is derived as by integrating the L.H.S of the following equation with limits 0 to W and R.H.S. from 0 to m (because when work is 0 then body has rest mass does not exist and when work W is done then all the particles make an object of rest mass m’).

∫dw = c^{2}∫dm

Thus W = m’c^{2}

Therefore, W will appear in the form of rest mass energy of the body, Thus rest mass energy will be

R = m’c^{2} (7)

The total energy of the object will be

E = kinetic energy + rest mass energy

Put equations (6) and (7) in this equation, we get

E = c^{2}(m – m’) + m’c^{2}

Or E = mc^{2}

This is the famous Einstein mass-energy equivalence relation.

**Significance**:

This equation represents that energy can neither be created nor be destroyed, but it can change its form.

**Example:**

Pair Annihilation:

In pair annihilation, electron and positron reacts to release photons.

e^{–} + e^{+} → γ

As electron and positron have mass but photon has energy but not mass. Therefore, here mass is changed into energy.

The opposite of this reaction is called pair production.

Why optical fibre is not used in the household phone application?

Because it costs more than the normal wire.

Is the copper wire is costlier than optical fibre? If yes why we prefer copper wire to optical fiber for landline phone application

according to me, fiber is costlier than wire. It may also depend upon the type of the fiber. For example, it is mono mode or multi mode