How is E = mc^2 derived?

Question

How is E = mc^2 derived?

Answer 1

The famous equation (E=mc²) was derived by Einstein as part of his 1905 paper 'On the Electrodynamics of Moving Bodies'. This theory is now better known as the ‘Theory of Special Relativity’. Its underlying concepts were that wherever an observer is located he or she must find the laws of physics to be the same and furthermore, whatever the observer’s velocity he or she must find the speed of light to have the same value. The speed of light is a universal upper limit to all velocities. To allow observers with different velocities to transform their results so that they agreed, Einstein provided a transform equation: -

x = x'/sqrt(1 -(v/c)²)

Thus measurement x for one observer, separating from another observer with a constant velocity v, can be transformed into the other observers measurement x’.

The famous equation is derived from the expression for work energy. Work equals force times distance.

dW = F.ds

Since force F equals d(mv)/dt and ds/dt = v we can expand the expression for work into: -

dW =(d(mv)/dt).ds

or, expanding further: -

dW=v²dm+mv.dv

Using the observers transform equation for masses (x = m, x’ = m’) we can obtain

m²c² = m²v² + m'²c²

Differentiating, with c and m' as constants, gives: -

2mc²dm = 2mv²dm + 2m²vdv

When the 2m is divided out, we are left with: -

dW = c²dm

Thus integrating between limits m and m' (the rest mass) with c constant gives:-

W = c²∫dm = c²(m - m')

Or (since the total work is the energy): -

E = mc² + m'c²

This is not Einstein's original derivation but I hope you find it perhaps a bit easier going than his briefer, but for all that, calculus rich approach.

  

Answer 2

It starts from the Lorentz Transformation. To derive the LT assume a linear spacetime transformation x'=a_ix_i (summed over i=1..4) between reference frames and solve for {a_i} which results in Mawell's Equations being reference frame invariant. Or it can be deduced directly from the assumption that the speed of light is constant in all reference frames by assuming a linear transformation in a thought experiment involving transverse and longitudinal light pulses making a round trip to and from reflectors in front of and to the side of a moving object.

Once you have the LT,  E^2=m^2c^4+P^2c^2 can be derived, where E and P are energy and momentum, by assuming that E and P are conserved in elastic scattering events. The subject equation is for P=0. m is rest mass, so the remaining term mc^2 is rest mass energy.

Answer 3

Time dilation

t' = t (1- v^2/c^2)^1/2

E = K + P

Rest mass:

m = m' (1- v^2/c^2)^1/2

Plug into the E equation and take the limits, it will give E = mc^2

Answer 4

Einstein himself invented a thought experiment to suggest that mass is associated with energy. Suppose that a photon of energy E is released from one end of a box of mass M and length L that is isolated from it's surroundings and initially at rest. The radiation carries momentum E/c and since the total momentum must remain 0 the box acquires a momentum -E/c. Therefore, using momentum = Mv, the box must recoil with a velocity v = -E/Mc. After travelling freely for a time dt (approximately L/c) the radiation hits the other end of the box and conveys an impluse equal and opposite to the one given initially and the box comes to a rest. The result of this is to move the box through a distance:

x = v*dt = (-E/Mc)(L/c) = -EL/(Mc^2)

But since this is a completely isolated box we find it difficult to believe that the center of mass of the box could move. We therefore postulate that the photon has carried with it an equivalent mass m such that the center of mass of the box remains where it was: This requires:

mL + Mx = 0

But using x from above:

mL = M[EL/(Mc^2)]

mL = EL/c^2

E = mc^2