Einstein’s famous equation of mass-energy equivalence (E = mc2) can be derived from the transformations so-far obtained for Special Relativity. We first need to look at how velocities transform between S and S' and, unfortunately, that will involve a bit of calculus.
Special Relativity defines a special time-like interval called the ‘proper time’ (τ) that is an invariant, i.e. the same for all inertial observers. S and S' will have different coordinates for two events (x,y,z,t) and (x0,y0,z0,t0) but they will both agree on the value of the following:
τ2 = (t – t0)2 – ((x – x0)2 + (y – y0)2 + (z – z0)2)/c2
Note that this is 0 for events along the path of a ray of light. Also, for events at a fixed location (i.e. x=x0 etc) then it is just the normal local time interval.
Let’s look at successive positions of a point moving relative to S. Let ds be the distance travelled between times t and t+dt, and let v be the speed ds/dt. Then:
dτ = √(dt2 – ds2/c2) = √(1 – v2/c2) dt
A velocity would normally be defined like vx = dx/dt but this would not transform as a vector in space-time using the Lorentz Transformation. Instead, we’ll start with the xi = (x,y,z,ict) that we used before, and differentiate it using the ‘proper time’ to get a new vector.
Vi = dxi / dτ = (dxi / dt) (dt / dτ) = (dxi / dt) / √(1 – v2/c2)
= (vx, vy, vz, ic) / √(1 – v2/c2)
The vector vi is called the 4-velocity because it has four components rather than the usual three.
So, if S’ is moving at velocity u relative to S in the x1 direction, where u = ic tan α, then we have:
V'1 = V1 cos α + V4 sin α
V'2 = V2
V'3 = V3
V'4 = -V1 sin α + V4 cos α
Expanding the values of the vector Vi, and dividing the first three equations by the fourth, we have:
v'x = (vx – u) / (1 – uvx/c2)
v'y = vy √(1 – u2/c2) / (1 – uvx/c2)
v'z = vz √(1 – u2/c2) / (1 – uvx/c2)
If all speeds are much less than c then this approximates to simply:
v'x = (vx – u), v'y = vy, v'z = vz
If we multiply our 4-velocity by the mass of an object to create a 4-momentum vector, how does that affect the ‘conservation of momentum’ law? Well, the quantity now being conserved is:
Pi = m Vi = m (vx, vy, vz, ic) / √(1 – v2/c2)
We can see that the first three components are not quite the same as the classical values for momentum as they are increased by a factor of 1/ √(1 – v2/c2). The fourth component, as we shall see later, is related to the conservation of both mass and energy. In other words, the four-component form will embrace all of the conservation laws for momentum, mass and energy.
Einstein modified the definition of mass to incorporate the square root term – also known as the ‘Lorentz factor’.
Pi = (m0 / √(1 – v2/c2)) (vx, vy, vz, ic)
m0 is termed the ‘rest mass’ of the object. The fact that the effective mass is measured as being greater for a moving object should not be confused with it “getting heavier”. Weight is only applicable when a mass is affected by a gravitational field, whereas mass is really a measure of the inertia of an object.
The 4-force vector is defined as:
Fi = dPi / dτ = m0 (dVi / dτ)
The 4-force can be expressed in terms of the classical force vector as follows:
Fi = (d / dτ) (px, py, pz, imc)
= (d / dt) (px, py, pz, imc) (dt / dτ)
= (fx, fy, fz, ic dm/dt) / √(1 – v2/c2)
The ‘dot product’ of the Vi vector with itself (Vi ∙ Vi) is just (–c2) so if we differentiate with respect to the proper time then we have:
Vi ∙ (dVi / dτ) => Vi ∙ Fi = 0
In other words, F and V are ‘orthogonal’. Expanding on F and V then gives:
(vx, vy, vz, ic) ∙ (fx, fy, fz, ic dm/dt) / √(1 – v2/c2) = 0
The work done by the force (i.e. the energy expended) is found by integrating the rate at which work is done (i.e. (vx, vy, vz) ∙ (fx, fy, fz)) over a given period of time (t1, t2).
⌠t2 c2 dm/dt dt = m2 c2 – m1 c2
This ‘work done’ is simply the increase in kinetic energy of the body, and the kinetic energy must, therefore, be given by:
T = mc2 + C
where C is a constant. When v=0 then T=0 so this sets the value of C at (– m0c2). Hence:
T = mc2 – m0c2 = m0c2 / √(1 – v2/c2) – m0c2
If v is much less than c then this approximates to ½ m0v2 which is the classical value. In effect, this is saying that the measured increase in the mass of a moving body is equivalent to the increase in its kinetic energy, and that the relation between them is given by the famous E= mc2.