E4 Energy has Mass. How Much Mass Does 1 Joule Have?
If we accelerate an object with rest mass m0, it gains not only velocity and kinetic energy, but its mass mv also becomes larger. We want to treat this computationally now, in order to determine quantitatively the additional mass gained by this energy input. We want to derive this important result exactly and therefore need integral calculus, albeit only to the extent that first year calculus students have at their disposal. Physics can also serve mathematics by showing how powerful formal mathematical methods can be !
The kinetic energy is equal to the accumulated work of acceleration, and we obtain this with the integral F • ds over the distance of the acceleration:
∆W = ∫ F • ds = ∆Ekin = Ekin (the last equal sign pertains only to the case where v0 = 0)
According to Newton force F is the change of momentum over time:
F = dp / dt = d(m • v) / dt
We use this to rewrite the integral over F • ds:
Instead of integrating over the distance of acceleration we can now integrate over the increase in velocity:
In order to gain some confidence in this procedure we first compute the classical case. In this case the accelerated mass is constant, and the derivative
of m • v over v simply gives m. Thus (1) yields:
We obtain the familiar expression for kinetic energy, which hopefully somewhat alleviates any distrust arising from all the juggling of dv's and dt's!
What value does the expression d(m • v) / dv have in the relativistic calculation? It is
This expression is occasionally called the ‘longitudinal mass’. We stress however that there is only one expression for the inertial mass of a body, i.e. mv = m0 / √ , and that expression is direction-independent. People spoke of the longitudinal and transversal mass before clarifying that force and acceleration transform differently for the directions parallel and perpendicular to v.
Thus we can calculate (1) in the relativistic case:
definition of mv in E1 !
We have found the connection between the energy gain ∆E (or performed work ∆W) and the mass increase ∆m we were looking for. The resulting formula is so simple that it has attained a strange popularity.
The illustration on the cover of the otherwise smart booklet  about the STR serves all the popular clichés. Of course, the formula E = mc2 is not
The US Navy even improved the formula in celebration of the 40th anniversary of nuclear-powered aircraft carriers. Sailors served as ‘pixels’:
Back to physics. We present in a red box, what Einstein himself later designated as the most meaningful result of the STR:
Thus 1 Joule of energy produces an increase in mass of 1 kg divided by c2. Here we are glad that we did not standardize c with the value 1, since otherwise this conversion factor between energy and mass would not be so obvious!
Our derivation also reveals the correct expression in the SRT for kinetic energy:
That this formula turns into the classical expression 0.5 • m0 • v2 for small speeds is not obvious. If the graphic on the following page does not suffice, then try the following: Develop the term 1/√ = (1 - (v/c)2) - 1/2 = (1 - x2) - 1/2 for x into a power series (find a collection of formulas or use a computer algebra program) and cancel out (for small values of x) the fourth and higher order terms.
By the way, Einstein found the relationship ∆E = ∆m • c2 shortly after the appearance of [09-123ff] and in the autumn of 1905 communicated it quasi as an addendum [09-161ff]. As early as 1901 Walter Kaufmann (1871-1947) had pondered a dependence of ‘transversal mass’ on velocity due to measurements he made of fast moving electrons. Due to its fundamental meaning the formula was experimentally examined again and again. In 2005 two groups of researchers in Canada and the USA were able to increase the accuracy to 1 part per million [ nature 438, p.1096-1097 ]. And, above all: None of the many experiments could prove any deviation from Einstein's formula! Theories cannot be confirmed by experiments, however they can be falsified.
We want to compare the kinetic energy according to the classical and the relativistic calculation in a diagram. We draw Ekin / E0 for values x = v/c from
0 to 1. The classical behavior corresponds to the blue curve with y = 0.5 • x2, while the relativistic to the red curve with y = 1/√ (1 - x2) - 1:
The curves deviate from each other only for larger velocities. Electrons can readily be accelerated to 0.8 • c and thereby show clear deviations from classical behavior (experiments by Kaufmann, see problem 4).
Very fast-moving particles (v ≈ c) also make possible a very fast derivation of the relationship between the mass increase and the energy gain. Assume a particle already has a velocity, which differs only fractionally from c (e.g., by a thousandth of c). Any added energy will serve practically only to increase the mass. As a good approximation p = mv • v ≈ mv • c and thus dp / dv = c • dm / dv. It follows immediately
dW = (dp / dv) • v • dv = c • (dm / dv) • c • dv = c2 • dm
and we are finished: The mass increase is proportional to the energy input, and the proportionality factor is the square of the speed of light! This is actually the counterpart to the classical calculation, which we made at the beginning of this section and where we additionally assumed m to be constant.
Thus the product of a mass with the square of a velocity represents (as we knew all along) an energy. We derived:
∆W = mv • c2 – m0 • c2 = ∆m • c2 = Ekin = m0 • c2 • (1 / √ – 1)
The expression m0 • c2 represents the quantity of energy corresponding to the rest mass, which the object already possessed before acceleration. Therefore one calls m0 • c2 the rest energy of the object and writes it as E0. The expression mv • c2 stands for the sum of the rest energy and the kinetic energy and is therefore the total energy of the object. We will write this as Etot.