There is a simple derivation of \(E = mc^2\) which is based on the Doppler effect and assumes the observer is moving slowly compared to the speed of light1. According to the Doppler Effect, the observed frequency of a wave increases by a factor of approximately v/c, where v is the speed of the observer moving toward the source of waves and the waves move at speed c. First we must derive this equation.
Section 1 – Derivation of Doppler shift equation
\(c = λ_S/T\)
c is the wave velocity
\(λ_S\) is the wavelength of the source
T is the period of the wave from the source, the time between the peaks of the wave.
\(T = λ_S/c\)
To show this is the case we can express the equation in terms of its units:
[Seconds] = [meters]/[meters/Second]
The observer is moving toward the source at velocity v. You could think of, for example, you are the observer and you are driving toward an ambulance or cop car at high speed. You already know from experience that it’s siren will sound higher pitched as you drive toward it and lower pitched as you drive away.
The change in distance between the observer and source which occurs in time T is:
\(∆d = vT\)
where ∆ indicates change.
v is the velocity of the observer toward the source.
When the observer is moving toward the source, the waves are becoming more frequent from its perspective and their observed wavelength is getting shorter.
\(λ_O = λ_S – ∆d\)
\(λ_O\) is the wavelength from the perspective of the observer
Next we substitute \(T = λ_S/c\) into \(∆d = -vT\)
Later in the derivation, this wave will be defined as light. You may recall that one result of the Theory of Relativity is that time slows down for the moving observer relative to the stationary source.
The above substitution of the left equation for the period of the wave, which is written from the perspective of the source, into the right equation for the distance traveled by the observer in one wave period, which is written from the perspective of the observer, assumes that the period of the wave, the time passed between its peaks, is approximately the same for the source and the observer. The assumption that the period of the light is approximately the same for the source and observer is approximately true when the observer velocity is small compared to the speed of light, as we assume in this derivation of \(E = mc^2\). Therefore, we can trust the result:
\(d = -v λ_S/c\)
Substituting d into \(λ_O = λ_S - ∆d\),
\(λ_O = λ_S + v λ_S/c\)
\(λ_O = λ_S(c-v)/c\)
\(f_O = f_S*c/(c-v)\)
We need to simplify this somewhat, which can be done by assuming that \(v_O^2\) << (is much less than) \(c^2\) so
\(f_O = f_S\frac{c(c+v)}{(c+v)(c-v)}\)
\(f_O = f_S\frac{c(c+v)}{c^2-v^2}\)
Since \(v_O^2 << c^2\)
\(f_O = f_S\frac{c(c+v)}{c^2} = f_S(1+v/c)\)
Here you may say, well then you are starting this derivation from an approximation! Well that is correct. Also, interestingly, \(E = mc^2\) is itself an approximation: http://www.askamathematician.com/2011/03/q-why-does-emc2/
Or as a physicist friend of mine told me, it is an exact equation for the energy contained by matter at 0 velocity.
One approximation is \(E ≈ mc^2 + (1/2)mv^2\)
You’ll notice the Physics 101 term for kinetic energy there, \(½mv^2\) which is negligible compared to the energy contained in the mass itself at low velocity. Then more accurate versions of the equation with many more terms exist.
Now let’s continue with the derivation:
Section 2 - Derivation of \(E = mc^2\) using the Doppler Effect
An observer is moving toward the source of waves, here light, when the source emits a photon, a unit of light, toward the observer and in the opposite direction as shown in the figure below.
According to the concept of relativity, the observer moving toward the source of light is equivalent to the source of light moving toward the observer as shown in the figure below. Now it stands to reason that the light moving toward the observer will be “blue shifted”, which means that the frequency of the light from the perspective of the observer is made greater by the movement of the source. Since blue is the color on the high end of the visible spectrum, we call this effect “blue shifting”. Basically the waves are squashed. The photon moving in the other direction is “red shifted”. That is the waves are stretched out from the perspective of the observer.
First we derive the momentum of the right-moving photon, which is its energy divided by its speed, c, here the speed of light. We define right-moving momentum as positive.
According to p. 378 of The Final Theory: Rethinking Our Scientific Legacy, this comes from experiments in which momentum is transferred to objects by shining light on them. For this experiment, objects which fully absorb light are used. (This derivation may be simpler.)
The extra momentum of the right-moving light is
\(∆Pright = \frac{(\frac{E}{2})(\frac{v}{c})}{c} = \frac{vE}{2c^2}\)
The momentum of the left moving light is shifted by the same amount with the opposite sign.
\(∆Pleft = -\frac{vE}{2c^2}\)
\(∆Psource = ∆Pright-∆Pleft = \frac{vE}{c^2}\)
Where did this surplus of momentum come from? The light waves left the source in opposite directions with the same energy so the source’s velocity is unaffected. It must have come from the mass. By conservation of momentum.
\(∆Mv + ∆Psource = 0\)
Where ∆m is the change in mass resulting from the light, having a total energy E, being emitted.
Rearranging…
\(∆Mv + \frac{vE}{c^2}= 0\)
\(-∆M = E/c^2\)
If we define m as the decrease in mass of the source, -∆M, then we get
\(E = mc^2\) !
1. Rohrlich, Fritz (1990), "An elementary derivation of E=mc2", American Journal of Physics58 (4): 348–349, Bibcode:1990AmJPh..58..348R, doi:10.1119/1.16168