Derivation of E=pc & E=MC2: Which Came First?

[Sagittarius A-Star]

Einstein's 1935 derivation of the equivalence of mass and energy:
https://projecteuclid.org/download/pdf_1/euclid.bams/1183498131

From an elastic eccentric collision scenario with 2 particles with equal mass, he derives (using a unit system with $c=1$), that the relativistic momentum $m\gamma \vec{v}$ and the relativistic kinetic energy $m(\gamma-1)$ of the system are conserved in any inertial reference frame. He makes use of the symmetry in the center-of-momentum frame and the Lorentz-transformation.

Then he considers an inelastic collision between 2 particles with equal mass and equal rest-energy $E_0$. He derives from the energy law for the system, that $E_0$ and $m$ of the particles change equally, and that therefore
$$E_0=m$$
in $E=E_0+m(\gamma -1)$.

He concludes:

If, from the beginning, we had provided the expression for the impulse with a mass-constant different from that of the energy, these considerations would show that the impulse-mass changes in an inelastic collision like the energy-mass. This is a partial justification for setting both mass-constants equal to each other.

The result of this consideration is therefore as follows. If for collisions of material points the conservation laws are to hold for an arbitrary (Lorentz) coordinate-system, the well known expressions for impulse and energy follow, as well as the validity of the principle of equivalence of mass and rest-energy.

 

[vanhees71]

The important point is that $E/c=(E_0+E_{\text{kin}})/c$ together with $\vec{p}$ forms a four-vector. The methodological breakthru has been Minkowski's formulation of SR spacetime as a 4D pseudo-Euclidean affine manifold!

 

[Sagittarius A-Star]

The important point is that $E/c=(E_0+E_{\text{kin}})/c$ together with $\vec{p}$ forms a four-vector. The methodological breakthru has been Minkowski's formulation of SR spacetime as a 4D pseudo-Euclidean affine manifold!

Yes. In the 1935 paper of Einstein, that I linked, he started with the definition of the four-momentum and did the momentum-calculations for the elastic collision for all 4 dimensions, see his 4 equation(s) number (1) on page 226.

We remark that the equations (1) can be derived more clearly if one considers directly the transformation for the sum of the four-vectors of the velocities of a particle-pair. I have chosen the above representation, however, because the conservation laws indicate the use of this 3-dimensionally inhomogeneous manner of writing.

Source:
https://projecteuclid.org/download/pdf_1/euclid.bams/1183498131

I think he did not want to give up completely the relation to the understanding of 3-momentum and energy in classical mechanics for didactical reasons.

 

[mitochan]

I have seen E=pc used to derive E=MC2. I have seen E=MC2 used to derive E=pc. This is circuitous. Which came first and how is E=pc derived?

From relativistic formula
E^2=p^2c^2+m^2c^4
we get $E=pc$ by making $m=0$ for photon and other massless particles.

Mass does not apply for a single photon, so let us prepare N photon gas of same wavelength and no total momentum in a theorotetical massless sphere vessel. In experiments done in the rest frame of this photon ball, it would show inertia and a source of gravity m, i.e.
m=\frac{Npc}{c^2}=\frac{E}{c^2}

 

[Sagittarius A-Star]

Mass does not apply for a single photon, so let us prepare N photon gas of same wavelength and no total momentum in a theorotetical massless sphere vessel.

Now assume instead a light-clock at rest in frame $S$ containing 2 photons, each with an energy of $L$, moving always in opposite direction between the mirrors at the top and at the bottom. Then the energy-content of this light-clock is $E_0 = 2L$.

In this reference frame $S$, the two mirrors are removed at the same time. One photon escapes now from the light clock in y-direction, the other in (-y)-direction.

Assume a frame $S'$, moving constantly with $v$ in positive x-direction. Two observers are at the same x'-coordinate, but at y'-coordinates with opposite signs.

Described in $S$, those oberserves in $S'$ (with time-dilated clocks) must receive the photons each blue-shifted with energy
$L' = L \gamma (1 - \frac{v}{c} * \cos{90°}) = \gamma L$, according to Einstein's Doppler-formula.
(In frame $S'$, the blue-shift comes from aberration.)

So in frame $S'$, the energy content of the light clock must have been $E = 2L' = \gamma E_0$ before loosing the photons. That means:

$$E_{kin} = mc^2(\gamma -1)= E -E_0 = E_0(\gamma -1)$$

Plausibility-check to classical mechanics (using for x << 1: $\frac{1}{\sqrt{1-x}}\approx 1+\frac{1}{2}x$):
$mc^2(\gamma -1) \approx \frac {1}{2}mv^2$

This is a similar, but simpified argumentation as in Einstein 1905:
https://www.fourmilab.ch/etexts/einstein/E_mc2/e_mc2.pdf

 

[Sagittarius A-Star]

Example from classical mechanics:
kinetic energy = $\int \vec F \cdot d\vec s = \int (m*\vec a) \, d\vec s = \int (m*\frac{d\vec v}{dt}) \, d\vec s = m\int \vec v \cdot d\vec v = \frac{1}{2}mv^2 + const_2$

With $a=\gamma^3 a_0$ (follows from relativistic velocity addition) and if the force is in direction of movement:

relativistic kinetic energy = $\int F \cdot ds = m \int \gamma^3 a_0 \cdot ds = m \int \gamma^3 \frac{dv}{dt} \cdot ds = m\int_0^v \gamma^3 v \cdot d v = mc^2(\gamma -1)$

Source (see after Einsteins definition of "longitudinal mass"):
https://en.wikisource.org/wiki/Translation:On_the_Electrodynamics_of_Moving_Bodies

 

[Sagittarius A-Star]

Correction to last posting:

With proper acceleration $\alpha=\gamma^3 a$ (follows from relativistic velocity addition) and if the force is in direction of movement:

relativistic kinetic energy = $\int F \cdot ds = \int m\alpha \cdot ds = m \int \gamma^3 a \cdot ds = m \int \gamma^3 \frac{dv}{dt} \cdot ds = m\int_0^v \gamma^3 v \cdot d v = mc^2(\gamma -1)$

Source (see after Einsteins definition of "longitudinal mass"):
https://en.wikisource.org/wiki/Translation:On_the_Electrodynamics_of_Moving_Bodies

The integral can be calculated via:
https://www.integral-calculator.com/

 

[Sagittarius A-Star]

Addition to above posting #67:

If the force is in direction of movement, the formula $F = m\gamma^3 a$ can also be derived from the relativistic momentum:

$F = \frac{d}{dt}(m\gamma v) = m \frac{d}{dv}(\gamma v) \frac{dv}{dt} = m\gamma^3 a$

The derivative $\frac{d}{dv}(\gamma v)$ can be calculated via:
https://www.derivative-calculator.net/