Below is how to understand the energy - momentum equation E² = p²c² + (m0c²)² in a natural way:

If the energy – momentum equation reflects the stationary situation, then, momentum p naturally equals to zero. Then, we got E² = 0 + (m0c²)², namely: E = m0c². It can be denoted exactly as E0 = m0c². This is the mass – energy equation in stationary situation;

If the energy – momentum equation reflects the dynamic situation, then, momentum p ≠ 0.

Transform the energy – momentum equation E² = p²c² + (m0c²)² into p² – E² / c² = - m0²c²,

- m0²c² = m0²v² / (1 – v² / c²) – m0²c² / (1 – v² / c²),

Because m² = m0² / (1 – v² / c²), then, - m0²c² = m²v² – m²c² = p² – E² / c²,

Because m²v² = p², then, – m²c² = – E² / c²,

Then E² = m²c ^4, namely: E = mc². This is the mass – energy equation in dynamic situation.

Since the energy – momentum equation E² = p²c² + (m0c²)² is generally applicable (to any particle), the stationary situation E0 = m0c² as well as the dynamic situation E = mc² is generally applicable (to any particle) too.

Liqiang Chen

Aug 18, 2020