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Next, just think it in 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 energy - mass 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 energy – mass equation in dynamic situation.

E = mc² is Dynamic mass – energy relationship in nature. It’s a basic natural property. (“Dynamic” means the particle is moving and hints it has momentum);

Next square both sides: E² = m²c ^4, namely: m²c² = E² / c², then, add a negative mark on both side: – m²c² = – E² / c²;

Add two “redundant” items m²v² on both sides: m²v² – m²c² = m²v² – E² / c²;

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

Because m = γm0, (pay attention here, it just means that you can consider the magnitude of the moving mass m is γm0, but not means the particle be rest), square both sides: m² = m0² / (1 – v² / c²);

Then, m0²v² / (1 – v² / c²) – m0²c² / (1 – v² / c²) = p² – E² / c²;

A mathematical calculation: m0²v² / (1 – v² / c²) – m0²c² / (1 – v² / c²) = - m0²c²;

Then, - m0²c² = p² – E² / c²;

Transform it in math, then, E² = p²c² + (m0c²)².

It’s the so called energy – momentum equation.

If it’s the stationary situation, v = 0, so, no math game can be played. Then, it’s just an energy - mass equation in stationary situation: E0 = m0c².

Liqiang Chen

Sept 19, 2020