E=mc²: Einstein explains his famous formula

Albert Einstein explaining the equivalence of energy and matter.

Category:

Science & Technology

Tags:

• physics
• science
• albert
• einstein
• theory
• relativity
• E=mc²

[podspd]

@SMGJohn what idiot....

[dcs002]

@LeconsdAnalyse Hmm... Now ya lost me. ;o)

[LeconsdAnalyse]

2/2

Hence, for an element of `work`, dW=Fdx=m₀γ³adx=m₀γ³(dv/dt)dx=m­₀γ³dv(dx/dt)=m₀γ³vdv=m₀c²β(1-β­²)^⁻3/2dβ. Therefore, E=W=m₀c² ∫β(1-β²)^⁻3/2dβ. Let u=1-β² so that du=-2βdβ or, -½du=βdβ. We get, E=W= -½m₀c²∫u^⁻3/2du = m₀γc² + constant = mc² + constant♦

[LeconsdAnalyse]

1/2

As a tribute to Einstein i wish to post the following proof (the simplest that i know of) of his most famous discovery:

F=d(mv)/dt=d(m₀γv)/dt=m₀d(γv)/­dt=m₀[γ(dv/dt)+v(dγ/dt)], where γ= 1/√(1-v²/c²). So dγ/dt=(dγ/dv)(dv/dt)=a(dγ/dv)=­a(dγ/dv)=ac⁻¹(dγ/dβ) (where, β=v/c)=ac⁻¹(d/dβ(1-β²)^⁻½)=ac⁻­¹[⁻½(-2β)(1-β²)^⁻3/2]=(av/c²)(­1-v²/c²)^⁻3/2.

Therefore F=m₀[γ(dv/dt)+v(dγ/dt)]=m₀γa + m₀aγv²/c²[c²/(c²-v²)]=m₀γa[(c²­-v²+v²)/(c²-v²)]=m₀γa[c²/(c²-v­²)]=m₀γ³a♦

[LeconsdAnalyse]

2/2

⁷₃Li + ¹₁H → ⁴₂He + ⁴₂He + energy.

High energy protons (¹₁H) making inelastic collisions with lithium (⁷₃Li). Producing helium (⁴₂He), and releasing energy.

[dcs002]

@dcs002 BTW LeconsdAnalyse, I don't yet understand your response completely, but I'm gonna think about it and work with it until I understand as much of it as I can. For the first time I can actually see mathematically just how mass approaches infinity (very rapidly) as v approaches c, though I don't have a sense of why that is. I'll need time to process this... This is actually kinda exciting!