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Both E's Are Numerically The Same

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Posted by Jim Bergquist on January 10, 2001 10:39:28 UTC

The two expressions for E are equivalent as the following argument will show:

Let: β = v/c γ = 1/sqrt(1-β^2) m0 = "rest mass"

m = γ m0 = "relativistic mass"

p = m v = γ m0 β c

E^2 = m0^2 c^4 + p^2 c^2 so that:

E^2 = m0^2 c^4 + γ^2 m0^2 β^2 c^4

= m0^2 c^4 (1 + γ^2 β^2 )

= m0^2 c^4 (1 + β^2 / (1-β^2) )

= m0^2 c^4 ( (1 - β^2 + β^2) / (1-β^2) )

= m0^2 c^4 ( 1 / (1-β^2) ) = m0^2 c^4 γ^2

= m^2 c^4

Or, finally, E = m c^2

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