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G.E. Ivey
science forum Guru

Joined: 29 Apr 2005
Posts: 308

Posted: Sun Jul 16, 2006 11:58 am    Post subject: Re: derivation of relativistic kinetic energy

Do you understand that "D" here means "the derivative with respect to t"? If you do, then the quotient rule (together with the chain rule), as Narcoleptic Insomniac said, gives the result.
Narcoleptic Insomniac
science forum Guru

Joined: 02 May 2005
Posts: 323

Posted: Sun Jul 16, 2006 11:06 am    Post subject: Re: derivation of relativistic kinetic energy

On Jul 15, 2006 7:22 PM CT, castertroy14@hotmail.com wrote:

 Quote: K = integral(vDp) = integral(vD(mv / (1-v^2/c^2)^0.5)) D(mv / (1-v^2/c^2)^0.5) = m(1 - v^2/c^2)^-1.5 Dv therefore K = integral(m(1-v^2/c^2)^-1.5 vDv) = mc^2(1/(1-(v^2/c^2)^0.5 - 1) so my question is how do you get from D(mv / (1-v^2/c^2)^0.5) to m(1 - v^2/c^2)^-1.5 Dv

We can use the the quotient rule to differentiate from
elementary calculus...

d[mv / (1 - v^2 / c^2)^(1/2)] / dv =

[m * (1 - v^2 / c^2)^(1/2) + (mv^2 / c^2) *
(1 - v^2 / c^2)^(-1/2)] / (1 - v^2 / c^2) =

m * [1 - v^2 / c^2 + v^2 / c^2] / (1 - v^2 / c^2)^(3/2) =

m * (1 - v^2 / c^2)^(-3/2).

Regards,
Kyle Czarnecki
castertroy14@hotmail.com1
science forum beginner

Joined: 15 Jul 2006
Posts: 5

 Posted: Sun Jul 16, 2006 12:22 am    Post subject: derivation of relativistic kinetic energy K = integral(vDp) = integral(vD(mv / (1-v^2/c^2)^0.5)) D(mv / (1-v^2/c^2)^0.5) = m(1 - v^2/c^2)^-1.5 Dv therefore K = integral(m(1-v^2/c^2)^-1.5 vDv) = mc^2(1/(1-(v^2/c^2)^0.5 - 1) so my question is how do you get from D(mv / (1-v^2/c^2)^0.5) to m(1 - v^2/c^2)^-1.5 Dv

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