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derivation of relativistic kinetic energy
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castertroy14@hotmail.com1
science forum beginner


Joined: 15 Jul 2006
Posts: 5

PostPosted: Sun Jul 16, 2006 12:22 am    Post subject: derivation of relativistic kinetic energy Reply with 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
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Narcoleptic Insomniac
science forum Guru


Joined: 02 May 2005
Posts: 323

PostPosted: Sun Jul 16, 2006 11:06 am    Post subject: Re: derivation of relativistic kinetic energy Reply with quote

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
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G.E. Ivey
science forum Guru


Joined: 29 Apr 2005
Posts: 308

PostPosted: Sun Jul 16, 2006 11:58 am    Post subject: Re: derivation of relativistic kinetic energy Reply with quote

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.
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