Re: derivation of kinetic energy


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N:dlzc D:aol T:com (dlzc) wrote:
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N:dlzc D:aol T:com (dlzc) wrote:
Dear castertroy14:

<castertroy14@xxxxxxxxxxx> wrote in message
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can anyone show me the process of getting
equation (2) from equation (1).
(1) D(mv / (1-v^2/c^2)^0.5)
(2) m(1 - v^2/c^2)^-1.5 Dv

I don't see how that relates to kinetic energy.

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4

David A. Smith

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

What text do you find this in? I have issues with the very
first:
K = integral(vDp)
I think this should be:
K = integral(pDv)

David A. Smith

K = integral(fDs) = integral([Dp/Dt]Dx) = integral(vDp)

correction:
K = integral(fDx) = integral([Dp/Dt]Dx) = integral(vDp)

.

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