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castertroy14@hotmail.com1
science forum beginner

Joined: 15 Jul 2006
Posts: 5

Posted: Sat Jul 15, 2006 12:19 am    Post subject: derivation of kinetic energy

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
N:dlzc D:aol T:com (dlzc)
science forum Guru

Joined: 25 Mar 2005
Posts: 2835

Posted: Sat Jul 15, 2006 1:22 am    Post subject: Re: derivation of kinetic energy

Dear castertroy14:

<castertroy14@hotmail.com> wrote in message
 Quote: 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
castertroy14@hotmail.com1
science forum beginner

Joined: 15 Jul 2006
Posts: 5

Posted: Sat Jul 15, 2006 9:27 pm    Post subject: Re: derivation of kinetic energy

N:dlzc D:aol T:com (dlzc) wrote:
 Quote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152922791.949560.235820@75g2000cwc.googlegroups.com... 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
N:dlzc D:aol T:com (dlzc)
science forum Guru

Joined: 25 Mar 2005
Posts: 2835

Posted: Sat Jul 15, 2006 11:30 pm    Post subject: Re: derivation of kinetic energy

Dear castertroy14:

<castertroy14@hotmail.com> wrote in message
 Quote: N:dlzc D:aol T:com (dlzc) wrote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152922791.949560.235820@75g2000cwc.googlegroups.com... 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:
 Quote: K = integral(vDp) I think this should be:

K = integral(pDv)

David A. Smith
castertroy14@hotmail.com1
science forum beginner

Joined: 15 Jul 2006
Posts: 5

Posted: Sun Jul 16, 2006 12:16 am    Post subject: Re: derivation of kinetic energy

N:dlzc D:aol T:com (dlzc) wrote:
 Quote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152998823.899220.19100@m79g2000cwm.googlegroups.com... N:dlzc D:aol T:com (dlzc) wrote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152922791.949560.235820@75g2000cwc.googlegroups.com... 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)
castertroy14@hotmail.com1
science forum beginner

Joined: 15 Jul 2006
Posts: 5

Posted: Sun Jul 16, 2006 12:19 am    Post subject: Re: derivation of kinetic energy

castertro...@hotmail.com wrote:
 Quote: N:dlzc D:aol T:com (dlzc) wrote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152998823.899220.19100@m79g2000cwm.googlegroups.com... N:dlzc D:aol T:com (dlzc) wrote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152922791.949560.235820@75g2000cwc.googlegroups.com... 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)
N:dlzc D:aol T:com (dlzc)
science forum Guru

Joined: 25 Mar 2005
Posts: 2835

Posted: Sun Jul 16, 2006 3:35 am    Post subject: Re: derivation of kinetic energy

Dear castertroy14:

<castertroy14@hotmail.com> wrote in message
 Quote: castertro...@hotmail.com wrote: N:dlzc D:aol T:com (dlzc) wrote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152998823.899220.19100@m79g2000cwm.googlegroups.com... N:dlzc D:aol T:com (dlzc) wrote: Dear castertroy14: castertroy14@hotmail.com> wrote in message news:1152922791.949560.235820@75g2000cwc.googlegroups.com... 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) K = integral(fDs) = integral([Dp/Dt]Dx) = integral(vDp) correction: K = integral(fDx) = integral([Dp/Dt]Dx) = integral(vDp)

I ask again, what text did you get this from?
http://scienceworld.wolfram.com/physics/KineticEnergy.html

Note that f, Dx, Dp, v, and p are vector quantities, and the
product represented is a dot product. If you are going to spend
time learning relativity, you will find that force is not well
defined. It is much easier to stick with momentum, where
possible.
K = integral(pDv)
.... has none of the issues you are trying to have with
derivatives, when you are intending to *integrate* anyway.

David A. Smith
Bilge
science forum Guru

Joined: 30 Apr 2005
Posts: 2816

Posted: Sun Jul 16, 2006 6:35 am    Post subject: Re: derivation of kinetic energy

 Quote: 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

Not unless you write your question in a more coherent fashion.
jem
science forum Guru

Joined: 08 May 2005
Posts: 616

Posted: Sun Jul 16, 2006 12:27 pm    Post subject: Re: derivation of kinetic energy

castertroy14@hotmail.com wrote:
 Quote: 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

Let g = 1/(1-v^2/c^2)^0.5
then Dg = vg^3 Dv, so

D(mvg) = m(g + v^2 g^3)Dv
= mg^3 Dv

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