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castertroy14@hotmail.com1
science forum beginner
Joined: 15 Jul 2006
Posts: 5
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 Posted: Sat Jul 15, 2006 12:19 am Post subject:
derivation of kinetic energy
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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 |
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N:dlzc D:aol T:com (dlzc)
science forum Guru
Joined: 25 Mar 2005
Posts: 2835
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| Posted: Sat Jul 15, 2006 1:22 am Post subject:
Re: derivation of kinetic energy
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Dear castertroy14:
<castertroy14@hotmail.com> wrote in message
news:1152922791.949560.235820@75g2000cwc.googlegroups.com...
| 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
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I don't see how that relates to kinetic energy.
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html#c4
David A. Smith |
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castertroy14@hotmail.com1
science forum beginner
Joined: 15 Jul 2006
Posts: 5
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| Posted: Sat Jul 15, 2006 9:27 pm Post subject:
Re: derivation of kinetic energy
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N:dlzc D:aol T:com (dlzc) wrote:
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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N:dlzc D:aol T:com (dlzc)
science forum Guru
Joined: 25 Mar 2005
Posts: 2835
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| Posted: Sat Jul 15, 2006 11:30 pm Post subject:
Re: derivation of kinetic energy
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Dear castertroy14:
<castertroy14@hotmail.com> wrote in message
news:1152998823.899220.19100@m79g2000cwm.googlegroups.com...
| 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
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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 |
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castertroy14@hotmail.com1
science forum beginner
Joined: 15 Jul 2006
Posts: 5
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| Posted: Sun Jul 16, 2006 12:16 am Post subject:
Re: derivation of kinetic energy
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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
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K = integral(fDs) = integral([Dp/Dt]Dx) = integral(vDp) |
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castertroy14@hotmail.com1
science forum beginner
Joined: 15 Jul 2006
Posts: 5
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| Posted: Sun Jul 16, 2006 12:19 am Post subject:
Re: derivation of kinetic energy
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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)
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correction:
K = integral(fDx) = integral([Dp/Dt]Dx) = integral(vDp) |
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N:dlzc D:aol T:com (dlzc)
science forum Guru
Joined: 25 Mar 2005
Posts: 2835
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| Posted: Sun Jul 16, 2006 3:35 am Post subject:
Re: derivation of kinetic energy
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Dear castertroy14:
<castertroy14@hotmail.com> wrote in message
news:1153009175.152183.305600@i42g2000cwa.googlegroups.com...
| 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)
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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 |
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Bilge
science forum Guru
Joined: 30 Apr 2005
Posts: 2816
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| Posted: Sun Jul 16, 2006 6:35 am Post subject:
Re: derivation of kinetic energy
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castertroy14@hotmail.com:
| 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
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Not unless you write your question in a more coherent fashion. |
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jem
science forum Guru
Joined: 08 May 2005
Posts: 616
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| Posted: Sun Jul 16, 2006 12:27 pm Post subject:
Re: derivation of kinetic energy
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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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