FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups 
 ProfileProfile   PreferencesPreferences   Log in to check your private messagesLog in to check your private messages   Log inLog in 
Forum index » Science and Technology » Physics » Relativity
derivation of kinetic energy
Post new topic   Reply to topic Page 1 of 1 [9 Posts] View previous topic :: View next topic
Author Message
castertroy14@hotmail.com1
science forum beginner


Joined: 15 Jul 2006
Posts: 5

PostPosted: Sat Jul 15, 2006 12:19 am    Post subject: derivation of kinetic energy Reply with 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
Back to top
N:dlzc D:aol T:com (dlzc)
science forum Guru


Joined: 25 Mar 2005
Posts: 2835

PostPosted: Sat Jul 15, 2006 1:22 am    Post subject: Re: derivation of kinetic energy Reply with quote

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

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

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

David A. Smith
Back to top
castertroy14@hotmail.com1
science forum beginner


Joined: 15 Jul 2006
Posts: 5

PostPosted: Sat Jul 15, 2006 9:27 pm    Post subject: Re: derivation of kinetic energy Reply with quote

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


Joined: 25 Mar 2005
Posts: 2835

PostPosted: Sat Jul 15, 2006 11:30 pm    Post subject: Re: derivation of kinetic energy Reply with quote

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

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
Back to top
castertroy14@hotmail.com1
science forum beginner


Joined: 15 Jul 2006
Posts: 5

PostPosted: Sun Jul 16, 2006 12:16 am    Post subject: Re: derivation of kinetic energy Reply with quote

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)
Back to top
castertroy14@hotmail.com1
science forum beginner


Joined: 15 Jul 2006
Posts: 5

PostPosted: Sun Jul 16, 2006 12:19 am    Post subject: Re: derivation of kinetic energy Reply with quote

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


Joined: 25 Mar 2005
Posts: 2835

PostPosted: Sun Jul 16, 2006 3:35 am    Post subject: Re: derivation of kinetic energy Reply with quote

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)

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
Back to top
Bilge
science forum Guru


Joined: 30 Apr 2005
Posts: 2816

PostPosted: Sun Jul 16, 2006 6:35 am    Post subject: Re: derivation of kinetic energy Reply with quote

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

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


Joined: 08 May 2005
Posts: 616

PostPosted: Sun Jul 16, 2006 12:27 pm    Post subject: Re: derivation of kinetic energy Reply with quote

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
Back to top
Google

Back to top
Display posts from previous:   
Post new topic   Reply to topic Page 1 of 1 [9 Posts] View previous topic :: View next topic
The time now is Fri Jun 17, 2011 1:34 am | All times are GMT
Forum index » Science and Technology » Physics » Relativity
Jump to:  

Similar Topics
Topic Author Forum Replies Last Post
No new posts Energy (not pollution) and global warming. PETER1929@TISCALI.CO.UK New Theories 3 Wed Jul 19, 2006 6:38 pm
No new posts How the public could understand dark energy? gb6724 New Theories 10 Wed Jul 19, 2006 1:55 pm
No new posts derivation of relativistic kinetic energy castertroy14@hotmail.com1 Math 2 Sun Jul 16, 2006 12:22 am
No new posts Study of gravity, dark energy and black holes gb7648 New Theories 1 Fri Jul 14, 2006 10:48 pm
No new posts Dark Energy gb6724 New Theories 2 Thu Jul 13, 2006 6:06 am

Copyright © 2004-2005 DeniX Solutions SRL
Other DeniX Solutions sites: Electronics forum |  Medicine forum |  Unix/Linux blog |  Unix/Linux documentation |  Unix/Linux forums  |  send newsletters
 


Powered by phpBB © 2001, 2005 phpBB Group
[ Time: 0.0890s ][ Queries: 16 (0.0506s) ][ GZIP on - Debug on ]