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I'm always anoyed when there is something that I know that I don't know. And one of these cases is when i read physics papers around the internet and they mention lagrange and Hamilton's discoverments, developments, etz.
Can anyone explain me what these two physicists (mathematicians?) did? And why they are som important and used? Equations are very welcome.
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Thread: Lagrange and Hamilton
- 10-12-2005, 05:57 AM #1The Thinker
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Lagrange and Hamilton
- 10-12-2005, 07:40 AM #2Moderator
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Joseph Louis Lagrange-1736-1813ad Guille young man,
Hope that this assists you,I think the man you are looking for is
Joseph-Louis-Lagrange,born in Turin-1736-died in 1813,regarded as the
founder of classical-mechanics.did also much study into mathamatics.
to find more information Amigo,go to the search engine,Ask Jeeves.
His birth name was-Giueppe-Lodovico-Lagrangia.
kind regards, michael.
- 10-12-2005, 01:29 PM #3Raider of the lost time
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About William Rowan Hamilton, it is more informative to visit the site at
http://www.maths.tcd.ie/pub/HistMath/People/Hamilton/Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
- 10-12-2005, 01:43 PM #4The Thinker
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Very thanks Michael and Antonio.Now, what is the "lagrange identity"?
- 10-12-2005, 01:53 PM #5Raider of the lost time
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This identity is usually mentioned in advanced calculus textbooks under the topic on vector analysis.
Originally Posted by GUILLE
It is a vector product composed of dot products and cross products. Supposed, A, B, C, D are vectors then the cross of A and B dot cross of C and D is equal to the difference of A dot D B dot C minus A dot C B dot D.Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
- 10-12-2005, 01:57 PM #6The Thinker
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Actually, I don't have calculus in school. Where I read the lagrange identity was in the paper you send me on Hadamard Matrices, remember: "However, the expansions of Hi in Lagrange's identity gives to different topologies...". Originally Posted by AntonioLao
About your second paragraph, I'm not sure if I understand it, it's that in language it's much harder, can you give it to me in equations?
- 10-12-2005, 02:15 PM #7Raider of the lost time
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A x B ∙ C x D = (A ∙ D)(B ∙ C) – (A ∙ C)(B ∙ D) Originally Posted by GUILLE
The cross product is the same definition as vector product or outer product. The dot product is the same definition as scalar product or inner product. Note the scalar product of vectors produces scalar quantities and the vector product produces vector quantities. In tensor analysis, the scalar product reduces the rank of the tensors.Time independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
- 10-13-2005, 02:02 AM #8The Thinker
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Thanks.
What about the hamiltonian?
- 10-13-2005, 01:28 PM #9Raider of the lost time
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Since the latex typesetting in this web forum is malfunctioning, the best I can do is to copy the equation from the site at scienceworld.com or better yet pleasee visit the site at Originally Posted by GUILLE
http://scienceworld.wolfram.com/phys...miltonian.htmlTime independence: [∂E(g)]²=[∂F(a)×∂r(a)]·[∂F(b)×∂r(b)] and Mass independence: ¶a(t)·¶r(t)=c²
- 10-13-2005, 01:39 PM #10The Thinker
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Thanks for the link. In the link, in the "see also" list, it says "Kamiltonian" but the by clicking in it it leads to an empty entry, so, can you tell me what the Kamiltonian is? Originally Posted by AntonioLao
Also, what does actually the hamiltonian of a system determine in physical meanings?
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