Deriving Lagrange's & Hamilton's Equations in 1-Dimension

[Ed Quanta]

Can someone direct me towards or provide me with a derivation of Lagrange's equations in one dimension? Are Hamilton's equations derived in a similar manner?

 

[StatMechGuy]

I think you could probably find both these things in any reasonable textbook on classical mechanics. Goldstein, Marion & Thornton, Corben & Stehle, Landau, any of those.

 

[Chris Hillman]

Landau and Lifschitz

I think you could probably find both these things in any reasonable textbook on classical mechanics. Goldstein, Marion & Thornton, Corben & Stehle, Landau, any of those.

I was just going to suggest Landau and Lifschitz, Mechanics, Pergamon, 1976. This classic textbook is short and sweet, highly readable, has many excellent problems, and is often cited and should be widely available e.g. via amazon.com

 

[Mindscrape]

Yeah, an eighth of the book is homage to Landau.

 

[vanesch]

Yes, Landau is brilliant ! Another book, for the more mathematically inclined, is Arnold.

 

[Marco_84]

i suggest you also "Mathematical methods for physics" Vladimir Arnold (i think this is the english translation.).
I think is the best because of its wide mathematical explanations. very good appendix.
To derive Lagrange's or hamilton's equation you just need the least action principle, or better the principle of stationary action.
applying this principle you get the right equations of motion even for fileds theory.
then if the hessian determinant of the Lagrangian or Hamiltonian is different form zero you can connect them via Legendre trasformation...
For a particle:

remeber that $$L=L(q,\dot{q})$$
THE PRINCIPLE STATES THAT:

$$\delta\int L dt=\int\delta L dt=0$$

$$\int\frac{\partial L}{\partial q}\delta q+ \frac{\partial L}{\partial \dot{q}}\delta\dot{q} dt=0$$

now integrating by parts the second term of this integral and making the assumption that the little variations

$$\deltaq$$

vanishes at the boundary. you get:

$$\int(\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}})\deltaq dt=0$$

so if it is zero for arbitrary $$\delta q$$ it must be:

$$\frac{\partial L}{\partial q} - \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=0$$

wich is the Eulero Lagrange equation for a single particle with a single degree of freedom. The equations for three degree of freedom are soon obtained using indices, the same for a system of particles.

bye MArco

 

[Marco_84]

I hope u can read i don't know what's going with latex,
i don't know if i understood right how to put symbols inside thge forum.

bye bye :-)

 

[Reshma]

You forgot to add a space after the code. Click on this symbol for the code.
$$\delta q$$