Comparing Lagrange's Equation of Motion and Euler-Lagrange Equations

[Niles]

Hi

What is the difference between Lagrange's equation of motion and the Euler-Lagrange equations? Don't they both yield the path which minimizes the action S?


Niles.

 

[vanhees71]

The Lagrange Equations of 2nd kind are the Euler-Lagrange equations of the Hamilton least-action principle in its Lagrangian formulation. In the Hamiltonian formulation you get the equivalent Hamilton Canonical Equations of Motion for configuration and conjugate momentum variables.

The Lagrange Equations of 1st kind are the Euler-Lagrange equations of the Hamilton least-action principle under constraints.

 

[Niles]

Just to be absolutely sure, then by the Lagrange Equations of 2nd kind you mean

$$\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial v_i }}\left( {\gamma (t),\mathop \gamma \limits^. (t),t} \right)} \right) - \frac{{\partial L}}{{\partial q_i }}(\gamma (t),\mathop \gamma \limits^. (t),t) = 0$$

where the corresponding Hamilton's equations are$$\begin{array}{l}<br /> \mathop q\limits^. _i (t) = \frac{{\partial H}}{{\partial p_i }}(q(t),p(t),t) \\ <br /> \mathop p\limits^. _i (t) = - \frac{{\partial H}}{{\partial p_i }}(q(t),p(t),t)<br /> \end{array}$$

?

 

[vanhees71]

Yep, but the last Eq. should read

$$\dot{p}_i=-\frac{\partial H}{\partial q^i}.$$

 

[Niles]

Great -- and for clarity, then by the Lagrange Eqations of the 1st kind you mean

$$\frac{d}{{dt}}\left( {\partial _{\mathop x\limits^. } L} \right) - \partial _x L = 0$$

?Niles.

 

[vanhees71]

No, these are the Lagrange equations of 2nd kind. Those of 1st kind are usually formulated in Cartesian coordinates taking into account constraints with help of Lagrange multipliers. These introduce the forces needed to fulfill the constraints explicitly into the equations of motion.

 

[Niles]

Ahh, I see. In that case I believe I have deciphered what my book means: The first EL-equations I wrote are for fields, where the ones I wrote in #5 are in 1D.