I'm trying to learn Lagrangian mechanics and have been reading a lot of articles on it. But many of the articles write the equations in different ways, probably for different purposes.

The Euler-Lagrange equation is:

$$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial{L}}{\partial{\dot{q_i}}}\right)-\frac{\partial{L}}{\partial{q_i}}=0$$

But then I've also seen a lot of other versions.

$$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial{T}}{\partial{\dot{x}}}\right)-\frac{\partial{T}}{\partial{x}}=Q$$

$$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial{L}}{\partial{\dot{q_i}}}\right)-\frac{\partial{L}}{\partial{q_i}}=Q_i$$

$$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial{L}}{\partial{\dot{q_i}}}\right)-\frac{\partial{L}}{\partial{q_i}}+\frac{\partial{P}}{\partial{\dot{q_i}}}=Q_i$$

My questions are as follows:

  1. In which cases do you only need to consider the kinetic energy as in the second equation?
  2. To include non-conservative forces as friction, do you just add the force as a generalized force in equation 3 or do you have to add a Rayleigh dissipation function as in equation 4?

1 Answer 1

  1. The first form is true whenever the Forces are derivable from a scalar, i.e when $Q_i=-\frac{\partial V}{\partial q_i}$

  2. The second equation however is true even when none of the forces can be derived from a scalar

  3. The third is true when some of the forces are derivable from a scalar and some are not, i.e. $L$ contains potential of the conservative forces and $Q_j$ represents forces not arising from the potential

  4. If the only non-conservative force is the Rayleigh dissipation function then $Q_i=-\frac{\partial P}{\partial \dot q_i}$ (in eq.3) if there are other/more non-conservative forces then eq.4 is valid


Goldstein: Classical Mechanics 3rd edition

  • $\begingroup$ Thanks, this is a very helpful answer. But could you elaborate on what you mean by 'forces that are derivable from a scalar' and 'forces not arising from the potential'? $\endgroup$
    – Kake_Fisk
    Commented Jun 19, 2016 at 11:33
  • $\begingroup$ @Kake_Fisk They mean the same, The scalar is the potential, Conservative forces obey the rule $\nabla \times F=0$ which implies that $F$ can be written as $F=-\nabla V$ , where $V$ is a potential which is a scalar $\endgroup$
    – Courage
    Commented Jun 19, 2016 at 13:16

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