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I'm learning Lagrangian mechanics, and more than a few times you find expressions of the form $$-\frac{\partial{Z}}{\partial{q_j}} + \frac{\mathrm d}{\mathrm dt}\left(\frac{\partial{Z}}{\partial{\dot{q_j}}}\right)=0.$$ Obviously it comes up where $Z=L$, and $Z=Q$ where it gives the components of generalised force, but I have trouble intuiting its meaning. I can obviously understand it term by term, but I don't fully understand how to interpret the whole.

The first part can be seen almost as being like $-\nabla Z$ except the derivative is with respect to state-space, not space proper, but the second half is conceptually mysterious to me. Can anyone give a good explanation of what this means in general?

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Intuitively, the second term is just the time derivative of the momentum.

For simple point mass in some potential, we have $\mathscr{L}=\frac{1}{2}m v^2-V(x)$. Then, the Euler-Lagrange equation becomes $$0=-\frac{\partial V}{\partial x} -\frac{\text{d}}{\text{d} t}\left(\frac{\partial }{\partial v}\frac{1}{2}mv^2\right)\,.$$ The first term you have already identified with the force as gradient of the potential. The second term becomes $$\frac{\text{d}}{\text{d} t}\left(mv\right)=\dot{p}\,.$$ In other words, we have $\dot{\vec p}=\vec F$, which is just Newton's second axiom.

For more complicated situations, e.g. motion with constraints, your second term is still interpreted as a "generalised momentum" conjugate to the respective coordinate, and the first term is the "generalised force".

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  • $\begingroup$ I was looking for how to interpret it for a generic function, not the lagrangian. While this is useful, it's not really what I was asking for. $\endgroup$ – user6873235 May 16 '18 at 1:05
  • $\begingroup$ Well, for a general function $Z\!\left(q_i(t),\dot{q}_i(t),t\right)$, it's just a derivative with respect to some variables which is then further differentiated with respect to time. The equation is derived by extremising some functional, the time integral of $Z$. It's not obvious to me what the intuition would be about that if you don't have some intuition about $Z$ in the first place. Do you assume that $q$ and $\dot q$ are (generalised) positions and momenta? What do you assume about your functional? Are you looking more for a physical or mathematical intuition? $\endgroup$ – Toffomat May 16 '18 at 7:59
  • $\begingroup$ A physical intuition, although not just classical mechanics. I intuit $q$ as effectively the state, and $\dot{q}$ as rate of change of state. I'm interested in it from a physics perspective, but not just a classical perspective. Given how much of modern physics is built on Lagrangians as first class entities, defined basically as any function which satisfies the euler-lagrange equation, with energy/momentum/angular momentum derived from them, it seems important to intuit the structure of the equation, much like understanding 4-divergence and its relation to conservation laws for any function. $\endgroup$ – user6873235 May 16 '18 at 14:13
  • $\begingroup$ The state really is given by both $q$ and $\dot q$ (in the usual cases), and so $\ddot q$ is more like the rate of change. For a function $Z(q,\dot q)$, the second term contains $q$, $\dot q$ and $\ddot q$, so you can think of that as a combination of accelerations, i.e. the change in the state. Then the Euler-Lagrange equations give a relation between these particular combinations of rate-of-change variables and some kind of generalised force. Is that clearer? $\endgroup$ – Toffomat May 17 '18 at 6:47
  • $\begingroup$ Is there a natural a-priori way of defining $q$? If the "state" of an arbitrary physical system is better defined by $q$ and $\dot{q}$, how do we distinguish between the two? Obviously in classical mechanics its easy to define in terms of concrete positions, but what about quantum mechanics or electromagnetism or other field theories? $\endgroup$ – user6873235 May 17 '18 at 9:05
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We can construct the Euler-Lagrange operator as follows:

$$ \bigg[\frac{d}{dt}\bigg(\frac{\partial}{\partial \dot q^i}\bigg) - \frac{\partial }{\partial q^i }\bigg] L = 0 $$

As you mention in your question, the second portion can be thought of as the force. We can recover this conceptually if we let $L = T(\dot q) - U(q)$ and hence: $$ -F_i = \frac{\partial U }{\partial q^i } $$ We then have: $$ \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot q^i}\bigg) - F_i = 0 $$ If we define the bracketed term as a momentum, $p = \partial L / \partial \dot q$, we recover Newton's second law: $$ \dot p^i = F_i $$ Therefore, we can interpret the first portion as the rate of change of momentum and the second portion as the force, if we must insist on paralleling the Newtonian formulation.

In addition, the Euler-Lagrange equation is a second order ODE in $q^i$ rather than a PDE.

Edit

In light of your comment, the first portion of Euler-Lagrange operator on some function $Z(q,\dot q)$ defines the rate of change of the conjugate momentum to $q^i$, where the definition of the momentum is: $$ p_i = \frac{\partial Z}{\partial \dot q^i} $$ It happens, that this then satisfies the Legendre transform relationship, linking $p_i$ and $\dot q^i$ as conjugate variables to a phase space description. As mentioned above and by @Toffomat, the rate of change of the conjugate momentum is by definition the generalised force $F(q,\dot q)$. Whether this coincides with the classical notion of force depends on whether $Z=L$.

In another framework, one cannot abstract further than this for a given $Z$. It is unclear, without further specification of $Z, q $ and $p$, how to interpret what the generalised momentum/force actually is.

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  • $\begingroup$ That's not really what I asked, I wanted to know how to interpret it for a general function, not just the Lagrangian. For instance when applied to T, you get generalised forces, so it can clearly be applied to more than just the Lagrangian function. $\endgroup$ – user6873235 May 15 '18 at 23:09

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