2
$\begingroup$

See this excerpt from Kinematic and Dynamic Simulation of Multibody Systems page 122-123:

Consider a system characterized by a set of $n$ independent coordiunates $q_i$. Let $L=T-V$ be the system Lagrangian, where $T$ and $V$ are the kinetic and potential energy, respectively, and $W_{nc}$ is the work done by the non-conservative forces. Hamilton's principle (Hamilton (1834)) establishes that the motion of the system from time $t_1$ to time $t_2$, at which the motion is specified, is such that the integral action $$A=\int^{t2}_{t1}L\text{ }dt + \int^{t2}_{t1}W_{nc}\text{ }dt \tag{4.6}$$ has a stationary value for the correct path of the motion. This means that the variation of the action $A$ has to vanish: $$\delta A = \int^{t2}_{t1}\delta L\text{ }dt + \int^{t2}_{t1}\delta W_{nc}\text{ }dt=0\tag{4.7}$$

My goal is to understand why (4.7) holds.

After doing some research and asking some questions in the chat, it appears to me that (4.7) looks very similar to the Principle of least Action, but they don't match exactly.

Can someone please help me to understand how (4.7) is motivated? A derivation or a reference to some work that does so is highly appreciated.

$\endgroup$
1
$\begingroup$

Recall how work is defined via the non-conservative force:

$W_{nc}[x, \dot{x}, t]=\int_{\vec{x}(t_1)}^{\vec{x}(t)}\vec{F}_{nc}(x', \dot{x}', t)\cdot d\vec{x'}$

Requiring $\int_{t_1}^{t_2}(\delta L + \delta W_{nc})dt=0$ gives, by a completely analogous argument to ordinary Euler-Lagrange,

$(-\frac{\partial}{\partial\vec{x}} +\frac{d}{dt}\frac{\partial}{\partial\vec{\dot{x}}})(L+W_{nc})=0$

Restrict to the familiar case of $L=T-V$. You know that the "Euler-Lagrange operator" acting on L gives $m\ddot{\vec{x}}+\nabla V(x)$.

How do we find the corresponding expression for $W_{nc}$? First, note that its functional dependence on $x, \dot{x}, t$ is only contained in the upper limit of the integral that defined $W_{nc}$, and moreover the dependence is only on $x$. Thus,

$(-\frac{\partial}{\partial\vec{x}} +\frac{d}{dt}\frac{\partial}{\partial\vec{\dot{x}}})(W_{nc})=-\frac{\partial}{\partial\vec{x}}W_{nc}$

And by the fundamental theorem of calculus,

$$=-\vec{F}_{nc}$$

Putting it all together, our "least action argument" has resulted in

$m\ddot{\vec{x}}+\nabla V(x) -\vec{F}_{nc}=0$

And you can rearrange that to get Newton's 2nd law incorporating both conservative and nonconservative forces.

As for how you motivate this whole business? One, the addition to the action to incorporate nonconservative forces is a quantity that is dimensionally appropriate, and "makes sense" as something to construct out of a nonconservative force. Two, if you have a good intuition for how these types of proofs work out, you could have guessed that this is the correct thing to do. Of course, the concrete reason is that this works, and I just mathematically proved it.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm going to edit my post to tighten up the notation. But in the definition of the work, the integration variable $d\vec{x}$ is different from the limit of integration. The $x(t)$ appearing in the limit of integration is the thing we treat as the dynamical variable when we take functional derivatives. $\endgroup$ – BRSTCohomology Mar 25 at 20:37
  • 1
    $\begingroup$ To be more precise, what do we mean when we take the functional variation $\delta W_{nc}$? Obviously, to make this whole system work, we need to specify that the only dynamical variable (i.e. what we're differentiating w.r.t. in the Euler-Lagrange eqn) is the $x(t)$ in the limit of integration. Do anything else and you'll get the wrong answer. I agree, this whole affair is kind of uncomfortable: this is because by nature, $W_{nc}$ is not well-defined in the 2N-dimensional phase space $(\vec{x}, \vec{\dot{x}})$, unlike $L$, which is well-defined in that phase space. $\endgroup$ – BRSTCohomology Mar 25 at 20:51
0
$\begingroup$
  1. The work $W_{\rm nc}(t)$ at instant $t$ of non-conservative (nc) forces in eq. (4.6) is ill-defined. Given a path $t\mapsto q^j(t)$, it is unclear what work $W_{\rm nc}(t)$ refers to. This dooms a variational principle of some action functional for nc systems, in agreement with standard lore.

  2. On the other hand, the infinitesimal virtual work $$\delta W_{\rm nc}(t)~=~\sum_{j=1}^nQ^{\rm nc}_j(q(t),\dot{q},t)~\delta q^j(t)$$ in eq. (4.7) does actually make sense. Here $Q^{\rm nc}_j$ are the generalized nc forces and $\delta q^j$ are infinitesimal virtual displacements. The second equality in eq. (4.7) is just a time-integrated version of d'Alembert's principle, cf. OP's question.

References:

  1. J. Garcia de Jalón & E. Bayo, Kinematic and Dynamic Simulation of Multibody Systems: The Real-Time Challenge, 1993; p.122-123.
| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.