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Let $f=f(q(t),\dot q(t),t)$, where $q(t)=\{q_1(t),...,q_N(t) \}=\{q_{a}\}_{a=1}^N$ and $\dot q:=\frac{dq}{dt}$. I want to show that if the following equations (Euler-Lagrange) are satisfied identically (independent of $q_a(t)$)

$$\frac{d}{dt}\frac{\partial f}{\partial \dot q_{a}}-\frac{\partial f}{\partial q_{a}}=0 \qquad(1)$$

then $f$ must have the following form

$$f(q(t),\dot q(t),t)=\frac{d \Lambda (q(t),t)}{dt} \qquad(2)$$

I read a book where they rewrite (1) as

$$\sum_{b=1}^N \left[ \frac{\partial^2f}{\partial \dot q_b \partial \dot q_a} \ddot q_b+\frac{\partial^2f}{\partial q_b \partial \dot q_a}\dot q_b \right]+\frac{\partial^2f}{\partial t \partial \dot q_a}=\frac{\partial f}{\partial q_a}\qquad(3)$$

Then the book says that, since we want (1) to be satisfied independent of $q_a$ then the coefficients of $\ddot q_a$ in (3) must be zero. So they find that $f$ must be a linear function of the $q_a$'s:

$$f(q,\dot q,t)=\sum_{b=1}^N A_b(q,t)\dot q_b+B(q,t)\qquad(4)$$

My question is: are they forgetting that the coefficients of the $\dot q_a$ must be zero as well?

The book inserts (4) into (1) and obtain a set of equations of the form

$$\sum_{b=1}^NC_{ab}\dot q_b+\frac{\partial A_a}{\partial t}=\frac{\partial B}{\partial q_a}\qquad(5)$$

Then they say again that since we want (1) to be satisfied independent of $q_a$ then the coefficients of $\dot q_a$ in (5) must be zero. And they find the functional form of $f$.

So my question can be restate as: Why do they have to do this step-by-step argument instead of making zero the coefficients of $\ddot q_a$ and $\dot q_a$ in (3) from the beginning?

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Here's my guess: none of the partials can generate a $\ddot{q}_b$ therefore, they must vanish, thus all of the $\frac{\partial^2f}{\partial \dot{q}_b \partial \dot{q}_a}$'s must vanish. However, both $\frac{\partial^2 f}{\partial t \partial \dot{q}_a}$ and $\frac{\partial f}{\partial q_a}$ can create $\dot{q}_b$'s (and $\dot{q}_a$) therefore it's not clear whether or not there is a way to write the function such that $\frac{\partial^2 f}{\partial q_b \partial \dot{q}_a} \neq 0$ (i.e. you cannot assume these are zero). In fact, I'm pretty sure it's not the case that that partial has to be $0$. So if you go through it, it's ugly, but here's what it looks like (keep in mind that $\Lambda$ has no dependence on $\dot{q}$).

\begin{align} f(q, \dot{q}, t) =& \frac{d\Lambda(q, t)}{dt} = \frac{\partial \Lambda}{\partial t} + \sum_1^N \frac{\partial \Lambda}{\partial q_b}\dot{q}_b & \qquad(A)\\ \frac{\partial f}{\partial \dot{q}_a} =& \frac{\partial \Lambda}{\partial q_a} &\qquad(B)\\ \frac{\partial^2f}{\partial q_b \partial \dot{q}_a} =& \frac{\partial^2\Lambda}{\partial q_b \partial q_a} \neq 0 &\qquad(C) \end{align}

Notice in equation (B), that most of the terms vanish because $\frac{\partial\Lambda}{\partial t}$ and all of the $\frac{\partial \Lambda}{\partial q_b}$'s have no dependence on any of the $\dot{q}$'s therefore $\frac{\partial^2 \Lambda}{\partial \dot{q}_i \partial t} = \frac{\partial^2 \Lambda}{\partial \dot{q}_i\partial q_j} = 0$. However, $\frac{\partial \Lambda}{\partial q_a}$ can be dependent on other $q_b$'s!

Just to go through the rest of the derivation, once you assume that $\frac{\partial^2 f}{\partial \dot{q}_b\partial \dot{q}_a} = 0$, this means that $\frac{\partial f}{\partial \dot{q}_a}$ must be independent of all of the $\dot{q}$'s which basically means they cannot be multiplied by each other, divided, put into functions, etc.--there can only be linear terms or no terms (since the linear terms will go away after the first partial). Which gives:

\begin{align} f(q, \dot{q}, t) =& B(q, t) + \sum_1^N A_b(q, t)\dot{q}_b & \qquad(D) \\ \frac{\partial f}{\partial q_a} =& \frac{\partial B}{\partial q_a} + \sum_1^N \frac{\partial A_b}{\partial q_a}\dot{q}_b & \qquad(E) \\ \frac{\partial f}{\partial \dot{q}_a} =& A_a(q, t) & \qquad(F)\\ \frac{d}{dt}\frac{\partial f}{\partial \dot{q}_a} =& \frac{\partial A_a}{\partial t} + \sum_1^N \frac{\partial A_a}{\partial q_b} \dot{q}_b & \qquad(G) \end{align}

Now the only way to generate the $\dot{q}$'s is with those summed partials, so they must be zero:

\begin{align} 0=&\sum_1^N\left(\frac{\partial A_a}{\partial q_b} - \frac{\partial A_b}{\partial q_a}\right)\dot{q}_b & \qquad(H) \\ \frac{\partial A_a}{\partial q_b} =& \frac{\partial A_b}{\partial q_a} & \qquad(I) \\ \frac{\partial B}{\partial q_a} =& \frac{\partial A_a}{\partial t} & \qquad(J) \end{align}

Other than through inspection, I can't come up with a great reason as to how they found the functional form. So here's something you should note (hopefully you remember this about partials of "well-behaved" functions):

$$ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} $$

So when you see two partials of something set equal, one way to enforce equality is to make the thing you are taking the partial of a partial derivative itself:

$$ \frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial y} \rightarrow f_1 = \frac{\partial F}{\partial y}, f_2 = \frac{\partial F}{\partial x} $$

Using that this would give:

$$ \frac{\partial A_a}{\partial q_b} = \frac{\partial A_b}{\partial q_a} \rightarrow A_a = \frac{\partial \Lambda}{\partial q_a}, A_b = \frac{\partial \Lambda}{\partial q_b} \\ \frac{\partial B}{\partial q_a} = \frac{\partial A_a}{\partial t} \rightarrow B = \frac{\partial \Gamma}{\partial t}, A_a = \frac{\partial \Gamma}{\partial q_a} $$

But we already said that $A_a = \frac{\partial \Lambda}{\partial q_a}$, so clearly $\Gamma = \Lambda$ and we finally arrive at:

\begin{align} f(q, \dot{q}, t) =& \frac{\partial \Lambda}{\partial t} + \sum_1^N \frac{\partial \Lambda}{\partial q_b}\dot{q}_b = \frac{d \Lambda(q, t)}{dt} & \qquad(K) \end{align}

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  • $\begingroup$ OK. I just have a question. What do you mean with "generate"? "none of the partials can generate a $\ddot q_b$" $\endgroup$ – Ana S. H. Mar 28 '14 at 9:16
  • $\begingroup$ Wait! Are you saying that due to the RHS of my (3) equation is a function of $(q,\dot q, t)$ then the coefficients of $\ddot q_b$ in the LHS of (3) must be zero? (i.e. we have to have a function of $(q,\dot q, t)$ in the LHS as well). And the same argument for my (5) equation? $\endgroup$ – Ana S. H. Mar 28 '14 at 9:19
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    $\begingroup$ Yes, the RHS of eq. (3) is a function of $(q,\dot{q}, t)$, but also $\frac{\partial^2 f}{\partial t \partial \dot{q}_a}$ and $\frac{\partial^2 f}{\partial q_b \partial \dot{q}_a} \dot{q}_b$. So the only place where $\ddot{q}$ appears is in that single term (which we say the coefficients of must be zero). Then, yes the argument is similar for eq. (5), except now the RHS and $\frac{\partial A_a}{\partial t}$ are functions of only $(q, t)$ and now only the $\sum C_{ab} \dot{q}_b$ contributes $\dot{q}$ terms and thus those coefficients (which are a difference of two partials) must be zero. $\endgroup$ – Jared Mar 28 '14 at 12:13

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