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Commutativity of Differential Operators in Lagrangian Mechanics

  1. Feb 1, 2010 #1

    I am having trouble realizing the following relation holds in Lagrangian Mechanics. It is used frequently in the derivation of the Euler-Lagrange equation but it is never elaborated on fully. I have looked at Goldstein, Hand and Finch, Landau, and Wikipedia and I still can't reason this. Could anybody elaborate or provide a proof? Thanks!

    Let [tex]\vec{r} = \vec{r} \left( q_1, q_2, ..., q_i \right)[/tex] be the position vector of a particle where [tex]q_1, q_2, ..., q_i[/tex] are the respective generalized coordinates and. Each [tex]q_i = q_i(t)[/tex], that is each coordinate is a function of time. In all derivations of the Euler-Lagrange equation, I see the following:

    [tex]\frac{d}{dt} \left( \frac{\partial \vec{r}}{\partial q_i} \right) = \frac{\partial}{\partial q_i} \left( \frac{d \vec{r}}{dt} \right) [/tex]

    Why is this so?
  2. jcsd
  3. Feb 2, 2010 #2
    It follows from the commutativity of differentiation with respect to different q's. Use the chain rule to write out the total time derivative explicitly as a sum of partial derivatives of r with respect to all the q's, each multiplied by the time derivative of the q-variable.

    Then both sides differ only in the order of differentiation with respect to the q's. If the function r is sufficiently regular it will therefore be true.

  4. Feb 2, 2010 #3
    I ended up working it out, quite simply really. I was getting held up with q dot, as I was treating it as a function of qs instead of strictly t.

    Working out the left side using the chain rule you get:

    [tex]\frac{d}{dt} \left( \frac{\partial \vec{r}}{\partial q_i} \right) \Rightarrow \frac{\partial^2 \vec{r}}{\partial q_i^2} \dot{q_i}[/tex]

    Working the right side out using the chain rule also give:
    [tex]\frac{\partial}{\partial q_i} \left( \frac{\partial \vec{r}}{\partial q_i} \dot{q_i} \right) \Rightarrow \frac{\partial^2 \vec{r}}{\partial q_i^2} \dot{q_i} [/tex]

    since [tex] q_i = q_i(t) [/tex].

    This shows both sides are equal, and the differentiation commutes.
    Last edited: Feb 2, 2010
  5. Feb 3, 2010 #4


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    campo, thanks for posting this. I remember also being confused by that, but I had forgotten how I convinced myself. Your explanation is very clear and comforting.
  6. Feb 4, 2010 #5
    Campo133, I think you forgot that [tex]\vec{r}[/tex] is a function of all the [tex]q[/tex]'s:

    \frac{d}{dt} \left( \frac{\partial \vec{r}}{\partial q_i} \right) =
    \sum_j \dot{q_j} \frac{\partial}{\partial q_j} \frac{\partial \vec{r}}{\partial q_i}

    On the other hand:

    \frac{\partial}{\partial q_i} \left( \frac{d \vec{r}}{dt} \right) =
    \frac{\partial}{\partial q_i}\sum_j \frac{\partial \vec{r}}{\partial q_j} \dot{q_j}=
    \sum_j \frac{\partial}{\partial q_i}\frac{\partial \vec{r}}{\partial q_j} \dot{q_j}

    Notice that the [tex]q_i[/tex] differentiation doesn't act upon [tex]\dot{q}_j[/tex] on the right hand side, even for the term [tex]j=i[/tex], because those are treated as inpedentent variables in the Lagrangian formalism.

    The difference between these two expressions is the order of the [tex]q[/tex]-differentiation. If [tex]\vec{r}[/tex] is sufficiently regular as a function of the [tex]q[/tex]'s, then this doesn't matter. This is what I expressed with words in my first reply.

  7. Feb 5, 2010 #6


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    Isn't that exactly what Campo did here:
    I'm confused.
  8. Feb 5, 2010 #7
    Well, when you do the time derivative on the first expression, you need to time-differentiate "through" all the different q_j-variables, not only q_i. The vector r was already differentiated once with respect to q_i (for a given i). So the result should be what I wrote in my post, right, where i is fixed, and there is a sum over j? My expression is not the same as campo's, so only one of us can be right :-)

  9. Feb 5, 2010 #8
    Yes torquil is correct in the most general sense. My explanation only works if r is a function of qi. Typically, r has many variables, so you need to sum over all of them. In my case, r just had one.
  10. Feb 5, 2010 #9


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    Oh, I thought that you were using an implicit summation convention for repeated indices. People don't do that anymore? I thought that it was standard practice.
  11. Feb 5, 2010 #10
    No they still use that convention, but it wouldn't be the same as my expression anyway. If I were to interpret it as a sum over the index i, the expression would still not contain any "cross-terms", i.e. the terms in my sum that have [tex]j \neq i[/tex], e.g. terms where r is differentiated with respect to e.g. both q_1 and q_2.

  12. Feb 5, 2010 #11


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    I totally missed that! Thanks, torquil.
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