# Commutativity of Differential Operators in Lagrangian Mechanics

1. Feb 1, 2010

### campo133

Hello.

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

$$\frac{d}{dt} \left( \frac{\partial \vec{r}}{\partial q_i} \right) = \frac{\partial}{\partial q_i} \left( \frac{d \vec{r}}{dt} \right)$$

Why is this so?

2. Feb 2, 2010

### torquil

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.

Torquil

3. Feb 2, 2010

### campo133

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:

$$\frac{d}{dt} \left( \frac{\partial \vec{r}}{\partial q_i} \right) \Rightarrow \frac{\partial^2 \vec{r}}{\partial q_i^2} \dot{q_i}$$

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

since $$q_i = q_i(t)$$.

This shows both sides are equal, and the differentiation commutes.

Last edited: Feb 2, 2010
4. Feb 3, 2010

### turin

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.

5. Feb 4, 2010

### torquil

Campo133, I think you forgot that $$\vec{r}$$ is a function of all the $$q$$'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 $$q_i$$ differentiation doesn't act upon $$\dot{q}_j$$ on the right hand side, even for the term $$j=i$$, because those are treated as inpedentent variables in the Lagrangian formalism.

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

Torquil

6. Feb 5, 2010

### turin

Isn't that exactly what Campo did here:
I'm confused.

7. Feb 5, 2010

### torquil

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 :-)

Torquil

8. Feb 5, 2010

### campo133

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.

9. Feb 5, 2010

### turin

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.

10. Feb 5, 2010

### torquil

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 $$j \neq i$$, e.g. terms where r is differentiated with respect to e.g. both q_1 and q_2.

Torquil

11. Feb 5, 2010

### turin

I totally missed that! Thanks, torquil.