# D'Alembert's Principle: Where does $-Q_j$ come from?

This is a follow-up question to D'Alembert's Principle and the term containing the reversed effective force.

From the second term of Eq. (1.45) \begin{align*} \sum_i{\dot{\mathbf{p}}_i \cdot \delta{\mathbf{r}_i}} &= \sum_i{m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j} \end{align*}

Starting from Eq. (1.50), I was able to follow that \begin{align*} \sum_i{m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j}} &= \sum_i{\left[ \frac{d}{dt} \left( m_i\mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial\dot{q}_j} \right) - m_i\mathbf{v}_i \cdot \frac{\partial\mathbf{v}_i}{\partial q_j} \right]} \end{align*}

Goldstein substituted the above equation to (1.45) by saying:

... and the second term on the left-hand side of Eq. (1.45) can be expanded to

By "second term", I understood it to be the very first equation I mentioned above. Therefore this is how I understood it:

\begin{align*} \sum_i{\dot{\mathbf{p}}_i \cdot \delta{\mathbf{r}_i}} &= \sum_i{m_i\ddot{\mathbf{r}}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j}\\ &= \sum_i\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) - Q_j \right]\delta q_j\\ &= \sum_i{\left[ \frac{d}{dt} \left( \frac{\partial T}{\partial\dot{q}_j} \right) - \frac{\partial T}{\partial q_j} -Q_j \right] \delta q_j} \end{align*}

I was able to follow \begin{align*} T &= \sum_i{m_i\mathbf{v}_i} \cdot \partial\mathbf{v}_i\\ T &=\frac{1}{2} \sum_i{m_iv_i^2} \end{align*}

But I am at a loss: Where does $-Q_j$ come from?

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There is an error in the question formulation (v1): The second equality in the third equation is not correct. –  Qmechanic Jul 17 '11 at 14:11

Similar to Newton's 2nd law, the D'Alembert's principle has both a dynamical and a kinetic term,

$$\sum_i (\mathbf{F}^{(a)}_i - \dot{\mathbf{p}}_i) \cdot \delta \mathbf{r}_i~=~0. \qquad (1.45)$$

On one hand, the dynamical term

$$\sum_i \mathbf{F}^{(a)}_i \cdot \delta \mathbf{r}_i = \sum_j Q_j \delta q_j \qquad (1.48)$$

contains the generalized force

$$Q_j=\sum_i\mathbf{F}^{(a)}_i\cdot \frac{\partial \mathbf{r}_i}{\partial q_j}.\qquad (1.49)$$

On the other hand, the kinetic term

$$\dot{\mathbf{p}}_i \cdot \delta \mathbf{r}_i~=~ \sum_j{\left[ \frac{d}{dt} \left( \frac{\partial T}{\partial\dot{q}_j} \right) - \frac{\partial T}{\partial q_j} \right] \delta q_j}$$

contains the kinetic energy $T =\frac{1}{2} \sum_i{m_iv_i^2}$.

Edit: It is true that the third edition of Goldstein wrongly says

and the second term on the left-hand side of Eq. (1.45) can be expanded into $$\sum_i\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) - Q_j \right]\delta q_j. \qquad$$

The second edition does not have the $-Q_j$ term, so an unfortunate mistake was introduced during the update to the third edition. This is not the first time that I have noticed that the second edition is often more carefully written than the third edition in what concerns old material. (The third edition contains a new chapter 11 about classical chaos.)

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Goldstein isn't wrong. He's multiplied 1.45 by -1 to keep + in front of the $\mathbf{\dot p}_i\cdot\mathbf{\delta r}_i$ –  Larry Harson Jul 19 '11 at 14:46

You've misinterpreted what Goldstein states:

and the second term on the left-hand side of Eq. (1.45) can be expanded into

To save confusion for some, it could be better expressed as:

and the second term on the left-hand side of Eq. (1.45) can be expanded, so that Eq. (1.45) becomes>

It looks repetitive, so the authors probably stuck with the current form, relying on the understanding of the student to see what is obviously meant. The relevant equations are

$$\sum_i (\mathbf{F}^{(a)}_i - \dot{\mathbf{p}}_i) \cdot \delta \mathbf{r}_i~=~0. \qquad (1.45)$$

$$\sum_i{\mathbf{F}_i \cdot \delta{\mathbf{r}_i}} = \sum_{i,j}{\mathbf{F}_i \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j = \sum_{j}{Q}_j \delta q_j}\qquad (1.48)$$

The second term in equation (1.45) is therefore expanded as in (1.48) and then combined with the expansion for the first term $\mathbf{\dot p}_i\cdot\delta\mathbf{r}_i$ elsewhere , to give $$\sum_j\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \sum_i{\frac{1}{2}m_iv_i^2} \right) - Q_j \right]\delta q_j$$

Note (1.45) is multiplied by -1 so the signs in the expression derived elsewhere for $\mathbf{\dot p}_i\cdot\delta\mathbf{r}_i$ remain the same, hence the $-Q_j$ term rather than $Q_j$. The outer summation is over $j$ and not $i$ as you've put down.

There's a link for correctons to this book if you're convinced you're right, http://astro.physics.sc.edu/goldstein/ although in this case it looks as if it's a misinterpretation on your part.

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Please explain why you down voted so i can correct my answer –  Larry Harson Jul 18 '11 at 22:00
\begin{align*} \sum_i{ (\dot{\mathbf{p}}_i-\mathbf{F}_i) \cdot \delta{\mathbf{r}_i}} &= \sum_{i,j}{(m_i\ddot{\mathbf{r}}_i-\mathbf{F}_i) \cdot \frac{\partial\mathbf{r}_i}{\partial q_j} \delta q_j}\\ &= \sum_{i,j}\left[ \frac{d}{dt}\left[ \frac{\partial}{\partial\dot{q}_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) \right] - \frac{\partial}{\partial q_j} \left( \frac{1}{2}\sum_i{m_iv_i^2} \right) - Q_j \right]\delta q_j\\ &= \sum_{i,j}{\left[ \frac{d}{dt} \left( \frac{\partial T}{\partial\dot{q}_j} \right) - \frac{\partial T}{\partial q_j} -Q_j \right] \delta q_j} \end{align*}
I'm using the 3rd ed. There must have been a mistake in the textbook. Equation (1.45) is $\sum_i{\left( \mathbf{F}_i^{(a)} - \dot{\mathbf{p}}_i \right) \cdot \delta\mathbf{r}_i} = 0$. Therefore, the "second term", as I understood it, is $\sum_i{\dot{\mathbf{p}}_i \cdot \delta\mathbf{r}_i}$. After working a bit more, I remembered that $\mathbf{F}_i^{(a)} \cdot \delta\mathbf{r}_i = Q_j\delta q_j$. That's when I realized that the $Q_j$ had something to do with $\mathbf{F}_i^{(a)}$. Therefore, it is a mistake by saying "...the second term expands to..." and including $-Q_j$. Am I right? –  Kit Jul 17 '11 at 14:36