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Let $S$ be a system with $g$ degrees of freedom, $\Sigma_g$ its configuration space and $\{q_i\}_{i=1}^g$ the lagrangian coordinates. If $P\in\Sigma_g$, we define a virtual displacement from $P$ as follows (using Einstein's convention)

\begin{equation} \delta P=\frac{\partial P}{\partial q_i}\delta q_i. \end{equation}

As far as I understand $\left\{\frac{\partial P}{\partial q_i}\right\}_{i=1}^g$ is a basis for the tangent space in $P$, i.e. $T_P\left(\Sigma_g\right)$.

What I don't understand is the definition of virtual work for a force $\boldsymbol{F}$:

\begin{equation} \delta W=\boldsymbol{F}\cdot\delta P. \end{equation}

$\boldsymbol{F}$ is a vector of ordinary space, while $\delta P$ is a vector of tangent space (that has dimension $g$). If the definition I wrote above are correct, why is dot product possible?

Please note that I am taking an undergrad course in Classical Mechanics and I've never taken any courses in differential geometry.

Qmechanic
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Mr. Feynman
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  • Related: https://physics.stackexchange.com/q/203600/2451 , https://physics.stackexchange.com/q/79533/2451 and links therein. – Qmechanic Dec 05 '20 at 11:58
  • I may be wrong, but I think this has something to do with evaluation of $F$ and $\delta P$ at a point (as opposed to evaluation at multiple points) within the base manifold. – PrawwarP Dec 05 '20 at 12:09

1 Answers1

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Because the $\delta P$ is a vector too: the virtual displacement vector that always lies in the tangent space.

Kksen
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