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.


1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.