# How is virtual work defined?

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)

$$$$\delta P=\frac{\partial P}{\partial q_i}\delta q_i.$$$$

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 $$\boldsymbol{F}$$:

$$$$\delta W=\boldsymbol{F}\cdot\delta P.$$$$

$$\boldsymbol{F}$$ 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.

• Related: physics.stackexchange.com/q/203600/2451 , physics.stackexchange.com/q/79533/2451 and links therein. Commented Dec 5, 2020 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. Commented Dec 5, 2020 at 12:09

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