# Meaning and Origin of an Expression which Involves Virtual Displacement

As an additional point of confusion related to the answer given here:

I have encountered the following expression in my study of virtual displacements.

$$\delta{\vec{r_i}}=\sum_j\frac{\partial\vec{r_i}}{\partial{q_j}}\delta{q_j}.$$

It is difficult for me to even parse this statement. On the left hand side we have an infinitesimal variation at $$\vec{r_i}$$ (which I now understand to be a tangent vector to the position manifold $$M$$ at that point) and it is being equated to the sum on the right. The partial derivative in the sum seems easy enough to understand; I assume that it denotes the usual component-wise partial derivative of $$\vec{r_i}$$ with respect to the coordinate $$q_j$$. The derivative then seems to be multiplied by an infinitesimal virtual displacement (which I now understand to be a tangent vector to the constraint submanifold $$C$$). The sum is obviously over all of the coordinates.

My confusion, then, is two-fold. First, what kind of multiplication is being carried out between $$\delta{q_j}$$ and $$\frac{\partial\vec{r_i}}{\partial{q_j}}$$? They don't even seem to be in the same spaces. Even if we took the image of $$\delta{q_j}$$ under some kind of inclusion map, it still isn't clear what operation we are performing. Is it the cross-product, the inner product, or something else?

Lastly, it is unclear to me where this expression even came from. In class, my professor practically pulled it out of a hat with the statement that "it is easy to see". Well, I fail to see how this is in any way obvious. In CuriousMind's comment here:

It is mentioned that this is actually the definition of an infinitesimal virtual displacement. However, I fail to see how this "definition" corresponds to my understanding that an infinitesimal virtual displacement is a tangent vector to a manifold. Any answer or clarification will be much appreciated.

• The derivative with respect to q does exist in the tangent space of he manifold defined by the {q}. So, they are dual to dq. – ggcg Jan 28 '19 at 1:15
• I'm not sure that I understand, can you elaborate? It's not clear to me that the derivative with respect to $q_j$ is in the tangent space. Saying that it is in the dual space of the tangent space doesn't help that case much, either. – J_Psi Jan 28 '19 at 2:18
• I am merely stating that, by definition, {dq} and {d/dq} are dual to each other. If that isn't meaningful then look up a diff geom text. The summation in your equation always makes sense. – ggcg Jan 28 '19 at 11:56
• Writing this out in more detail as an answer could be beneficial. – J_Psi Jan 28 '19 at 21:28