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

Confusion with Virtual Displacement

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:

Mathematics of the Virtual Displacement

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.

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  • $\begingroup$ 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. $\endgroup$
    – user196418
    Commented Jan 28, 2019 at 1:15
  • $\begingroup$ 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. $\endgroup$
    – J_Psi
    Commented Jan 28, 2019 at 2:18
  • $\begingroup$ 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. $\endgroup$
    – user196418
    Commented Jan 28, 2019 at 11:56
  • $\begingroup$ Writing this out in more detail as an answer could be beneficial. $\endgroup$
    – J_Psi
    Commented Jan 28, 2019 at 21:28

1 Answer 1

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Before you talk about dq or dr you need to talk about q and r.

These are coordinates on the manifolds. A typical use of the mechanism you have cited is found in expressing differential elements in curvilinear coordinates. The {q} are a set of independent parameters and {r} the usual Cartesian coordinates in R^(N). It should be noted, because I have seen a lot of confusion on this point, that r is NOT a vector, and likewise {q} is not a vector in any space. These are coordinates and are to be thought of as a set of scalar functions {r_i}, i = 1...N. When dealing with Euclidean space it is easy to develop this confusion because the full Manifold can in fact be covered with a single copy of R^(N) so much of what we learn in differential geometry is redundant when applied to Euclidean space (or superfluous).

When dealing with an arbitrary manifold, M, one maps open sets in R^(N) into open sets in M and then uses the standard coordinate in R^(N) to label points in M. This labeling does not have the structure of a vector space. Also, one cannot in general cover M with a single open set of R^(N). While {r_i} is not a vector in M, differential displacements {dr_i} form a vector space. They obey the axioms of a vector space. Now, to make a differential displacement one needs a base point. On a Manifold, M, we build this construction at each point of M. The space made of all differential displacements at the point p is called the dual tangent space at p, T*(M, p) is one notation for this. I say dual because the vector space at p is defined by the gradient of a scalar function on M evaluated at p, {df/dx_i}. These quantities obey the general coordinate transform for the components of a vector in a vector space. The tangent space of M at p is denoted T(M, p). Using the standard coordinates from the mappings of open sets of R^(N) into M creates what we can a coordinate basis for T and T*. We have a complete set of linearly independent elements generated by the {dx_i} and the {df/dx_i}. Again, each point of M has its own tangent space.

One can only do vector algebra operations on vectors in the spaces at the same point. There is a one to one correspondence between elements of T and T* at each point of M. The linear map between the two spaces is defined in terms of the metric tensor associated with M, g_ij. With this one can define inner products (dot product generalization) as g_ij * V^i * U^j, sum over i and j, between two elements of T(M, p) or T*(M, p). Typically, once we make the distinction between these two spaces we index elements of of one space with a lower index (subscript) and the other with an upper index (superscript). This is equivalent to taking v_i * U^j, v_i = g_ik * V^k. In other words we can project an element of T onto an element of T*. If the underlying manifold is "flat", i.e. Euclidean with the typical metric = diag(1, 1, ..., 1) then T and T* are identical copies of each other, one cannot really decipher the difference between the two except as a formality.

Now, in your case you state that the {q} are coordinates in the constraint manifold (or constrained manifold). I can relate this to looking a systems in mechanics and mechanical engineering like a linkage, where there are objects connected by frictionless ideal joints that act as holonomic constraints. As a simple example, a spherical pendulum would have a coordinate in R^(3), {x, y, z} but is constrained by r = constant. Here the {q} would be {theta, phi}. Your dq would be {d(theta), d(phi)}. The Cartesian coordinates {x, y, z} are scalar functions of theta and phi. Hence don't think of them as vectors in R^3 but as as set of scalar functions on S^2 (the 2-sphere). Hence the derivatives, dx/d(theta) etc are elements of T(S^2, p), and dq are elements of T*(S^2, p). Note that the indices line up properly, you are summing over the index of the {q} coordinates, for each r_i. the d(r_i) on the l.h.s is also a dual vector. Think of the dq as a basis for T*(S^2, p) and the partials of {r_i} as coefficients of d(r_i) in that basis.

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  • $\begingroup$ Consider reformatting the answer using MathJax, as it hard to read your notation in plain text. $\endgroup$
    – J_Psi
    Commented Jan 30, 2019 at 0:09

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