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.


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.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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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