Vector spaces and bundles in classical mechanics My understanding so far is that if we have a manifold of coordinates, the velocity field is more generally called a tangent space $T_xQ$, further the space described by position and momentum is a co-tangent space $T_x^*Q$. The union of all the tangent spaces is itself another manifold: the tangent bundle $TQ={T_xQ|x\in Q}$ and the same for the co tangent space. 
While I can comprehend the tangent spaces to a manifold .. What is the difference between the tangent space and the tangent bundle in mechanics? 
Is it like taking the gradient at each point of the manifold and "smooshing" into another manifold itself? 
 A: The tangent bundle is not a manifold in general, it is a fibre bundle.
Just as a manifold is a topological space that looks locally like $ \mathbb{R}^n $, a fibre bundle is a topological space that looks locally like a direct product (a.k.a. Cartesian product) of two spaces.
A fibre bundle is formally defined as the quadruple $ ( E, B, F, p ) $ where:


*

*$ E $ is the total space - the bundle itself

*$ B $ is the base space - in your example the manifold you started with

*$ F $ is the fibre - the space you're 'attaching' to $ B $

*$ p : E \rightarrow B $ is the projection  - a function which takes a point in the bundle and gets rid of the fibre, leaving you with the point it was attached to.
So, in the phrase "a fibre bundle is a topological space that looks locally like a direct product (a.k.a. Cartesian product) of two spaces" from above, the space $ E $ looks locally like $ B \times F $, but not globally. As you move around the bundle (which is described by other functions known as sections) you have to do work to "stay in the bundle". In your example of the tangent bundle to a manifold, that work would be prescribed by the connection.
Two useful examples:


*

*An hollow cylinder is a trivial bundle i.e. it is globally the direct product $ S^1 \times I $ for some $ I = [ a, b ] \subset \mathbb{R} $.

*The Möbius strip, however, is a nontrivial bundle. If you look locally at it, it looks like $ S^1 \times I $, but it is not that simple globally. It has some twisting to the way the copy of $ I $ has been attached.
In the specific example of classical mechanics, we take the manifold of initial positions of the particles - in general some subset of $ \mathbb{R}^n $ - and, as you say, we look at the cotangent space at a point i.e. the space of all functions from the tangent space to $ \mathbb{R} $. We often call such functions $1$-forms. To fully see that the contangent space is the proper way to model the space of all position and momenta takes a little bit more work than we can do here, but some pointers are:


*

*learn about $1$-,$2$-, and $p$-forms on a manifold

*learn about the exterior derivative and the exterior algebra

*see how a basis for the cotangent space can be formed by the exterior derivatives of the co-ordinate functions of the base manifold (a.k.a configuration space)
from this we can see that the cotangent space has a basis written in terms of the differentials of the co-ordinate functions on the manifold, which is exactly what we want to encode the positions (the co-ordinate functions on configuration space) and the momenta (a function of the derivative of position).
