I am studying a book called "A Student's Guide to Lagrangians and Hamiltonians".
There, they show a proof for this identity:
$$\frac{\partial v_i}{\partial \dot{q_j}} = \frac{\partial x_i}{\partial q_j}$$
where:
- $x_i$ are Cartesian coordinates,
- $q_j$ are generalized coordinates,
- $v_i$ is velocity along $x_i$; that is $v_i = \frac{dx_i}{dt}$
- $\dot{q_j}$ is velocity along $q_j$; that is $v_i = \frac{dq_j}{dt}$
The proof is here and my two questions are placed in between the proof:
... $x_i$ is a function of the $q$s, so using the chain rule of differentiation we have $$v_i = \sum_{k} \frac{\partial x_i}{\partial q_k} \frac{dq_k}{dt} + \frac{\partial x_i}{\partial t} = \sum_{k} \frac{\partial x_i}{\partial q_k} \dot{q_k} + \frac{\partial x_i}{\partial t}$$ We now obtain a very useful relation involving the partial derivative of $v_i$ with respect to $\dot{q_j}$ . Since a mixed second-order partial derivative does not depend on the order in which the derivatives are taken, we can write $$\frac{\partial}{\partial \dot{q_j}} \frac{\partial x_i}{\partial t} = \frac{\partial}{\partial t} \frac{\partial x_i}{\partial \dot{q_j}}$$ But $\frac{\partial x_i}{\partial \dot{q_j}} = 0$ because $x$ does not depend on the generalized velocity $\dot{q}$.
Question 1: How is it that $x$ not dependent on $\dot{q}$? Cartesian coordinates are tied by their transformation to generalized coordinates, and generalized velocity is time derivative of generalized coordinates. They have a mathematical relationship. First, transform $x$ which must depend on time (that's why $v$ exists) , like $q=T(x(t))$ then take time derivative of it, and you have $\dot{q}$. Why does the book say they are independent? Is it because it is "partial" derivative and would not equal to zero for total derivative?
Let us take the partial derivative of $v_i$ with respect to $\dot{q_j}$ ... This yields $$\frac{\partial v_i}{\partial \dot{q_j}} = \frac{\partial}{\partial \dot{q_j}} \sum_{k} \frac{\partial x_i}{\partial q_k} \dot{q_k} + \frac{\partial}{\partial \dot{q_j}} \frac{\partial x_i}{\partial t}$$ $$= \frac{\partial}{\partial \dot{q_j}} \sum_{k} \frac{\partial x_i}{\partial q_k} \dot{q_k} + 0$$ $$= \sum_{k} \left( \frac{\partial}{\partial \dot{q_j}} \frac{\partial x_i}{\partial q_k} \right) \dot{q_k} + \sum_{k} \frac{\partial x_i}{\partial q_k} \frac{\partial \dot{q_k}}{\partial \dot{q_j}} = 0 + \sum_{k} \frac{\partial x_i}{\partial q_k} \delta _{ij}$$
Question 1.5: What does $\delta _{ij}$ mean? Delta generally means some difference but what does this specific use mean?
$$ = \frac{\partial x_i}{\partial q_j}$$ Thus, we conclude that $$ \frac{\partial v_i}{\partial \dot{q_j}} = \frac{\partial x_i}{\partial q_j}$$ ... it is easy to remember because it states that the Cartesian velocity is related to the generalized velocity in the same way as the Cartesian coordinate is related to the generalized coordinate.
Question 2: I agree with the book on that this equality shows a nice relationship and also I think is intuitive. However, when I look at the same equality from this perspective $$\frac{\partial \left( \frac{dx_i}{dt} \right)}{\partial \left( \frac{dq_j}{dt} \right)} = \frac{\partial x_i}{\partial q_j}$$ I feel like this shouldn't require any proof for it to work. I want to "cancel" $\frac{}{dt}$ parts of the equation. Because it feels like the derivative of parametric equations. But this uses partial derivatives so I am not sure.
Does my intuition make sense? Did we really needed that long proof?
