Why is generalised force $Q_j = \frac{\partial L}{\partial q_j}$ when kinetic energy is independent of position?

just a heads up - to solve this problem we must know about Lagrange's equation and expressions for generalised velocity, kinetic energy and generalised forces. It's not purely mathematics

Show that if kinetic energy does not depend on generalised coordinates, $q_j$ but only on time or $\dot{q_j},$ then $Q_j = \frac{\partial L}{\partial q_j},$ where $L$ is the Lagrangian.

I've worked out that

$$\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial T}{\partial \dot{q_j}} = \sum_i m_i \ddot{x_i}\frac{\partial x_i}{\partial q_j} + m_i \dot{x_i} \frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial x_i}{\partial q_j}$$

by $$\frac{\partial \dot{x_i}}{\partial \dot{q_j}} = \frac{\partial x_i}{\partial q_j}$$

where $T$ is kinetic energy. But by the chain rule, the second term in the first expression is

$$\frac{\partial T}{\partial q_j} = m_i\dot{x_i}\frac{\partial \dot{x_i}}{\partial q_j} = m_i\dot{x_i}\frac{\partial }{\partial q_j} \left (\ \sum_{k}\frac{\partial x_i}{\partial q_k}\dot{q_k} + \frac{\partial x_i}{\partial t} \right )$$

and the first term is $Q_j$. So the expression equals $Q_j + \frac{\partial T}{\partial q_j}$. Now Lagrange's equation also has a $-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}}$ term and a $-\frac{\partial L}{\partial q_j}$ term on the LHS. Because we have that kinetic energy does not depend on coordinates, $Q_j -\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}} + \frac{\partial V}{\partial q_j} = 0$

I may well have messed up somewhere and I'm not sure if this method is even going to work. I don't think I really have the intuition to be able to solve this yet. Could anyone help?

The Lagrangian $\mathcal{L}$ is built up from the sum of the kinetik energy $T$ and the potential energy $V$ with its sign reversed, $i.e.$ $$\mathcal{L}=T-V\tag{*}$$

The equations of motion come from the minimum principle of the action. They are derived from: $$\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial\dot{q}}\right)=\frac{\partial \mathcal{L}}{\partial q}\tag{**}$$ where $q$ and $\dot{q}$ are the generalised coodinates.

If $T=T(\dot{q})$ and $V=V(q)$ then equation $(*)$ introduced in $(**)$ reads: $$\frac{d}{dt}\left(\frac{\partial T}{\partial\dot{q}}\right)=-\frac{\partial V}{\partial q}\tag{***}$$ The latter equation $(***)$ is the Newton second law, and it is clear that the generalised forces $Q$ are given by: $$Q=-\frac{\partial V}{\partial q}=\frac{\partial L}{\partial q}$$

• so is $\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial V}{\partial \dot{q_j}} = 0$ Or was my working wrong? (and this equation is true by some other reason) Sep 2 '17 at 15:11
• i think i'm on the right lines because you claim the lhs = Q, which agrees with my working Sep 2 '17 at 15:14
• Formally $V$ is independent of the velocity $\dot{q}$ because not all forces can be derived from potentials. Instead of this you can put a generalised force to the RHS of $(***)$. For example the frictional force $-c\dot{q}$. See: physics.stackexchange.com/questions/20929/…
– HBR
Sep 2 '17 at 15:20
• wait. if it's true generally then why do we need that term for equation two? Sep 2 '17 at 15:21
• I meant, not all $\textbf{forces that depend on the velocity}$ can be derived from potentials.
– HBR
Sep 2 '17 at 15:30