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I've got a system where I know that the derivative of one of the generalized coordinates is constant. So to find the Hamiltonian of the system I need to take the partial derivative with respect to that derivative to get my conjugate momentum. The computation is not particularly complicated, I'm just wondering if this is allowable. If for instance, I know that $\dot q =$ constant, then does $\dfrac {\partial \mathcal L}{\partial \dot q}$ make sense?

  • Just think of $\mathcal{L}$ as being any other function. If you had $f(x,y) = x(y-1)$ then $f(0,y)$ is a constant, while $\frac {\partial f} {\partial x}|_{x=0}$ still makes sense and is not constant. – Arthur Suvorov Apr 13 '15 at 02:18

2 Answers2

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  1. If e.g. we consider a 1D non-relativistic free particle with kinetic energy $$\tag{1} T(q,v,t)~=~\frac{1}{2}mv^2$$ the information that the velocity is a constant$^1$ $$\tag{2} v~\approx~\text{constant},$$ only came afterwards from solving the Lagrange eqs. $$\tag{3} \left(\frac{d}{dt}\frac{\partial T}{\partial v}-\frac{\partial T}{\partial q}-Q\right)_{v=\dot{q}} ~\approx~0,$$ or equivalently, Newton's 2nd law.

  2. Echoing NeuroFuzzy's answer, if the reader is familiar with the principle of stationary action $S=\int \! dt~L$, then the Euler-Lagrange (EL) equations $$\tag{4} \frac{\delta S}{\delta q} ~=~ \left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial v}\right)_{v=\dot{q}} ~\approx~0,$$ is then the ODE for the stationary path. Needless to say, one is not supposed to use information about the solution to the EL eqs. during the construction phase of the EL eqs. For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian $L$ but dependent variables in the action $S$, etc.; see e.g. this related Phys.SE post and links therein.

  3. There are situations where the system from the onset has a constraint. Here OP might have a valid point. If the constraint is holonomic, say the position $x=0$ is fixed, it can be dealt with implicitly (via d'Alemberts principle), so that Lagrange equations (3) continues to hold. For semi-holonomic constraints, it can be dealt with explicitly via Lagrange multipliers.

References:

  1. Goldstein, Classical Mechanics, Chapter 1 & 2.

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$^1$ The $\approx$ symbol means equality modulo eqs. of motion.

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Yes. The action functional $S[q]=\int \mathcal{L} dt$ is something that can be applied to ANY [differentiable] path. So $\mathcal{L}$ is to be understood as an object which can really take n any value of $q$ and $\dot{q}$ independently. This makes it so that $\mathcal{L}$ really is a function of two variables, and so that it really does have to be defined for the values you have to plug in.

In the derivation you refer to, you look at a virtual variation of a path about the path physically taken. The key word being "virtual". You could phrase it as follows: you're considering MANY paths, not all of which have $\dot{q}=\mbox{constant}$, even though the physical path does have $\dot{q}=\mbox{constant}$.