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Consider a system whose generalized co-ordinates are $q_i$ and is under the constraints $\dot{q_i} = K_i \forall i = 1,2,3,...$ where $K_i$ are constants. I have a problem in writing the Lagrange's equation for this system as $\dot{q_i}$ is a constant. What would be $\frac{\partial L}{\partial \dot{q_i}}$ ?

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This problem is in general quite hard and has been much explored in the context of quantization of electromagnetism and other gauge theories which are also singular. Check out the wikipedia page on Dirac bracket: – Marek Jun 19 '11 at 9:39
up vote 3 down vote accepted

I will assume that we are talking about a classical (as opposed to a quantum) system.

Well, Karsus Ren is of course right that $\dot{q}^i=K^i$ implies that the solution $q^i(t)$ is an affine function of time $t$.

I guess the heart of the question (v2) is not so much the solution itself, but more that if the Lagrangian $L=L(q,\dot{q},t)$ is given, and we have constraints $\dot{q}^i=K^i$, should we substitute $\dot{q}^i\to K^i$ in (1) none, (2) all, or (3) some of the $\dot{q}$ appearances in $L=L(q,\dot{q},t)$ before differentiating wrt. $\dot{q}^i$?

The answer is that it doesn't matter. Normally we treat constraints by introducing Lagrange multipliers $\lambda_i$ and a new Lagrangian

$$ \tilde{L}(q,\dot{q},\lambda,t) = L(q,\dot{q},t) + \lambda_i (\dot{q}^i-K^i). $$

The Lagrange eqs. wrt. $\lambda_i$ yield the constraints $\dot{q}^i=K^i$. This is the evolution equation for $q^i$.

On the other hand, the Lagrange eqs. wrt. $q^i$ yield

$$\frac{\partial L}{\partial q^i} = \frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{q}^i} + \lambda_i \right]. $$

This is the evolution equation for $\lambda_i$. We see from the last equation that a change in the definition of the derivative $\frac{\partial L}{\partial \dot{q}^i}$ leads to a corresponding change in $\lambda_i$, but this has no consequences for $q^i$, as expected.

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Hm, true, this is the end of story in classical mechanics. But if one wanted to quantize this system (either by path integral or by going to Hamiltonian picture) then one would need to deal with the singularity due to constraints. – Marek Jun 19 '11 at 13:30
Thank you very much for the answer. – Rajesh Dachiraju Jun 20 '11 at 3:40

You already have the relation of $q$ in terms of $t$, what are you solving?

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I see what you're getting at but this is too brief to be an answer. Perhaps you could change it to being a comment or edit your answer to elaborate. -1 until then I'm afraid. – qftme Jun 19 '11 at 8:38
I'd suggest that this is not good as an answer... – Waffle's Crazy Peanut Oct 26 '12 at 15:36
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Manishearth Dec 2 '12 at 2:07

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