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Found Lagrange Equation $$E = 2F\ddot q + \frac{dF}{dq} + \frac{dv}{dq} $$

knowing that Equation for Total energy is E = T + v $$E = 2F\ddot q + \frac{dF}{dq} - \frac{dv}{dq} $$ Is that correct? ( change of signs? ) How would i find Time derivative $$ \frac{DE}{dt}? $$

Thanks in advice!

Unknownstarz
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  • Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. You have also defined $T$ and $V$ two times. – Winther Oct 30 '15 at 00:24
  • @Winther Basically i solved the lagrange Equation, not really sure how to name the derivative of the T and V, and just have no idea how to find time derivative – Unknownstarz Oct 30 '15 at 00:33
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    There are so many weird variables in your question. Do you mean $\frac{De}{dt}$ or $\dfrac{dE}{dt}$? What is $V = -\dfrac{dv}{dq}$? Your question is very confusing. – cgo Oct 30 '15 at 00:36
  • @cgo im Baffed myself, so basically i got T = F(q)q'^2 and V = V(q) i found the derivatives of F and V which i dont know what to name? Dv/Dt? Df/Dt? and its dE/dt for lagrange equation – Unknownstarz Oct 30 '15 at 00:45
  • I am not sure if we are thinking about the same Lagrangian equation. (Euler-Lagrange?) Because Euler Lagrange Equation uses $T-V$, not $T+V$. The latter expression is for Hamiltonian, not Lagrangian. – cgo Oct 30 '15 at 01:23

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I have tried to decipher what you mean. It seems you are working with a Lagrangian

$$L(q,\dot{q}) = T-V = F(q) \dot{q}^2 - V(q)$$

The Euler-Lagrange equations coming from this Lagrangian, $\frac{dL}{dq} - \frac{d}{dt}\frac{dL}{d\dot{q}} = 0$, is

$$0 = 2F(q)\ddot{q} + \frac{dF(q)}{dq}\dot{q}^2 + \frac{dV(q)}{dq}$$

This is the same equation as you have, but in your question this equation is equated with the energy which is incorrect. The total energy is the sum of kinetic and potential energy

$$E = T + V = F(q) \dot{q}^2 + V(q)$$

To compute the time-derivative of the total energy $\frac{dE}{dt}$ use the chain-rule

$$\frac{dV(q)}{dt} = \frac{dV(q)}{dq} \frac{dq}{dt}$$

and similar for the other quantities. If you then use the Euler-Lagrange equation derived above you should be able to simplify the result and show that $\frac{dE}{dt} = 0$.

Winther
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