E-L equations only hold for independent variables?

AI Thread Summary
The discussion centers on the assumption of independence among variables in the derivation of the Euler-Lagrange (E-L) equations. It is confirmed that when variables are dependent, the standard E-L equations cannot be applied directly. Instead, the method of Lagrange multipliers should be utilized to account for constraints between the variables. The conversation also touches on the implications of time-dependent potentials in the context of Lagrangian mechanics. Ultimately, if the variables are not independent, a modified approach to the E-L equations is necessary.
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I'm talking about this:

http://www.cs.cornell.edu/courses/cs6650/2008fa/images/thumb_EL.jpg

In the derivation when you minimize action you assume that all the variations in coordinates are independent and thus conclude that each term has to be zero. When this isn't the case anymore one doesn't reach this conclusion.

Question 1: Is the above correct,

If yes, here follows the real reason for this post:

Question 2: One can also derive the E-L equations from d'Alemberts principle of virtual work. One arrives at the same equations. However it seems that no explicit assumption of the variables being independent was ever made. Where does this assumption hide in this derivation?
 
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Often if there is a constraint relating the variables so they aren't independent, then the method of Lagrange multipliers may be used in conjunction with the Lagrangian at the unconstrained coordinates. For example if we have a constraint relating the variables ##\int dt\; g(\mathbf{q}(t), \dot{\mathbf{q}}(t)) = 0## Then solutions may be found by using
$$\left[\frac{\partial}{\partial q_i} - \frac{d}{dt}\frac{\partial }{\partial \dot{q}_i}\right]\left[ \mathcal{L} + \lambda g\right] = 0 $$ This does relate to the the variations not being independent. You can see an explanation of that on page 156 here.
 
MisterX said:
Often if there is a constraint relating the variables so they aren't independent, then the method of Lagrange multipliers may be used in conjunction with the Lagrangian at the unconstrained coordinates. For example if we have a constraint relating the variables ##\int dt\; g(\mathbf{q}(t), \dot{\mathbf{q}}(t)) = 0## Then solutions may be found by using
$$\left[\frac{\partial}{\partial q_i} - \frac{d}{dt}\frac{\partial }{\partial \dot{q}_i}\right]\left[ \mathcal{L} + \lambda g\right] = 0 $$ This does relate to the the variations not being independent. You can see an explanation of that on page 156 here.

Hey, I might just being slow but this doesn't answer my question directly or does it? You are talking about constraints here between the coordinates, what I'm talking about is the potential as used in ##L=T-V## being a function of the time. Or more correct, the forces in space being the gradient of a time dependent function. Can I then still use the standard E-L equations?
 
Da
MisterX said:
Often if there is a constraint relating the variables so they aren't independent, then the method of Lagrange multipliers may be used in conjunction with the Lagrangian at the unconstrained coordinates. For example if we have a constraint relating the variables ##\int dt\; g(\mathbf{q}(t), \dot{\mathbf{q}}(t)) = 0## Then solutions may be found by using
$$\left[\frac{\partial}{\partial q_i} - \frac{d}{dt}\frac{\partial }{\partial \dot{q}_i}\right]\left[ \mathcal{L} + \lambda g\right] = 0 $$ This does relate to the the variations not being independent. You can see an explanation of that on page 156 here.
Sorry for my previous reply, I mixed thus thread up with another question I posted. So to conclude, if I have dependent variables in the lagrangian I can't just use the standard method?
 
Yes if the ##q_i## and ##\dot{q}_i## are not independent you must use a modified E-L equation.
 
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