Independence of generalized coordinates and generalized velocities

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VVS2000
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I was studying the derivation of the lagrangian formalism from Goldstein's textbook for mechanics and at one point they made a claim that generalized co-ordinates and velocities are independent and the derivative of one with respect to the other is zero.
How can I make sense of this and further how to think of this in the context of phase space diagrams?
 
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This question has been asked a number of times. Lagrangian mechanics analyses the functional form of the Lagrangian in order to derive the Euler Lagrange equations. At which point the coordinates and derivatives revert to their usual role, with one the time derivative of the other.

If you search my recent posts for the text in italics you should find a more complete answer.
 
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PeroK said:
This question has been asked a number of times. Lagrangian mechanics analyses the functional form of the Lagrangian in order to derive the Euler Lagrange equations. At which point the coordinates and derivatives revert to their usual role, with one the time derivative of the other.

If you search my recent posts for the text in italics you should find a more complete answer.
Well you're apparently a P-F galaxy so can you provide the link to your post regarding this topic?
 
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PeroK said:
I've never found a way to do it from my phone.
Ok, if not too much trouble, can you explain how they're independent here itself?
 
VVS2000 said:
Ok, if not too much trouble, can you explain how they're independent here itself?
You treat them as independent variables and study the functional form of the Lagrangian. In the simplest case you have$$L(x,\dot x) = \frac 1 2 m\dot x^2 - V(x)$$Effectively what you do is say, okay, let's study a function of the form$$L(X,Y) = \frac 1 2 mY^2 - V(X)$$That's perfectly legitimate mathematically.